Symmetric rational and nonrational inequalities [Volume 2]

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Vasile Cîrtoaje C j

MATHEM MATICAL INEQUA Q ALITIES Volume 2

SYMM METRIC RATIONAL AND NONRATIONAL RATIONAL AND  ALITIES INEQUA

UNIVERSITY OF PLLOIESTI, ROMANIA 2015

Contents 1 Symmetric Rational Inequalities 1 1.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2 Symmetric Nonrational Inequalities 257 2.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 2.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 3 Symmetric Power-Exponential Inequalities 419 3.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 3.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 4 Bibliography

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Chapter 1

Symmetric Rational Inequalities 1.1

Applications

1.1. If a, b are nonnegative real numbers, then 1 1 1 + ≥ . (1 + a)2 (1 + b)2 1 + ab

1.2. If a, b, c are nonnegative real numbers, then a2 − bc b2 − ca c2 − a b + + ≥ 0. 3a + b + c 3b + c + a 3c + a + b

1.3. If a, b, c are positive real numbers, then 4a2 − b2 − c 2 4b2 − c 2 − a2 4c 2 − a2 − b2 + + ≤ 3. a(b + c) b(c + a) c(a + b)

1.4. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a) (b) (c)

a2 2a2 a2

1 1 1 3 + 2 + 2 ≥ ; + bc b + ca c + a b a b + bc + ca 1 1 1 2 + 2 + 2 ≥ . + bc 2b + ca 2c + a b a b + bc + ca

1 1 1 2 + 2 + 2 > . + 2bc b + 2ca c + 2a b a b + bc + ca 1

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1.5. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a(b + c) b(c + a) c(a + b) + 2 + 2 ≥ 2. a2 + bc b + ca c + ab

1.6. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 b2 c2 a b c + + ≥ + + . 2 2 2 2 2 2 b +c c +a a +b b+c c+a a+b

1.7. Let a, b, c be positive real numbers. Prove that 1 1 1 a b c + + ≥ 2 + 2 + 2 . b+c c+a a+b a + bc b + ca c + a b

1.8. Let a, b, c be positive real numbers. Prove that 1 1 1 2a 2b 2c + + ≥ 2 + 2 + 2 . b+c c+a a+b 3a + bc 3b + ca 3c + a b

1.9. Let a, b, c be positive real numbers. Prove that (a)

(b)

b c 13 2(a b + bc + ca) a + + ≥ − ; b+c c+a a+b 6 3(a2 + b2 + c 2 )  ‹ p a b c 3 a b + bc + ca + + − ≥ ( 3 − 1) 1 − 2 . b+c c+a a+b 2 a + b2 + c 2

1.10. Let a, b, c be positive real numbers. Prove that  ‹2 1 1 1 a+b+c . + + ≤ a2 + 2bc b2 + 2ca c 2 + 2a b a b + bc + ca

1.11. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 (b + c) b2 (c + a) c 2 (a + b) + 2 + 2 ≥ a + b + c. b2 + c 2 c + a2 a + b2

Symmetric Rational Inequalities 1.12. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 3(a2 + b2 + c 2 ) a2 + b2 b2 + c 2 c 2 + a2 + + ≤ . a+b b+c c+a a + b + c)

1.13. Let a, b, c be positive real numbers. Prove that a2

1 1 9 1 + 2 + 2 ≥ . 2 2 2 + ab + b b + bc + c c + ca + a (a + b + c)2

1.14. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 b2 c2 1 + + ≤ . (2a + b)(2a + c) (2b + c)(2b + a) (2c + a)(2c + b) 3

1.15. Let a, b, c be positive real numbers. Prove that P

(a)

P

(b)

a 1 ≤ ; (2a + b)(2a + c) a+b+c

a3 1 ≤ . 2 2 2 2 (2a + b )(2a + c ) a+b+c

1.16. If a, b, c are positive real numbers, then X

1 2 1 ≥ + . 2 (a + 2b)(a + 2c) (a + b + c) 3(a b + bc + ca)

1.17. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 1 1 1 4 + + ≥ ; 2 2 2 (a − b) (b − c) (c − a) a b + bc + ca

(a)

(b) (c)

a2

1 1 1 3 + 2 + 2 ≥ ; 2 2 2 − ab + b b − bc + c c − ca + a a b + bc + ca a2

1 1 1 5 + 2 + 2 ≥ . 2 2 2 +b b +c c +a 2(a b + bc + ca)

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1.18. Let a, b, c be positive real numbers, no two of which are zero. Prove that (a2 + b2 )(a2 + c 2 ) (b2 + c 2 )(b2 + a2 ) (c 2 + a2 )(c 2 + b2 ) + + ≥ a2 + b2 + c 2 . (a + b)(a + c) (b + c)(b + a) (c + a)(c + b) 1.19. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that a2

1 1 1 + 2 + 2 ≤ 1. +b+c b +c+a c +a+b

1.20. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that a2 − bc b2 − ca c 2 − a b + 2 + 2 ≥ 0. a2 + 3 b +3 c +3 1.21. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 − bc 1 − ca 1 − a b + + ≥ 0. 5 + 2a 5 + 2b 5 + 2c 1.22. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that a2

1 1 1 3 + 2 + 2 ≤ . 2 2 2 + b +2 b +c +2 c +a +2 4

1.23. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that 1 1 1 1 + + ≤ . 4a2 + b2 + c 2 4b2 + c 2 + a2 4c 2 + a2 + b2 2 1.24. Let a, b, c be nonnegative real numbers such that a + b + c = 2. Prove that bc ca ab + 2 + 2 ≤ 1. +1 b +1 c +1

a2

1.25. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that bc ca ab 1 + + ≤ . a+1 b+1 c+1 4

Symmetric Rational Inequalities

5

1.26. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that 1 1 3 1 + + ≤ . 2 2 2 a(2a + 1) b(2b + 1) c(2c + 1) 11a bc 1.27. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that 1 1 1 + + ≤ 1. a3 + b + c b3 + c + a c 3 + a + b 1.28. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that a2 b2 c2 + + ≥ 1. 1 + b3 + c 3 1 + c 3 + a3 1 + a3 + b3 1.29. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 1 1 3 + + ≤ . 6 − a b 6 − bc 6 − ca 5 1.30. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 1 1 1 + 2 + 2 ≤ . + 7 2b + 7 2c + 7 3

2a2

1.31. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ 1 ≥ c and a + b + c = 3. Prove that 1 1 1 3 + 2 + 2 ≤ . 2 a +3 b +3 c +3 4 1.32. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 1 1 3 + + ≥ . 2a2 + 3 2b2 + 3 2c 2 + 3 5 1.33. Let a, b, c be nonnegative real numbers such that a ≥ 1 ≥ b ≥ c and a + b + c = 3. Prove that 1 1 1 + 2 + 2 ≥ 1. 2 a +2 b +2 c +2

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1.34. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that 1 1 a+b+c 3 1 + + ≥ + . a+b b+c c+a 6 a+b+c

1.35. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that a2

1 1 1 3 + 2 + 2 ≥ . +1 b +1 c +1 2

1.36. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that a2 b2 c2 + + ≥ 1. a2 + b + c b2 + c + a c 2 + a + b

1.37. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that bc + 4 ca + 4 a b + 4 bc + 2 ca + 2 a b + 2 + 2 + 2 ≤3≤ 2 + + . 2 a +4 b +4 c +4 a + 2 b2 + 2 c 2 + 2 p 1.38. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. If k ≥ 2 + 3, then 1 1 1 3 + + ≤ . a+k b+k c+k 1+k

1.39. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 3. Prove that a(b + c) b(c + a) c(a + b) + + ≤ 3. 1 + bc 1 + ca 1 + ab

1.40. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that a2 + b2 b2 + c 2 c 2 + a2 + + ≤ 3. a+b b+c c+a

Symmetric Rational Inequalities 1.41. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that ab bc ca 7 + + + 2 ≤ (a + b + c). a+b b+c c+a 6

1.42. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that (a)

1 1 3 1 + + ≤ ; 3 − a b 3 − bc 3 − ca 2

(b)

1 1 1 3 +p +p ≤p . p 6 − ab 6 − bc 6 − ca 6−1

1.43. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that 1 1 3 1 + + ≥ . 1 + a5 1 + b5 1 + c 5 2 1.44. Let a, b, c be positive real numbers such that a bc = 1. Prove that 1 1 1 + + ≥ 1. a2 + a + 1 b2 + b + 1 c 2 + c + 1 1.45. Let a, b, c be positive real numbers such that a bc = 1. Prove that a2

1 1 1 + 2 + 2 ≤ 3. −a+1 b − b+1 c −c+1

1.46. Let a, b, c be positive real numbers such that a bc = 1. Prove that 3+ b 3+c 3+a + + ≥ 3. 2 2 (1 + a) (1 + b) (1 + c)2

1.47. Let a, b, c be positive real numbers such that a bc = 1. Prove that 7 − 6a 7 − 6b 7 − 6c + + ≥ 1. 2 + a2 2 + b2 2 + c 2

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1.48. Let a, b, c be positive real numbers such that a bc = 1. Prove that b6 c6 a6 + + ≥ 1. 1 + 2a5 1 + 2b5 1 + 2c 5 1.49. Let a, b, c be positive real numbers such that a bc = 1. Prove that a b c 1 + + ≤ . a2 + 5 b2 + 5 c 2 + 5 2 1.50. Let a, b, c be positive real numbers such that a bc = 1. Prove that 1 1 1 2 + + + ≥ 1. 2 2 2 (1 + a) (1 + b) (1 + c) (1 + a)(1 + b)(1 + c)

1.51. Let a, b, c be nonnegative real numbers such that 1 1 1 3 + + = . a+b b+c c+a 2 Prove that

3 2 1 ≥ + 2 . a+b+c a b + bc + ca a + b2 + c 2

1.52. Let a, b, c be nonnegative real numbers such that 7(a2 + b2 + c 2 ) = 11(a b + bc + ca). Prove that

51 a b c ≤ + + ≤ 2. 28 b+c c+a a+b

1.53. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2

1 1 1 10 + 2 + 2 ≥ . 2 2 2 +b b +c c +a (a + b + c)2

1.54. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2

1 1 1 3 + 2 + 2 ≥ . 2 2 2 − ab + b b − bc + c c − ca + a max{a b, bc, ca}

Symmetric Rational Inequalities 1.55. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a(2a + b + c) b(2b + c + a) c(2c + a + b) + + ≥ 6. b2 + c 2 c 2 + a2 a2 + b2

1.56. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 (b + c)2 b2 (c + a)2 c 2 (a + b)2 + 2 + ≥ 2(a b + bc + ca). b2 + c 2 c + a2 a2 + b2 1.57. If a, b, c are real numbers such that a bc > 0, then  ‹  ‹ X a b c 1 1 1 a +5 + + ≥8 + + . 3 b2 − bc + c 2 bc ca a b a b c

1.58. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that ‹  1 1 1 + + + a2 + b2 + c 2 ≥ 2(a b + bc + ca); (a) 2a bc a+b b+c c+a (b)

a2 b2 c2 3(a2 + b2 + c 2 ) + + ≤ . a+b b+c c+a 2(a + b + c)

1.59. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

a2 − bc b2 − ca c 2 − a b 3(a b + bc + ca) + 2 + + ≥ 3; b2 + c 2 c + a2 a2 + b2 a2 + b2 + c 2

(b)

a2 b2 c2 a b + bc + ca 5 + + + 2 ≥ ; b2 + c 2 c 2 + a2 a2 + b2 a + b2 + c 2 2

(c)

a2 + bc b2 + ca c 2 + a b a b + bc + ca + 2 + 2 ≥ 2 + 2. 2 2 2 2 b +c c +a a +b a + b2 + c 2

1.60. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 b2 c2 (a + b + c)2 + + ≥ . b2 + c 2 c 2 + a2 a2 + b2 2(a b + bc + ca)

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1.61. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 5 2a b 2bc 2ca a2 + b2 + c 2 ≥ . + + + (a + b)2 (b + c)2 (c + a)2 a b + bc + ca 2

1.62. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that bc ca 1 a b + bc + ca ab + + + ≥ 2 . (a + b)2 (b + c)2 (c + a)2 4 a + b2 + c 2

1.63. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 3a b 3bc 3ca a b + bc + ca 5 + + ≤ 2 + . (a + b)2 (b + c)2 (c + a)2 a + b2 + c 2 4

1.64. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

a3 + a bc b3 + a bc c 3 + a bc + + ≥ a2 + b2 + c 2 ; b+c c+a a+b

(b)

a3 + 2a bc b3 + 2a bc c 3 + 2a bc 1 + + ≥ (a + b + c)2 ; b+c c+a a+b 2

(c)

a3 + 3a bc b3 + 3a bc c 3 + 3a bc + + ≥ 2(a b + bc + ca). b+c c+a a+b

1.65. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a3 + 3a bc b3 + 3a bc c 3 + 3a bc + + ≥ a + b + c. (b + c)2 (c + a)2 (a + b)2

1.66. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

a3 + 3a bc b3 + 3a bc c 3 + 3a bc 3 + + ≥ ; 3 3 3 (b + c) (c + a) (a + b) 2

(b)

3a3 + 13a bc 3b3 + 13a bc 3c 3 + 13a bc + + ≥ 6. (b + c)3 (c + a)3 (a + b)3

Symmetric Rational Inequalities

11

1.67. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

b3 c3 3 a3 + + + a b + bc + ca ≥ (a2 + b2 + c 2 ); b+c c+a a+b 2

(b)

2a2 + bc 2b2 + ca 2c 2 + a b 9(a2 + b2 + c 2 ) + + ≥ . b+c c+a a+b 2(a + b + c)

1.68. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a(b + c) b(c + a) c(a + b) + 2 + 2 ≥ 2. 2 2 + bc + c c + ca + a a + a b + b2

b2

1.69. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that Y  a − b ‹2 a(b + c) b(c + a) c(a + b) + + ≥2+4 . b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2 a+b

1.70. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a b − bc + ca bc − ca + a b ca − a b + bc 3 + + ≥ . 2 2 2 2 2 2 b +c c +a a +b 2 1.71. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then X a b + (k − 1)bc + ca b2

+ k bc

+ c2

≥

3(k + 1) . k+2

1.72. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then X 3bc − a(b + c) b2

+ k bc

+ c2

≤

3 . k+2

1.73. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that ab + 1 bc + 1 ca + 1 4 + 2 + 2 ≥ . 2 2 2 2 a +b b +c c +a 3

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1.74. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that 5a b + 1 5bc + 1 5ca + 1 + + ≥ 2. (a + b)2 (b + c)2 (c + a)2

1.75. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that b2 − ca c2 − a b a2 − bc + + ≥ 0. 2b2 − 3bc + 2c 2 2c 2 − 3ca + 2a2 2a2 − 3a b + 2b2

1.76. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 2a2 − bc 2b2 − ca 2c 2 − a b + + ≥ 3. b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2

1.77. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 b2 c2 + + ≥ 1. 2b2 − bc + 2c 2 2c 2 − ca + 2a2 2a2 − a b + 2b2

1.78. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 4b2

1 1 9 1 + 2 + 2 ≥ . 2 2 2 2 − bc + 4c 4c − ca + 4a 4a − a b + 4b 7(a + b2 + c 2 )

1.79. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 2a2 + bc 2b2 + ca 2c 2 + a b 9 + 2 + 2 ≥ . 2 2 2 2 b +c c +a a +b 2

1.80. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 2b2 + 3ca 2c 2 + 3a b 2a2 + 3bc + + ≥ 5. b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2

Symmetric Rational Inequalities

13

1.81. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 21 2a2 + 5bc 2b2 + 5ca 2c 2 + 5a b + + ≥ . (b + c)2 (c + a)2 (a + b)2 4

1.82. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then X 2a2 + (2k + 1)bc b2

+ k bc

+ c2

≥

3(2k + 3) . k+2

1.83. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then X

3bc − 2a2 3 ≤ . 2 2 b + k bc + c k+2

1.84. If a, b, c are nonnegative real numbers, no two of which are zero, then a2 + 16bc b2 + 16ca c 2 + 16a b + 2 + 2 ≥ 10. b2 + c 2 c + a2 a + b2

1.85. If a, b, c are nonnegative real numbers, no two of which are zero, then a2 + 128bc b2 + 128ca c 2 + 128a b + + ≥ 46. b2 + c 2 c 2 + a2 a2 + b2

1.86. If a, b, c are nonnegative real numbers, no two of which are zero, then a2 + 64bc b2 + 64ca c 2 + 64a b + + ≥ 18. (b + c)2 (c + a)2 (a + b)2

1.87. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ −1, then X a2 (b + c) + ka bc b2 + k bc + c 2

≥ a + b + c.

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1.88. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ then

X a3 + (k + 1)a bc b2 + k bc + c 2

−3 , 2

≥ a + b + c.

1.89. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > 0, then 2a k − b k − c k 2b k − c k − a k 2c k − a k − b k + 2 + 2 ≥ 0. b2 − bc + c 2 c − ca + a2 a − a b + b2 1.90. If a, b, c are the lengths of the sides of a triangle, then (a)

c+a−b a+b−c 2(a + b + c) b+c−a + + ≥ ; b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2 a2 + b2 + c 2

(b)

a2 − 2bc b2 − 2ca c 2 − 2a b + + ≤ 0. b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2

1.91. If a, b, c are nonnegative real numbers, then a2 b2 c2 1 + + ≤ . 2 2 2 2 2 2 5a + (b + c) 5b + (c + a) 5c + (a + b) 3 1.92. If a, b, c are nonnegative real numbers, then b2 + c 2 − a2 c 2 + a2 − b2 a2 + b2 − c 2 1 + + ≥ . 2 2 2 2 2 2 2a + (b + c) 2b + (c + a) 2c + (a + b) 2 1.93. Let a, b, c be positive real numbers. If k > 0, then 3a2 − 2bc 3b2 − 2ca 3c 2 − 2a b 3 + + ≤ . 2 2 2 2 2 2 ka + (b − c) k b + (c − a) kc + (a − b) k 1.94. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ 3 + then (a) (b)

a2 ka2

a b c 9 + 2 + 2 ≥ ; + k bc b + kca c + ka b (1 + k)(a + b + c)

1 1 1 9 + 2 + 2 ≥ . + bc k b + ca kc + a b (k + 1)(a b + bc + ca)

p 7,

Symmetric Rational Inequalities

15

1.95. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that

2a2

1 1 1 6 + 2 + 2 ≥ 2 . 2 2 + bc 2b + ca 2c + a b a + b + c + a b + bc + ca

1.96. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that

22a2

1 1 1 1 + + ≥ . 2 2 + 5bc 22b + 5ca 22c + 5a b (a + b + c)2

1.97. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 1 1 1 8 + + ≥ . 2a2 + bc 2b2 + ca 2c 2 + a b (a + b + c)2

1.98. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 1 1 12 1 + + ≥ . a2 + bc b2 + ca c 2 + a b (a + b + c)2

1.99. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

(b)

a2

1 1 1 1 2 + 2 + 2 ≥ 2 + ; 2 2 + 2bc b + 2ca c + 2a b a +b +c a b + bc + ca a(b + c) b(c + a) c(a + b) a b + bc + ca + 2 + 2 ≥1+ 2 . 2 a + 2bc b + 2ca c + 2a b a + b2 + c 2

1.100. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

(b)

a2

a b c a+b+c + 2 + 2 ≤ ; + 2bc b + 2ca c + 2a b a b + bc + ca

a(b + c) b(c + a) c(a + b) a2 + b2 + c 2 + + ≤ 1 + . a2 + 2bc b2 + 2ca c 2 + 2a b a b + bc + ca

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1.101. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

2a2

b c a+b+c a + 2 + 2 ≥ 2 ; + bc 2b + ca 2c + a b a + b2 + c 2

b+c c+a a+b 6 + 2 + 2 ≥ . 2 2a + bc 2b + ca 2c + a b a+b+c

(b)

1.102. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a(b + c) b(c + a) c(a + b) (a + b + c)2 + + ≥ . a2 + bc b2 + ca c2 + a b a2 + b2 + c 2 1.103. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > 0, then p p p p 3(2 + 3) b2 + c 2 + 3bc c 2 + a2 + 3ca a2 + b2 + 3a b + + ≥ . a2 + k bc b2 + kca c 2 + ka b 1+k

1.104. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2

1 1 1 8 6 + 2 + 2 + 2 ≥ . 2 2 2 2 2 +b b +c c +a a +b +c a b + bc + ca

1.105. If a, b, c are the lengths of the sides of a triangle, then a(b + c) b(c + a) c(a + b) + 2 + 2 ≤ 2. 2 a + 2bc b + 2ca c + 2a b

1.106. If a, b, c are real numbers, then a2 − bc b2 − ca c2 − a b + + ≥ 0. 2a2 + b2 + c 2 2b2 + c 2 + a2 2c 2 + a2 + b2

1.107. If a, b, c are nonnegative real numbers, then 3a2 − bc 3b2 − ca 3c 2 − a b 3 + + ≤ . 2 2 2 2 2 2 2 2 2 2a + b + c 2b + c + a 2c + a + b 2

Symmetric Rational Inequalities 1.108. If a, b, c are nonnegative real numbers, then (b + c)2 (c + a)2 (a + b)2 + + ≥ 2. 4a2 + b2 + c 2 4b2 + c 2 + a2 4c 2 + a2 + b2 1.109. If a, b, c are positive real numbers, then P

(a) (b)

11a2 P 4a2

3 1 ≤ ; 2 2 + 2b + 2c 5(a b + bc + ca)

1 1 1 ≤ + . 2 2 2 2 2 +b +c 2(a + b + c ) a b + bc + ca

1.110. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then p p p a b c 3 + + ≥ . b+c c+a a+b 2 1.111. If a, b, c are nonnegative real numbers such that a b + bc + ca ≥ 3, then 1 1 1 1 1 1 + + ≥ + + . 2+a 2+ b 2+c 1+ b+c 1+c+a 1+a+ b 1.112. If a, b, c are the lengths of the sides of a triangle, then (a)

a2 − bc b2 − ca c2 − a b + + ≤ 0; 3a2 + b2 + c 2 3b2 + c 2 + a2 3c 2 + a2 + b2

(b)

a4 − b2 c 2 b4 − c 2 a2 c 4 − a2 b2 + + ≤ 0. 3a4 + b4 + c 4 3b4 + c 4 + a4 3c 4 + a4 + b4

1.113. If a, b, c are the lengths of the sides of a triangle, then 4a2

bc ca ab 1 + 2 + 2 ≥ . 2 2 2 2 2 2 +b +c 4b + c + a 4c + a + b 2

1.114. If a, b, c are the lengths of the sides of a triangle, then b2

1 1 1 9 + 2 + 2 ≤ . 2 2 2 +c c +a a +b 2(a b + bc + ca)

17

18

Vasile Cîrtoaje

1.115. If a, b, c are the lengths of the sides of a triangle, then a + b b + c c + a (a) a − b + b − c + c − a > 5; 2 a + b2 b2 + c 2 c 2 + a2 (b) a2 − b2 + b2 − c 2 + c 2 − a2 ≥ 3. 1.116. If a, b, c are the lengths of the sides of a triangle, then  ‹ b+c c+a a+b a b c + + +3≥6 + + . a b c b+c c+a a+b 1.117. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that X 3a(b + c) − 2bc 3 ≥ . (b + c)(2a + b + c) 2

1.118. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that X

a(b + c) − 2bc ≥ 0. (b + c)(3a + b + c)

1.119. Let a, b, c be positive real numbers such that a2 + b2 + c 2 ≥ 3. Prove that b5 − b2 c5 − c2 a5 − a2 + + ≥ 0. a5 + b2 + c 2 b5 + c 2 + a2 c 5 + a2 + b2 1.120. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = a3 + b3 + c 3 . Prove that a2 b2 c2 3 + + ≥ . b+c c+a a+b 2 1.121. If a, b, c ∈ [0, 1], then (a)

a b c + + ≤ 1; bc + 2 ca + 2 a b + 2

(b)

ab bc ca + + ≤ 1. 2bc + 1 2ca + 1 2a b + 1

Symmetric Rational Inequalities

19

1.122. Let a, b, c be positive real numbers such that a + b + c = 2. Prove that  ‹ 1 1 1 + + 5(1 − a b − bc − ca) + 9 ≥ 0. 1 − a b 1 − bc 1 − ca

1.123. Let a, b, c be nonnegative real numbers such that a + b + c = 2. Prove that 2 − c2 2 − a2 2 − b2 + + ≤ 3. 2 − bc 2 − ca 2 − a b

1.124. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 3 + 5a2 3 + 5b2 3 + 5c 2 + + ≥ 12. 3 − bc 3 − ca 3 − ab

1.125. Let a, b, c be nonnegative real numbers such that a + b + c = 2. If 7 −1 ≤m≤ , 7 8 then

a2 + m b2 + m c2 + m 3(4 + 9m) + + ≥ . 3 − 2bc 3 − 2ca 3 − 2a b 19

1.126. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 47 − 7a2 47 − 7b2 47 − 7c 2 + + ≥ 60. 1 + bc 1 + ca 1 + ab

1.127. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 26 − 7a2 26 − 7b2 26 − 7c 2 57 + + ≤ . 1 + bc 1 + ca 1 + ab 2

1.128. Let a, b, c be nonnegative real numbers, no all are zero. Prove that X 5a(b + c) − 6bc ≤ 3. a2 + b2 + c 2 + bc

20

Vasile Cîrtoaje

1.129. Let a, b, c be nonnegative real numbers, no two of which are zero, and let x=

a2 + b2 + c 2 . a b + bc + ca

Prove that a b c 1 1 + + + ≥x+ ; b+c c+a a+b 2 x ‹  b c 4 a + + ≥ 5x + ; 6 b+c c+a a+b x

(a)

(b)

 ‹ a b c 3 1 1 + + − ≥ x− . b+c c+a a+b 2 3 x

(c)

1.130. If a, b, c are real numbers, then a2

1 1 1 9 + 2 + 2 ≤ . 2 2 2 2 2 2 + 7(b + c ) b + 7(c + a ) c + 7(a + b ) 5(a + b + c)2

1.131. If a, b, c are real numbers, then ca ab 3 bc + + ≤ . 3a2 + b2 + c 2 3b2 + c 2 + a2 3c 2 + a2 + b2 5

1.132. If a, b, c are real numbers such that a + b + c = 3, then (a)

1 1 1 3 + + ≤ ; 2 2 2 2 2 2 2+ b +c 2+c +a 2+a + b 4

(b)

1 1 1 1 + + ≤ . 2 2 2 2 2 2 8 + 5(b + c ) 8 + 5(c + a ) 8 + 5(a + b ) 6

1.133. If a, b, c are real numbers, then (a + b)(a + c) (b + c)(b + a) (c + a)(c + b) 4 + 2 + 2 ≤ . 2 2 2 2 2 2 2 a + 4(b + c ) b + 4(c + a ) c + 4(a + b ) 3

Symmetric Rational Inequalities

21

1.134. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that X

1 1 ≤ . (b + c)(7a + b + c) 2(a b + bc + ca)

1.135. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that X b2

+ c2

9 1 ≤ . + 4a(b + c) 10(a b + bc + ca)

1.136. If a, b, c are nonnegative real numbers such that a + b + c = 3, then 1 1 1 9 + + ≤ . 3 − a b 3 − bc 3 − ca 2(a b + bc + ca)

1.137. If a, b, c are nonnegative real numbers such that a + b + c = 3, then a2

bc ca ab 3 + 2 + 2 ≤ . +a+6 b + b+6 c +c+6 8

1.138. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then 1 1 1 1 + + ≥ . 8a2 − 2bc + 21 8b2 − 2ca + 21 8c 2 − 2a b + 21 9

1.139. Let a, b, c be real numbers, no two of which are zero. Prove that (a)

a2 + bc b2 + ca c 2 + a b (a + b + c)2 + + ≥ ; b2 + c 2 c 2 + a2 a2 + b2 a2 + b2 + c 2

(b)

a2 + 3bc b2 + 3ca c 2 + 3a b 6(a b + bc + ca) + 2 + 2 ≥ . b2 + c 2 c + a2 a + b2 a2 + b2 + c 2

1.140. Let a, b, c be real numbers such that a b + bc + ca ≥ 0 and no two of which are zero. Prove that a(b + c) b(c + a) c(a + b) 3 + 2 + 2 ≥ . 2 2 2 2 b +c c +a a +b 10

22

Vasile Cîrtoaje

1.141. If a, b, c are positive real numbers such that a bc > 1, then 1 4 1 + ≥ . a + b + c − 3 a bc − 1 a b + bc + ca − 3

1.142. Let a, b, c be positive real numbers, no two of which are zero. Prove that X (4b2 − ac)(4c 2 − a b) b+c

≤

27 a bc. 2

1.143. Let a, b, c be nonnegative real numbers, no two of which are zero, such that a + b + c = 3. Prove that

b c 2 a + + ≥ . 3a + bc 3b + ca 3c + a b 3

1.144. Let a, b, c be positive real numbers such that  ‹ 1 1 1 (a + b + c) + + = 10. a b c Prove that

19 a b c 5 ≤ + + ≤ . 12 b+c c+a a+b 3

1.145. Let a, b, c be nonnegative real numbers, no two of which are zero, such that a + b + c = 3. Prove that 9 a b c < + + ≤ 1. 10 2a + bc 2b + ca 2c + a b

1.146. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a3 b3 c3 a3 + b3 + c 3 + + ≤ . 2a2 + bc 2b2 + ca 2c 2 + a b a2 + b2 + c 2

Symmetric Rational Inequalities

23

1.147. Let a, b, c be positive real numbers, no two of which are zero. Prove that b3 c3 a+b+c a3 + + ≥ . 4a2 + bc 4b2 + ca 4c 2 + a b 5

1.148. If a, b, c are positive real numbers, then 1 1 3 1 + + ≥ . (2 + a)2 (2 + b)2 (2 + c)2 6 + a b + bc + ca

1.149. If a, b, c are positive real numbers, then 1 1 1 3 + + ≥ . 1 + 3a 1 + 3b 1 + 3c 3 + a bc

1.150. Let a, b, c be real numbers, no two of which are zero. If 1 ≤ k ≤ 3, then  ‹ ‹ ‹ 2a b 2bc 2ca k+ 2 k+ 2 k+ 2 ≥ (k − 1)(k2 − 1). a + b2 b + c2 c + a2

1.151. If a, b, c are non-zero and distinct real numbers, then • ˜  ‹ 1 1 1 1 1 1 1 1 1 + + +3 + + ≥4 + + . a2 b2 c 2 (a − b)2 (b − c)2 (c − a)2 a b bc ca

1.152. Let a, b, c be positive real numbers, and let A=

a b + + k, b a

B=

b c + + k, c b

C=

c a + + k, a b

where −2 < k ≤ 4. Prove that 1 1 1 1 4 + + ≤ + . A B C k + 2 A + B + C − (k + 2)

1.153. If a, b, c are nonnegative real numbers, no two of which are zero, then b2

1 1 1 1 1 1 + 2 + 2 ≥ 2 + 2 + 2 . 2 2 2 + bc + c c + ca + a a + ab + b 2a + bc 2b + ca 2c + a b

24

Vasile Cîrtoaje

1.154. If a, b, c are nonnegative real numbers such that a + b + c = 3, then 1 1 1 1 1 1 + + ≥ 2 + 2 + 2 . 2a b + 1 2bc + 1 2ca + 1 a +2 b +2 c +2 1.155. If a, b, c are nonnegative real numbers such that a + b + c = 4, then 1 1 1 1 1 1 + + ≥ 2 + 2 + 2 . a b + 2 bc + 2 ca + 2 a +2 b +2 c +2

1.156. If a, b, c are nonnegative real numbers, no two of which are zero, then a b + bc + ca (a − b)2 (b − c)2 (c − a)2 + ≤ 1; a2 + b2 + c 2 (a2 + b2 )(b2 + c 2 )(c 2 + a2 )

(a)

a b + bc + ca (a − b)2 (b − c)2 (c − a)2 + ≤ 1. a2 + b2 + c 2 (a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 )

(b)

1.157. If a, b, c are nonnegative real numbers, no two of which are zero, then a2

1 1 45 1 + 2 + 2 ≥ . 2 2 2 2 2 2 +b b +c c +a 8(a + b + c ) + 2(a b + bc + ca)

1.158. If a, b, c are real numbers, no two of which are zero, then a2 − 7bc b2 − 7ca c 2 − 7a b 9(a b + bc + ca) + 2 + 2 + ≥ 0. b2 + c 2 a + b2 a + b2 a2 + b2 + c 2 1.159. If a, b, c are real numbers such that a bc 6= 0, then (b + c)2 (c + a)2 (a + b)2 10(a + b + c)2 + + ≥ 2 + . a2 b2 c2 3(a2 + b2 + c 2 )

1.160. If a, b, c are nonnegative real numbers, no two of which are zero, then a2 − 4bc b2 − 4ca c 2 − 4a b 9(a b + bc + ca) 9 + 2 + 2 + ≥ . b2 + c 2 a + b2 a + b2 a2 + b2 + c 2 2

Symmetric Rational Inequalities 1.161. If a, b, c are nonnegative real numbers, no two of which are zero, then 9(a − b)2 (b − c)2 (c − a)2 a2 + b2 + c 2 ≥1+ . a b + bc + ca (a + b)2 (b + c)2 (c + a)2

1.162. If a, b, c are nonnegative real numbers, no two of which are zero, then p a2 + b2 + c 2 (a − b)2 (b − c)2 (c − a)2 ≥ 1 + (1 + 2)2 2 . a b + bc + ca (a + b2 )(b2 + c 2 )(c 2 + a2 )

1.163. If a, b, c are nonnegative real numbers, no two of which are zero, then 2 2 2 5 5 5 + + ≥ + + . a+b b+c c+a 3a + b + c 3b + c + a 3c + a + b

1.164. If a, b, c are real numbers, no two of which are zero, then (a)

8a2 + 3bc 8b2 + 3ca 8c 2 + 3a b + + ≥ 11; b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2

(b)

8a2 − 5bc 8b2 − 5ca 8c 2 − 5a b + + ≥ 9. b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2

1.165. If a, b, c are real numbers, no two of which are zero, then 4b2 + ca 4c 2 + a b 4a2 + bc + + ≥ 1. 4b2 + 7bc + 4c 2 4c 2 + 7ca + 4a2 4a2 + 7a b + 4b2

1.166. If a, b, c are real numbers, no two of which are equal, then 1 1 1 27 + + ≥ . 2 2 2 2 2 2 (a − b) (b − c) (c − a) 4(a + b + c − a b − bc − ca)

1.167. If a, b, c are real numbers, no two of which are zero, then 1 1 1 14 + + ≥ . a2 − a b + b2 b2 − bc + c 2 c 2 − ca + a2 3(a2 + b2 + c 2 )

25

26

Vasile Cîrtoaje

1.168. Let a, b, c be real numbers such that a b + bc + ca ≥ 0 and no two of which are zero. Prove that b c 3 a + + ≥ ; b+c c+a a+b 2

(a) (b) i f a b ≤ 0, then

a b c + + ≥ 2. b+c c+a a+b

1.169. If a, b, c are nonnegative real numbers, then a b c a b + bc + ca + + ≥ . 7a + b + c 7b + c + a 7c + a + b (a + b + c)2

1.170. If a, b, c are the lengths of the sides of a triangle, then a2 b2 c2 1 + + ≥ . 2 2 2 4a + 5bc 4b + 5ca 4c + 5a b 3

1.171. If a, b, c are the lengths of the sides of a triangle, then 1 1 1 3 + + ≥ . 7a2 + b2 + c 2 7b2 + c 2 + a2 7c 2 + a2 + b2 (a + b + c)2

1.172. Let a, b, c be the lengths of the sides of a triangle. If k > −2, then X a(b + c) + (k + 1)bc b2

+ k bc

+ c2

≤

3(k + 3) . k+2

1.173. Let a, b, c be the lengths of the sides of a triangle. If k > −2, then X 2a2 + (4k + 9)bc b2

+ k bc

+ c2

≤

3(4k + 11) . k+2

1.174. If a ≥ b ≥ c ≥ d such that a bcd = 1, then 1 1 1 3 + + ≥ . p 3 1+a 1+ b 1+c 1 + a bc

Symmetric Rational Inequalities

27

1.175. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that X

1 ≤ 1. 1 + a b + bc + ca

1.176. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that 1 1 1 1 + + + ≥ 1. 2 2 2 (1 + a) (1 + b) (1 + c) (1 + d)2

1.177. Let a, b, c, d 6=

1 be positive real numbers such that a bcd = 1. Prove that 3

1 1 1 1 + + + ≥ 1. 2 2 2 (3a − 1) (3b − 1) (3c − 1) (3d − 1)2

1.178. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that 1 1 1 1 + + + ≥ 1. 1 + a + a2 + a3 1 + b + b2 + b3 1 + c + c 2 + c 3 1 + d + d 2 + d 3

1.179. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that 1 1 1 1 + + + ≥ 1. 2 2 2 1 + a + 2a 1 + b + 2b 1 + c + 2c 1 + d + 2d 2

1.180. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that 1 1 1 1 9 25 + + + + ≥ . a b c d a+b+c+d 4

1.181. If a, b, c, d are real numbers such that a + b + c + d = 0, then (a − 1)2 (b − 1)2 (c − 1)2 (d − 1)2 + + 2 + ≤ 4. 3a2 + 1 3b2 + 1 3c + 1 3d 2 + 1

28

Vasile Cîrtoaje

1.182. If a, b, c, d ≥ −5 such that a + b + c + d = 4, then 1− b 1−c 1−d 1−a + + + ≥ 0. 2 2 2 (1 + a) (1 + b) (1 + c) (1 + d)2

1.183. Let a1 , a2 , . . . , an be positive real numbers such that a1 + a2 + · · · + an = n. Prove that X 1 1 ≤ . 2 2 2 (n + 1)a1 + a2 + · · · + an2

1.184. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = 0. Prove that (a1 + 1)2 a12 + n − 1

+

(a2 + 1)2 a22 + n − 1

+ ··· +

(an + 1)2 n ≥ . 2 an + n − 1 n−1

1.185. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that 1 1 1 + + ··· + ≥ 1. 1 + (n − 1)a1 1 + (n − 1)a2 1 + (n − 1)an

1.186. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that 1 1 − a1 + na12

+

1 1 − a2 + na22

+ ··· +

1 ≥ 1. 1 − an + nan2

1.187. Let a1 , a2 , . . . , an be positive real numbers such that a1 , a2 , . . . , an ≥

k(n − k − 1) , kn − k − 1

k>1

and a1 a2 · · · an = 1. Prove that

1 1 1 n + + ··· + ≤ . a1 + k a2 + k an + k 1+k

Symmetric Rational Inequalities

29

1.188. Let a1 , a2 , . . . , an be positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , Prove that

1 − a1 3 + a12

+

1 − a2 3 + a22

a1 a2 · · · an = 1.

+ ··· +

1 − an ≥ 0. 3 + an2

1.189. If a1 , a2 , . . . , an ≥ 0, then 1 1 1 n + + ··· + ≥ . 1 + na1 1 + na2 1 + nan n + a1 a2 · · · an

1.190. If a1 , a2 , . . . , an are positive real numbers, then bn an a1 a2 b1 b2 + + ··· + ≥ + + ··· + , a1 a2 an b1 b2 bn where bi =

1 X aj, n − 1 j6=i

i = 1, 2, · · · , n.

1.191. If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an =

1 1 1 + + ··· + , a1 a2 an

then (a)

1 1 1 + + ··· + ≥ 1; 1 + (n − 1)a1 1 + (n − 1)a2 1 + (n − 1)an

(b)

1 1 1 + + ··· + ≤ 1. n − 1 + a1 n − 1 + a2 n − 1 + an

30

Vasile Cîrtoaje

Symmetric Rational Inequalities

1.2

31

Solutions

P 1.1. If a, b are nonnegative real numbers, then 1 1 1 + ≥ . 2 2 (1 + a) (1 + b) 1 + ab (Vasile Cîrtoaje, 1994) First Solution. Use the Cauchy-Schwarz inequality as follows: 1 1 1 (b + a)2 1 + − ≥ − 2 2 2 2 2 2 (1 + a) (1 + b) 1 + ab b (1 + a) + a (1 + b) 1 + ab a b[a2 + b2 − 2(a + b) + 2] = (1 + a b)[b2 (1 + a)2 + a2 (1 + b)2 ] a b[(a − 1)2 + (b − 1)2 ] = ≥ 0. (1 + a b)[b2 (1 + a)2 + a2 (1 + b)2 ] The equality holds for a = b = 1. Second Solution. By the Cauchy-Schwarz inequality, we have ‹  1 ≥ (a + 1)2 , (a + b) a + b

(a + b)



‹ 1 + b ≥ (1 + b)2 , a

hence 1 1 1 1 1 + ≥ + = . 2 2 (1 + a) (1 + b) (a + b)(a + 1/b) (a + b)(1/a + b) 1 + a b Third Solution. The desired inequality follows from the identity 1 1 a b(a − b)2 + (1 − a b)2 1 + − = . (1 + a)2 (1 + b)2 1 + a b (1 + a)2 (1 + b)2 (1 + a b)

P 1.2. If a, b, c are nonnegative real numbers, then a2 − bc b2 − ca c2 − a b + + ≥ 0. 3a + b + c 3b + c + a 3c + a + b

32

Vasile Cîrtoaje

Solution. We use the SOS method. Without loss of generality, assume that a ≥ b ≥ c. We have 2

X (a − b)(a + c) + (a − c)(a + b) X a2 − bc = 3a + b + c 3a + b + c X (a − b)(a + c) X (b − a)(b + c) + = 3a + b + c 3b + c + a X (a − b)2 (a + b − c) = (3a + b + c)(3b + c + a)

Since a + b − c ≥ 0, it suffices to show that (b − c)2 (b + c − a)(3a + b + c) + (c − a)2 (c + a − b)(3b + c + a) ≥ 0; that is, (a − c)2 (c + a − b)(3b + c + a) ≥ (b − c)2 (a − b − c)(3a + b + c). This inequality is trivial for a ≤ b + c. Otherwise, we can get it by multiplying the obvious inequalities c + a − b ≥ a − b − c, b2 (a − c)2 ≥ a2 (b − c)2 , a(3b + c + a) ≥ b(3a + b + c), a ≥ b. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.3. If a, b, c are positive real numbers, then 4a2 − b2 − c 2 4b2 − c 2 − a2 4c 2 − a2 − b2 + + ≤ 3. a(b + c) b(c + a) c(a + b) (Vasile Cîrtoaje, 2006) Solution. We use the SOS method. Write the inequality as follows:  X 4a2 − b2 − c 2 1− ≥ 0, a(b + c) X b2 + c 2 − 4a2 + a(b + c) ≥ 0, a(b + c)

Symmetric Rational Inequalities

33

X (b2 − a2 ) + a(b − a) + (c 2 − a2 ) + a(c − a) a(b + c) X (b − a)(2a + b) + (c − a)(2a + c) a(b + c) X (b − a)(2a + b) a(b + c) X

+

≥ 0,

≥ 0,

X (a − b)(2b + a) b(c + a)

≥ 0,

c(a + b)(a − b)2 (bc + ca − a b) ≥ 0.

Without loss of generality, assume that a ≥ b ≥ c. Since ca + a b − bc > 0, it suffices to show that b(c + a)(c − a)2 (a b + bc − ca) + c(a + b)(a − b)2 (bc + ca − a b) ≥ 0, that is, b(c + a)(a − c)2 (a b + bc − ca) ≥ c(a + b)(a − b)2 (a b − bc − ca). For the nontrivial case a b − bc − ca > 0, this inequality follows by multiplying the inequalities a b + bc − ca > a b − bc − ca, (a − c)2 ≥ (a − b)2 , b(c + a) ≥ c(a + b). The equality holds for a = b = c

P 1.4. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

(b)

(c)

a2

2a2

a2

1 1 1 3 + 2 + 2 ≥ ; + bc b + ca c + a b a b + bc + ca 1 1 1 2 + 2 + 2 ≥ . + bc 2b + ca 2c + a b a b + bc + ca

1 1 1 2 + 2 + 2 > . + 2bc b + 2ca c + 2a b a b + bc + ca (Vasile Cîrtoaje, 2005)

34

Vasile Cîrtoaje

Solution. (a) Since

a(b + c − a) a b + bc + ca =1+ , 2 a + bc a2 + bc we can write the inequality as a(b + c − a) b(c + a − b) c(a + b − c) + + ≥ 0. a2 + bc b2 + ca c2 + a b Without loss of generality, assume that a = min{a, b, c}. Since b + c − a > 0, it suffices to show that b(c + a − b) c(a + b − c) + ≥ 0. b2 + ca c2 + a b This is equivalent to each of the following inequalities (b2 + c 2 )a2 − (b + c)(b2 − 3bc + c 2 )a + bc(b − c)2 ≥ 0, (b − c)2 a2 − (b + c)(b − c)2 a + bc(b − c)2 + a bc(2a + b + c) ≥ 0, (b − c)2 (a − b)(a − c) + a bc(2a + b + c) ≥ 0. The last inequality is obviously true. The equality holds for a = 0 and b = c (or any cyclic permutation thereof). (b) According to the identities 2a2 + bc = a(2a − b − c) + a b + bc + ca, 2b2 + ca = b(2b − c − a) + a b + bc + ca, 2c 2 + a b = c(2c − a − b) + a b + bc + ca, we can write the inequality as 1 1 1 + + ≥ 2, 1+ x 1+ y 1+z where x=

a(2a − b − c) , a b + bc + ca

y=

b(2b − c − a) c(2c − a − b) , z= . a b + bc + ca a b + bc + ca

Without loss of generality, assume that a = min{a, b, c}. Since x ≤ 0 and suffices to show that

1 1 + ≥ 1. 1+ y 1+z

This is equivalent to each of the following inequalities 1 ≥ yz,

1 ≥ 1, it 1+ x

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35

(a b + bc + ca)2 ≥ bc(2b − c − a)(2c − a − b), a2 (b2 + bc + c 2 ) + 3a bc(b + c) + 2bc(b − c)2 ≥ 0. The last inequality is obviously true. The equality holds for a = 0 and b = c (or any cyclic permutation thereof). (c) According to the identities a2 + 2bc = (a − b)(a − c) + a b + bc + ca, b2 + 2ca = (b − c)(b − a) + a b + bc + ca, c 2 + 2a b = (c − a)(c − b) + a b + bc + ca, we can write the inequality as 1 1 1 + + > 2, 1+ x 1+ y 1+z where x=

(a − b)(a − c) , a b + bc + ca

y=

(b − c)(b − a) (c − a)(c − b) , z= . a b + bc + ca a b + bc + ca

Since x y + yz + z x = 0 and x yz = we have

−(a − b)2 (b − c)2 (c − a)2 ≤ 0, (a b + bc + ca)3

1 − 2x yz 1 1 1 + + −2= > 0. 1+ x 1+ y 1+z (1 + x)(1 + y)(1 + z)

P 1.5. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a(b + c) b(c + a) c(a + b) + 2 + 2 ≥ 2. a2 + bc b + ca c + ab (Pham Kim Hung, 2006) Solution. Without loss of generality, assume that a ≥ b ≥ c and write the inequality as b(c + a) (a − b)(a − c) (a − c)(b − c) ≥ + . b2 + ca a2 + bc c2 + a b

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Since

a−b (a − b)(a − c) (a − b)a ≤ 2 ≤ 2 a + bc a + bc a

and

(a − c)(b − c) a(b − c) b−c ≤ 2 ≤ , 2 c + ab c + ab b

it suffices to show that

b(c + a) a−b b−c ≥ + . b2 + ca a b

This inequality is equivalent to b2 (a − b)2 − 2a bc(a − b) + a2 c 2 + a b2 c ≥ 0, (a b − b2 − ac)2 + a b2 c ≥ 0. The equality holds for for a = 0 and b = c (or any cyclic permutation).

P 1.6. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 b2 c2 a b c + + ≥ + + . 2 2 2 2 2 2 b +c c +a a +b b+c c+a a+b (Vasile Cîrtoaje, 2002) Solution. We have X

a2 a − b2 + c 2 b + c



=

X a b(a − b) + ac(a − c) (b2 + c 2 )(b + c)

X a b(a − b) ba(b − a) + 2 2 2 (b + c )(b + c) (c + a2 )(c + a) X a b(a − b)2 = (a2 + b2 + c 2 + a b + bc + ca) ≥ 0. (b2 + c 2 )(c 2 + a2 )(b + c)(c + a) =

X

The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 1.7. Let a, b, c be positive real numbers. Prove that 1 1 1 a b c + + ≥ 2 + 2 + 2 . b+c c+a a+b a + bc b + ca c + a b

Symmetric Rational Inequalities First Solution. Without loss of generality, assume that a = min{a, b, c}. Since ‹ X X 1 X 1 a a − = − b+c a2 + bc b + c a2 + bc X (a − b)(a − c) = (b + c)(a2 + bc) and (a − b)(a − c) ≥ 0, it suffices to show that (b − c)(b − a) (c − a)(c − b) + ≥ 0. 2 (c + a)(b + ca) (a + b)(c 2 + a b) This inequality is equivalent to (b − c)[(b2 − a2 )(c 2 + a b) + (a2 − c 2 )(b2 + ca)] ≥ 0, a(b − c)2 (b2 + c 2 − a2 + a b + bc + ca) ≥ 0, and is clearly true for a = min{a, b, c}. The equality holds for a = b = c. Second Solution. Since X

˜ X • ˜ X• b c 1 1 1 = + = a + , b+c (b + c)2 (b + c)2 (a + b)2 (a + c)2

we can write the inequality as ˜ X • 1 1 1 a + − ≥ 0. (a + b)2 (a + c)2 a2 + bc This is true, since 1 1 1 bc(b − c)2 + (a2 − bc)2 + − = ≥ 0. (a + b)2 (a + c)2 a2 + bc (a + b)2 (a + c)2 (a2 + bc) We can also prove this inequality using the Cauchy-Schwarz inequality, as follows 1 1 1 (c + b)2 1 + − ≥ − (a + b)2 (a + c)2 a2 + bc c 2 (a + b)2 + b2 (a + c)2 a2 + bc bc[2a2 − 2a(b + c) + b2 + c 2 ] = 2 (a + bc)[c 2 (a + b)2 + b2 (a + c)2 ] bc[(2a − b − c)2 + (b − c)2 ] = ≥ 0. 2(a2 + bc)[c 2 (a + b)2 + b2 (a + c)2 ]

37

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P 1.8. Let a, b, c be positive real numbers. Prove that 1 1 2a 2b 2c 1 + + ≥ 2 + 2 + 2 . b+c c+a a+b 3a + bc 3b + ca 3c + a b (Vasile Cîrtoaje, 2005) Solution. Since X

‹ X X 1 1 2a 2a − = − b+c 3a2 + bc b + c 3a2 + bc X (a − b)(a − c) + a(2a − b − c) = , (b + c)(3a2 + bc)

it suffices to show that X

(a − b)(a − c) ≥0 (b + c)(3a2 + bc)

X

a(2a − b − c) ≥ 0. (b + c)(3a2 + bc)

and

In order to prove the first inequality, assume that a = min{a, b, c}. Since (a − b)(a − c) ≥ 0, it is enough to show that (b − c)(b − a) (c − a)(c − b) + ≥ 0. (c + a)(3b2 + ca) (a + b)(3c 2 + a b) This is equivalent to the obvious inequality a(b − c)2 (b2 + c 2 − a2 + 3a b + bc + 3ca) ≥ 0. The second inequality is also true, since X

X a(a − b) + a(a − c) a(2a − b − c) = (b + c)(3a2 + bc) (b + c)(3a2 + bc) X X a(a − b) b(b − a) + = 2 (b + c)(3a + bc) (c + a)(3b2 + ca) • ˜ X a b = (a − b) − (b + c)(3a2 + bc) (c + a)(3b2 + ca) X c(a − b)2 [(a − b)2 + c(a + b)] = ≥ 0. (b + c)(c + a)(3a2 + bc)(3b2 + ca)

The equality holds for a = b = c.

Symmetric Rational Inequalities

39

P 1.9. Let a, b, c be positive real numbers. Prove that (a) (b)

b c 13 2(a b + bc + ca) a + + ≥ − ; b+c c+a a+b 6 3(a2 + b2 + c 2 )  ‹ p a b c 3 a b + bc + ca + + − ≥ ( 3 − 1) 1 − 2 . b+c c+a a+b 2 a + b2 + c 2 (Vasile Cîrtoaje, 2006)

Solution. (a) We use the SOS method. Rewrite the inequality as  ‹ a a b + bc + ca b c 3 2 1− 2 . + + − ≥ b+c c+a a+b 2 3 a + b2 + c 2 Since ‹ X X a (a − b) + (a − c) 1 − = b+c 2 2(b + c) X a−b X b−a = + 2(b + c) 2(c + a) ‹ X a−b  1 1 = − 2 b+c c+a X (a − b)2 = 2(b + c)(c + a) and

 ‹ X (a − b)2 2 a b + bc + ca 1− 2 = , 3 a + b2 + c 2 3(a2 + b2 + c 2 ) the inequality can be restated as • ˜ X 1 1 − ≥ 0. (a − b)2 2(b + c)(c + a) 3(a2 + b2 + c 2 ) This is true, since 3(a2 + b2 + c 2 ) − 2(b + c)(c + a) = (a + b − c)2 + 2(a − b)2 ≥ 0. The equality holds for a = b = c. (b) Let p = a + b + c,

q = a b + bc + ca,

r = a bc.

We have X

 X a X 1 a = +1 −3= p −3 b+c b+c b+c p(p2 + q) = − 3. pq − r

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Vasile Cîrtoaje

According to P 3.57-(a) in Volume 1, for fixed p and q, the product r is minimal when a = 0 or b = c. Therefore, it suffices to prove the inequality for a = 0 and for b = c = 1. Case 1: a = 0. The original inequality can be written as  ‹ p bc b c 3 + − ≥ ( 3 − 1) 1 − 2 . c b 2 b + c2 It suffices to show that

bc b c 3 + − ≥1− 2 . c b 2 b + c2

Denoting t=

b2 + c 2 , bc

t ≥ 2,

this inequality becomes t−

3 1 ≥1− , 2 t

(t − 2)(2t − 1) ≥ 0. Case 2: b = c = 1. The original inequality becomes as follows:  ‹ p a 2 3 2a + 1 + − ≥ ( 3 − 1) 1 − 2 , 2 a+1 2 a +2 p (a − 1)2 ( 3 − 1)(a − 1)2 ≥ , 2(a + 1) a2 + 2 p (a − 1)2 (a − 3 + 1)2 ≥ 0. a The equality holds for a = b = c, and for p = b = c (or any cyclic permutation). 3−1

P 1.10. Let a, b, c be positive real numbers. Prove that  ‹2 1 1 a+b+c 1 . + + ≤ a2 + 2bc b2 + 2ca c 2 + 2a b a b + bc + ca (Vasile Cîrtoaje, 2006) First Solution. Write the inequality as ‹ X  a b + bc + ca (a + b + c)2 −3≥ − 1 , a b + bc + ca a2 + 2bc

Symmetric Rational Inequalities

41

(a − b)2 + (b − c)2 + (a − b)(b − c) X (a − b)(a − c) + ≥ 0. a b + bc + ca a2 + 2bc Without loss of generality, assume that a ≥ b ≥ c. Since (a − b)(a − c) ≥ 0 and (c − a)(c − b) ≥ 0, it suffices to show that (a − b)2 + (b − c)2 + (a − b)(b − c) −

(a b + bc + ca)(a − b)(b − c) ≥ 0. b2 + 2ca

This inequality is equivalent to (a − b)2 + (b − c)2 −

(a − b)2 (b − c)2 ≥ 0, b2 + 2ca

or

c(a − b)2 (2a + 2b − c) ≥ 0, b2 + 2ca which is clearly true. The equality holds for a = b = c. Second Solution. Assume that a ≥ b ≥ c and write the desired inequality as ‹ X  a b + bc + ca (a + b + c)2 −3≥ −1 , a b + bc + ca a2 + 2bc X X (a − b)(a − c) 1 (a − b)(a − c) + ≥ 0, a b + bc + ca a2 + 2bc  ‹ X a b + bc + ca 1+ (a − b)(a − c) ≥ 0. a2 + 2bc Since (c − a)(c − b) ≥ 0 and a − b ≥ 0, it suffices to prove that ‹  ‹  a b + bc + ca a b + bc + ca (a − c) + 1 + (c − b) ≥ 0. 1+ a2 + 2bc b2 + 2ca (b − c)2 +

Write this inequality as ‹ a−c c−b a − b + (a b + bc + ca) 2 + ≥ 0, a + 2bc b2 + 2ca   (a b + bc + ca)(3ac + 3bc − a b − 2c 2 (a − b) 1 + ≥ 0. (a2 + 2bc)(b2 + 2ca) 

Since a − b ≥ 0 and 2ac + 3bc − 2c 2 > 0, it is enough to show that 1+

(a b + bc + ca)(ac − a b) ≥ 0. (a2 + 2bc)(b2 + 2ca)

We have 1+

(a b + bc + ca)(ac − a b) (a b + bc + ca)(ac − a b) ≥1+ (a2 + 2bc)(b2 + 2ca) a2 (b2 + ca) 2 (a + b)c + (a2 − b2 )c = > 0. a(b2 + ca)

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P 1.11. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 (b + c) b2 (c + a) c 2 (a + b) + 2 + 2 ≥ a + b + c. b2 + c 2 c + a2 a + b2 (Darij Grinberg, 2004) First Solution. We use the SOS method. We have  X a2 (b + c) X X  a2 (b + c) − a= −a b2 + c 2 b2 + c 2 X a b(a − b) + ac(a − c) = b2 + c 2 X a b(a − b) X ba(b − a) = + b2 + c 2 c 2 + a2 X a b(a + b)(a − b)2 = ≥ 0. (b2 + c 2 )(c 2 + a2 ) The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation). Second Solution. By virtue of the Cauchy-Schwarz inequality, we have P X a2 (b + c) [ a2 (b + c)]2 ≥P . b2 + c 2 a2 (b + c)(b2 + c 2 ) Then, it suffices to show that X X X [ a2 (b + c)]2 ≥ ( a)[ a2 (b + c)(b2 + c 2 )]. Let p = a + b + c and q = a b + bc + ca. Since X [ a2 (b + c)]2 = (pq − 3a bc)2 = p2 q2 − 6a bc pq + 9a2 b2 c 2 and X

X (b + c)[(a2 b2 + b2 c 2 + c 2 a2 ) − b2 c 2 ] X = 2p(a2 b2 + b2 c 2 + c 2 a2 ) − b2 c 2 (p − a)

a2 (b + c)(b2 + c 2 ) =

= p(a2 b2 + b2 c 2 + c 2 a2 ) + a bcq = p(q2 − 2a bc p) + a bcq, the inequality can be written as a bc(2p3 + 9a bc − 7pq) ≥ 0.

Symmetric Rational Inequalities

43

Using Schur’s inequality p3 + 9a bc − 4pq ≥ 0, we have 2p3 + 9a bc − 7pq ≥ p(p2 − 3q) ≥ 0.

P 1.12. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 3(a2 + b2 + c 2 ) a2 + b2 b2 + c 2 c 2 + a2 + + ≤ . a+b b+c c+a a + b + c) Solution. We use the SOS method. First Solution. Multiplying by 2(a + b + c), the inequality successively becomes X a  2 1+ (b + c 2 ) ≤ 3(a2 + b2 + c 2 ), b+c X X a (b2 + c 2 ) ≤ a2 , b+c  X  b2 + c 2 a a− ≥ 0, b+c X a b(a − b) − ac(c − a) ≥ 0, b+c X a b(a − b) X ba(a − b) − ≥ 0, b+c c+a X a b(a − b)2 ≥ 0. (b + c)(c + a) The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation). Second Solution. Subtracting a + b + c from the both sides, the desired inequality becomes as follows: X  a2 + b2 a + b  3(a2 + b2 + c 2 ) − (a + b + c) ≥ − , a+b+c a+b 2 X (a − b)2 X (a − b)2 ≥ , a+b+c 2(a + b) X (a + b − c)(a − b)2 ≥ 0. a+b

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Without loss of generality, assume that a ≥ b ≥ c. Since a + b − c ≥ 0, it suffices to prove that (a − b − c)(b − c)2 (a + c − b)(a − c)2 ≥ . a+c b+c a−c b−c This inequality is true because a + c − b ≥ a − b − c, a − c ≥ b − c and ≥ . a+c b+c The last inequality reduces to c(a − b) ≥ 0. Third Solution. Write the inequality as follows  X  3(a2 + b2 ) a2 + b2 − ≥ 0, 2(a + b + c) a+b X (a2 + b2 )(a + b − 2c)

≥ 0, a+b X (a2 + b2 )(a − c) X (a2 + b2 )(b − c) + ≥ 0, a+b a+b X (a2 + b2 )(a − c) X (b2 + c 2 )(c − a) + ≥ 0, a+b b+c X (a − c)2 (a b + bc + ca − b2 ) ≥ 0. (a + b)(b + c) It suffices to prove that X (a − c)2 (a b + bc − ca − b2 ) (a + b)(b + c)

≥ 0.

Since a b + bc − ca − b2 = (a − b)(b − c), this inequality is equivalent to (a − b)(b − c)(c − a) which is true because X

X

c−a ≥ 0, (a + b)(b + c)

c−a = 0. (a + b)(b + c)

P 1.13. Let a, b, c be positive real numbers. Prove that a2

1 1 1 9 + 2 + 2 ≥ . 2 2 2 + ab + b b + bc + c c + ca + a (a + b + c)2 (Vasile Cîrtoaje, 2000)

Symmetric Rational Inequalities

45

First Solution. Due to homogeneity, we may assume that a + b + c = 1. Let q = a b + bc + ca. Since b2 + bc + c 2 = (a + b + c)2 − a(a + b + c) − (a b + bc + ca) = 1 − a − q, we can write the inequality as X

1 ≥ 9, 1−a−q

9q3 − 6q2 − 3q + 1 + 9a bc ≥ 0. From Schur’s inequality (a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca), we get 1 + 9a bc − 4q ≥ 0. Therefore, 9q3 − 6q2 − 3q + 1 + 9a bc = (1 + 9a bc − 4q) + q(3q − 1)2 ≥ 0. The equality holds for a = b = c. Second Solution. Multiplying by a2 + b2 +c 2 +a b+ bc +ca, the inequality can be written as X a 9(a b + bc + ca) (a + b + c) + ≥ 6. 2 2 b + bc + c (a + b + c)2 By the Cauchy-Schwarz inequality, we have X

a (a + b + c)2 a+b+c P ≥ = . 2 2 2 2 b + bc + c a b + bc + ca a(b + bc + c )

Then, it suffices to show that (a + b + c)2 9(a b + bc + ca) + ≥ 6. a b + bc + ca (a + b + c)2 This follows immediately from the AM-GM inequality.

P 1.14. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 b2 c2 1 + + ≤ . (2a + b)(2a + c) (2b + c)(2b + a) (2c + a)(2c + b) 3 (Tigran Sloyan, 2005)

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Vasile Cîrtoaje

First Solution. The inequality is equivalent to each of the inequalities  X a a2 − ≤ 0, (2a + b)(2a + c) 3(a + b + c) X a(a − b)(a − c) ≥ 0. (2a + b)(2a + c) Due to symmetry, we may consider that a ≥ b ≥ c. Since c(c − a)(c − b) ≥ 0, it suffices to prove that b(b − c)(b − a) a(a − b)(a − c) + ≥ 0. (2a + b)(2a + c) (2b + c)(2b + a) This is equivalent to the obvious inequality (a − b)2 [(a + b)(2a b − c 2 ) + c(a2 + b2 + 5a b)] ≥ 0. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation). Second Solution (by Vo Quoc Ba Can). Apply the Cauchy-Schwarz inequality in the following manner 9a2 (2a + a)2 2a a2 = ≤ + . (2a + b)(2a + c) 2a(a + b + c) + (2a2 + bc) a + b + c 2a2 + bc Then, X

X a2 9a2 ≤2+ , (2a + b)(2a + c) 2a2 + bc

and from the known inequality X

a2 ≤ 1, 2a2 + bc

the conclusion follows. The last inequality is equivalent to X

bc ≥ 1, 2a2 + bc

and can be obtained using the Cauchy-Schwarz inequality, as follows P X ( bc)2 bc ≥P = 1. 2a2 + bc bc(2a2 + bc) Remark. From the inequality in P 1.14 and Hölder’s inequality X  ”X Æ —2 a2 a(2a + b)(2a + c) ≥ (a + b + c)3 , (2a + b)(2a + c)

Symmetric Rational Inequalities

47

we get the following result: • If a, b, c are nonnegative real numbers such that a + b + c = 3, then Æ Æ Æ a(2a + b)(2a + c) + b(2b + c)(2b + a) + c(2c + a)(2c + bc) ≥ 9, 3 3 with equality for a = b = c = 1, and for (a, b, c) = (0, , ) or any cyclic permutation. 2 2

P 1.15. Let a, b, c be positive real numbers. Prove that (a)

(b)

P

P

a 1 ≤ ; (2a + b)(2a + c) a+b+c

1 a3 ≤ . (2a2 + b2 )(2a2 + c 2 ) a+b+c (Vasile Cîrtoaje, 2005)

Solution. (a) Write the inequality as ˜ X•1 a(a + b + c) − ≥ 0, 3 (2a + b)(2a + c) X (a − b)(a − c) ≥ 0. (2a + b)(2a + c) Assume that a ≥ b ≥ c. Since (a − b)(a − c) ≥ 0, it suffices to prove that (b − c)(b − a) (a − c)(b − c) + ≥ 0. (2b + c)(2b + a) (2c + a)(2c + b) Since b − c ≥ 0 and a − c ≥ a − b ≥ 0, it is enough to show that 1 1 ≥ . (2c + a)(2c + b) (2b + c)(2b + a) This is equivalent to the obvious inequality (b − c)(a + 4b + 4c) ≥ 0. The equality holds for a = b = c. (b) We obtain the desired inequality by summing the inequalities a3 a ≥ , 2 2 2 2 (2a + b )(2a + c ) (a + b + c)2

48

Vasile Cîrtoaje b b3 ≥ , 2 2 2 2 (2b + c )(2b + a ) (a + b + c)2 c c3 ≥ , 2 2 2 2 (2c + a )(2c + b ) (a + b + c)2

which are consequences of the Cauchy-Schwarz inequality. For example, from (a2 + a2 + b2 )(c 2 + a2 + a2 ) ≥ (ac + a2 + ba)2 , the first inequality follows. The equality holds for a = b = c.

P 1.16. If a, b, c are positive real numbers, then X

1 1 2 ≥ + . (a + 2b)(a + 2c) (a + b + c)2 3(a b + bc + ca)

Solution. Write the inequality as follows ˜ X• 1 1 2 2 − ≥ − , 2 (a + 2b)(a + 2c) (a + b + c) 3(a b + bc + ca) (a + b + c)2 X (b − c)2 (b − c)2 ≥ , (a + 2b)(a + 2c) 3(a b + bc + ca) X b−c (a − b)(b − c)(c − a) ≥ 0. (a + 2b)(a + 2c) X

Since X

˜ X• b−c b−c b−c = − (a + 2b)(a + 2c) (a + 2b)(a + 2c) 3(a b + bc + ca) (a − b)(b − c)(c − a) X 1 = , 3(a b + bc + ca) (a + 2b)(a + 2c)

the desired inequality is equivalent to the obvious inequality (a − b)2 (b − c)2 (c − a)2

X

1 ≥ 0. (a + 2b)(a + 2c)

The equality holds for a = b, or b = c, or c = a.

Symmetric Rational Inequalities

49

P 1.17. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 1 1 4 1 + + ≥ ; 2 2 2 (a − b) (b − c) (c − a) a b + bc + ca

(a)

(b)

1 1 1 3 + + ≥ ; a2 − a b + b2 b2 − bc + c 2 c 2 − ca + a2 a b + bc + ca

(c)

a2

1 1 1 5 + 2 + 2 ≥ . 2 2 2 +b b +c c +a 2(a b + bc + ca)

Solution. Let Ek (a, b, c) =

a b + bc + ca a b + bc + ca a b + bc + ca + 2 + 2 . 2 2 2 a − ka b + b b − k bc + c c − kca + a2

We will prove that Ek (a, b, c) ≥ αk , where

 5 − 2k  , 0≤k≤1  2−k αk = .  1≤k≤2  2 + k,

To show this, assume that a ≤ b ≤ c and prove that Ek (a, b, c) ≥ Ek (0, b, c) ≥ αk . For the nontrivial case a > 0, the left inequality is true because Ek (a, b, c) − Ek (0, b, c) = a b2 + (1 + k)bc − ac c 2 + (1 + k)bc − a b b+c = + + b(a2 − ka b + b2 ) b2 − k bc + c 2 c(c 2 − kca + a2 ) bc − ac b+c bc − a b > + 2 + > 0. 2 2 2 2 b(a − ka b + b ) b − k bc + c c(c − kca + a2 ) In order to prove the right inequality Ek (0, b, c) ≥ αk , where Ek (0, b, c) =

b2

bc b c + + , 2 − k bc + c c b

by virtue of the AM-GM inequality, we have Ek (0, b, c) =

bc b2 − k bc + c 2 + + k ≥ 2 + k. b2 − k bc + c 2 bc

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Vasile Cîrtoaje

Thus, it remains to consider the case 0 ≤ k ≤ 1. We have Ek (0, b, c) =

b2 − k bc + c 2 bc + b2 − k bc + c 2 (2 − k)2 bc • ˜ ‹ 1 b c k + 1− + + 2 (2 − k) c b (2 − k)2 • ˜ 2 1 k 5 − 2k ≥ +2 1− + = . 2−k (2 − k)2 (2 − k)2 2−k

b c + = 1 + k (or any cyclic c b permutation). For 0 ≤ k ≤ 1, the equality holds when a = 0 and b = c (or any cyclic permutation).

For 1 ≤ k ≤ 2, the equality holds when a = 0 and

P 1.18. Let a, b, c be positive real numbers, no two of which are zero. Prove that (a2 + b2 )(a2 + c 2 ) (b2 + c 2 )(b2 + a2 ) (c 2 + a2 )(c 2 + b2 ) + + ≥ a2 + b2 + c 2 . (a + b)(a + c) (b + c)(b + a) (c + a)(c + b) (Vasile Cîrtoaje, 2011) Solution. Using the identity (a2 + b2 )(a2 + c 2 ) = a2 (a2 + b2 + c 2 ) + b2 c 2 , we can write the inequality as follows X

  X b2 c 2 a2 2 2 2 ≥ (a + b + c ) 1 − , (a + b)(a + c) (a + b)(a + c) X b2 c 2 (b + c) ≥ 2a bc(a2 + b2 + c 2 ), X X a3 (b2 + c 2 ) ≥ 2 a3 bc, X a3 (b − c)2 ≥ 0.

Since the last form is obvious, the proof is completed. The equality holds for a = b = c.

P 1.19. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that a2

1 1 1 + 2 + 2 ≤ 1. +b+c b +c+a c +a+b

Symmetric Rational Inequalities

51

First Solution. By virtue of the Cauchy-Schwarz inequality, we have (a2 + b + c)(1 + b + c) ≥ (a + b + c)2 . Therefore, X

X 1+ b+c 1 3 + 2(a + b + c) ≤ = = 1. 2 2 a +b+c (a + b + c) (a + b + c)2

The equality occurs for a = b = c = 1. Second Solution. Rewrite the inequality as 1 1 1 + + ≤ 1. a2 − a + 3 b2 − b + 3 c 2 − c + 3 We see that the equality holds for a = b = c = 1. Thus, if there exists a real number k such that  ‹ 1 1 ≤k+ −k a a2 − a + 3 3 for all a ∈ [0, 3], then X

 ‹ ˜  ‹X X• 1 1 1 ≤ k+ − k a = 3k + −k a = 1. a2 − a + 3 3 3

We have ‹ 1 (a − 1)[(1 − 3k)a2 + 3ka + 3(1 − 3k)] 1 −k a− 2 = . k+ 3 a −a+3 3(a2 − a + 3) 

Setting k = 4/9, we get k+



‹ 1 (a − 1)2 (3 − a) 1 −k a− 2 = ≥ 0. 3 a −a+3 9(a2 − a + 3)

P 1.20. Let a, b, c be real numbers such that a + b + c = 3. Prove that a2 − bc b2 − ca c 2 − a b + 2 + 2 ≥ 0. a2 + 3 b +3 c +3 (Vasile Cîrtoaje, 2005)

52

Vasile Cîrtoaje

Solution. Apply the SOS method. We have 2

X a2 − bc a2 + 3

=

X (a − b)(a + c) + (a − c)(a + b)

a2 + 3 X (a − b)(a + c) X (b − a)(b + c) = + a2 + 3 b2 + 3  ‹ X a+c b+c − 2 = (a − b) 2 a +3 b +3 X (a − b)2 = (3 − a b − bc − ca) ≥ 0. (a2 + 3)(b2 + 3)

Thus, it suffices to show that 3 − a b − bc − ca ≥ 0. This follows immediately from the known inequality (a + b + c)2 ≥ 3(a b + bc + ca), which is equivalent to (a − b)2 + (b − c)2 + (c − a)2 ≥ 0. The equality holds for a = b = c = 1.

P 1.21. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 − bc 1 − ca 1 − a b + + ≥ 0. 5 + 2a 5 + 2b 5 + 2c Solution. Since 9(1 − bc) = (a + b + c)2 − 9bc, we can write the inequality as X a2 + b2 + c 2 + 2a(b + c) − 7bc 5 + 2a

≥ 0.

From (a − b)(a + k b + mc) + (a − c)(a + kc + mb) = = 2a2 − k(b2 + c 2 ) + (k + m − 1)a(b + c) − 2mbc, choosing k = −2 and m = 7, we get (a − b)(a − 2b + 7c) + (a − c)(a − 2c + 7b) = 2[a2 + b2 + c 2 + 2a(b + c) − 7bc]. Therefore, the desired inequality becomes as follows: X (a − b)(a − 2b + 7c) 5 + 2a X (a − b)(a − 2b + 7c) 5 + 2a

+ +

X (a − c)(a − 2c + 7b) 5 + 2a X (b − a)(b − 2a + 7c) 5 + 2b

≥ 0, ≥ 0,

Symmetric Rational Inequalities

53

X (a − b)(5 + 2c)[(5 + 2b)(a − 2b + 7c) − (5 + 2a)(b − 2a + 7c)] ≥ 0, X (a − b)2 (5 + 2c)(15 + 4a + 4b − 14c) ≥ 0, X (a − b)2 (5 + 2c)(a + b − c) ≥ 0. Without loss of generality, assume that a ≥ b ≥ c. Clearly, it suffices to show that (a − c)2 (5 + 2b)(a + c − b) ≥ (b − c)2 (5 + 2a)(a − b − c). Since a − c ≥ b − c ≥ 0 and a + c − b ≥ a − b − c, it suffices to prove that (a − c)(5 + 2b) ≥ (b − c)(5 + 2a). Indeed, (a − c)(5 + 2b) − (b − c)(5 + 2a) = (a − b)(5 + 2c) ≥ 0. The equality holds for a = b = c = 1, and for c = 0 and a = b = 3/2 (or any cyclic permutation).

P 1.22. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that a2

1 1 1 3 + 2 + 2 ≤ . 2 2 2 + b +2 b +c +2 c +a +2 4 (Vasile Cîrtoaje, 2006)

Solution. Since

1 a2 + b2 1 = − , a2 + b2 + 2 2 a2 + b2 + 2

we write the inequality as a2 + b2 b2 + c 2 c 2 + a2 3 + + ≥ . 2 2 2 2 2 2 a + b +2 b +c +2 c +a +2 2 By the Cauchy-Schwarz inequality, we have Pp P Pp X a2 + b2 ( a2 + b2 )2 2 a2 + 2 (a2 + b2 )(a2 + c 2 ) P P ≥ = a2 + b2 + 2 (a2 + b2 + 2) 2 a2 + 6 P 2 P 2 P 2 2 a + 2 (a + bc) 3 a + 9 3 P ≥ = P = . 2 a2 + 6 2 a2 + 6 2 The equality holds for a = b = c = 1

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Vasile Cîrtoaje

P 1.23. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that 4a2

1 1 1 1 + 2 + 2 ≤ . 2 2 2 2 2 2 +b +c 4b + c + a 4c + a + b 2 (Vasile Cîrtoaje, 2007)

Solution. According to the Cauchy-Schwarz inequality, we have 9 (a + b + c)2 = 4a2 + b2 + c 2 2a2 + (a2 + b2 ) + (a2 + c 2 ) 1 b2 c2 ≤ + 2 + . 2 a + b2 a2 + c 2 Therefore, X

  b2 9 3 X c2 ≤ + + 4a2 + b2 + c 2 2 a2 + b2 a2 + c 2   b2 3 X a2 3 9 = + + = +3= . 2 a2 + b2 b2 + a2 2 2

The equality holds for a = b = c = 1.

P 1.24. Let a, b, c be nonnegative real numbers such that a + b + c = 2. Prove that ca ab bc + 2 + 2 ≤ 1. +1 b +1 c +1

a2

(Pham Kim Hung, 2005) Solution. Let p = a + b + c and q = a b + bc + ca, q ≤ p2 /3 = 4/3. If one of a, b, c is zero, then the inequality is true. Otherwise, write the inequality as 1 1 ≤ , + 1) a bc ‹ X1 a 1 − 2 ≤ , a a +1 a bc X a 1 1 1 1 ≥ + + − , 2 a +1 a b c a bc X a q−1 ≥ , 2 a +1 a bc X

a(a2

Symmetric Rational Inequalities

55

Using the inequality a2

2 ≥ 2 − a, +1

which is equivalent to a(a − 1)2 ≥ 0, we get X

X a(2 − a) X a(b + c) a ≥ = = q. a2 + 1 2 2

Therefore, it suffices to prove that 1 + a bcq ≥ q. Since a4 + b4 + c 4 = (a2 + b2 + c 2 )2 − 2(a2 b2 + b2 c 2 + c 2 a2 ) = (p2 − 2q)2 − 2(q2 − 2a bc p) = p4 − 4p2 q + 2q2 + 4a bc p by Schur’s inequality of degree four a4 + b4 + c 4 + 2a bc(a + b + c) ≥ (a b + bc + ca)(a2 + b2 + c 2 ), we get a bc ≥

(p2 − q)(4q − p2 ) , 6p

a bc ≥

(4 − q)(q − 1) . 3

Thus 1 + a bcq − q ≥ 1 +

q(4 − q)(q − 1) (3 − q)(q − 1)2 −q = ≥ 0. 3 3

The equality holds if a = 0 and b = c = 1 (or any cyclic permutation).

P 1.25. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that bc ca ab 1 + + ≤ . a+1 b+1 c+1 4 (Vasile Cîrtoaje, 2009)

56

Vasile Cîrtoaje

First Solution. We have X X bc bc = a+1 (a + b) + (c + a)  ‹ 1X 1 1 ≤ bc + 4 a+b c+a 1 X bc 1 X bc + = 4 a+b 4 c+a 1 X bc 1 X ca = + 4 a+b 4 a+b 1X 1 X bc + ca 1 = = c= . 4 a+b 4 4 The equality holds for a = b = c = 1/3, and for a = 0 and b = c = 1/2 (or any cyclic permutation). Second Solution. It is easy to check that the inequality is true if one of a, b, c is zero. Otherwise, write the inequality as 1 1 1 1 + + ≤ . a(a + 1) b(b + 1) c(c + 1) 4a bc Since

1 1 1 = − , a(a + 1) a a+1

we can write the required inequality as 1 1 1 1 1 1 1 + + ≥ + + − . a+1 b+1 c+1 a b c 4a bc In virtue of the Cauchy-Schwarz inequality, 1 1 1 9 9 + + ≥ = . a + 1 b + 1 c + 1 (a + 1) + (b + 1) + (c + 1) 4 Therefore, it suffices to prove that 9 1 1 1 1 ≥ + + − . 4 a b c 4a bc This is equivalent to Schur’s inequality (a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca).

Symmetric Rational Inequalities

57

P 1.26. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that 1 1 3 1 + + ≤ . 2 2 2 a(2a + 1) b(2b + 1) c(2c + 1) 11a bc (Vasile Cîrtoaje, 2009) Solution. Since

1 2a 1 = − 2 , 2 a(2a + 1) a 2a + 1

we can write the inequality as X 2a 1 1 1 3 ≥ + + − . 2a2 + 1 a b c 11a bc By the Cauchy-Schwarz inequality, we have P X 2a 2( a)2 2 ≥P = . 2 3 3 2 2a + 1 a(2a + 1) 2(a + b + c 3 ) + 1 Therefore, it suffices to show that 2(a3 where q = a b + bc + ca, q ≤

11q − 3 2 ≥ , 3 + +c )+1 11a bc b3

1 1 (a + b + c)2 = . Since 3 3

a3 + b3 + c 3 = 3a bc + (a + b + c)3 − 3(a + b + c)(a b + bc + ca) = 3a bc + 1 − 3q, we need to prove that 22a bc ≥ (11q − 3)(6a bc + 3 − 6q), or, equivalently, 2(20 − 33q)a bc ≥ 3(11q − 3)(1 − 2q). From Schur’s inequality (a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca), we get 9a bc ≥ 4q − 1. Thus, 2(20 − 33q)a bc − 3(11q − 3)(1 − 2q) ≥ 2(20 − 33q)(4q − 1) ≥ − 3(11q − 3)(1 − 2q) 9 330q2 − 233q + 41 (1 − 3q)(41 − 110q) = = ≥ 0. 9 9 This completes the proof. The equality holds for a = b = c = 1/3.

58

Vasile Cîrtoaje

P 1.27. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that a3

1 1 1 + 3 + 3 ≤ 1. +b+c b +c+a c +a+b (Vasile Cîrtoaje, 2009)

Solution. Write the inequality in the form 1 1 1 + + ≤ 1. a3 − a + 3 b3 − b + 3 c 3 − c + 3 Assume that a ≥ b ≥ c. There are two cases to consider. Case 1: c ≤ b ≤ a ≤ 2. The desired inequality follows by adding the inequalities a3

1 5 − 2a 1 5 − 2b 1 5 − 2c ≤ , 3 ≤ , 3 ≤ . −a+3 9 b − b+3 9 c −c+3 9

Indeed, we have 1 5 − 2a (a − 1)2 (a − 2)(2a + 3) − = ≤ 0. a3 − a + 3 9 9(a3 − a + 3) Case 2: a > 2. From a + b + c = 3, we get b + c < 1. Since X a3

1 1 1 1 1 1 1 < 3 + + < + + , −a+3 a −a+3 3− b 3−c 9 3− b 3−c

it suffices to prove that 1 1 8 + ≤ . 3− b 3−c 9 We have

1 1 8 −3 − 15(1 − b − c) − 8bc + − = < 0. 3− b 3−c 9 9(3 − b)(3 − c)

The equality holds for a = b = c = 1.

P 1.28. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that a2 b2 c2 + + ≥ 1. 1 + b3 + c 3 1 + c 3 + a3 1 + a3 + b3

Symmetric Rational Inequalities

59

Solution. Using the Cauchy-Schwarz inequality, we have P X ( a2 )2 a2 ≥P , 1 + b3 + c 3 a2 (1 + b3 + c 3 ) and it remains to show that (a2 + b2 + c 2 )2 ≥ (a2 + b2 + c 2 ) +

X

a2 b2 (a + b).

Let p = a + b + c and q = a b + bc + ca, q ≤ 3. Since a2 + b2 + c 2 = 9 − 2q and X X X a2 b2 (a + b) = a2 b2 (3 − c) = 3 a2 b2 − qa bc = 3q2 − (q + 18)a bc, the desired inequality can be written as q2 − 34q + 72 + (q + 18)a bc ≥ 0. This inequality is clearly true for q ≤ 2. Consider further that 2 < q ≤ 3. Since a4 + b4 + c 4 = (a2 + b2 + c 2 )2 − 2(a2 b2 + b2 c 2 + c 2 a2 ) = (p2 − 2q)2 − 2(q2 − 2a bc p) = p4 − 4p2 q + 2q2 + 4a bc p by Schur’s inequality of degree four a4 + b4 + c 4 + 2a bc(a + b + c) ≥ (a b + bc + ca)(a2 + b2 + c 2 ), we get a bc ≥

(p2 − q)(4q − p2 ) (9 − q)(4q − 9) = . 6p 18

Therefore (q + 18)(9 − q)(4q − 9) 18 (3 − q)(4q2 + 21q − 54) = ≥ 0. 18

q2 − 34q + 72 + (q + 18)a bc ≥ q2 − 34q + 72 +

The equality holds for a = b = c = 1.

P 1.29. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 1 1 3 + + ≤ . 6 − a b 6 − bc 6 − ca 5

60

Vasile Cîrtoaje

Solution. Rewrite the inequality as 108 − 48(a b + bc + ca) + 13a bc(a + b + c) − 3a2 b2 c 2 ≥ 0, 4[9 − 4(a b + bc + ca) + 3a bc] + a bc(1 − a bc) ≥ 0. By the AM-GM inequality, a+b+c 3 Consequently, it suffices to show that 1=



‹3

≥ a bc.

9 − 4(a b + bc + ca) + 3a bc ≥ 0. We see that the homogeneous form of this inequality is just Schur’s inequality of third degree (a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca). The equality holds for a = b = c = 1, as well as for a = 0 and b = c = 3/2 (or any cyclic permutation).

P 1.30. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 1 1 1 + 2 + 2 ≤ . + 7 2b + 7 2c + 7 3

2a2

(Vasile Cîrtoaje, 2005) Solution. Assume that a = max{a, b, c} and prove that E(a, b, c) ≤ E(a, s, s) ≤ where s= and E(a, b, c) =

1 , 3

b+c , 0 ≤ s ≤ 1, 2

1 1 1 + + . 2a2 + 7 2b2 + 7 2c 2 + 7

We have ‹  ‹ 1 1 1 1 − + − 2s2 + 7 2b2 + 7 2s2 + 7 2c 2 + 7 • ˜ 1 (b − c)(b + s) (c − b)(c + s) = 2 + 2s + 7 2b2 + 7 2c 2 + 7 2 2 (b − c) (7 − 4s − 2bc) = . (2s2 + 7)(2b2 + 7)(2c 2 + 7)

E(a, s, s) − E(a, b, c) =



Symmetric Rational Inequalities

61

Since bc ≤ s2 ≤ 1, it follows that 7 − 4s2 − 2bc > 0, and hence E(a, s, s) ≥ E(a, b, c). Also, 1 1 4(s − 1)2 (2s − 1)2 − E(a, s, s) = − E(3 − 2s, s, s) = ≥ 0. 3 3 3(2a2 + 7)(2s2 + 7) The equality holds for a = b = c = 1, as well as for a = 2 and b = c = 1/2 (or any cyclic permutation).

P 1.31. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ 1 ≥ c and a + b + c = 3. Prove that 1 1 3 1 + + ≤ . a2 + 3 b2 + 3 c 2 + 3 4 (Vasile Cîrtoaje, 2005) First Solution (by Nguyen Van Quy). Write the inequality as follows:  ‹  ‹  ‹ 1 3−a 1 3− b 3−c 1 − + − ≤ − 2 , a2 + 3 8 b2 + 3 8 8 c +3 (a − 1)3 (b − 1)3 (1 − c)3 + ≤ . a2 + 3 b2 + 3 c2 + 3 Indeed, we have (1 − c)3 (a − 1 + b − 1)3 (a − 1)3 + (b − 1)3 (a − 1)3 (b − 1)3 = ≥ ≥ + 2 . c2 + 3 c2 + 3 c2 + 3 a2 + 3 b +3 The proof is completed. The equality holds for a = b = c = 1. Second Solution. Let d be a positive number such that c + d = 2. We have a + b = 1 + d, In addition, we claim that c2 Indeed,

d ≥ a ≥ b ≥ 1.

1 1 1 + 2 ≤ . +3 d +3 2

1 1 1 (cd − 1)2 − 2 − 2 = ≥ 0. 2 c + 3 d + 3 2(c 2 + 3)(d 2 + 3)

Thus, it suffices to show that 1 1 1 1 + ≤ 2 + . a2 + 3 b2 + 3 d +3 4

62

Vasile Cîrtoaje

Since a2

1 1 (d − a)(d + a) − 2 = 2 , + 3 d + 3 (a + 3)(d 2 + 3)

we need to prove that (a2

1 1 (b − 1)(b + 1) − 2 = , 4 b +3 4(b2 + 3)

d+a b+1 ≤ . 2 + 3)(d + 3) 4(b2 + 3)

We can get this inequality by multiplying the inequalities a+1 d+a ≤ , 2 d +3 4 a+1 b+1 ≤ 2 . 2 a +3 b +3 We have

a+1 d+a (d − 1)(ad + a + d − 3) − 2 = ≥ 0, 4 d +3 4(d 2 + 3) b+1 a+1 (a − b)(a b + a + b − 3) − 2 = ≥ 0. 2 b +3 a +3 (a2 + 3)(b2 + 3)

P 1.32. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 1 1 3 + 2 + 2 ≥ . + 3 2b + 3 2c + 3 5

2a2

(Vasile Cîrtoaje, 2005) First Solution (by Nguyen Van Quy). Write the inequality as ‹ X1 1 2 − 2 ≤ , 3 2a + 3 5 a2 3 ≤ . 2 2a + 5 5 Using the Cauchy-Schwarz inequality gives X

25 25 = 3(2a2 + 3) 6a2 + (a + b + c)2 (2 + 2 + 1)2 = 2(2a2 + bc) + 2a(a + b + c) + a2 + b2 + c 2 22 22 1 ≤ + + 2 , 2 2(2a + bc) 2a(a + b + c) a + b2 + c 2

Symmetric Rational Inequalities

63

hence X

X 2a2 X X 25a2 2a a2 ≤ + + 3(2a2 + 3) 2a2 + bc a+b+c a2 + b2 + c 2 X 2a2 + 3. = 2a2 + bc

Therefore, it suffices to show that X

a2 ≤ 1, 2a2 + bc

which is equivalent to  a2 1 − 2 ≥ , 2 2a + bc 2 X bc ≥ 1. 2 2a + bc Using again the Cauchy-Schwarz inequality, we get P X ( bc)2 bc = 1. ≥P 2a2 + bc bc(2a2 + bc) X1

The equality holds for a = b = c = 1, as well as for a = 0 and b = c = 3/2 (or any cyclic permutation). Second Solution. First, we can check that the desired inequality becomes an equality for a = b = c = 1, and for a = 0 and b = c = 3/2. Consider then the inequality f (x) ≥ 0, where 1 f (x) = − A − B x. 2 2x + 3 We have −4x − B. f 0 (x) = (2x 2 + 3)2 From the conditions f (1) = 0 and f 0 (1) = 0, we get A = 9/25 and B = −4/25. Also, from the conditions f (3/2) = 0 and f 0 (3/2) = 0, we get A = 22/75 and B = −8/75. From these values of A and B, we obtain the identities 1 9 − 4x 2(x − 1)2 (4x − 1) − = , 2x 2 + 3 25 25(2x 2 + 3) 1 22 − 8x (2x − 3)2 (4x + 1) − = , 2x 2 + 3 75 75(2x 2 + 3) and the inequalities 1 9 − 4x ≥ , 2x 2 + 3 25

x≥

1 , 4

64

Vasile Cîrtoaje 22 − 8x 1 ≥ , 2x 2 + 3 75

x ≥ 0.

Without loss of generality, assume that a ≥ b ≥ c. 1 . By summing the inequalities 4

Case 1: a ≥ b ≥ c ≥

9 − 4a 1 ≥ , 2a2 + 3 25 we get

1 9 − 4b ≥ , 2b2 + 3 25

1 9 − 4c ≥ , 2c 2 + 3 25

1 1 27 − 4(a + b + c) 3 1 + 2 + 2 ≥ = . + 3 2b + 3 2c + 3 25 5

2a2 Case 2: a ≥ b ≥ X

1 ≥ c. We have 4 1 22 − 8a 22 − 8b 1 ≥ + + 2 +3 75 75 2c + 3 44 − 8(a + b) 1 20 + 8c 1 = + 2 = + 2 . 75 2c + 3 75 2c + 3

2a2

Therefore, it suffices to show that 1 3 20 + 8c + 2 ≥ , 75 2c + 3 5 which is equivalent to the obvious inequality c(8c 2 − 25c + 12) ≥ 0.

Case 3: a ≥

1 ≥ b ≥ c. We have 4 X 1 2a2

+3

>

1 1 + 2 ≥ + 3 2c + 3

2b2

2 1 8

+3

>

3 . 5

P 1.33. Let a, b, c be nonnegative real numbers such that a ≥ 1 ≥ b ≥ c and a + b + c = 3. Prove that 1 1 1 + 2 + 2 ≥ 1. 2 a +2 b +2 c +2 (Vasile Cîrtoaje, 2005)

Symmetric Rational Inequalities

65

First Solution. Let b1 and c1 be positive numbers such that b + b1 = 2,

c + c1 = 2.

We have b1 + c1 = 1 + a, 1 ≤ b1 ≤ c1 ≤ a. In addition, we claim that b2

1 1 2 + 2 ≥ , + 2 b1 + 2 3

c2

1 1 2 + 2 ≥ . + 2 c1 + 2 3

Indeed, 2b b1 (1 − b b1 ) b b1 (b − b1 )2 1 1 2 + = − = ≥ 0, b2 + 2 b12 + 2 3 3(b2 + 2)(b12 + 2) 6(b2 + 2)(b12 + 2) cc1 (c − c1 )2 1 2 1 + − = ≥ 0. c 2 + 2 c12 + 2 3 6(c 2 + 2)(c12 + 2) Using these inequalities, it suffices to show that a2

1 1 1 1 + ≥ 2 + 2 . +2 3 b1 + 2 c1 + 2

Since (b1 − 1)(b1 + 1) 1 1 = , − 2 3 b1 + 2 3(b12 + 2) we need to prove that

b1 + 1 3(b12

+ 2)

≥

1 c12

+2

−

a2

(a − c1 )(a + c1 ) 1 , = + 2 (a2 + 2)(c12 + 2)

a + c1 (a2

+ 2)(c12 + 2)

.

We can get this inequality by multiplying the inequalities b1 + 1 b12

+2

≥

c1 + 1 c12 + 2

,

c1 + 1 a + c1 ≥ 2 . 3 a +2 We have

b1 + 1 b12

+2

−

c1 + 1 c12

+2

=

(c1 − b1 )(b1 c1 + b1 + c1 − 2) (b12 + 2)(c12 + 2)

≥ 0,

c1 + 1 a + c1 (a − 1)(ac1 + a + c1 − 2) − 2 = ≥ 0. 3 a +2 3(a2 + 2)

66

Vasile Cîrtoaje

The proof is completed. The equality holds for a = b = c = 1, as well as for a = 2, b = 1 and c = 0. Second Solution. First, we can check that the desired inequality becomes an equality for a = b = c = 1, and also for a = 2, b = 1, c = 0. Consider then the inequality f (x) ≥ 0, where 1 f (x) = 2 − A − B x. x +2 We have −2x − B. f 0 (x) = 2 (x + 2)2 From the conditions f (1) = 0 and f 0 (1) = 0, we get A = 5/9 and B = −2/9. Also, from the conditions f 2) = 0 and f 0 (2) = 0, we get A = 7/18 and B = −1/9. From these values of A and B, we obtain the identities 1 5 − 2x (x − 1)2 (2x − 1) − = , x2 + 2 9 9(x 2 + 2) 7 − 2x (x − 2)2 (2x + 1) 1 − = , x2 + 2 18 18(x 2 + 2) and the inequalities 1 5 − 2x ≥ , x2 + 2 9 1 7 − 2x ≥ , 2 x +2 18 Let us define g(x) =

x≥

1 , 2

x ≥ 0.

1 . x2 + 2

p Notice that for d ∈ (0, 2] and x ∈ [0, d], we have g(x) ≥ g(0) + because g(x) − g(0) −

g(d) − g(0) x, d

g(d) − g(0) x(d − x)(2 − d x) x= ≥ 0. d 2(d 2 + 2)(x 2 + 2)

For d = 1/2 and d = 1, we get the inequalities x2

1 9 − 2x ≥ , +2 18

0≤ x ≤

1 , 2

1 3− x ≥ , 0 ≤ x ≤ 1. x2 + 2 6 Consider further two cases: c ≥ 1/2 and c ≤ 1/2.

Symmetric Rational Inequalities

Case 1: c ≥

67

1 . By summing the inequalities 2 1 5 − 2a ≥ , a2 + 2 9

1 5 − 2b ≥ , b2 + 2 9

1 5 − 2c ≥ , c2 + 2 9

we get a2 Case 2: c ≤

1 1 15 − 2(a + b + c) 1 + 2 + 2 ≥ = 1. +2 b +2 c +2 9

1 . We have 2 a2

1 7 − 2a ≥ , +2 18

1 3− b 8 − 2b ≥ ≥ , b2 + 2 6 18 9 − 2c 1 ≥ . c2 + 2 18 Therefore,

a2

1 1 1 7 − 2a 8 − 2b 9 − 2c + 2 + 2 ≥ + + = 1. +2 b +2 c +2 18 18 18

P 1.34. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that 1 1 a+b+c 3 1 + + ≥ + . a+b b+c c+a 6 a+b+c (Vasile Cîrtoaje, 2007) First Solution. Denoting x = a + b + c, we have 1 1 1 (a + b + c)2 + a b + bc + ca x2 + 3 + + = = . a+b b+c c+a (a + b + c)(a b + bc + ca) − a bc 3x − a bc Then, the inequality becomes x2 + 3 x 3 ≥ + , 3x − a bc 6 x or 3(x 3 + 9a bc − 12x) + a bc(x 2 − 9) ≥ 0.

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Vasile Cîrtoaje

This inequality is true since from (a + b + c)2 ≥ 3(a b + bc + ca), we get x 2 − 9 ≥ 0, and from Schur’s inequality of degree three (a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca), 3 we get xp + 9a bc − 12 ≥ 0. The equality holds for a = b = c = 1, and for a = 0 and b = c = 3 (or any cyclic permutation).

Second Solution. We apply the SOS method. Write the inequality as follows: 1 1 1 a+b+c 3 + + ≥ + , a+b b+c c+a 2(a b + bc + ca) a + b + c ‹ 1 1 (a + b + c)2 1 + + ≥ + 6, a+b b+c c+a a b + bc + ca  ‹ 1 1 1 (a + b + c)2 [(a + b) + (b + c) + (c + a)] + + −9≥ − 3, a+b b+c c+a a b + bc + ca X X 1 (b − c)2 ≥ (b − c)2 , (a + b)(c + a) 2(a b + bc + ca) X a b + bc + ca − a2 (b − c)2 ≥ 0. (a + b)(c + a) 2(a + b + c)



Without loss of generality, assume that a ≥ b ≥ c. Since a b + bc + ca − c 2 ≥ 0, it suffices to show that a b + bc + ca − b2 a b + bc + ca − a2 (b − c)2 + (c − a)2 ≥ 0. (a + b)(c + a) (b + c)(a + b) Since a b + bc + ca − b2 ≥ b(a − b) ≥ 0 and (c − a)2 ≥ (b − c)2 , it is enough to prove that a b + bc + ca − a2 a b + bc + ca − b2 (b − c)2 + (b − c)2 ≥ 0. (a + b)(c + a) (b + c)(a + b) This is true if

a b + bc + ca − a2 a b + bc + ca − b2 + ≥ 0, . (a + b)(c + a) (b + c)(a + b)

which is equivalent to 3 − a2 3 − b2 + ≥ 0, 3 + a2 3 + b2 Indeed,

2(9 − a2 b2 ) 2c(a + b)(3 + a b) 3 − a2 3 − b2 + = = ≥ 0. 2 2 2 2 3+a 3+ b (3 + a )(3 + b ) (3 + a2 )(3 + b2 )

Symmetric Rational Inequalities

69

P 1.35. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that a2

1 1 1 3 + 2 + 2 ≥ . +1 b +1 c +1 2 (Vasile Cîrtoaje, 2005)

First Solution. After expanding, the inequality can be restated as a2 + b2 + c 2 + 3 ≥ a2 b2 + b2 c 2 + c 2 a2 + 3a2 b2 c 2 . From (a + b + c)(a b + bc + ca) − 9a bc = a(b − c)2 + b(c − a)2 + c(a − b)2 ≥ 0, we get a + b + c ≥ 3a bc. So, it suffices to show that a2 + b2 + c 2 + 3 ≥ a2 b2 + b2 c 2 + c 2 a2 + a bc(a + b + c). This is equivalent to the homogeneous inequalities (a b + bc + ca)(a2 + b2 + c 2 ) + (a b + bc + ca)2 ≥ 3(a2 b2 + b2 c 2 + c 2 a2 ) + 3a bc(a + b + c), a b(a2 + b2 ) + bc(b2 + c 2 ) + ca(c 2 + a2 ) ≥ 2(a2 b2 + b2 c 2 + c 2 a2 ), a b(a − b)2 + bc(b − c)2 + ca(c − a)2 ≥ 0. The equality holds for a = b = c = 1, and for a = 0 and b = c = permutation).

p 3 (or any cyclic

Second Solution. Without loss of generality, assume that a = min{a, b, c}. From a b + bc + ca = 3, we get bc ≥ 1. Also, from (a + b + c)(a b + bc + ca) − 9a bc = a(b − c)2 + b(c − a)2 + c(a − b)2 ≥ 0, we get a + b + c ≥ 3a bc. The desired inequality follows by summing the inequalities b2

1 1 2 + 2 ≥ , +1 c +1 bc + 1 a2

1 2 3 + ≥ . + 1 bc + 1 2

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We have b2

1 2 b(c − b) c(b − c) 1 + 2 − = 2 + 2 + 1 c + 1 bc + 1 (b + 1)(bc + 1) (c + 1)(bc + 1) (b − c)2 (bc − 1) = 2 ≥0 (b + 1)(c 2 + 1)(bc + 1)

and 2 3 a2 − bc + 3 − 3a2 bc a(a + b + c − 3a bc) 1 + − = = ≥ 0. 2 2 a + 1 bc + 1 2 2(a + 1)(bc + 1) 2(a2 + 1)(bc + 1) Third Solution. Since 1 a2 1 b2 1 c2 = 1 − , = 1 − , = 1 − , a2 + 1 a2 + 1 b2 + 1 b2 + 1 c 2 + 1 c2 + 1 we can rewrite the inequality as a2 b2 c2 3 + + ≤ , 2 2 2 a +1 b +1 c +1 2 or, in the homogeneous form X

a2 1 ≤ . 2 3a + a b + bc + ca 2

According to the Cauchy-Schwarz inequality, we have 4a2 (a + a)2 a a2 = ≤ + . 3a2 + a b + bc + ca a(a + b + c) + (2a2 + bc) a + b + c 2a2 + bc Then, X

X 4a2 a2 ≤ 1 + , 3a2 + a b + bc + ca 2a2 + bc

and it suffices to show that X

a2 ≤ 1, 2a2 + bc

or, equivalently, X

bc ≥ 1. + bc

2a2

This follows from the Cauchy-Schwarz inequality as follows: P P 2 2 P X b c + 2a bc a ( bc)2 bc P P ≥P = = 1. 2a2 + bc bc(2a2 + bc) 2a bc a + b2 c 2

Symmetric Rational Inequalities

71

Remark. We can write the inequality in P 1.35 in the homogeneous form 1 1+

2

3a a b + bc + ca

Substituting a, b, c by

1

+ 1+

2

3b a b + bc + ca

1

+ 1+

2

3c a b + bc + ca

≥

3 . 2

1 1 1 , , , respectively, we get x y z

y x z 3 + + ≥ . 3 yz 3z x 3x y 2 y+ x+ z+ x + y +z x + y +z x + y +z So, we find the following result. • If x, y, z are positive real numbers such that x + y + z = 3, then y x z 3 + + ≥ . x + yz y + zx z + x y 2

P 1.36. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that b2 c2 a2 + + ≥ 1. a2 + b + c b2 + c + a c 2 + a + b (Vasile Cîrtoaje, 2005) Solution. We apply the Cauchy-Schwarz inequality in the following way
2 P 3 P X a3/2 + b3/2 + c 3/2 a + 2 (a b)3/2 a2 P ≥ P = . a2 + b + c a(a2 + b + c) a3 + 6 Then, we still have to show that (a b)3/2 + (bc)3/2 + (ca)3/2 ≥ 3. By the AM-GM inequality, (a b)3/2 = and hence

(a b)3/2 + (a b)3/2 + 1 1 3a b 1 − ≥ − , 2 2 2 2

X 3X 3 (a b)3/2 ≥ a b − = 3. 2 2 The equality holds for a = b = c = 1.

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P 1.37. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that bc + 2 ca + 2 a b + 2 bc + 4 ca + 4 a b + 4 + 2 + 2 ≤3≤ 2 + + . 2 a +4 b +4 c +4 a + 2 b2 + 2 c 2 + 2 (Vasile Cîrtoaje, 2007) Solution. More general, using the SOS method, we will show that ‹  bc + k ca + k a b + k + + 2 −3 ≤0 (k − 3) 2 a + k b2 + k c +k for k > 0. This inequality is equivalent to (k − 3)

X a2 − bc a2 + k

≥ 0.

Since 2

X a2 − bc a2 + k

we have 2(k − 3)

=

X (a − b)(a + c) + (a − c)(a + b)

a2 + k X (a − b)(a + c) X (b − a)(b + c) = + a2 + k b2 + k X (a − b)2 = (k − a b − bc − ca) (a2 + p)(b2 + p) 2 X (a − b) = (k − 3) , 2 (a + p)(b2 + p)

X a2 − bc a2 + k

= (k − 3)2

X

(a − b)2 ≥ 0. (a2 + k)(b2 + k)

Equality in both inequalities holds for a = b = c = 1.

p P 1.38. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. If k ≥ 2 + 3, then 1 1 1 3 + + ≤ . a+k b+k c+k 1+k (Vasile Cîrtoaje, 2007) Solution. Let us denote p = a + b + c, p ≥ 3. By expanding, the inequality becomes k(k − 2)p + 3a bc ≥ 3(k − 1)2 .

Symmetric Rational Inequalities

73

Since this inequality is true for p ≥ 3(k − 1)2 /(k2 − 2k), consider further that p≤

3(k − 1)2 . k(k − 2)

From Schur’s inequality (a + b + c)3 + 9a bc ≥ 4(a b + bc + ca)(a + b + c), we get 9a bc ≥ 12p − p3 . Therefore, it suffices to prove that 3k(k − 2)p + 12p − p3 ≥ 9(k − 1)2 , or, equivalently, (p − 3)[(3(k − 1)2 − p2 − 3p] ≥ 0. Thus, it remains to prove that 3(k − 1)2 − p2 − 3p ≥ 0. p Since p ≤ 3(k − 1)2 /(k2 − 2k) and k ≥ 2 + 3, we have 9(k − 1)4 9(k − 1)2 − k2 (k − 2)2 k(k − 2) 2 2 2 3(k − 1) (k − 3)(k − 4k + 1) = ≥ 0. k2 (k − 2)2 p The equality holds for p a = b = c = 1. In the case k = 2 + 3, the equality holds again for a = 0 and b = c = 3 (or any cyclic permutation). 3(k − 1)2 − p2 − 3p ≥ 3(k − 1)2 −

P 1.39. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 3. Prove that a(b + c) b(c + a) c(a + b) + + ≤ 3. 1 + bc 1 + ca 1 + ab (Vasile Cîrtoaje, 2010) Solution. Write the inequality in the homogeneous form X a2 or

X•

a(b + c) ≤ 1, + b2 + c 2 + 3bc

˜ a(b + c) a − ≤ 0, a2 + b2 + c 2 + 3bc a + b + c

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Vasile Cîrtoaje

X a(a − b)(a − c) ≥ 0. a2 + b2 + c 2 + 3bc Without loss of generality, assume that a ≥ b ≥ c. Then, it suffices to prove that b(b − c)(b − a) a(a − b)(a − c) + 2 ≥ 0. 2 2 + b + c + 3bc a + b2 + c 2 + 3ca

a2 This is true if

a(a − c) b(b − c) ≥ 2 . 2 2 + b + c + 3bc a + b2 + c 2 + 3ca Since a(a − c) ≥ b(b − c) and a2

a2

+

b2

1 1 ≥ 2 , 2 2 + c + 3bc a + b + c 2 + 3ca

the conclusion follows. The equality holds for a = b = c = 1, and for a = 0 and p b = c = 3/2 (or any cyclic permutation).

P 1.40. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that a2 + b2 b2 + c 2 c 2 + a2 + + ≤ 3. a+b b+c c+a (Cezar Lupu, 2005) First Solution. We apply the SOS method. Write the inequality in the homogeneous form X  b2 + c 2 b + c  Æ − ≥ 3(a2 + b2 + c 2 ) − a − b − c, b+c 2 or P X (b − c)2 (b − c)2 ≥p . 2(b + c) 3(a2 + b2 + c 2 ) + a + b + c p Since 3(a2 + b2 + c 2 ) + a + b + c ≥ 2(a + b + c) > 2(b + c), the conclusion follows. The equality holds for a = b = c = 1. Second Solution. By virtue of the Cauchy-Schwarz inequality, we have Pp P Pp X a2 + b2 ( a2 + b2 )2 2 a2 + 2 (a2 + b2 )(a2 + c 2 ) P P ≥ = a+b (a + b) 2 a P 2 P 2 P 2 P 2 a + 2 (a + bc) 3 a + ( a)2 P P ≥ = 2 a 2 a P P 9 + ( a)2 ( a − 3)2 P P = =3+ ≥ 3. 2 a 2 a

Symmetric Rational Inequalities

75

P 1.41. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that bc ca 7 ab + + + 2 ≤ (a + b + c). a+b b+c c+a 6 (Vasile Cîrtoaje, 2011) Solution. We apply the SOS method. Write the inequality as ‹ X 4bc b+c− ≥ 8(3 − a − b − c). 3 b+c Since b+c−

4bc (b − c)2 = b+c b+c

and 3−a− b−c =

3(a2 + b2 + c 2 ) − (a + b + c)2 9 − (a + b + c)2 = 3+a+ b+c 3+a+ b+c X 1 = (b − c)2 , 3+a+ b+c

we can write the inequality as Sa (b − c)2 + S b (c − a)2 + Sc (a − b)2 ≥ 0, where

8 3 − . b+c 3+a+ b+c Without loss of generality, assume that a ≥ b ≥ c. Since Sa ≥ S b ≥ Sc , it suffices to show that S b + Sc ≥ 0. Indeed, if this condition is fulfilled, then Sa ≥ S b ≥ (S b + Sc )/2 ≥ 0, and hence Sa =

Sa (b − c)2 + S b (c − a)2 + Sc (a − b)2 ≥ S b (c − a)2 + Sc (a − b)2 ≥ S b (a − b)2 + Sc (a − b)2 = (a − b)2 (S b + Sc ) ≥ 0. Since S b + Sc =

4(9 − 5a − b − c) , (a + b + 2c)(3 + a + b + c)

we need to show that 9 ≥ 5a+ b+c. This follows immediately from the Cauchy-Schwarz inequality (25 + 1 + 1)(a2 + b2 + c 2 ) ≥ (5a + b + c)2 . Thus, the proof is completed. The equality holds for a = b = c = 1, and for a = 5/3 and b = c = 1/3 (or any cyclic permutation).

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P 1.42. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that (a)

1 1 3 1 + + ≤ ; 3 − a b 3 − bc 3 − ca 2

(b)

1 1 1 3 +p +p ≤p . p 6 − ab 6 − bc 6 − ca 6−1 (Vasile Cîrtoaje, 2005)

Solution. (a) Since ab 2a b 3 =1+ =1+ 2 2 3 − ab 3 − ab a + b + 2c 2 + (a − b)2 2a b (a + b)2 ≤1+ 2 ≤ 1 + , a + b2 + 2c 2 2(a2 + b2 + 2c 2 ) it suffices to prove that (b + c)2 (c + a)2 (a + b)2 + + ≤ 3. a2 + b2 + 2c 2 b2 + c 2 + 2a2 c 2 + a2 + 2b2 By the Cauchy-Schwarz inequality, we have (a + b)2 (a + b)2 a2 b2 = ≤ + . a2 + b2 + 2c 2 (a2 + c 2 ) + (b2 + c 2 ) a2 + c 2 b2 + c 2 Thus, X

X a2 X b2 X a2 X c2 (a + b)2 ≤ + = + = 3. a2 + b2 + 2c 2 a2 + c 2 b2 + c 2 a2 + c 2 c 2 + a2

The equality holds for a = b = c. (b) According to P 1.29, the following inequality holds 1 1 1 3 + + ≤ . 2 2 2 2 2 2 6−a b 6− b c 6−c a 5 Since

p 2 6 1 1 =p +p , 2 2 6−a b 6 − ab 6 + ab

this inequality becomes X

1

+ p 6 − ab

X

p 6 6 . ≤ p 5 6 + ab 1

Symmetric Rational Inequalities

77

Thus, it suffices to show that X

3 1 ≥p . p 6 + ab 6+1

Since a b + bc + ca ≤ a2 + b2 + c 2 = 3, by the Cauchy-Schwarz inequality, we have X

1 9 9 3 ≥Pp = p ≥p . p 6 + ab ( 6 + a b) 3 6 + a b + bc + ca 6+1

The equality holds for a = b = c = 1.

P 1.43. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = 3. Prove that 1 1 1 3 + + ≥ . 5 5 5 1+a 1+ b 1+c 2 (Vasile Cîrtoaje, 2007) Solution. Let a = min{a, b, c}. There are two cases to consider Case 1: a ≥ 1/2. The desired inequality follows by summing the inequalities 8 ≥ 9 − 5a2 , 1 + a5

8 ≥ 9 − 5b2 , 1 + b5

8 ≥ 9 − 5c 2 ; 1 + c5

8 ≥ p + q x 2 , whose 1 + x5 coefficients p and q will be determined such that the polynomial P(x) = 8−(1+ x 5 )(p + q x 2 ) divides by (x − 1)2 . It is easy to check that P(1) = 0 involves p + q = 4, when To obtain these inequalities, we start from the inequality

P(x) = 4(2 − x 2 − x 7 ) − p(1 − x 2 + x 5 − x 7 ) = (1 − x)Q(x), where Q(x) = 4(2 + 2x + x 2 + x 3 + x 4 + x 5 + x 6 ) − p(1 + x + x 5 + x 6 ). In addition, Q(1) = 0 involves p = 9. In this case, P(x) = (1 − x)2 (5x 5 + 10x 4 + 6x 3 + 2x 2 − 2x − 1) = (1 − x)2 [x 5 + (2x − 1)(2x 4 + 6x 3 + 6x 2 + 4x + 1)] ≥ 0.

Case 2: a ≥ 1/2. Write the inequality as 1 1 b5 c 5 − 1 − ≥ . 1 + a5 2 (1 + b5 )(1 + c 5 )

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Vasile Cîrtoaje

Since

1 1 32 1 31 − ≥ − = 5 1+a 2 33 2 66

and (1 + b5 )(1 + c 5 ) ≥ (1 + it suffices to show that 31(1 +

p

p

b5 c 5 )2 ,

b5 c 5 )2 ≥ 66(b5 c 5 − 1),

which is equivalent to bc ≤ (97/35)2/5 . Indeed, from 3 = a2 + b2 + c 2 > b2 + c 2 ≥ 2bc, we get bc < 3/2 < (97/35)2/5 . This completes the proof. The equality holds for a = b = c = 1.

P 1.44. Let a, b, c be positive real numbers such that a bc = 1. Prove that 1 1 1 + + ≥ 1. a2 + a + 1 b2 + b + 1 c 2 + c + 1 First Solution. Using the substitutions a = yz/x 2 , b = z x/ y 2 , c = x y/z 2 , where x, y, z are positive real numbers, the inequality becomes X

x4 ≥ 1. x 4 + x 2 yz + y 2 z 2

By the Cauchy-Schwarz inequality, we have P P 4 P X ( x 2 )2 x + 2 y 2z2 x4 P P ≥P =P . x 4 + x 2 yz + y 2 z 2 (x 4 + x 2 yz + y 2 z 2 ) x 4 + x yz x + y 2 z 2 Therefore, it suffices to show that X which is equivalent to

P

y 2 z 2 ≥ x yz

X

x,

x 2 ( y − z)2 ≥ 0. The equality holds for a = b = c = 1.

Second Solution. Using the substitutions a = y/x, b = z/ y, c = x/z, where x, y, z > 0, we need to prove that y2 x2 z2 + + ≥ 1. x2 + x y + y2 y 2 + yz + z 2 z 2 + z x + z 2

Symmetric Rational Inequalities Since

79

x 2 z(x + y + z) x 2 (x 2 + y 2 + z 2 + x y + yz + z x) 2 = x + , x2 + x y + y2 x2 + x y + y2

multiplying by x 2 + y 2 + z 2 + x y + yz + z x, the inequality can be written as X

x y + yz + z x x 2z ≥ . x2 + x y + y2 x + y +z

By the Cauchy-Schwarz inequality, we have P X ( xz)2 x y + yz + z x x 2z ≥P . = 2 2 2 2 x +xy+ y x + y +z z(x + x y + y ) Remark. The inequality in P 1.44 is a particular case of the following more general inequality: (Vasile Cîrtoaje, 2009). • Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If p and q are nonnegative real numbers satisfying p + q = n − 1, then i=n X

1

i=1

1 + pai + qai2

≥ 1.

P 1.45. Let a, b, c be positive real numbers such that a bc = 1. Prove that a2

1 1 1 + 2 + 2 ≤ 3. −a+1 b − b+1 c −c+1

First Solution. Since 1 1 2(a2 + 1) 2a4 + = = 2 − , a2 − a + 1 a2 + a + 1 a4 + a2 + 1 a4 + a2 + 1 we can rewrite the inequality as X

X 1 a4 + 2 ≥ 3. a2 + a + 1 a4 + a2 + 1

Thus, it suffices to show that X and X

1 ≥1 a2 + a + 1

a4 ≥ 1. a4 + a2 + 1

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Vasile Cîrtoaje

The first inequality is just the inequality in P 1.44, while the second follows from the first by substituting a, b, c with a−2 , b−2 , c −2 , respectively. The equality holds for a = b = c = 1. Second Solution. Write the inequality as ‹ X4 1 − 2 ≥ 1, 3 a −a+1 X (2a − 1)2 ≥ 3. a2 − a + 1 Let p = a + b + c and q = a b + bc + ca. By the Cauchy-Schwarz inequality, we have P X (2a − 1)2 (2 a − 3)2 (2p − 3)2 P ≥ . = a2 − a + 1 p2 − 2q − p + 3 (a2 − a + 1) Thus, it suffices to show that (2p − 3)2 ≥ 3(p2 − 2q − p + 3), which is equivalent to p2 + 6q − 9p ≥ 0. From the known inequality (a b + bc + ca)2 ≥ 3a bc(a + b + c), we get q2 ≥ 3p. Using this inequality and the AM-GM inequality, we get Æ Æ 3 3 p2 + 6q = p2 + 3q + 3q ≥ 3 9p2 q2 ≥ 3 9p2 (3p) = 9p.

P 1.46. Let a, b, c be positive real numbers such that a bc = 1. Prove that 3+a 3+ b 3+c + + ≥ 3. 2 2 (1 + a) (1 + b) (1 + c)2 Solution. Using the inequality in P 1.1, we have X 3+a X X 1 2 = + (1 + a)2 (1 + a)2 1+a ˜ X X• 1 1 1 = + + 2 2 (1 + a) (1 + b) 1+c X 1 X ab ≥ + = 3. 1 + ab 1 + ab The equality holds for a = b = c = 1.

Symmetric Rational Inequalities

81

P 1.47. Let a, b, c be positive real numbers such that a bc = 1. Prove that 7 − 6a 7 − 6b 7 − 6c + + ≥ 1. 2 + a2 2 + b2 2 + c 2 (Vasile Cîrtoaje, 2008) Solution. Write the inequality as  ‹  ‹  ‹ 7 − 6a 7 − 6b 7 − 6c +1 + +1 + + 1 ≥ 4, 2 + a2 2 + b2 2 + c2 (3 − a)2 (3 − b)2 (3 − c)2 + + ≥ 4. 2 + a2 2 + b2 2 + c2 Substituting a, b, c by 1/a, 1/b, 1/c, respectively, we need to prove that a bc = 1 involves (3a − 1)2 (3b − 1)2 (3c − 1)2 + + ≥ 4. 2a2 + 1 2b2 + 1 2c 2 + 1 By the Cauchy-Schwarz inequality, we have P P P P X (3a − 1)2 (3 a − 3)2 9 a2 + 18 a b − 18 a + 9 P ≥ P = . 2a2 + 1 (2a2 + 1) 2 a2 + 3 Thus, it suffices to prove that f (a) + f (b) + f (c) ≥ 3, where

‹ 1 f (x) = x + 18 −x . x In order to do this, we use the mixing method. Without loss of generality, assume that a = max{a, b, c}, a ≥ 1, bc ≤ 1. Since ‹  p p p 1 f (b) + f (c) − 2 f ( bc) = (b − c)2 + 18( b − c)2 − 1 ≥ 0, bc 2

it suffices to show that



p f (a) + 2 f ( bc) ≥ 3.

Write this inequality as  ‹ 1 f (x ) + 2 f ≥ 3, x p where x = a. It is equivalent to 2

x 6 − 18x 4 + 36x 3 − 3x 2 − 36x + 20 ≥ 0, (x − 1)2 (x − 2)2 (x + 1)(x + 5) ≥ 0. Since the last inequality is true, the proof is completed. The equality holds for a = b = c = 1, and also for a = 1/4 and b = c = 2 (or any cyclic permutation).

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P 1.48. Let a, b, c be positive real numbers such that a bc = 1. Prove that b6 c6 a6 + + ≥ 1. 1 + 2a5 1 + 2b5 1 + 2c 5 (Vasile Cîrtoaje, 2008) Solution. Using the substitutions a=

v t 2 3 x yz

,

b=

v t 2 3 y zx

,

v t 2 3 z c= , xy

the inequality becomes X

x4 ≥ 1. p y 2 z 2 + 2x 3 3 x yz

By the Cauchy-Schwarz inequality, we have P P X ( x 2 )2 ( x 2 )2 x4 P . ≥P =P p p p y 2 z 2 + 2x 3 3 x yz ( y 2 z 2 + 2x 3 3 x yz) x 2 y 2 + 2 3 x yz x 3 Therefore, we need to show that X X X p ( x 2 )2 ≥ x 2 y 2 + 2 3 x yz x 3. p Since x + y + z ≥ 3 3 x yz, it suffices to prove that X X X X 3( x 2 )2 ≥ 3 x 2 y 2 + 2( x)( x 3 ); that is,

X

x4 + 3

X

x2 y2 ≥ 2

X

x y(x 2 + y 2 ),

or, equivalently, X (x − y)4 ≥ 0. The equality holds for a = b = c = 1.

P 1.49. Let a, b, c be positive real numbers such that a bc = 1. Prove that a b c 1 + + ≤ . a2 + 5 b2 + 5 c 2 + 5 2 (Vasile Cîrtoaje, 2008)

Symmetric Rational Inequalities

83

Solution. Let F (a, b, c) =

a2

a b c + 2 + 2 . +5 b +5 c +5

Without loss of generality, assume that a = min{a, b, c}. Case 1: a ≤ 1/5. We have a 1 b c 1 1 + p + p ≤ +p < . 5 2 5b2 2 5c 2 25 5 2 p p Case 2: a > 1/5. Let x = bc, a = 1/x 2 , x < 5. We will show that F (a, b, c)
0. The right inequality, F (a, x, x) ≤ 1/2, is equivalent to (x − 1)2 (5x 4 − 10x 3 − 2x 2 + 6x + 5) ≥ 0. It is also true since 5x 4 − 10x 3 − 2x 2 + 6x + 5 = 5(x − 1)4 + 2x(5x 2 − 16x + 13) and 5x 2 + 13 ≥ 2

p

65x 2 > 16x.

The equality holds for a = b = c = 1.

P 1.50. Let a, b, c be positive real numbers such that a bc = 1. Prove that 1 1 1 2 + + + ≥ 1. 2 2 2 (1 + a) (1 + b) (1 + c) (1 + a)(1 + b)(1 + c) (Pham Van Thuan, 2006)

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First Solution. There are two of a, b, c either greater than or equal to 1, or less than or equal to 1. Let b and c be these numbers; that is, (1 − b)(1 − c) ≥ 0. Since 1 1 1 + ≥ 2 2 (1 + b) (1 + c) 1 + bc (see P 1.1), it suffices to show that 1 1 2 + + ≥ 1. 2 (1 + a) 1 + bc (1 + a)(1 + b)(1 + c) This inequality is equivalent to b2 c 2 1 2bc + + ≥ 1, 2 (1 + bc) 1 + bc (1 + bc)(1 + b)(1 + c) which can be written in the obvious form bc(1 − b)(1 − c) ≥ 0. (1 + bc)(1 + b)(1 + c) The equality holds for a = b = c = 1. Second Solution. Setting a = yz/x 2 , b = z x/ y 2 , c = x y/z 2 , where x, y, z > 0, the inequality becomes X

2x 2 y 2 z 2 x4 + ≥ 1. (x 2 + yz)2 (x 2 + yz)( y 2 + z x)(z 2 + x y)

Since (x 2 + yz)2 ≤ (x 2 + y 2 )(x 2 + z 2 ), we have X

X 2x 2 y 2 z 2 x4 x4 ≥ = 1 − . (x 2 + yz)2 (x 2 + y 2 )(x 2 + z 2 ) (x 2 + y 2 )( y 2 + z 2 )(z 2 + x 2 )

Then, it suffices to show that (x 2 + y 2 )( y 2 + z 2 )(z 2 + x 2 ) ≥ (x 2 + yz)( y 2 + z x)(z 2 + x y). This inequality follows by multiplying the inequalities (x 2 + y 2 )(x 2 + z 2 ) ≥ (x 2 + yz)2 , ( y 2 + z 2 )( y 2 + x 2 ) ≥ ( y 2 + z x)2 , (z 2 + x 2 )(z 2 + y 2 ) ≥ (z 2 + x y)2 . Third Solution. We make the substitutions 1+ y 1 1+ x 1 1 1+z = , = , = ; 1+a 2 1+ b 2 1+c 2

Symmetric Rational Inequalities that is, a=

85

1− x , 1+ x

b=

1− y , 1+ y

c=

1−z , 1+z

where −1 < x, y, z < 1. Since a bc = 1 involves x + y + z + x yz = 0, we need to prove that x + y + z + x yz = 0 implies (1 + x)2 + (1 + y)2 + (1 + z)2 + (1 + x)(1 + y)(1 + z) ≥ 4. This inequality is equivalent to x 2 + y 2 + z 2 + (x + y + z)2 + 4(x + y + z) ≥ 0. By virtue of the AM-GM inequality, we have x 2 + y 2 + z 2 + (x + y + z)2 + 4(x + y + z) = x 2 + y 2 + z 2 + x 2 y 2 z 2 − 4x yz Æ 4 ≥ 4 x 4 y 4 z 4 − 4x yz = 4|x yz| − 4x yz ≥ 0.

P 1.51. Let a, b, c be nonnegative real numbers such that 1 1 1 3 + + = . a+b b+c c+a 2 Prove that

3 2 1 ≥ + 2 . a+b+c a b + bc + ca a + b2 + c 2

Solution. Write the inequality in the homogeneous form  ‹ 1 2 1 1 2 1 + + ≥ + 2 . a+b+c a+b b+c c+a a b + bc + ca a + b2 + c 2 Denote q = a b + bc + ca and assume that a + b + c = 1. From the known inequality (a + b + c)2 ≥ 3(a b + bc + ca), we get 1 − 3q ≥ 0. Rewrite the desired inequality as follows  ‹ 1 1 1 2 1 2 + + ≥ + , 1−c 1−a 1− b q 1 − 2q 2(q + 1) 2 − 3q ≥ , q − a bc q(1 − 2q) q2 (1 − 4q) + (2 − 3q)a bc ≥ 0.

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By Schur’s inequality, we have (a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca), 1 − 4q ≥ −9a bc. Then, q2 (1 − 4q) + (2 − 3q)a bc ≥ −9q2 a bc + (2 − 3q)a bc = (1 − 3q)(2 + 3q)a bc ≥ 0. The equality holds for a = b = c = 1, and for a = 0 and b = c = 5/3 (or any cyclic permutation).

P 1.52. Let a, b, c be nonnegative real numbers such that 7(a2 + b2 + c 2 ) = 11(a b + bc + ca). Prove that

a b c 51 ≤ + + ≤ 2. 28 b+c c+a a+b

Solution. Due to homogeneity, we may assume that b + c = 2. Let us denote x = bc, 0 ≤ x ≤ 1. By the hypothesis 7(a2 + b2 + c 2 ) = 11(a b + bc + ca), we get x=

7a2 − 22a + 28 . 25

Then, x ≤ 1 involves 1/7 ≤ a ≤ 3. Since a b c a a(b + c) + (b + c)2 − 2bc + + = + b+c c+a a+b b+c a2 + (b + c)a + bc a 2(a + 2 − x) 4a3 + 27a + 11 = + 2 = , 2 a + 2a + x 8a2 + 7a + 7 the required inequalities become 51 4a3 + 27a + 11 ≤ ≤ 2. 28 8a2 + 7a + 7 We have

4a3 + 27a + 11 51 (7a − 1)(4a − 7)2 − = ≥0 8a2 + 7a + 7 28 28(8a2 + 7a + 7)

Symmetric Rational Inequalities

87

and

4a3 + 27a + 11 (3 − a)(2a − 1)2 = ≥ 0. 8a2 + 7a + 7 8a2 + 7a + 7 This completes the proof. The left inequality becomes an equality for 7a = b = c (or any cyclic permutation), while the right inequality is an equality for a/3 = b = c (or any cyclic permutation). 2−

P 1.53. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2

1 1 1 10 + 2 + 2 ≥ . 2 2 2 +b b +c c +a (a + b + c)2

Solution. Assume that a = min{a, b, c}, and denote a x = b+ , 2

a y =c+ . 2

Since a2 + b2 ≤ x 2 , b2 + c 2 ≤ x 2 + y 2 , c 2 + a2 ≤ y 2 , (a + b + c)2 = (x + y)2 ≥ 4x y, it suffices to show that

1 1 1 5 + 2 + 2≥ . 2 2 x x +y y 2x y

We have 1 1 1 5 + 2 + 2− = 2 2 x x +y y 2x y

‹  ‹ 1 1 2 1 1 + 2− + − x2 y xy x 2 + y 2 2x y (x − y)2 (x − y)2 − = x2 y2 2x y(x 2 + y 2 ) 2 (x − y) (2x 2 − x y + 2 y 2 ) = ≥ 0. 2x 2 y 2 (x 2 + y 2 ) 

The equality holds for a = 0 and b = c (or any cyclic permutation).

P 1.54. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2

1 1 1 3 + 2 + 2 ≥ . 2 2 2 − ab + b b − bc + c c − ca + a max{a b, bc, ca}

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Solution. Assume that a = min{a, b, c}, hence bc = max{a b, bc, ca}. Since a2

1 1 1 1 1 1 + 2 + 2 ≥ 2+ 2 + 2, 2 2 2 2 − ab + b b − bc + c c − ca + a b b − bc + c c

it suffices to show that

We have

1 1 3 1 . + + ≥ b2 b2 − bc + c 2 c 2 bc

1 1 1 3 (b − c)4 + + − = ≥ 0. b2 b2 − bc + c 2 c 2 bc b2 c 2 (b2 − bc + c 2 )

The equality holds for a = b = c, and also a = 0 and b = c (or any cyclic permutation).

P 1.55. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a(2a + b + c) b(2b + c + a) c(2c + a + b) + + ≥ 6. b2 + c 2 c 2 + a2 a2 + b2 Solution. By the Cauchy-Schwarz inequality, we have P X a(2a + b + c) [ a(2a + b + c)]2 ≥P . b2 + c 2 a(2a + b + c)(b2 + c 2 ) Thus, we still need to show that X X X 2( a2 + a b)2 ≥ 3 a(2a + b + c)(b2 + c 2 ), which is equivalent to X X X X a+ a b(a2 + b2 ) ≥ 6 a2 b2 . 2 a4 + 2a bc We can obtain this inequality by adding Schur’s inequality of degree four X X X a4 + a bc a≥ a b(a2 + b2 ) and

X

a b(a2 + b2 ) ≥ 2

X

a2 b2 ,

multiplied by 2 and 3, respectively. The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

Symmetric Rational Inequalities

89

P 1.56. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 (b + c)2 b2 (c + a)2 c 2 (a + b)2 + 2 + ≥ 2(a b + bc + ca). b2 + c 2 c + a2 a2 + b2 Solution. We apply the SOS method. Since a2 (b + c)2 ) 2a2 bc 2 = a + , b2 + c 2 b2 + c 2 we can write the inequality as X X X  2( a2 − a b) − a2 1 −

2bc b2 + c 2

‹

≥ 0,

X X a2 (b − c)2 ≥ 0, (b − c)2 − b2 + c 2  X a2 1− 2 (b − c)2 ≥ 0. b + c2 Without loss of generality, assume that a ≥ b ≥ c. Since 1 − prove that

c2 > 0, it suffices to a2 + b2

   a2 b2 2 1− 2 (b − c) + 1 − 2 (a − c)2 ≥ 0, b + c2 c + a2



which is equivalent to (a2 − b2 + c 2 )(a − c)2 (a2 − b2 − c 2 )(b − c)2 ≥ . a2 + c 2 b2 + c 2 This inequality follows by multiplying the inequalities a2 − b2 + c 2 ≥ a2 − b2 − c 2 ,

(a − c)2 (b − c)2 ≥ . a2 + c 2 b2 + c 2

The latter inequality is true since (a − c)2 (b − c)2 2bc 2ac 2c(a − b)(a b − c 2 ) − = − = ≥ 0. a2 + c 2 b2 + c 2 b2 + c 2 a2 + c 2 (b2 + c 2 )(a2 + c 2 ) The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

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P 1.57. If a, b, c are real numbers such that a bc > 0, then  ‹  ‹ X a b c a 1 1 1 3 + 5 + + ≥ 8 + + . b2 − bc + c 2 bc ca a b a b c (Vasile Cîrtoaje, 2011) Solution. In order to apply the SOS method, we multiply the inequality by a bc and write it as follows: ‹ X X X  bc 2 2 ≥ 0, 8( a − bc) − 3 a 1− 2 b − bc + c 2 X X a2 (b − c)2 (b − c)2 − 3 ≥ 0, b2 − bc + c 2 X (b − c)2 (4b2 − 4bc + 4c 2 − 3a2 ) ≥ 0. b2 − bc + c 2 Without loss of generality, assume that a ≥ b ≥ c. Since 4

4a2 − 4a b + 4b2 − 3c 2 = (2a − b)2 + 3(b2 − c 2 ) ≥ 0, it suffices to prove that (a − c)2 (4a2 − 4ac + 4c 2 − 3b2 ) (b − c)2 (3a2 − 4b2 + 4bc − 4c 2 ) ≥ . a2 − ac + c 2 b2 − bc + c 2 Notice that 4a2 − 4ac + 4c 2 − 3b2 = (a − 2c)2 + 3(a2 − b2 ) ≥ 0. Thus, the desired inequality follows by multiplying the inequalities 4a2 − 4ac + 4c 2 − 3b2 ≥ 3a2 − 4b2 + 4bc − 4c 2 and

(a − c)2 (b − c)2 ≥ . a2 − ac + c 2 b2 − bc + c 2 The first inequality is equivalent to (a − 2c)2 + (b − 2c)2 ≥ 0. Also, we have (a − c)2 (b − c)2 bc ac − = 2 − 2 2 2 2 2 2 a − ac + c b − bc + c b − bc + c a − ac + c 2 c(a − b)(a b − c 2 ) = 2 ≥ 0. (b − bc + c 2 )(a2 − ac + c 2 ) The equality occurs for a = b = c, and for 2a = b = c (or any cyclic permutation).

Symmetric Rational Inequalities

91

P 1.58. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that  ‹ 1 1 1 (a) 2a bc + + + a2 + b2 + c 2 ≥ 2(a b + bc + ca); a+b b+c c+a b2 c2 3(a2 + b2 + c 2 ) a2 + + ≤ . a+b b+c c+a 2(a + b + c)

(b)

Solution. (a) First Solution. We have 2a bc

X

X X a(2bc + a b + ac) 1 + a2 = b+c b+c X a b(a + c) X ac(a + b) = + b+c b+c X a b(a + c) X ba(b + c) + = b+ c ‹c + a X X a+c b+c = ab + ≥2 a b. b+c a+c

The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation). Second Solution. Write the inequality as ‹ X  2a bc 2 + a − a b − ac ≥ 0. b+c We have X  2a bc b+c

2

‹

+ a − a b − ac =

X a b(a − b) + ac(a − c)

b+c X a b(a − b) X ba(b − a) = + b+c c+a X a b(a − b)2 = ≥ 0. (b + c)(c + a)

(b) Since

‹ X a2 X X ab ab = a− =a+b+c− , a+b a+b a+b we can write the desired inequality as X ab 3(a2 + b2 + c 2 ) + ≥ a + b + c. a+b 2(a + b + c) Multiplying by 2(a + b + c), the inequality can be written as X a  2 1+ bc + 3(a2 + b2 + c 2 ) ≥ 2(a + b + c)2 , b+c

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or

1 + a2 + b2 + c 2 ≥ 2(a b + bc + ca), b+c which is just the inequality in (a). 2a bc

X

P 1.59. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

a2 − bc b2 − ca c 2 − a b 3(a b + bc + ca) + 2 + + ≥ 3; b2 + c 2 c + a2 a2 + b2 a2 + b2 + c 2

(b)

a2 b2 c2 a b + bc + ca 5 + + + 2 ≥ ; 2 2 2 2 2 2 2 2 b +c c +a a +b a +b +c 2

(c)

a b + bc + ca a2 + bc b2 + ca c 2 + a b + 2 + 2 ≥ 2 + 2. 2 2 2 2 b +c c +a a +b a + b2 + c 2 (Vasile Cîrtoaje, 2014)

Solution. (a) Write the inequality as follows:  X ‹  ‹ X  2a2 2bc a b + bc + ca − 1 + 1 − − 6 1 − ≥ 0, b2 + c 2 b2 + c 2 a2 + b2 + c 2 X 2a2 − b2 − c 2 b2 + c 2

+

X (b − c)2 b2 + c 2

−3

X

(b − c)2 ≥ 0. a2 + b2 + c 2

Since X 2a2 − b2 − c 2 b2 + c 2

= =

X a2 − b2 X

+

X a2 − c 2

=

X a2 − b2

+

X b2 − a2 c 2 + a2

b2 + c 2 b2 + c 2 b2 + c 2 2 2 2 X (a − b ) (b2 − c 2 )2 = , (b2 + c 2 )(c 2 + a2 ) (a2 + b2 )(a2 + c 2 )

we can write the inequality as X (b − c)2 Sa ≥ 0, where Sa =

(b + c)2 1 3 + 2 − 2 . 2 2 2 2 2 (a + b )(a + c ) b + c a + b2 + c 2

It suffices to show that Sa ≥ 0 for all nonnegative real numbers a, b, c, no two of which are zero. Denoting x 2 = b2 + c 2 , we have Sa =

x 2 + 2bc 1 3 + 2− 2 , 4 2 2 2 2 a +a x +b c x a + x2

Symmetric Rational Inequalities

93

and the inequality Sa ≥ 0 becomes (a2 − 2x 2 )b2 c 2 + 2x 2 (a2 + x 2 )bc + (a2 + x 2 )(a2 − x 2 )2 ≥ 0. Clearly, this is true if −2x 2 b2 c 2 + 2x 4 bc ≥ 0. Indeed, −2x 2 b2 c 2 + 2x 4 bc = 2x 2 bc(x 2 − bc) = 2bc(b2 + c 2 )(b2 + c 2 − bc) ≥ 0. The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation). (b) First Solution. We get the desired inequality by summing the inequality in (a) and the inequality b2

ca ab 1 2(a b + bc + ca) bc + 2 + 2 + ≥ . 2 2 2 +c c +a a +b 2 a2 + b2 + c 2

This inequality is equivalent to ‹ X  2bc 4(a b + bc + ca) + 1 ≥ + 2, b2 + c 2 a2 + b2 + c 2 X (b + c)2

≥

b2 + c 2 By the Cauchy-Schwarz inequality, we have X (b + c)2 b2 + c 2

2(a + b + c)2 . a2 + b2 + c 2

P 2 (b + c) 2(a + b + c)2 ≥ P . = 2 a + b2 + c 2 (b2 + c 2 )

The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation). Second Solution. Let p = a + b + c,

q = a b + bc + ca,

r = a bc.

By the Cauchy-Schwarz inequality, we have X

P 2
2 a (p2 − 2q)2 a2 P ≥ . = b2 + c 2 a2 (b2 + c 2 ) 2(q2 − 2pr)

Therefore, it suffices to show that (p2 − 2q)2 2q + 2 ≥ 5. 2 q − 2pr p − 2q

(*)

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Consider the following cases: p2 ≥ 4q and 3q ≤ p2 < 4q. Case 1: p2 ≥ 4q. The inequality (*) is true if 2q (p2 − 2q)2 + 2 ≥ 5, 2 q p − 2q which is equivalent to the obvious inequality   (p2 − 4q) (p2 − q)2 − 2q2 ≥ 0. Case 2: 3q ≤ p2 < 4q. Using Schur’s inequality of degree four 6pr ≥ (p2 − q)(4q − p2 ), the inequality (*) is true if 3(p2 − 2q)2 2q + ≥ 5, 3q2 − (p2 − q)(4q − p2 ) p2 − 2q which is equivalent to the obvious inequality (p2 − 3q)(p2 − 4q)(2p2 − 5q) ≤ 0. Third Solution (by Nguyen Van Quy). Write the inequality (*) from the preceding solution as follows: (a2 + b2 + c 2 )2 2(a b + bc + ca) + ≥ 5, 2 2 2 2 2 2 a b +b c +c a a2 + b2 + c 2 (a2 + b2 + c 2 )2 2(a b + bc + ca) −3≥2− , 2 2 2 2 2 2 a b +b c +c a a2 + b2 + c 2 a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2 2(a2 + b2 + c 2 − a b − bc − ca) ≥ . a2 b2 + b2 c 2 + c 2 a2 a2 + b2 + c 2 Since 2(a2 b2 + b2 c 2 + c 2 a2 ) ≤

X

a b(a2 + b2 ) ≤ (a b + bc + ca)(a2 + b2 + c 2 ),

it suffices to prove that a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2 ≥ a2 + b2 + c 2 − a b − bc − ca, a b + bc + ca which is just Schur’s inequality of degree four a4 + b4 + c 4 + a bc(a + b + c) ≥ a b(a2 + b2 ) + bc(b2 + c 2 ) + ca(c 2 + a2 ).

Symmetric Rational Inequalities

95

(c) We get the desired inequality by summing the inequality in (a) and the inequality

2ca 2a b 4(a b + bc + ca) 2bc + 2 + 2 +1≥ , 2 2 2 +c c +a a +b a2 + b2 + c 2 which is proved at (b). The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation). b2

P 1.60. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 b2 c2 (a + b + c)2 + + ≥ . b2 + c 2 c 2 + a2 a2 + b2 2(a b + bc + ca) Solution. Applying the Cauchy-Schwarz inequality, we have X

P 2
2 a (a2 + b2 + c 2 )2 a2 P ≥ = . b2 + c 2 a2 (b2 + c 2 ) 2(a2 b2 + b2 c 2 + c 2 a2 )

Therefore, it suffices to show that (a2 + b2 + c 2 )2 (a + b + c)2 ≥ , 2(a2 b2 + b2 c 2 + c 2 a2 ) 2(a b + bc + ca) which is equivalent to (a2 + b2 + c 2 )2 (a + b + c)2 − 3 ≥ − 3, a2 b2 + b2 c 2 + c 2 a2 a b + bc + ca a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2 a2 + b2 + c 2 − a b − bc − ca ≥ . a2 b2 + b2 c 2 + c 2 a2 a b + bc + ca Since a2 b2 + b2 c 2 + c 2 a2 ≤ (a b + bc + ca)2 , it suffices to show that a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2 ≥ (a2 + b2 + c 2 − a b − bc − ca)(a b + bc + ca), which is just Schur’s inequality of degree four a4 + b4 + c 4 + a bc(a + b + c) ≥ a b(a2 + b2 ) + bc(b2 + c 2 ) + ca(c 2 + a2 ). The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

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P 1.61. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 2bc 2ca a2 + b2 + c 2 5 2a b + + + ≥ . (a + b)2 (b + c)2 (c + a)2 a b + bc + ca 2 (Vasile Cîrtoaje, 2006) First Solution. We use the SOS method. Write the inequality as follows ‹ X1 a2 + b2 + c 2 2bc −1≥ − , a b + bc + ca 2 (b + c)2 X

X (b − c)2 (b − c)2 ≥ , ab + bc + ca (b + c)2

(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0, where Sa = 1 −

a b + bc + ca a b + bc + ca a b + bc + ca , Sb = 1 − , Sc = 1 − . 2 2 (b + c) (c + a) (a + b)2

Without loss of generality, assume that a ≥ b ≥ c. We have Sc > 0 and Sb ≥ 1 −

(c + a)(c + b) a−b = ≥ 0. 2 (c + a) c+a

If b2 Sa + a2 S b ≥ 0, then X a2 (b − c)2 Sa ≥ (b − c)2 Sa + (c − a)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b b 2 2 2 (b − c) (b Sa + a S b ) ≥ 0. = b2 We have 2

2

2



‹2

 a 2  b Sa + a S b = a + b − (a b + bc + ca) + c+a   ‹2  b a 2 2 2 ≥ a + b − (b + c)(c + a)) + b+c c+a ‹    c+a b+c = a2 1 − + b2 1 − c+a b+c 2 (a − b) (a b + bc + ca) = ≥ 0. (b + c)(c + a) 2

b b+c

The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

Symmetric Rational Inequalities

97

Second Solution. Multiplying by a b + bc + ca, the inequality becomes X 1 X 2a2 b2 5 + 2a bc + a2 + b2 + c 2 ≥ (a b + bc + ca), 2 (a + b) a+b 2 X 1 X1 • X 4a b ˜ 2a bc + a2 + b2 + c 2 − 2(a b + bc + ca) − ab 1 − ≥ 0. a+b 2 (a + b)2 According to the second solution of P 1.58-(a), we can write the inequality as follows X a b(a − b)2 X a b(a − b)2 − ≥ 0, (b + c)(c + a) 2(a + b)2 (b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0, where

bc [2(b + c)2 − (a + b)(a + c)]. b+c Without loss of generality, assume that a ≥ b ≥ c. We have Sc > 0 and Sa =

ac ac [2(a + c)2 − (a + b)(b + c)] ≥ [2(a + c)2 − (2a)(a + c)] a+c a+c 2ac 2 (a + c) = ≥ 0. a+c

Sb =

If Sa + S b ≥ 0, then X (b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 (Sa + S b ) ≥ 0. The inequality Sa + S b ≥ 0 is true if bc ac [2(a + c)2 − (a + b)(b + c)] ≥ [(a + b)(a + c) − 2(b + c)2 ]. a+c b+c Since

ac bc ≥ , a+c b+c

it suffices to show that 2(a + c)2 − (a + b)(b + c) ≥ (a + b)(a + c) − 2(b + c)2 . This is true since is equivalent to (a − b)2 + 2c(a + b) + 4c 2 ≥ 0.

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P 1.62. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that bc ca 1 a b + bc + ca ab + + + ≥ 2 . 2 2 2 (a + b) (b + c) (c + a) 4 a + b2 + c 2 (Vasile Cîrtoaje, 2011) First Solution. We use the SOS method. Write the inequality as follows 1−

• ˜ bc a b + bc + ca X 1 − ≥ , a2 + b2 + c 2 4 (b + c)2

X (b − c)2 (b − c)2 ≥ , a2 + b2 + c 2 (b + c)2   X a2 + b2 + c 2 2 (b − c) 2 − ≥ 0. (b + c)2 2

Since 2−

X

 a 2 2bc − a2 a2 + b2 + c 2 = 1 + ≥ 1 − , (b + c)2 (b + c)2 b+c

it suffices to show that (b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0, where

 ‹2  c 2  a 2 b , Sb = 1 − , Sc = 1 − . Sa = 1 − b+c c+a a+b

Without loss of generality, assume that a ≥ b ≥ c. Since S b ≥ 0 and Sc > 0, if b2 Sa + a2 S b ≥ 0, then X a2 (b − c)2 Sa ≥ (b − c)2 Sa + (c − a)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b b (b − c)2 (b2 Sa + a2 S b ) = ≥ 0. b2 We have ‹  ‹ ab 2 ab 2 b Sa + a S b = a + b − − b+c c+a    ‹2 •  a 2 ˜ b = a2 1 − + b2 1 − ≥ 0. b+c c+a 2

2

2

2



The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

Symmetric Rational Inequalities

99

Second Solution. Since (a + b)2 ≤ 2(a2 + b2 ), it suffices to prove that ab 1 a b + bc + ca + ≥ 2 , 2 +b ) 4 a + b2 + c 2

X

2(a2 which is equivalent to

X 2a b 4(a b + bc + ca) +1≥ , 2 2 a +b a2 + b2 + c 2 X (a + b)2

≥2+

+ X (a + b)2 a2

b2

4(a b + bc + ca) , a2 + b2 + c 2

2(a + b + c)2 . a2 + b2 a2 + b2 + c 2 The last inequality follows immediately by the Cauchy-Schwarz inequality P X (a + b)2 [ (a + b)]2 ≥ P . a2 + b2 (a2 + b2 ) ≥

Remark. The following generalization of the inequalities in P 1.61 and P 1.62 holds: • Let a, b, c be nonnegative real numbers, no two of which are zero. If 0 ≤ k ≤ 2, then X

4a b a2 + b2 + c 2 a b + bc + ca + k ≥ 3k − 1 + 2(2 − k) 2 . 2 (a + b) a b + bc + ca a + b2 + c 2

with equality for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.63. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 3a b 3bc 3ca a b + bc + ca 5 + + ≤ 2 + . 2 2 2 (a + b) (b + c) (c + a) a + b2 + c 2 4 (Vasile Cîrtoaje, 2011) Solution. We use the SOS method. Write the inequality as follows ˜ X•1 bc a b + bc + ca 3 − ≥1− 2 , 2 4 (b + c) a + b2 + c 2 3

X (b − c)2 (b + c)2

≥2

X

(b − c)2 , a2 + b2 + c 2

(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0,

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Vasile Cîrtoaje

where Sa =

3(a2 + b2 + c 2 ) 3(a2 + b2 + c 2 ) 3(a2 + b2 + c 2 ) − 2, S = − 2, S = − 2. b c (b + c)2 (c + a)2 (a + b)2

Without loss of generality, assume that a ≥ b ≥ c. Since Sa > 0 and Sb =

a2 + 3b2 + c 2 − 4ac (a − 2c)2 + 3(b2 − c 2 ) = > 0, (c + a)2 (c + a)2

if S b + Sc ≥ 0, then X (b − c)2 Sa ≥ (c − a)2 S b + (a − b)2 Sc ≥ (a − b)2 (S b + Sc ) ≥ 0. Using the Cauchy-Schwarz Inequality, we have ˜ • 1 1 + −4 S b + Sc = 3(a2 + b2 + c 2 ) (c + a)2 (a + b)2 12(a2 + b2 + c 2 ) 4(a − b − c)2 + 4(b − c)2 ≥ − 4 = ≥ 0. (c + a)2 + (a + b)2 (c + a)2 + (a + b)2 The equality occurs for a = b = c, and for

a = b = c (or any cyclic permutation). 2

P 1.64. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

a3 + a bc b3 + a bc c 3 + a bc + + ≥ a2 + b2 + c 2 ; b+c c+a a+b

(b)

a3 + 2a bc b3 + 2a bc c 3 + 2a bc 1 + + ≥ (a + b + c)2 ; b+c c+a a+b 2

(c)

a3 + 3a bc b3 + 3a bc c 3 + 3a bc + + ≥ 2(a b + bc + ca). b+c c+a a+b

Solution. (a) First Solution. Write the inequality as  X  a3 + a bc 2 − a ≥ 0, b+c X a(a − b)(a − c) b+c

≥ 0.

Symmetric Rational Inequalities

101

Assume that a ≥ b ≥ c. Since (c − a)(c − b) ≥ 0 and a(a − b)(a − c) b(b − c)(b − a) (a − b)2 (a2 + b2 + c 2 + a b) + = ≥ 0, b+c b+c (b + c)(c + a) the conclusion follows. The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation). (b) Taking into account the inequality in (a), it suffices to show that a bc a bc a bc 1 + + + a2 + b2 + c 2 ≥ (a + b + c)2 , b+c c+a a+b 2 which is just the inequality (a) from P 1.58. The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation). (c) The desired inequality follows by adding the inequality in (a) to the inequality (a) from P 1.58. The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.65. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a3 + 3a bc b3 + 3a bc c 3 + 3a bc + + ≥ a + b + c. (b + c)2 (c + a)2 (a + b)2 (Vasile Cîrtoaje, 2005) Solution. We use the SOS method. We have  X 3 X a3 + 3a bc X X  a3 + 3a bc a − a(b2 − bc + c 2 ) − a = − a = (b + c)2 (b + c)2 (b + c)2 = =

X a3 (b + c) − a(b3 + c 3 )

X a b(a2 − b2 )

+

=

(b + c)3 X ba(b2 − a2 )

X a b(a2 − b2 ) + ac(a2 − c 2 )

=

(b + c)3 X a b(a2 − b2 )[(c + a)3 − (b + c)3 ]

(b + c)3 (c + a)3 (b + c)3 (c + a)3 X a b(a + b)(a − b)2 [(c + a)2 + (c + a)(b + c) + (b + c)2 ] = ≥ 0. (b + c)3 (c + a)3

The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

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P 1.66. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

a3 + 3a bc b3 + 3a bc c 3 + 3a bc 3 + + ≥ ; 3 3 3 (b + c) (c + a) (a + b) 2

(b)

3a3 + 13a bc 3b3 + 13a bc 3c 3 + 13a bc + + ≥ 6. (b + c)3 (c + a)3 (a + b)3 (Vasile Cîrtoaje and Ji Chen, 2005)

Solution. (a) First Solution. Use the SOS method. We have X a3 + 3a bc (b + c)3

= =

X a(b + c)2 + a(a2 + bc − b2 − c 2 )

3 2 3 = 2 3 = 2 3 = 2 ≥

(b + c)3 X a3 − a(b2 − bc + c 2 )

a + b+c (b + c)3 X a3 (b + c) − a(b3 + c 3 ) + (b + c)4 X a b(a2 − b2 ) + ac(a2 − c 2 ) + (b + c)4 X a b(a2 − b2 ) X ba(b2 − a2 ) + + (b + c)4 (c + a)4 X a b(a + b)(a − b)[(c + a)4 − (b + c)4 ] + ≥ 0. (b + c)4 (c + a)4

X

The equality occurs for a = b = c. Second Solution. Assume that a ≥ b ≥ c. Since a3 + 3a bc b3 + 3a bc c 3 + 3a bc ≥ ≥ b+c c + ac a+b and

1 1 1 ≥ ≥ , 2 2 (b + c) (c + a) (a + b)2

by Chebyshev’s inequality, we get X a3 + 3a bc (b + c)3 Thus, it suffices to show that X

  1 X a3 + 3a bc X 1 ≥ . 3 b+c (b + c)2

 1 9 a3 + 3a bc X ≥ . 2 b+c (b + c) 2

Symmetric Rational Inequalities

103

We can obtain this inequality by multiplying the known inequality (Iran-1996) X 9 1 ≥ 2 (b + c) 4(a b + bc + ca) and the inequality (c) from P 1.64. (b) We have X 3a3 + 13a bc (b + c)3 =

=

(b + c)3

X 3a X X a3 − a(b2 − bc + c 2 ) 1 + 4a bc + 3 . b+c (b + c)3 (b + c)3

Since X and

X 3a(b + c)2 + 4a bc + 3a(a2 + bc − b2 − c 2 )

1 3 ≥ (b + c)3 (a + b)(b + c)(c + a)

X a3 − a(b2 − bc + c 2 ) (b + c)3

=

X a3 (b + c) − a(b3 + c 3 )

(b + c)4 X a b(a2 − b2 ) + ac(a2 − c 2 ) X a b(a2 − b2 ) X ba(b2 − a2 ) = = + (b + c)4 (b + c)4 (c + a)4 X a b(a + b)(a − b)[(c + a)4 − (b + c)4 ] ≥ 0, = (b + c)4 (c + a)4 it suffices to prove that X 3a 12a bc + ≥ 6. b + c (a + b)(b + c)(c + a) This inequality is equivalent to the third degree Schur’s inequality X a3 + b3 + c 3 + 3a bc ≥ a b(a + b). The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.67. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

(b)

a3 b3 c3 3 + + + a b + bc + ca ≥ (a2 + b2 + c 2 ); b+c c+a a+b 2 2a2 + bc 2b2 + ca 2c 2 + a b 9(a2 + b2 + c 2 ) + + ≥ . b+c c+a a+b 2(a + b + c) (Vasile Cîrtoaje, 2006)

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Vasile Cîrtoaje

Solution. (a) We apply the SOS method. Write the inequality as  X X  2a3 2 (a − b)2 . −a ≥ b+c Since

 X 2 X  2a3 a (a − b) + a2 (a − c) 2 −a = b+c b+c =

X a2 (a − b) b+c

+

X b2 (b − a) X (a − b)2 (a2 + b2 + a b + bc + ca) = , c+a (b + c)(c + a)

we can write the inequality as (b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0, where Sa = (b + c)(b2 + c 2 − a2 ), S b = (c + a)(c 2 + a2 − b2 ), Sc = (a + b)(a2 + b2 − c 2 ). Without loss of generality, assume that a ≥ b ≥ c. Since S b ≥ 0, Sc ≥ 0, and Sa + S b = (a + b)(a − b)2 + c 2 (a + b + 2c) ≥ 0, we have X (b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 (Sa + S b ) ≥ 0. The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation). (b) Multiplying by a + b + c, the inequality can be written as X 1+ or

a  9 (2a2 + bc) ≥ (a2 + b2 + c 2 ), b+c 2

X 2a3 + a bc

5 + a b + bc + ca ≥ (a2 + b2 + c 2 ). b+c 2 This inequality follows using the inequality in (a) and the first inequality from P 1.58. The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.68. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a(b + c) b(c + a) c(a + b) + 2 + 2 ≥ 2. 2 2 + bc + c c + ca + a a + a b + b2

b2

Symmetric Rational Inequalities

105

First Solution. Apply the SOS method. We have ˜ X• ˜ •X a(b + c)(a + b + c) a(b + c) −2 = − 2a (a + b + c) b2 + bc + c 2 b2 + bc + c 2 =

X a(a b + ac − b2 − c 2 )

=

X a b(a − b) − ca(c − a)

b2 + bc + c 2 b2 + bc + c 2 X a b(a − b) X a b(a − b) = − b2 + bc + c 2 c 2 + ca + a2 X a b(a − b)2 = (a + b + c) ≥ 0. (b2 + bc + c 2 )(c 2 + ca + a2 )

The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation). Second Solution. By the AM-GM inequality, we have 4(b2 + bc + c 2 )(a b + bc + ca) ≤ (b2 + bc + c 2 + a b + bc + ca)2 = (b + c)2 (a + b + c)2 . Thus,

X a(b + c)(a b + bc + ca) a(b + c) = 2 2 2 b + bc + c (b + bc + c 2 )(a b + bc + ca) X 4a(a b + bc + ca) 4(a b + bc + ca) X a ≥ = , (b + c)(a + b + c)2 (a + b + c)2 b+c X

and it suffices to show that X

(a + b + c)2 a ≥ . b+c 2(a b + bc + ca)

This follows immediately from the Cauchy-Schwarz inequality X

a (a + b + c)2 ≥ P . b+c a(b + c)

Third Solution. By the Cauchy-Schwarz inequality, we have X

a(b + c) (a + b + c)2 ≥ . P a(b2 + bc + c 2 ) b2 + bc + c 2 b+c

Thus, it is enough to show that (a + b + c)2 ≥ 2 Since

X a(b2 + bc + c 2 ) b+c

.

 ‹ a(b2 + bc + c 2 ) bc a bc =a b+c− = a b + ca − , b+c b+c b+c

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Vasile Cîrtoaje

this inequality is equivalent to  ‹ 1 1 1 2a bc + + + a2 + b2 + c 2 ≥ 2(a b + bc + ca), b+c c+a a+b which is just the inequality (a) from P 1.58. Fourth Solution. By direct calculation, we can write the inequality as X X a b(a4 + b4 ) ≥ a2 b2 (a2 + b2 ), which is equivalent to the obvious inequality X a b(a − b)(a3 − b3 ) ≥ 0.

P 1.69. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that Y  a − b ‹2 a(b + c) b(c + a) c(a + b) + + ≥2+4 . b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2 a+b (Vasile Cîrtoaje, 2011) Solution. For b = c = 1, the inequality reduces to a(a − 1)2 ≥ 0. Assume further that a < b < c. According to the first solution of P 1.68, we have X

X a(b + c) bc(b − c)2 − 2 = . b2 + bc + c 2 (a2 + a b + b2 )(a2 + ac + c 2 )

Therefore, it remains to show that X

Y  a − b ‹2 bc(b − c)2 ≥4 . (a2 + a b + b2 )(a2 + ac + c 2 ) a+b

Since (a2 + a b + b2 )(a2 + ac + c 2 ) ≤ (a + b)2 (a + c)2 , it suffices to show that X

Y  a − b ‹2 bc(b − c)2 ≥4 , (a + b)2 (a + c)2 a+b

which is equivalent to X

bc(b + c)2 ≥ 4. (a − b)2 (a − c)2

Symmetric Rational Inequalities

107

We have X

bc(b + c)2 bc(b + c)2 ≥ (a − b)2 (a − c)2 (a − b)2 (a − c)2 bc(b + c)2 (b + c)2 ≥ 4. ≥ = b2 c 2 bc

The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.70. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a b − bc + ca bc − ca + a b ca − a b + bc 3 + + ≥ . b2 + c 2 c 2 + a2 a2 + b2 2 Solution. Use the SOS method. We have X  a b − bc + ca 1 ‹ X (b + c)(2a − b − c) − = b2 + c 2 2 2(b2 + c 2 ) = =

X (b + c)(a − b) 2(b2 + c 2 ) X (b + c)(a − b)

+ +

X (b + c)(a − c) 2(b2 + c 2 ) X (c + a)(b − a)

2(b2 + c 2 ) 2(c 2 + a2 ) X (a − b)2 (a b + bc + ca − c 2 ) = . 2(b2 + c 2 )(c 2 + a2 )

Since a b + bc + ca − c 2 = (b − c)(c − a) + 2a b ≥ (b − c)(c − a), it suffices to show that X (a2 + b2 )(a − b)2 (b − c)(c − a) ≥ 0. This inequality is equivalent to (a − b)(b − c)(c − a)

X (a − b)(a2 + b2 ) ≥ 0,

or (a − b)2 (b − c)2 (c − a)2 ≥ 0. The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

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Vasile Cîrtoaje

P 1.71. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then X a b + (k − 1)bc + ca b2

+ k bc

≥

+ c2

3(k + 1) . k+2 (Vasile Cîrtoaje, 2005)

First Solution. Apply the SOS method. Write the inequality as X • a b + (k − 1)bc + ca b2 + k bc + c 2 X b2

˜ k+1 − ≥ 0, k+2

A ≥ 0, + k bc + c 2

where A = (b + c)(2a − b − c) + k(a b + ac − b2 − c 2 ). Since A =(b + c)[(a − b) + (a − c)] + k[b(a − b) + c(a − c)] = (a − b)[(k + 1)b + c] + (a − c)[(k + 1)c + b], the inequality is equivalent to X (a − b)[(k + 1)b + c] b2 + k bc + c 2 X (a − b)[(k + 1)b + c]

+ +

X (a − c)[(k + 1)c + b] b2 + k bc + c 2 X (b − a)[(k + 1)a + c]

b2 + k bc + c 2 c 2 + kca + a2 X (b − c)2 R a Sa ≥ 0,

≥ 0, ≥ 0,

where R a = b2 + k bc + c 2 , Sa = a(b + c − a) + (k + 1)bc. Without loss of generality, assume that a ≥ b ≥ c. Case 1: k ≥ −1. Since Sa ≥ a(b + c − a), it suffices to show that X a(b + c − a)(b − c)2 R a ≥ 0. We have X

a(b + c − a)(b − c)2 R a ≥ a(b + c − a)(b − c)2 R a + b(c + a − b)(c − a)2 R b ≥ (b − c)2 [a(b + c − a)R a + b(c + a − b)R b ].

Symmetric Rational Inequalities

109

Thus, it is enough to prove that a(b + c − a)R a + b(c + a − b)R b ≥ 0. Since b + c − a ≥ −(c + a − b), we have a(b + c − a)R a + b(c + a − b)R b ≥ (c + a − b)(bR b − aR a ) = (c + a − b)(a − b)(a b − c 2 ) ≥ 0. Case 2: −2 < k ≤ 1. Since Sa = (a − b)(c − a) + (k + 2)bc ≥ (a − b)(c − a), we have

X X (b − c)2 R a Sa ≥ (a − b)(b − c)(c − a) (b − c)R a .

From X X (b − c)R a = (b − c)[b2 + bc + c 2 − (1 − k)bc] X X = (b3 − c 3 ) − (1 − k) bc(b − c) = (1 − k)(a − b)(b − c)(c − a), we get (a − b)(b − c)(c − a)

X (b − c)R a = (1 − k)(a − b)2 (b − c)2 (c − a)2 ≥ 0.

This completes the proof. The equality occurs for a = b = c, and for a = 0 and b = c (or any cyclic permutation). Second Solution. Let p = a + b + c, q = a b + bc + ca, r = a bc. Write the inequality in the form f6 (a, b, c) ≥ 0, where X [a(b + c) + (k − 1)bc](a2 + ka b + b2 )(a2 + kac + c 2 ) f6 (a, b, c) = (k + 2) −3(k + 1) = (k + 2)

Y (b2 + k bc + c 2 )

X [(k − 2)bc + q](ka b − c 2 + p2 − 2q)(kac − b2 + p2 − 2q) Y −3(k + 1) (k bc − a2 + p2 − 2q).

Thus, f6 (a, b, c) has the same highest coefficient A as (k + 2)(k − 2)P2 (a, b, c) − 3(k + 1)P3 (a, b, c), where P2 (a, b, c) =

X

bc(ka b − c 2 )(kac − b2 ),

110

Vasile Cîrtoaje P3 (a, b, c) =

Y (k bc − a2 ).

According to Remark 2 from the proof of P 2.75 in Volume 1, A = (k + 2)(k − 2)P2 (1, 1, 1) − 3(k + 1)P3 (1, 1, 1) = 3(k + 2)(k − 2)(k − 1)2 − 3(k + 1)(k − 1)3 = −9(k − 1)2 . Since A ≤ 0, according to P 3.76-(a) in Volume 1, it suffices to prove the original inequality for b = c = 1, and for a = 0. For these cases, this inequality has the obvious forms (k + 2)a(a − 1)2 ≥ 0 and (b − c)2 [(k + 2)(b2 + c 2 ) + (k2 + k + 1)bc] ≥ 0, respectively. Remark. For k = 1 and k = 0, from P 1.71, we get the inequalities in P 1.68 and P 1.70, respectively. Besides, for k = 2, we get the well-known inequality (Iran 1996): 1 1 1 9 + + ≥ . 2 2 2 (a + b) (b + c) (c + a) 4(a b + bc + ca)

P 1.72. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then X 3bc − a(b + c) b2 + k bc + c 2

≤

3 . k+2 (Vasile Cîrtoaje, 2011)

Solution. Write the inequality in P 1.71 as ˜ X• a b + (k − 1)bc + ca 3 1− ≤ , 2 2 b + k bc + c k+2 X b2 + c 2 + bc − a(b + c) 3 ≤ . b2 + k bc + c 2 k+2 Since b2 + c 2 ≥ 2bc, we get X 3bc − a(b + c) b2

+ k bc

+ c2

≤

3 , k+2

which is just the desired inequality. The equality occurs for a = b = c.

Symmetric Rational Inequalities

111

P 1.73. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that bc + 1 ca + 1 4 ab + 1 + 2 + 2 ≥ . 2 2 2 2 a +b b +c c +a 3 Solution. Write the inequality in the homogeneous form E(a, b, c) ≥ 4, where E(a, b, c) =

4a b + bc + ca 4bc + ca + a b 4ca + a b + bc + + . a2 + b2 b2 + c 2 c 2 + a2

Without loss of generality, assume that a = min{a, b, c}. We will show that E(a, b, c) ≥ E(0, b, c) ≥ 4. For a = 0, we have E(a, b, c) = E(0, b, c), and for a > 0, we have E(a, b, c) − E(0, b, c) 4b2 + c(b − a) b+c 4c 2 + b(c − a) = + + > 0. a b(a2 + b2 ) b2 + c 2 c(c 2 + a2 ) Also,

b 4bc c (b − c)4 + 2 + − 4 = ≥ 0. c b + c2 b bc(b2 + c 2 ) p The equality holds for a = 0 and b = c = 3 (or any cyclic permutation). E(0, b, c) − 4 =

P 1.74. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that 5a b + 1 5bc + 1 5ca + 1 + + ≥ 2. (a + b)2 (b + c)2 (c + a)2 Solution. Write the inequality as E(a, b, c) ≥ 6, where E(a, b, c) =

16a b + bc + ca 16bc + ca + a b 16ca + a b + bc + + . (a + b)2 (b + c)2 (c + a)2

Without loss of generality, assume that a ≤ b ≤ c. Case 1: 16b2 ≥ c(a + b). We will show that E(a, b, c) ≥ E(0, b, c) ≥ 6. For a = 0, we have E(a, b, c) = E(0, b, c), and for a > 0, we have E(a, b, c) − E(0, b, c) 16b2 − c(a + b) 1 16c 2 − b(a + c) = + + > 0. a b(a + b)2 b+c c(c + a)2

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Vasile Cîrtoaje

Also, E(0, b, c) − 6 =

16bc c (b − c)4 b + + − 6 = ≥ 0. c (b + c)2 b bc(b + c)2

Case 2: 16b2 < c(a + b). We have 16a b + 16b2 2(5b − 3a) 16a b + bc + ca − 6 > −6= > 0. 2 2 (a + b) (a + b) a+b p The equality holds for a = 0 and b = c = 3 (or any cyclic permutation). E(a, b, c) − 6 >

P 1.75. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 − bc b2 − ca c2 − a b + + ≥ 0. 2b2 − 3bc + 2c 2 2c 2 − 3ca + 2a2 2a2 − 3a b + 2b2 (Vasile Cîrtoaje, 2005) Solution. The hint is applying the Cauchy-Schwarz inequality after we made the numerators of the fractions to be nonnegative and as small as possible. Thus, we write the inequality as  X a2 − bc + 1 ≥ 3, 2b2 − 3bc + 2c 2 X a2 + 2(b − c)2 ≥ 3. 2b2 − 3bc + 2c 2 Using the Cauchy-Schwarz inequality P P X a2 + 2(b − c)2 (5 a2 − 4 a b)2 ≥P , 2b2 − 3bc + 2c 2 [a2 + 2(b − c)2 ](2b2 − 3bc + 2c 2 ) it suffices to prove that X X X (5 a2 − 4 a b)2 ≥ 3 [a2 + 2(b − c)2 ](2b2 − 3bc + 2c 2 ). This inequality is equivalent to X X X X a4 + a bc a+2 a b(a2 + b2 ) ≥ 6 a2 b2 . We can obtain it by summing Schur’s inequality of degree four X X X a4 + a bc a≥ a b(a2 + b2 ) to the obvious inequality 3

X

a b(a2 + b2 ) ≥ 6

X

a2 b2 .

The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

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113

P 1.76. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 2b2 − ca 2c 2 − a b 2a2 − bc + + ≥ 3. b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2 (Vasile Cîrtoaje, 2005) Solution. Write the inequality such that the numerators of the fractions are nonnegative and as small as possible:  X  2a2 − bc + 1 ≥ 6, b2 − bc + c 2 X 2a2 + (b − c)2

≥ 6. b2 − bc + c 2 Applying the Cauchy-Schwarz inequality, we get P P X 2a2 + (b − c)2 4(2 a2 − a b)2 ≥P . b2 − bc + c 2 [2a2 + (b − c)2 ](b2 − bc + c 2 ) Thus, we still have to prove that X X X 2(2 a2 − a b)2 ≥ 3 [2a2 + (b − c)2 ](b2 − bc + c 2 ). This inequality is equivalent to X X X X 2 a4 + 2a bc a+ a b(a2 + b2 ) ≥ 6 a2 b2 . We can obtain it by summing up Schur’s inequality of degree four X X X a4 + a bc a≥ a b(a2 + b2 ) and

X

a b(a2 + b2 ) ≥ 2

X

a2 b2 ,

multiplied by 2 and 3, respectively. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.77. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 b2 c2 + + ≥ 1. 2b2 − bc + 2c 2 2c 2 − ca + 2a2 2a2 − a b + 2b2 (Vasile Cîrtoaje, 2005)

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Solution. By the Cauchy-Schwarz inequality, we have P X ( a2 )2 a2 ≥P . 2b2 − bc + 2c 2 a2 (2b2 − bc + 2c 2 ) Therefore, it suffices to show that X X ( a2 )2 ≥ a2 (2b2 − bc + 2c 2 ), which is equivalent to X

a4 + a bc

X

a≥2

X

a2 b2 .

This inequality follows by adding Schur’s inequality of degree four X X X a4 + a bc a≥ a b(a2 + b2 ) and

X

a b(a2 + b2 ) ≥ 2

X

a2 b2 .

The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.78. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 4b2

1 1 1 9 + 2 + 2 ≥ . 2 2 2 2 − bc + 4c 4c − ca + 4a 4a − a b + 4b 7(a + b2 + c 2 ) (Vasile Cîrtoaje, 2005)

Solution. We use the SOS method. Without loss of generality, assume that a ≥ b ≥ c. Write the inequality as  X  7(a2 + b2 + c 2 ) − 3 ≥ 0, 4b2 − bc + 4c 2 X 7a2 − 5b2 − 5c 2 + 3bc ≥ 0, 4b2 − bc + 4c 2 X 5(2a2 − b2 − c 2 ) − 3(a2 − bc) ≥ 0. 4b2 − bc + 4c 2 Since 2(a2 − bc) = (a − b)(a + c) + (a − c)(a + b), we have 10(2a2 − b2 − c 2 ) − 6(a2 − bc) = = 10(a2 − b2 ) − 3(a − b)(a + c) + 10(a2 − c 2 ) − 3(a − c)(a + b) = (a − b)(7a + 10b − 3c) + (a − c)(7a + 10c − 3b).

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Thus, we can write the desired inequality as follows X (a − b)(7a + 10b − 3c) 4b2 − bc + 4c 2 X (a − b)(7a + 10b − 3c) 4b2 − bc + 4c 2

+ +

X (a − c)(7a + 10c − 3b) 4b2 − bc + 4c 2

≥ 0,

X (b − a)(7b + 10a − 3c) 4c 2 − ca + 4a2

≥ 0,

X (a − b)2 (28a2 + 28b2 − 9c 2 + 68a b − 19bc − 19ca) (4b2 − bc + 4c 2 )(4c 2 − ca + 4a2 )

,

(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0, where Sa = (4b2 − bc + 4c 2 )[(b − a)(28b + 9a) + c(−19a + 68b + 28c)], S b = (4c 2 − ca + 4a2 )[(a − b)(28a + 9b) + c(−19b + 68a + 28c)], Sc = (4a2 − a b + 4b2 )[(b − c)(28b + 9c) + a(68b − 19c + 28a)]. Since S b ≥ 0 and Sc > 0, it suffices to show that Sa + S b ≥ 0. We have Sa ≥ (4b2 − bc + 4c 2 )[(b − a)(28b + 9a) − 19ac], S b ≥ (4c 2 − ca + 4a2 )[(a − b)(28a + 9b) + 19ac], 19ac[(4c 2 − ca + 4a2 ) − (4b2 − bc + 4c 2 ] = 19ac(a − b)(4a + 4b − c) ≥ 0, and hence Sa + S b ≥ (a − b)[−(4b2 − bc + 4c 2 )(28b + 9a) + (4c 2 − ca + 4a2 )(28a + 9b)] = (a − b)2 [112(a2 + a b + b2 ) + 76c 2 − 28c(a + b)] ≥ 0. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.79. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 2a2 + bc 2b2 + ca 2c 2 + a b 9 + + ≥ . b2 + c 2 c 2 + a2 a2 + b2 2 (Vasile Cîrtoaje, 2005)

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First Solution. We apply the SOS method. Since  X 2a2 − b2 − c 2 X (b − c)2 X  2(2a2 + bc) − 3 = 2 − b2 + c 2 b2 + c 2 b2 + c 2 and X 2a2 − b2 − c 2 b2 + c 2

=

X a2 − b2 b2 + c 2

+

X a2 − c 2

=

X a2 − b2

+

X b2 − a2 c 2 + a2

b2 + c 2 b2 + c 2 ‹ X  X 1 (a2 − b2 )2 1 − = = (a2 − b2 ) 2 b + c 2 c 2 + a2 (b2 + c 2 )(c 2 + a2 ) ≥

X (a − b)2 (a2 + b2 ) (b2 + c 2 )(c 2 + a2 )

we can write the inequality as 2

X (b − c)2 (b2 + c 2 ) X (b − c)2 ≥ , (c 2 + a2 )(a2 + b2 ) b2 + c 2

or (b − c)2 Sa + (c − a)2 S b + (c − a)2 Sc ≥ 0, where Sa = 2(b2 + c 2 )2 − (c 2 + a2 )(a2 + b2 ). Assume that a ≥ b ≥ c. We have S b = 2(c 2 + a2 )2 − (a2 + b2 )(b2 + c 2 ) ≥ 2(c 2 + a2 )(c 2 + b2 ) − (a2 + b2 )(b2 + c 2 ) = (b2 + c 2 )(a2 − b2 + 2c 2 ) ≥ 0, Sc = 2(a2 + b2 )2 − (b2 + c 2 )(c 2 + a2 ) > 0, Sa + S b = (a2 − b2 )2 + 2c 2 (a2 + b2 + 2c 2 ) ≥ 0. Therefore, (b − c)2 Sa + (c − a)2 S b + (c − a)2 Sc ≥ (b − c)2 Sa + (c − a)2 S b ≥ (b − c)2 (Sa + S b ) ≥ 0. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation). Second Solution. Since bc ≥

2b2 c 2 , b2 + c 2

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we have X 2a2 + bc b2 + c 2

≥

2b2 c 2 b2 +c 2 b2 + c 2

X 2a2 +

= 2(a2 b2 + b2 c 2 + c 2 a2 )

X

1 . (b2 + c 2 )2

Therefore, it suffices to show that X

1 9 ≥ , (b2 + c 2 )2 4(a2 b2 + b2 c 2 + c 2 a2 )

which is just the known Iran-1996 inequality (see Remark from P 1.71). Third Solution. We get the desired inequality by summing the inequality in P 1.59-(a), namely 2a2 − 2bc 2b2 − 2ca 2c 2 − 2a b 6(a b + bc + ca) + 2 + 2 + ≥ 6, b2 + c 2 c + a2 a + b2 a2 + b2 + c 2 and the inequality 3bc 3ca 3a b 3 6(a b + bc + ca) + 2 + 2 + ≥ . 2 2 2 +c c +a a +b 2 a2 + b2 + c 2

b2

This inequality is equivalent to ‹ X  2bc 4(a b + bc + ca) +1 ≥ + 2, 2 2 b +c a2 + b2 + c 2 X (b + c)2 b2 + c 2

≥

2(a + b + c)2 . a2 + b2 + c 2

By the Cauchy-Schwarz inequality, we have X (b + c)2 b2 + c 2

P 2 (b + c) 2(a + b + c)2 ≥ P = 2 . a + b2 + c 2 (b2 + c 2 )

P 1.80. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 2a2 + 3bc 2b2 + 3ca 2c 2 + 3a b + + ≥ 5. b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2 (Vasile Cîrtoaje, 2005)

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Solution. We apply the SOS method. Write the inequality as  X  3(2a2 + 3bc) − 5 ≥ 0, b2 + bc + c 2 or

X 6a2 + 4bc − 5b2 − 5c 2 b2 + bc + c 2

≥ 0.

Since 2(a2 − bc) = (a − b)(a + c) + (a − c)(a + b), we have 6a2 + 4bc − 5b2 − 5c 2 = 5(2a2 − b2 − c 2 ) − 4(a2 − bc) = 5(a2 − b2 ) − 2(a − b)(a + c) + 5(a2 − c 2 ) − 2(a − c)(a + b) = (a − b)(3a + 5b − 2c) + (a − c)(3a + 5c − 2b). Thus, we can write the desired inequality as follows X (a − b)(3a + 5b − 2c) b2 + bc + c 2 X (a − b)(3a + 5b − 2c)

+ +

X (a − c)(3a + 5c − 2b) b2 + bc + c 2 X (b − a)(3b + 5a − 2c)

b2 + bc + c 2 c 2 + ca + a2 X (a − b)2 (3a2 + 3b2 − 4c 2 + 8a b + bc + ca) (b2 + bc + c 2 )(c 2 + ca + a2 )

≥ 0, ≥ 0,

,

(b − c)2 Sa + (c − a)2 S b + (a − b)2 Sc ≥ 0, where Sa = (b2 + bc + c 2 )(3b2 + 3c 2 − 4a2 + a b + 8bc + ca). Assume that a ≥ b ≥ c. Since S b = (c 2 + ca + a2 )[(a − b)(3a + 4b) + c(8a + b + c)] ≥ 0 and Sc = (a2 + a b + b2 )[(b − c)(3b + 4c) + a(8b + c + a)] > 0, it suffices to show that Sa + S b ≥ 0. We have Sa + S b ≥ (b2 + bc + c 2 )(b − a)(3b + 4a) + (c 2 + ca + a2 )(a − b)(3a + 4b) = (a − b)2 [3(a + b)(a + b + c) + a b − c 2 ] ≥ 0. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

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P 1.81. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 21 2a2 + 5bc 2b2 + 5ca 2c 2 + 5a b + + ≥ . (b + c)2 (c + a)2 (a + b)2 4 (Vasile Cîrtoaje, 2005) Solution. Use the SOS method.Write the inequality as follows X  2a2 + 5bc 7  ≥ 0, − (b + c)2 4 X 4(a2 − b2 ) + 4(a2 − c 2 ) − 3(b − c)2 (b + c)2 X c 2 − b2

≥ 0,

X (b − c)2 X b2 − c 2 + 4 − 3 ≥ 0, (c + a)2 (a + b)2 (b + c)2 X (b − c)2 (b + c)(2a + b + c) X (b − c)2 4 − 3 ≥ 0. (c + a)2 (a + b)2 (b + c)2 Substituting b + c = x, c + a = y and a + b = z, we can rewrite the inequality in the form ( y − z)2 S x + (z − x)2 S y + (x − y)2 Sz ≥ 0, 4

where S x = 4x 3 ( y + z) − 3 y 2 z 2 , S y = 4 y 3 (z + x) − 3z 2 x 2 , Sz = 4z 3 (x + y) − 3x 2 y 2 . Without loss of generality, assume that 0 < x ≤ y ≤ z, z ≤ x + y. We have Sz > 0 and S y ≥ 4x 2 y(z + x) − 3x 2 z(x + y) = x 2 [4x y + z( y − 3x)] ≥ 0, since for the nontrivial case y − 3x < 0, we get 4x y + z( y − 3x) ≥ 4x y + (x + y)( y − 3x) = x 2 (3x + y)( y − x) ≥ 0. Thus, it suffices to show that S x + S y ≥ 0. Since S x + S y = 4x y(x 2 + y 2 ) + 4(x 3 + y 3 )z − 3(x 2 + y 2 )z 2 ≥ 4x y(x 2 + y 2 ) + 4(x 3 + y 3 )z − 3(x 2 + y 2 )(x + y)z = 4x y(x 2 + y 2 ) + (x 2 − 4x y + y 2 )(x + y)z, for the nontrivial case x 2 − 4x y + y 2 < 0, we get S x + S y ≥ 4x y(x 2 + y 2 ) + (x 2 − 4x y + y 2 )(x + y)2 ≥ 2x y(x + y)2 + (x 2 − 4x y + y 2 )(x + y)2 = (x − y)2 (x + y)2 . The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

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P 1.82. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then X 2a2 + (2k + 1)bc b2 + k bc + c 2

≥

3(2k + 3) . k+2 (Vasile Cîrtoaje, 2005)

First Solution. There are two cases to consider. Case 1: −2 < k ≤ −1/2. Write the inequality as X  2a2 + (2k + 1)bc 2k + 1  6 − ≥ , 2 2 b + k bc + c k+2 k+2 X 2(k + 2)a2 − (2k + 1)(b − c)2 b2 + k bc + c 2

≥ 6.

Since 2(k + 2)a2 − (2k + 1)(b − c)2 ≥ 0 for −2 < k ≤ −1/2, we can apply the CauchySchwarz inequality. Thus, it suffices to show that P P [2(k + 2) a2 − (2k + 1) (b − c)2 ]2 P ≥ 6, [2(k + 2)a2 − (2k + 1)(b − c)2 ](b2 + k bc + c 2 ) which is equivalent to each of the following inequalities P P 2[(1 − k) a2 + (2k + 1) a b]2 P ≥ 3, [2(k + 2)a2 − (2k + 1)(b − c)2 ](b2 + k bc + c 2 ) X X X X 2(k + 2) a4 + 2(k + 2)a bc a − (2k + 1) a b(a2 + b2 ) ≥ 6 a2 b2 , X X X X 2(k + 2)[ a4 + a bc a− a b(a2 + b2 )] + 3 a b(a − b)2 ≥ 0. The last inequality is true since, by Schur’s inequality of degree four, we have X X X a4 + a bc a− a b(a2 + b2 ) ≥ 0. Case 2: k ≥ −9/5. Use the SOS method. Without loss of generality, assume that a ≥ b ≥ c. Write the inequality as X  2a2 + (2k + 1)bc 2k + 3  − ≥ 0, b2 + k bc + c 2 k+2 X 2(k + 2)a2 − (2k + 3)(b2 + c 2 ) + 2(k + 1)bc b2 + k bc + c 2 X (2k + 3)(2a2 − b2 − c 2 ) − 2(k + 1)(a2 − bc) b2 + k bc + c 2

≥ 0, ≥ 0.

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Since 2(a2 − bc) = (a − b)(a + c) + (a − c)(a + b), we have (2k + 3)(2a2 − b2 − c 2 ) − 2(k + 1)(a2 − bc) = = (2k + 3)(a2 − b2 ) − (k + 1)(a − b)(a + c) + (2k + 3)(a2 − c 2 ) − (k + 1)(a − c)(a + b) = (a − b)[(k + 2)a + (2k + 3)b − (k + 1)c] + (a − c)[(k + 2)a + (2k + 3)c − (k + 1)b). Thus, we can write the desired inequality as X (a − b)[(k + 2)a + (2k + 3)b − (k + 1)c] b2 + k bc + c 2 + or

X (a − c)[(k + 2)a + (2k + 3)c − (k + 1)b] b2 + k bc + c 2

+ ≥ 0,

X (a − b)[(k + 2)a + (2k + 3)b − (k + 1)c]

+ b2 + k bc + c 2 X (b − a)[(k + 2)b + (2k + 3)a − (k + 1)c] ≥ 0, c 2 + kca + a2 or (b − c)2 R a Sa + (c − a)2 R b S b + (a − b)2 R c Sc ≥ 0, where R a = b2 + k bc + c 2 , R b = c 2 + kca + a2 , R c = a2 + ka b + b2 , Sa = (k + 2)(b2 + c 2 ) − (k + 1)2 a2 + (3k + 5)bc + (k2 + k − 1)a(b + c) = −(a − b)[(k + 1)2 a + (k + 2)b] + c[(k2 + k − 1)a + (3k + 5)b + (k + 2)c], S b = (k + 2)(c 2 + a2 ) − (k + 1)2 b2 + (3k + 5)ca + (k2 + k − 1)b(c + a) = (a − b)[(k + 2)a + (k + 1)2 b] + c[(3k + 5)a + (k2 + k − 1)b + (k + 2)c], Sc = (k + 2)(a2 + b2 ) − (k + 1)2 c 2 + (3k + 5)a b + (k2 + k − 1)c(a + b) = (k + 2)(a2 + b2 ) + (3k + 5)a b + c[(k2 + k − 1)(a + b) − (k + 1)2 c] ≥ (5k + 9)a b + c[(k2 + k − 1)(a + b) − (k + 1)2 c]. We have S b ≥ 0, since for the nontrivial case (3k + 5)a + (k2 + k − 1)b + (k + 2)c < 0, we get S b ≥ (a − b)[(k + 2)a + (k + 1)2 b] + b[(3k + 5)a + (k2 + k − 1)b + (k + 2)c] = (k + 2)(a2 − b2 ) + (k + 2)2 a b + (k + 2)bc > 0.

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Also, we have Sc ≥ 0 for k ≥ −9/5, since (5k + 9)a b + c[(k2 + k − 1)(a + b) − (k + 1)2 c] ≥ ≥ (5k + 9)ac + c[(k2 + k − 1)(a + b) − (k + 1)2 c] = (k + 2)(k + 4)ac + (k2 + k − 1)bc − (k + 1)2 c 2 ≥ (2k2 + 7k + 7)bc − (k + 1)2 c 2 ≥ (k + 2)(k + 3)c 2 ≥ 0. Therefore, it suffices to show that R a Sa + R b S b ≥ 0. Since bR b − aR a = (a − b)(a b − c 2 ) ≥ 0 and R a Sa + R b S b ≥ R a (Sa + it suffices to show that Sa +

a S b ), b

a S b ≥ 0. b

We have bSa + aS b = (k + 2)(a + b)(a − b)2 + c f (a, b, c) ≥ 2(k + 2)b(a − b)2 + c f (a, b, c), Sa +

c a S b ≥ 2(k + 2)(a − b)2 + f (a, b, c), b b

where f (a, b, c) = b[(k2 + k − 1)a + (3k + 5)b] + a[(3k + 5)a + (k2 + k − 1)b] +(k + 2)c(a + b) = (3k + 5)(a2 + b2 ) + 2(k2 + k − 1)a b + (k + 2)c(a + b). For the nontrivial case f (a, b, c) < 0, we have Sa +

a S b ≥ 2(k + 2)(a − b)2 + f (a, b, c) b

≥ 2(k + 2)(a − b)2 + (3k + 5)(a2 + b2 ) + 2(k2 + k − 1)a b = (5k + 9)(a2 + b2 ) + 2(k2 − k − 5)a b ≥ 2(k + 2)2 a b ≥ 0. The proof is completed. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation). Second Solution. Let p = a + b + c,

q = a b + bc + ca.

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Write the inequality as f6 (a, b, c) ≥ 0, where X f6 (a, b, c) = (k + 2) [2a2 + (2k + 1)bc](a2 + ka b + b2 )(a2 + kac + c 2 ) −3(2k + 3)

Y (b2 + k bc + c 2 ).

Since (a2 + ka b + b2 )(a2 + kac + c 2 ) = (p2 − 2q + ka b − c 2 )(p2 − 2q + kac − b2 ), b2 + k bc + c 2 = p2 − 2q + k bc − a2 , f6 (a, b, c has the same highest coefficient A as (k + 2)P2 (a, b, c) − 3(2k + 3)P3 (a, b, c), where P2 (a, b, c) =

X Y [2a2 +(2k +1)bc](ka b − c 2 )(kac − b2 ), P3 (a, b, c) = (b2 + k bc + c 2 ).

According to Remark 2 from the proof of P 2.75 in Volume 1, we have A = (k + 2)P2 (1, 1, 1) − 3(2k + 3)P3 (1, 1, 1) = 9(2k + 3)(k − 1)2 . On the other hand, f6 (a, 1, 1) = 2(k + 2)a(a2 + ka + 1)(a − 1)2 (a + k + 2) ≥ 0,   f6 (0, b, c) = (b − c)2 2(k + 2)(b2 + c 2 )2 + 2(k + 2)2 bc(b2 + c 2 ) + (4k2 + 6k − 1)b2 c 2 . For −2 < k ≤ −3/2, we have A ≤ 0. According to P 3.76-(a) in Volume 1, it suffices to show that f6 (a, 1, 1) ≥ 0 and f6 (0, b, c) ≥ 0 for all a, b, c ≥ 0. The first condition is clearly satisfied. The second condition is also satisfied since 2(k + 2)(b2 + c 2 )2 + (4k2 + 6k − 1)b2 c 2 ≥ [8(k + 2) + 4k2 + 6k − 1]b2 c 2 = (4k2 + 14k + 15)b2 c 2 ≥ 0. For k > −3/2, when A ≥ 0, we will apply the highest coefficient cancellation method. Consider two cases: p2 ≤ 4q and p2 > 4q. Case 1: p2 ≤ 4q. Since f6 (1, 1, 1) = f6 (0, 1, 1) = 0, define the homogeneous function P(a, b, c) = a bc + B(a + b + c)3 + C(a + b + c)(a b + bc + ca)

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such that P(1, 1, 1) = P(0, 1, 1) = 0; that is, 4 1 P(a, b, c) = a bc + (a + b + c)3 − (a + b + c)(a b + bc + ca). 9 9 We will prove the sharper inequality g6 (a, b, c) ≥ 0, where g6 (a, b, c) = f6 (a, b, c) − 9(2k + 3)(k − 1)2 P 2 (a, b, c). Clearly, g6 (a, b, c) has the highest coefficient A = 0. Then, according to Remark 1 from the proof of P 3.76 in Volume 1, it suffices to prove that g6 (a, 1, 1) ≥ 0 for 0 ≤ a ≤ 4. We have a(a − 1)2 P(a, 1, 1) = , 9 hence g6 (a, 1, 1) = f6 (a, 1, 1) − 9(2k + 3)(k − 1)2 P 2 (a, 1, 1) =

a(a − 1)2 g(a) , 9

where g(a) = 18(k + 2)(a2 + ka + 1)(a + k + 2) − (2k + 3)(k − 1)2 a(a − 1)2 . Since a2 + ka + 1 ≥ (k + 2)a, it suffices to show that 18(k + 2)2 (a + k + 2) ≥ (2k + 3)(k − 1)2 (a − 1)2 . Also, since (a − 1)2 ≤ 2a + 1, it is enough to prove that h(a) ≥ 0, where h(a) = 18(k + 2)2 (a + k + 2) − (2k + 3)(k − 1)2 (2a + 1). Since h(a) is a linear function, the inequality h(a) ≥ 0 is true if h(0) ≥ 0 and h(4) ≥ 0. Setting x = 2k + 3, x > 0, we get h(0) = 18(k + 2)3 − (2k + 3)(k − 1)2 =

1 (8x 3 + 37x 2 + 2x + 9) > 0. 4

Also, 1 h(4) = 2(k + 2)2 (k + 6) − (2k + 3)(k − 1)2 = 3(7k2 + 20k + 15) > 0. 9 Case 2: p2 > 4q. We will prove the sharper inequality g6 (a, b, c) ≥ 0, where g6 (a, b, c) = f6 (a, b, c) − 9(2k + 3)(k − 1)2 a2 b2 c 2 .

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125

We see that g6 (a, b, c) has the highest coefficient A = 0. According to Remark 1 from the proof of P 3.76 in Volume 1, it suffices to prove that g6 (a, 1, 1) ≥ 0 for a > 4 and g6 (0, b, c) ≥ 0 for all b, c ≥ 0. We have g6 (a, 1, 1) = f6 (a, 1, 1) − 9(2k + 3)(k − 1)2 a2 = a[2(k + 2)(a2 + ka + 1)(a − 1)2 (a + k + 2) − 9(2k + 3)(k − 1)2 a]. Since a2 + ka + 1 > (k + 2)a,

(a − 1)2 > 9,

it suffices to show that 2(k + 2)2 (a + k + 2) ≥ (2k + 3)(k − 1)2 . Indeed, 2(k + 2)2 (a + k + 2) − (2k + 3)(k − 1)2 > 2(k + 2)2 (k + 6) − (2k + 3)(k − 1)2 = 3(7k2 + 20k + 15) > 0. Also, g6 (0, b, c) = f6 (0, b, c) ≥ 0.

P 1.83. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > −2, then X

3bc − 2a2 3 ≤ . b2 + k bc + c 2 k+2 (Vasile Cîrtoaje, 2011)

First Solution. Write the inequality as  X  2a2 − 3bc 3 6 + ≥ , b2 + k bc + c 2 k + 2 k+2 X 2(k + 2)a2 + 3(b − c)2 b2 + k bc + c 2

≥ 6.

Applying the Cauchy-Schwarz inequality, it suffices to show that P P [2(k + 2) a2 + 3 (b − c)2 ]2 P ≥ 6, [2(k + 2)a2 + 3(b − c)2 ](b2 + k bc + c 2 )

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which is equivalent to each of the following inequalities P P 2[(k + 5) a2 − 3 a b]2 P ≥ 3, [2(k + 2)a2 + 3(b − c)2 ](b2 + k bc + c 2 ) X X X X 2(k + 8) a4 + 2(2k + 19) a2 b2 ≥ 6(k + 2)a bc a + 21 a b(a2 + b2 ), 2(k + 2) f (a, b, c) + 3g(a, b, c) ≥ 0, where

X X X f (a, b, c) = a4 + 2 a2 b2 − 3a bc a, X X X a4 + 10 a2 b2 − 7 a b(a2 + b2 ). g(a, b, c) = 4

We need to show that f (a, b, c) ≥ 0 and g(a, b, c) ≥ 0. Indeed, X X X X f (a, b, c) = ( a2 )2 − 3a bc a≥( a b)2 − 3a bc a≥0 and X [2(a4 + b4 ) + 10a2 b2 − 7a b(a2 + b2 )] X = (a − b)2 (2a2 − 3a b + 2b2 ) ≥ 0.

g(a, b, c) =

The equality occurs for a = b = c. Second Solution. Write the inequality in P 1.82 as  X 2a2 + (2k + 1)bc 3 2− ≤ , b2 + k bc + c 2 k+2 X 2(b2 + c 2 ) − bc − 2a2 b2

+ k bc

+ c2

≤

3 . k+2

Since b + c ≥ 2bc, we get 2

2

X

3bc − 2a2 3 ≤ , b2 + k bc + c 2 k+2

which is just the desired inequality.

P 1.84. If a, b, c are nonnegative real numbers, no two of which are zero, then a2 + 16bc b2 + 16ca c 2 + 16a b + 2 + 2 ≥ 10. b2 + c 2 c + a2 a + b2 (Vasile Cîrtoaje, 2005)

Symmetric Rational Inequalities

127

Solution. Let a ≤ b ≤ c, and E(a, b, c) =

a2 + 16bc b2 + 16ca c 2 + 16a b + 2 + 2 . b2 + c 2 c + a2 a + b2

Consider two cases. Case 1: 16b3 ≥ ac 2 . We will show that E(a, b, c) ≥ E(0, b, c) ≥ 10. We have E(a, b, c) − E(0, b, c) =

a2 a(16c 3 − a b2 ) a(16b3 − ac 2 ) + + 2 2 ≥ 0, b2 + c 2 c 2 (c 2 + a2 ) b (a + b2 )

since c 3 − a b2 ≥ 0 and 16b3 − ac 2 ≥ 0. Also, 16bc b2 c 2 + + 2 − 10 b2 + c 2 c 2 b (b − c)4 (b2 + c 2 + 4bc) ≥ 0. = b2 c 2 (b2 + c 2 )

E(0, b, c) − 10 =

Case 2: 16b3 ≤ ac 2 . It suffices to show that c 2 + 16a b ≥ 10. a2 + b2 Indeed, 16b3 + 16a b c + 16a b a − 10 ≥ − 10 a2 + b2 a2 + b2 16b = − 10 ≥ 16 − 10 > 0. a 2

This completes the proof. The equality holds for a = 0 and b = c (or any cyclic permutation).

P 1.85. If a, b, c are nonnegative real numbers, no two of which are zero, then a2 + 128bc b2 + 128ca c 2 + 128a b + + ≥ 46. b2 + c 2 c 2 + a2 a2 + b2 (Vasile Cîrtoaje, 2005)

128

Vasile Cîrtoaje

Solution. Let a ≤ b ≤ c, and E(a, b, c) =

a2 + 128bc b2 + 128ca c 2 + 128a b + + . b2 + c 2 c 2 + a2 a2 + b2

Consider two cases. Case 1: 128b3 ≥ ac 2 . We will show that E(a, b, c) ≥ E(0, b, c) ≥ 46. We have E(a, b, c) − E(0, b, c) =

a(128c 3 − a b2 ) a(128b3 − ac 2 ) a2 + + ≥ 0, b2 + c 2 c 2 (c 2 + a2 ) b2 (a2 + b2 )

since c 3 − a b2 ≥ 0 and 128b3 − ac 2 ≥ 0. Also, 128bc b2 c 2 + + 2 − 46 b2 + c 2 c 2 b 2 2 2 2 (b + c − 4bc) (b + c 2 + 8bc) ≥ 0. = b2 c 2 (b2 + c 2 )

E(0, b, c) − 46 =

Case 2: 128b3 ≤ ac 2 . It suffices to show that c 2 + 128a b ≥ 46. a2 + b2 Indeed, 128b3 + 128a b c + 128a b a − 46 ≥ − 46 a2 + b2 a2 + b2 128b − 46 ≥ 128 − 46 > 0. = a 2

This completes the proof. The equality holds for a = 0 and permutation).

c b + = 4 (or any cyclic c b

P 1.86. If a, b, c are nonnegative real numbers, no two of which are zero, then a2 + 64bc b2 + 64ca c 2 + 64a b + + ≥ 18. (b + c)2 (c + a)2 (a + b)2 (Vasile Cîrtoaje, 2005)

Symmetric Rational Inequalities

129

Solution. Let a ≤ b ≤ c, and E(a, b, c) =

a2 + 64bc b2 + 64ca c 2 + 64a b + + . (b + c)2 (c + a)2 (a + b)2

Consider two cases. Case 1: 64b3 ≥ c 2 (a + 2b). We will show that E(a, b, c) ≥ E(0, b, c) ≥ 18. We have E(a, b, c) − E(0, b, c) = ≥

a2 a[64c 3 − b2 (a + 2c)] a[64b3 − c 2 (a + 2b)] + + (b + c)2 c 2 (c + a)2 b2 (a + b)2

a[64c 3 − b2 (a + 2c]) a[64b2 c − b2 (c + 2c)] 61a b2 c ≥ = ≥ 0. c 2 (c + a)2 c 2 (c + a)2 c 2 (c + a)2

Also, 64bc b2 c 2 + + 2 − 18 (b + c)2 c 2 b 4 2 2 (b − c) (b + c + 6bc) = ≥ 0. b2 c 2 (b + c)2

E(0, b, c) − 18 =

Case 2: 64b3 ≤ c 2 (a + 2b). It suffices to show that c 2 + 64a b ≥ 18. (a + b)2 Indeed, 64b3 + 64a b c + 64a b a + 2b − 18 ≥ − 18 (a + b)2 (a + b)2 64b 64 = − 18 ≥ − 18 > 0. a + 2b 3 2

This completes the proof. The equality holds for a = 0 and b = c (or any cyclic permutation).

P 1.87. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ −1, then X a2 (b + c) + ka bc b2 + k bc + c 2

≥ a + b + c.

130

Vasile Cîrtoaje

Solution. We apply the SOS method. Write the inequality as follows X  a2 (b + c) + ka bc b2 + k bc + c 2

 − a ≥ 0,

X a(a b + ac − b2 − c 2 )

≥ 0, b2 + k bc + c 2 X a b(a − b) X ac(a − c) + ≥ 0, b2 + k bc + c 2 b2 + k bc + c 2 X a b(a − b) X ba(b − a) + ≥ 0, b2 + k bc + c 2 c 2 + kca + a2 X a b(a2 + ka b + b2 )(a + b + kc)(a − b)2 ≥ 0. Without loss of generality, assume that a ≥ b ≥ c. Since a + b + kc ≥ a + b − c > 0, it suffices to show that b(b2 + k bc + c 2 )(b + c + ka)(b − c)2 + a(c 2 + kca + a2 )(c + a + k b)(c − a)2 ≥ 0. Since c + a + k b ≥ c + a − b ≥ 0 and c 2 + kca + a2 ≥ b2 + k bc + c 2 , it is enough to prove that b(b + c + ka)(b − c)2 + a(c + a + k b)(c − a)2 ≥ 0. We have b(b + c + ka)(b − c)2 + a(c + a + k b)(c − a)2 ≥ ≥ [b(b + c + ka) + a(c + a + k b)](b − c)2 = [a2 + b2 + 2ka b + c(a + b)](b − c)2 ≥ [(a − b)2 + c(a + b)](b − c)2 ≥ 0. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.88. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ X a3 + (k + 1)a bc b2 + k bc + c 2

−3 , then 2

≥ a + b + c. (Vasile Cîrtoaje, 2009)

Symmetric Rational Inequalities

131

Solution. Use the SOS method. Write the inequality as follows  X  a3 + (k + 1)a bc X a3 − a(b2 − bc + c 2 ) − a ≥ 0, ≥ 0, b2 + k bc + c 2 b2 + k bc + c 2 X (b + c)a3 − a(b3 + c 3 ) (b + c)(b2 + k bc + c 2 ) X

≥ 0,

X a b(a2 − b2 ) + ac(a2 − c 2 ) (b + c)(b2 + k bc + c 2 )

≥ 0,

X ba(b2 − a2 ) a b(a2 − b2 ) + ≥ 0, (b + c)(b2 + k bc + c 2 ) (c + a)(c 2 + kca + a2 )

X (a2 − b2 )2 a b(a2 + ka b + b2 )[a2 + b2 + a b + (k + 1)c(a + b + c)] ≥ 0, X (b2 − c 2 )2 bc(b2 + k bc + c 2 )Sa ≥ 0, where Sa = b2 + c 2 + bc + (k + 1)a(a + b + c). Since Sc > 0, it suffices to show that (b2 − c 2 )2 b(b2 + k bc + c 2 )Sa + (c 2 − a2 )2 a(c 2 + kca + a2 )S b ≥ 0. Since (c 2 − a2 )2 ≥ (b2 − c 2 )2 , a ≥ b, c 2 + kca + a2 − (b2 + k bc + c 2 ) = (a − b)(a + b + kc) ≥ 0, and S b = a2 + c 2 + ac + (k + 1)b(a + b + c) ≥ a2 + c 2 + ac − =

(a − b)(2a + b) + c(2a + 2c − b) ≥ 0, 2

1 b(a + b + c) 2

it is enough to show that Sa + S b ≥ 0. Indeed, Sa + S b = a2 + b2 + 2c 2 + c(a + b) + (k + 1)(a + b)(a + b + c) 1 ≥ a2 + b2 + 2c 2 + c(a + b) − (a + b)(a + b + c) 2 (a − b)2 + c(a + b + 4c) = ≥ 0. 2 This completes the proof. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

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Vasile Cîrtoaje

P 1.89. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > 0, then 2a k − b k − c k 2b k − c k − a k 2c k − a k − b k + 2 + 2 ≥ 0. b2 − bc + c 2 c − ca + a2 a − a b + b2 (Vasile Cîrtoaje, 2004) Solution. Let X = bk − c k ,

Y = c k − ak ,

Z = ak − bk ,

A = b2 − bc + c 2 , B = c 2 − ca + a2 , C = a2 − a b + b2 . Without loss of generality, assume that a ≥ b ≥ c. This involves A ≤ B, A ≤ C, X ≥ 0, and Z ≥ 0. Since X 2a k − b k − c k b2

− bc

+ c2

X + 2Z X − Z 2X + Z + − B‹ A C ‹ 1 1 2 2 1 1 =X + − +Z − − , A B C A B C =

it suffices to prove that 1 1 2 + − ≥ 0. A B C Write this inequality as 1 1 1 1 − ≥ − , A C C B that is, (a − c)(a + c − b)(a2 − ac + c 2 ) ≥ (b − c)(a − b − c)(b2 − bc + c 2 ). For the nontrivial case a > b + c, we can obtain this inequality from a + c − b ≥ a − b − c, a − c ≥ b − c, a2 − ac + c 2 > b2 − bc + c 2 . This completes the proof. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

Symmetric Rational Inequalities

133

P 1.90. If a, b, c are the lengths of the sides of a triangle, then c+a−b a+b−c 2(a + b + c) b+c−a + 2 + 2 ≥ 2 ; 2 2 2 − bc + c c − ca + a a − ab + b a + b2 + c 2

(a)

b2

a2 − 2bc b2 − 2ca c 2 − 2a b + + ≤ 0. b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2

(b)

(Vasile Cîrtoaje, 2009) Solution. (a) By the Cauchy-Schwarz inequality, we get P X b+c−a [ (b + c − a)]2 ≥P b2 − bc + c 2 (b + c − a)(b2 − bc + c 2 ) P ( a)2 P = P . 2 a3 − a2 (b + c) + 3a bc On the other hand, from (b + c − a)(c + a − b)(a + b − c) ≥ 0, we get 2a bc ≤

X

a2 (b + c) −

X

a3 ,

and hence 2

X

3

a −

X

P 2

a (b + c) + 3a bc ≤

Therefore, X

a3 +

P

a2 (b + c)

2

=

P P ( a)( a2 ) 2

.

P 2 a b+c−a ≥ P . b2 − bc + c 2 a2

The equality holds for a degenerate triangle with a = b + c (or any cyclic permutation). (b) Since

a2 − 2bc (b − c)2 + (b + c)2 − a2 = 2 − , b2 − bc + c 2 b2 − bc + c 2 we can write the inequality as X

X

X b+c−a (b − c)2 + (a + b + c) ≥ 6. b2 − bc + c 2 b2 − bc + c 2

Using the inequality in (a), it suffices to prove that X

(b − c)2 2(a + b + c)2 + ≥ 6. b2 − bc + c 2 a2 + b2 + c 2

134

Vasile Cîrtoaje

Write this inequality as X

X 2(b − c)2 (b − c)2 ≥ , b2 − bc + c 2 a2 + b2 + c 2

or, equivalently, X (b − c)2 (a − b + c)(a + b − c)

≥ 0, b2 − bc + c 2 which is true. The equality holds for degenerate triangles with either a/2 = b = c (or any cyclic permutation), or a = 0 and b = c (or any cyclic permutation). Remark. The following generalization of the inequality in (b) holds (Vasile Cîrtoaje, 2009): • Let a, b, c be the lengths of the sides of a triangle. If k ≥ −1, then X a2 − 2(k + 2)bc b2 + k bc + c 2

≤ 0.

with equality for a = 0 and b = c (or any cyclic permutation).

P 1.91. If a, b, c are nonnegative real numbers, then a2 b2 c2 1 + + ≤ . 2 2 2 2 2 2 5a + (b + c) 5b + (c + a) 5c + (a + b) 3 (Vo Quoc Ba Can, 2009) Solution. Apply the Cauchy-Schwarz inequality in the following manner (1 + 2)2 1 2 9 = ≤ 2 + 2 . 2 2 2 2 2 2 2 2 5a + (b + c) (a + b + c ) + 2(2a + bc) a +b +c 2a + bc Then, X

X X 2a2 X 9a2 a2 bc ≤ + = 4 − , 5a2 + (b + c)2 a2 + b2 + c 2 2a2 + bc 2a2 + bc

and it remains to show that

bc ≥ 1. + bc This is a known inequality, which can be proved by the Cauchy-Schwarz inequality, as follows P X ( bc)2 bc ≥P = 1. 2a2 + bc bc(2a2 + bc) X

2a2

The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

Symmetric Rational Inequalities

135

P 1.92. If a, b, c are nonnegative real numbers, then c 2 + a2 − b2 a2 + b2 − c 2 1 b2 + c 2 − a2 + + ≥ . 2 2 2 2 2 2 2a + (b + c) 2b + (c + a) 2c + (a + b) 2 (Vasile Cîrtoaje, 2011) Solution. We apply the SOS method. Write the inequality as follows  X  b2 + c 2 − a2 1 − ≥ 0, 2a2 + (b + c)2 6 X 5(b2 + c 2 − 2a2 ) + 2(a2 − bc) 2a2 + (b + c)2

≥ 0;

X 5(b2 − a2 ) + 5(c 2 − a2 ) + (a − b)(a + c) + (a − c)(a + b)

≥ 0; 2a2 + (b + c)2 X (b − a)[5(b + a) − (a + c)] X (c − a)[5(c + a) − (a + b)] + ≥ 0; 2a2 + (b + c)2 2a2 + (b + c)2 X (b − a)[5(b + a) − (a + c)] X (a − b)[5(a + b) − (b + c)] + ≥ 0; 2a2 + (b + c)2 2b2 + (c + a)2 X (a − b)2 [2c 2 + (a + b)2 ][2(a2 + b2 ) + c 2 + 3a b − 3c(a + b)] ≥ 0, X (b − c)2 R a Sa ≥ 0, where R a = 2a2 + (b + c)2 ,

Sa = a2 + 2(b2 + c 2 ) + 3bc − 3a(b + c).

Without loss of generality, assume that a ≥ b ≥ c. We have S b = b2 + 2(c 2 + a2 ) + 3ca − 3b(c + a) = (a − b)(2a − b) + 2c 2 + 3c(a − b) ≥ 0, Sc = c 2 + 2(a2 + b2 ) + 3a b − 3c(a + b) ≥ 7a b − 3c(a + b) ≥ 3a(b − c) + 3b(a − c) ≥ 0, Sa + S b = 3(a − b)2 + 4c 2 ≥ 0. Since

X (b − c)2 R a Sa ≥ (b − c)2 R a Sa + (c − a)2 R b S b = (b − c)2 R a (Sa + S b ) + [(c − a)2 R b − (b − c)2 R a ]S b ,

it suffices to prove that (a − c)2 R b ≥ (b − c)2 R a . We can get this by multiplying the inequalities b2 (a − c)2 ≥ a2 (b − c)2

136

Vasile Cîrtoaje

and a2 R b ≥ b2 R a . The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.93. Let a, b, c be positive real numbers. If k > 0, then 3a2 − 2bc 3b2 − 2ca 3c 2 − 2a b 3 + + ≤ . 2 2 2 2 2 2 ka + (b − c) k b + (c − a) kc + (a − b) k (Vasile Cîrtoaje, 2011) Solution. Use the SOS method. Write the inequality as follows  X1 3a2 − 2bc − ≥ 0, k ka2 + (b − c)2 X b2 + c 2 − 2a2 + 2(k − 1)(bc − a2 ) ≥ 0; ka2 + (b − c)2 X (b2 − a2 ) + (c 2 − a2 ) + (k − 1)[(a + b)(c − a) + (a + c)(b − a)]

≥ 0; ka2 + (b − c)2 X (b − a)[b + a + (k − 1)(a + c)] X (c − a)[c + a + (k − 1)(a + b)] + ≥ 0; ka2 + (b − c)2 ka2 + (b − c)2 X (b − a)[b + a + (k − 1)(a + c)] X (a − b)[a + b + (k − 1)(b + c)] + ≥ 0; ka2 + (b − c)2 k b2 + (c − a)2 X (a − b)2 [kc 2 + (a − b)2 ][(k − 1)c 2 + 2c(a + b) + (k2 − 1)(a b + bc + ca)] ≥ 0. For k ≥ 1, the inequality is clearly true. Consider further that 0 < k < 1. Since (k − 1)c 2 + 2c(a + b) + (k2 − 1)(a b + bc + ca) > > −c 2 + 2c(a + b) − (a b + bc + ca) = (b − c)(c − a), it suffices to prove that (a − b)(b − c)(c − a)

X (a − b)[kc 2 + (a − b)2 ] ≥ 0.

Since X X X (a − b)[kc 2 + (a − b)2 ] = k (a − b)c 2 + (a − b)3 = (3 − k)(a − b)(b − c)(c − a),

Symmetric Rational Inequalities

137

we have (a − b)(b − c)(c − a)

X (a − b)[kc 2 + (a − b)2 ] =

= (3 − k)(a − b)2 (b − c)2 (c − a)2 ≥ 0. This completes the proof. The equality holds for a = b = c.

P 1.94. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ 3 + then (a)

(b)

a2 ka2

p 7,

b c 9 a + 2 + 2 ≥ ; + k bc b + kca c + ka b (1 + k)(a + b + c)

1 1 1 9 + 2 + 2 ≥ . + bc k b + ca kc + a b (k + 1)(a b + bc + ca) (Vasile Cîrtoaje, 2005)

Solution. (a) Assume that a = max{a, b, c}. Setting t = (b + c)/2, t ≤ a, by the Cauchy-Schwarz inequality, we get b c (b + c)2 4t 2 + ≥ = b2 + kca c 2 + ka b b(b2 + kca) + c(c 2 + ka b) 8t 3 − 6bc t + 2ka bc 2t 2 2 2t 2 ≥ 3 = . = 3 2 4t + (ka − 3t)bc 4t + (ka − 3t)t t + ka On the other hand, a2

a a ≥ 2 . + k bc a + kt 2

Therefore, it suffices to prove that 2 9 a + ≥ , a2 + kt 2 t + ka (k + 1)(a + 2t) which is equivalent to (a − t)2 [(k2 − 6k + 2)a + k(4k − 5)t] ≥ 0. This inequality is true, since k2 − 6k + 2 ≥ 0 and 4k − 5 > 0. The equality holds for a = b = c. (b) For a = 0, the inequality becomes 1 1 k(8 − k) + 2≥ . 2 b c (k + 1)bc

138 We have

Vasile Cîrtoaje

1 k(8 − k) 2 k(8 − k) k2 − 6k + 2 1 + − ≥ − = ≥ 0. b2 c 2 (k + 1)bc bc (k + 1)bc (k + 1)bc

For a, b, c > 0, the desired inequality follows from the inequality in (a) by substituting a, b, c with p 1/a, 1/b, 1/c, respectively. The equality holds for a = b = c. In the case k = 3 + 7, the equality also holds for a = 0 and b = c (or any cyclic permutation).

P 1.95. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 1 1 6 1 + + ≥ 2 . 2a2 + bc 2b2 + ca 2c 2 + a b a + b2 + c 2 + a b + bc + ca (Vasile Cîrtoaje, 2005) Solution. Applying the Cauchy-Schwarz inequality, we have X

X 1 (b + c)2 4(a + b + c)2 P = ≥ . 2a2 + bc (b + c)2 (2a2 + bc) (b + c)2 (2a2 + bc)

Thus, it suffices to show that 2(a + b + c)2 (a2 + b2 + c 2 + a b + bc + ca) ≥ 3

X (b + c)2 (2a2 + bc),

which is equivalent to X X X X 2 a4 + 3 a b(a2 + b2 ) + 2a bc a ≥ 10 a2 b2 . This follows by adding Schur’s inequality X X X 2 a4 + 2a bc a≥2 a b(a2 + b2 ) to the inequality 5

X

a b(a2 + b2 ) ≥ 10

X

a2 b2 .

The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 1.96. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 22a2

1 1 1 1 + + ≥ . 2 2 + 5bc 22b + 5ca 22c + 5a b (a + b + c)2 (Vasile Cîrtoaje, 2005)

Symmetric Rational Inequalities

139

Solution. Applying the Cauchy-Schwarz inequality, we have X

X (b + c)2 4(a + b + c)2 1 P = ≥ . 22a2 + 5bc (b + c)2 (22a2 + 5bc) (b + c)2 (22a2 + 5bc)

Thus, it suffices to show that 4(a + b + c)4 ≥

X (b + c)2 (22a2 + 5bc),

which is equivalent to X X X X 4 a4 + 11 a b(a2 + b2 ) + 4a bc a ≥ 30 a2 b2 . This follows by adding Schur’s inequality X X X 4 a4 + 4a bc a≥4 a b(a2 + b2 ) to the inequality 15

X

a b(a2 + b2 ) ≥ 30

X

a2 b2 .

The equality holds for a = b = c.

P 1.97. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 1 1 8 1 + + ≥ . 2a2 + bc 2b2 + ca 2c 2 + a b (a + b + c)2 (Vasile Cîrtoaje, 2005) First Solution. Applying the Cauchy-Schwarz inequality, we have X

X 1 (b + c)2 4(a + b + c)2 P = ≥ . 2a2 + bc (b + c)2 (2a2 + bc) (b + c)2 (2a2 + bc)

Thus, it suffices to show that (a + b + c)4 ≥ 2

X (b + c)2 (2a2 + bc),

which is equivalent to X X X X a4 + 2 ab(a2 + b2 ) + 4a bc a≥6 a2 b2 . We will prove the sharper inequality X X X X a4 + 2 a b(a2 + b2 ) + a bc a≥6 a2 b2 .

140

Vasile Cîrtoaje

This follows by adding Schur’s inequality X X X a4 + a bc a≥ a b(a2 + b2 ) to the inequality 3

X

a b(a2 + b2 ) ≥ 6

X

a2 b2 .

The equality holds for a = 0 and b = c (or any cyclic permutation). Second Solution. Without loss of generality, we may assume that a ≥ b ≥ c. Since the equality holds for c = 0 and a = b, write the inequality as 2a2

1 1 1 8 1 + 2 + 2 + 2 ≥ + bc 2b + ca 4c + 2a b 4c + 2a b (a + b + c)2

and then apply the Cauchy-Schwarz inequality. It suffices to prove that (2a2

+

bc) + (2b2

8 16 ≥ , 2 2 + ca) + (4c + 2a b) + (4c + 2a b) (a + b + c)2

which is equivalent to the obvious inequality c(a + b − 2c) ≥ 0.

P 1.98. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 1 1 1 12 + + ≥ . a2 + bc b2 + ca c 2 + a b (a + b + c)2 (Vasile Cîrtoaje, 2005) Solution. Due to homogeneity, we may assume that a + b + c = 1. On this assumption, write the inequality as ‹ X 1 − 1 ≥ 9, a2 + bc X 1 − a2 − bc ≥ 9. a2 + bc Since 1 − a2 − bc = (a + b + c)2 − a2 − bc > 0, by the Cauchy-Schwarz inequality, we have X 1 − a2 − bc a2 + bc

[

P (1 − a2 − bc)]2

≥P . (1 − a2 − bc)(a2 + bc)

Symmetric Rational Inequalities

141

Then, it suffices to prove that X X X X (3 − a2 − bc)2 ≥ 9 (a2 + bc) − 9 (a2 + bc)2 , which is equivalent to (1 − 4q)(4 − 7q) + 36a bc ≥ 0, where q = a b + bc + ca. For q ≤ 1/4, this inequality is clearly true. Consider further that q > 1/4. By Schur’s inequality of degree three (a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca), we get 1 + 9a bc ≥ 4q, and hence 36a bc ≥ 16q − 4. Thus, (1 − 4q)(4 − 7q) + 36a bc ≥ (1 − 4q)(4 − 7q) + 16q − 4 = 7q(4q − 1) > 0. The equality holds for a = 0 and b = c (or any cyclic permutation).

P 1.99. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

(b)

a2

1 1 1 1 2 + 2 + 2 ≥ 2 + ; 2 2 + 2bc b + 2ca c + 2a b a +b +c a b + bc + ca a(b + c) b(c + a) c(a + b) a b + bc + ca + 2 + 2 ≥1+ 2 . 2 a + 2bc b + 2ca c + 2a b a + b2 + c 2 (Darij Grinberg and Vasile Cîrtoaje, 2005)

Solution. (a) Write the inequality as P 2 (b + 2ca)(c 2 + 2a b) (a2 + 2bc)(b2 + 2ca)(c 2 + 2a b)

≥

a b + bc + ca + 2a2 + 2b2 + 2c 2 . (a2 + b2 + c 2 )(a b + bc + ca)

Since X (b2 + 2ca)(c 2 + 2a b) = (a b + bc + ca)(a b + bc + ca + 2a2 + 2b2 + 2c 2 ), it suffices to show that (a2 + b2 + c 2 )(a b + bc + ca)2 ≥ (a2 + 2bc)(b2 + 2ca)(c 2 + 2a b), which is just the inequality (a) in P 2.16 in Volume 1. The equality holds for a = b, or b = c, or c = a.

142

Vasile Cîrtoaje

(b) Write the inequality in (a) as X a b + bc + ca a2

+ 2bc

≥2+

a b + bc + ca , a2 + b2 + c 2

or

X a(b + c) X bc a b + bc + ca + ≥2+ 2 . 2 2 a + 2bc a + 2bc a + b2 + c 2 The desired inequality follows by adding this inequality to bc . + 2bc

1≥

X

X

a2 ≥ 1, a2 + 2bc

a2

The last inequality is equivalent to

which follows by applying the AM-GM inequality as follows X

X a2 a2 ≥ = 1. a2 + 2bc a2 + b2 + c 2

The equality holds for a = b = c.

P 1.100. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a)

(b)

a2

b c a+b+c a + 2 + 2 ≤ ; + 2bc b + 2ca c + 2a b a b + bc + ca

a(b + c) b(c + a) c(a + b) a2 + b2 + c 2 + + ≤ 1 + . a2 + 2bc b2 + 2ca c 2 + 2a b a b + bc + ca (Vasile Cîrtoaje, 2008)

Solution. (a) Use the SOS method. Write the inequality as ‹ X  a b + bc + ca a 1− ≥ 0, a2 + 2bc X a(a − b)(a − c)

≥ 0. a2 + 2bc Assume that a ≥ b ≥ c. Since (c − a)(c − b) ≥ 0, it suffices to show that a(a − b)(a − c) b(b − a)(b − c) + ≥ 0. a2 + 2bc b2 + 2ca

Symmetric Rational Inequalities

143

This inequality is equivalent to c(a − b)2 [2a(a − c) + 2b(b − c) + 3a b] ≥ 0, which is clearly true. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation). (b) Since

a(b + c) a(a + b + c) a2 = − , a2 + 2bc a2 + 2bc a2 + 2bc we can write the inequality as (a + b + c)

X

a2 + b2 + c 2 X a2 a ≤ 1 + + . a2 + 2bc a b + bc + ca a2 + 2bc

According to the inequality in (a), it suffices to show that a2 (a + b + c)2 a2 + b2 + c 2 X ≤1+ + , a b + bc + ca a b + bc + ca a2 + 2bc which is equivalent to X

a2 ≥ 1. a2 + 2bc

Indeed,

X a2 a2 ≥ = 1. a2 + 2bc a2 + b2 + c 2 The equality holds for a = b = c. X

P 1.101. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (a) (b)

2a2

b c a+b+c a ; + 2 + 2 ≥ 2 + bc 2b + ca 2c + a b a + b2 + c 2

b+c c+a a+b 6 + 2 + 2 ≥ . 2 2a + bc 2b + ca 2c + a b a+b+c (Vasile Cîrtoaje, 2008)

Solution. Assume that a ≥ b ≥ c. (a) Multiplying by a + b + c, we can write the inequality as follows X a(a + b + c) 2a2 + bc

≥

(a + b + c)2 , a2 + b2 + c 2

144

Vasile Cîrtoaje • ˜ (a + b + c)2 X a(a + b + c) 3− 2 1− ≥ , a + b2 + c 2 2a2 + bc X X (a − b)(a − c) , 2 (a − b)(a − c) ≥ (a2 + b2 + c 2 ) 2a2 + bc X 3a2 − (b − c)2 (a − b)(a − c) ≥ 0, 2a2 + bc 3 f (a, b, c) + (a − b)(b − c)(c − a)g(a, b, c) ≥ 0,

where

X a2 (a − b)(a − c)

b−c . + bc 2a2 + bc It suffices to show that f (a, b, c) ≥ 0 and g(a, b, c) ≤ 0. We have f (a, b, c) =

2a2

,

g(a, b, c) =

X

a2 (a − b)(a − c) b2 (b − a)(b − c) + 2a2 + bc 2b2 + ca 2 2 a (a − b)(b − c) b (b − a)(b − c) ≥ + 2a2 + bc 2b2 + ca 2 2 2 a c(a − b) (b − c)(a + a b + b2 ) = ≥ 0. (2a2 + bc)(2b2 + ca)

f (a, b, c) ≥

Also, (a − b) + (b − c) a−b b−c − + 2 2 2 2a + bc 2b + ca  ‹ 2c + a b  ‹ 1 1 1 1 = (a − b) − + (b − c) − 2c 2 + a b 2b2 + ca 2a2 + bc 2b2 + ca • ˜ (a − b)(b − c) 2b + 2c − a 2b + 2a − c = − = 2b2 + ca 2c 2 + a b 2a2 + bc 2(a − b)(b − c)(c − a)(a2 + b2 + c 2 − a b − bc − ca) ≤ 0. = (2a2 + bc)(2b2 + ca)(2c 2 + a b)

g(a, b, c) =

The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation). (b) We apply the SOS method. Write the inequality as follows ˜ X • (b + c)(a + b + c) − 2 ≥ 0, 2a2 + bc X (b2 + a b − 2a2 ) + (c 2 + ca − 2a2 )

≥ 0, 2a2 + bc X (b − a)(b + 2a) + (c − a)(c + 2a) ≥ 0, 2a2 + bc X (b − a)(b + 2a) X (a − b)(a + 2b) + ≥ 0, 2a2 + bc 2b2 + ca

Symmetric Rational Inequalities

145

 ‹ X b + 2a a + 2b (a − b) − ≥ 0, 2b2 + ca 2a2 + bc X (a − b)2 (2c 2 + a b)(a2 + b2 + 3a b − ac − bc) ≥ 0. Since a2 + b2 + 3a b − ac − bc ≥ a2 + b2 + 2a b − ac − bc = (a + b)(a + b − c), it suffices to show that X (a − b)2 (2c 2 + a b)(a + b)(a + b − c) ≥ 0. This inequality is true if (b − c)2 (2a2 + bc)(b + c)(b + c − a) + (c − a)2 (2b2 + ca)(c + a)(c + a − b) ≥ 0; that is, (a − c)2 (2b2 + ca)(a + c)(a + c − b) ≥ (b − c)2 (2a2 + bc)(b + c)(a − b − c). Since a + c ≥ b + c and a + c − b ≥ a − b − c, it is enough to prove that (a − c)2 (2b2 + ca) ≥ (b − c)2 (2a2 + bc). We can obtain this inequality by multiplying the inequalities b2 (a − c)2 ≥ a2 (b − c)2 and a2 (2b2 + ca) ≥ b2 (2a2 + bc). The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.102. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a(b + c) b(c + a) c(a + b) (a + b + c)2 + + ≥ . a2 + bc b2 + ca c2 + a b a2 + b2 + c 2 (Pham Huu Duc, 2006)

146

Vasile Cîrtoaje

Solution. Assume that a ≥ b ≥ c and write the inequality as follows 3−

 ‹ (a + b + c)2 X a b + ac ≥ 1 − , a2 + b2 + c 2 a2 + bc

X X (a − b)(a − c) (a − b)(a − c) ≥ (a2 + b2 + c 2 ) , a2 + bc X (a − b)(a − c)(a + b − c)(a − b + c) ≥ 0. a2 + bc It suffices to show that 2

(b − c)(b − a)(b + c − a)(b − c + a) (c − a)(c − b)(c + a − b)(c − a + b) + ≥ 0, b2 + ca c2 + a b which is equivalent to the obvious inequality (b − c)2 (c − a + b)2 (a2 + bc) ≥ 0. (b2 + ca)(c 2 + a b) The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.103. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > 0, then p p p p 3(2 + 3) b2 + c 2 + 3bc c 2 + a2 + 3ca a2 + b2 + 3a b + + ≥ . a2 + k bc b2 + kca c 2 + ka b 1+k (Vasile Cîrtoaje, 2013) Solution. Write the inequality in the form f6 (a, b, c) ≥ 0, where X p f6 (a, b, c) = (1 + k) (b2 + c 2 + 3bc)(b2 + kca)(c 2 + ka b) −3(2 +

p 3 )(a2 + k bc)(b2 + kca)(c 2 + ka b).

Clearly, f6 (a, b, c) has the same highest coefficient as f (a, b, c), where Xp f (a, b, c) = (1 + k) ( 3bc − a2 )(b2 + kca)(c 2 + ka b) −3(2 +

p 3 )(a2 + k bc)(b2 + kca)(c 2 + ka b));

therefore, p p A = 3(1 + k)3 ( 3 − 1) − 3(2 + 3 )(1 + k)3 = −9(1 + k)3 .

Symmetric Rational Inequalities

147

Since A ≤ 0, according to P 3.76-(a) in Volume 1, it suffices to prove the original inequality for b = c = 1 and for a = 0. In the first case, this inequality is equivalent to  p  p 2 ™ 1 + 3 3 (k + 1)a2 − 1 + (k − 2)a + k − ≥ 0. 2 2

– (a − 1)2



For the nontrivial case k > 2, we have  p 2 p  p 1+ 3 1+ 3 (k + 1)a + k − a ≥2 k+1 k− 2 2 2



  p  p  p 1+ 3 3 ≥2 3 k− a ≥ 1+ (k − 2)a. 2 2 In the second case (a = 0), the original inequality can be written as  p ‹  2  1 b c p b c2 3(2 + 3) + + 3 + 2 + 2 ≥ , k c b c b 1+k and is true if

p p 2+ 3 3(2 + 3) +2≥ , k 1+k

which is equivalent to

p 2 1+ 3 ≥ 0. k− 2 p 1+ 3 The equality holds for a = b = c. If k = , then the equality holds also for a = 0 2 and b = c (or any cyclic permutation). 

P 1.104. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that

a2

1 1 8 6 1 + 2 + 2 + 2 ≥ . 2 2 2 2 2 +b b +c c +a a +b +c a b + bc + ca (Vasile Cîrtoaje, 2013)

Solution. Multiplying by a2 + b2 + c 2 , the inequality becomes a2 b2 c2 6(a2 + b2 + c 2 ) + + + 11 ≥ . b2 + c 2 c 2 + a2 a2 + b2 a b + bc + ca

148

Vasile Cîrtoaje

Since

 b2 c2 a2 (a2 b2 + b2 c 2 + c 2 a2 ) = + + b2 + c 2 c 2 + a2 a2 + b2  ‹ 1 1 1 4 4 4 2 2 2 =a +b +c +a b c + + ≥ a4 + b4 + c 4 , a2 + b2 b2 + c 2 c 2 + a2 

it suffices to show that 6(a2 + b2 + c 2 ) a4 + b4 + c 4 + 11 ≥ , a2 b2 + b2 c 2 + c 2 a2 a b + bc + ca which is equivalent to (a2 + b2 + c 2 )2 6(a2 + b2 + c 2 ) + 9 ≥ . a2 b2 + b2 c 2 + c 2 a2 a b + bc + ca Clearly, it is enough to prove that 

a2 + b2 + c 2 a b + bc + ca

2

+9≥

6(a2 + b2 + c 2 ) , a b + bc + ca

which is 

a2 + b2 + c 2 −3 a b + bc + ca

The equality holds for a = 0 and

2 ≥ 0.

b c + = 3 (or any cyclic permutation). c b

P 1.105. If a, b, c are the lengths of the sides of a triangle, then a(b + c) b(c + a) c(a + b) + 2 + 2 ≤ 2. 2 a + 2bc b + 2ca c + 2a b (Vo Quoc Ba Can and Vasile Cîrtoaje, 2010) Solution. Write the inequality as ‹ X a b + ac) ≥ 1, 1− 2 a + 2bc X a2 + 2bc − a b − ac a2 + 2bc

≥ 1.

Since a2 + 2bc − a b − ac = bc − (a − c)(b − a) ≥ |a − c||b − a| − (a − c)(b − a) ≥ 0,

Symmetric Rational Inequalities

149

by the Cauchy-Schwarz inequality, we have X a2 + 2bc − a b − ac a2 + 2bc

[

P 2 (a + 2bc − a b − ac)]2

≥P . (a2 + 2bc)(a2 + 2bc − a b − ac)

Thus, it suffices to prove that (a2 + b2 + c 2 )2 ≥

X (a2 + 2bc)(a2 + 2bc − a b − ac),

which reduces to the obvious inequality a b(a − b)2 + bc(b − c)2 + ca(c − a)2 ≥ 0. The equality holds for an equilateral triangle, and for a degenerate triangle with a = 0 and b = c (or any cyclic permutation).

P 1.106. If a, b, c are real numbers, then b2 − ca c2 − a b a2 − bc + + ≥ 0. 2a2 + b2 + c 2 2b2 + c 2 + a2 2c 2 + a2 + b2 (Nguyen Anh Tuan, 2005) First Solution. Rewrite the inequality as  X1 a2 − bc 3 − ≤ , 2 2a2 + b2 + c 2 2 (b + c)2 ≤ 3. 2a2 + b2 + c 2 If two of a, b, c are zero, then the inequality is trivial. Otherwise, applying the CauchySchwarz inequality, we get X

X (b + c)2 (b + c)2 = 2a2 + b2 + c 2 (a2 + b2 ) + (a2 + c 2 )  X X  b2 X a2 c2 b2 ≤ + = + = 3. a2 + b2 a2 + c 2 a2 + b2 b2 + a2 X

The equality holds for a = b = c. Second Solution. Use the SOS method. We have 2

X

X (a − b)(a + c) + (a − c)(a + b) a2 − bc = 2a2 + b2 + c 2 2a2 + b2 + c 2

150

Vasile Cîrtoaje =

X (a − b)(a + c)

+

X (b − a)(b + c)

2a2 + b2 + c 2 2b2 + c 2 + a2 ‹  X a+c b+c = (a − b) − 2a2 + b2 + c 2 2b2 + c 2 + a2 X (a − b)2 = (a2 + b2 + c 2 − a b − bc − ca) ≥ 0. (2a2 + b2 + c 2 )(2b2 + c 2 + a2 )

P 1.107. If a, b, c are nonnegative real numbers, then 3a2 − bc 3b2 − ca 3c 2 − a b 3 + + ≤ . 2 2 2 2 2 2 2 2 2 2a + b + c 2b + c + a 2c + a + b 2 (Vasile Cîrtoaje, 2008) First Solution. Write the inequality as  X3 3a2 − bc − ≥ 3, 2 2a2 + b2 + c 2 X 8bc + 3(b − c)2 2a2 + b2 + c 2 By the Cauchy-Schwarz inequality, we have 8bc + 3(b − c)2 ≥

≥ 6.

2(b + c)4 3[4bc + (b − c)2 ]2 = . 6bc + (b − c)2 b2 + c 2 + 4bc

Therefore, it suffices to prove that X

(b + c)4 ≥ 2. (2a2 + b2 + c 2 )(b2 + c 2 + 4bc)

Using again the Cauchy-Schwarz inequality, we get X

P [ (b + c)2 ]2 (b + c)4 ≥P = 2. (2a2 + b2 + c 2 )(b2 + c 2 + 4bc) (2a2 + b2 + c 2 )(b2 + c 2 + 4bc)

The equality holds for a = b = c, for a = 0 and b = c, and for b = c = 0 (or any cyclic permutation). Second Solution. Write the inequality as  X1 3a2 − bc − ≥ 0, 2 2a2 + b2 + c 2

Symmetric Rational Inequalities

151

X (b + c + 2a)(b + c − 2a)

≥ 0, 2a2 + b2 + c 2 X (b + c + 2a)(b − a) + (b + c + 2a)(c − a) ≥ 0, 2a2 + b2 + c 2 X (b + c + 2a)(b − a) X (c + a + 2b)(a − b) + ≥ 0, 2a2 + b2 + c 2 2b2 + c 2 + a2 ‹  X b + c + 2a c + a + 2b − ≥ 0, (a − b) 2b2 + c 2 + a2 2a2 + b2 + c 2 X (3a b + bc + ca − c 2 )(2c 2 + a2 + b2 )(a − b)2 ≥ 0. Since 3a b + bc + ca − c 2 ≥ c(a + b − c), it suffices to show that X c(a + b − c)(2c 2 + a2 + b2 )(a − b)2 ≥ 0. Assume that a ≥ b ≥ c. It is enough to prove that a(b + c − a)(2a2 + b2 + c 2 )(b − c)2 + b(c + a − b)(2b2 + c 2 + a2 )(c − a)2 ≥ 0; that is, b(c + a − b)(2b2 + c 2 + a2 )(a − c)2 ≥ a(a − b − c)(2a2 + b2 + c 2 )(b − c)2 . Since c + a − b ≥ a − b − c, it suffices to prove that b(2b2 + c 2 + a2 )(a − c)2 ≥ a(2a2 + b2 + c 2 )(b − c)2 . We can obtain this inequality by multiplying the inequalities b2 (a − c)2 ≥ a2 (b − c)2 and a(2b2 + c 2 + a2 ) ≥ b(2a2 + b2 + c 2 ). The last inequality is equivalent to (a − b)[(a − b)2 + a b + c 2 ] ≥ 0.

P 1.108. If a, b, c are nonnegative real numbers, then (b + c)2 (c + a)2 (a + b)2 + + ≥ 2. 4a2 + b2 + c 2 4b2 + c 2 + a2 4c 2 + a2 + b2 (Vasile Cîrtoaje, 2005)

152

Vasile Cîrtoaje

Solution. By the Cauchy-Schwarz inequality, we have P X (b + c)2 [ (b + c)2 ]2 ≥P 4a2 + b2 + c 2 (b + c)2 (4a2 + b2 + c 2 ) P P P P 2[ a4 + 3 a2 b2 + 4a bc a + 2 a b(a2 + b2 )] P P P = P ≥ 2, a4 + 5 a2 b2 + 4a bc a + a b(a2 + b2 ) since

X

a b(a2 + b2 ) ≥ 2

X

a2 b2 .

The equality holds for a = b = c, and for b = c = 0 (or any cyclic permutation).

P 1.109. If a, b, c are positive real numbers, then P

(a)

(b)

11a2 P

3 1 ≤ ; 2 2 + 2b + 2c 5(a b + bc + ca)

1 1 1 ≤ + . 4a2 + b2 + c 2 2(a2 + b2 + c 2 ) a b + bc + ca (Vasile Cîrtoaje, 2008)

Solution. We will prove that X ka2

k+2 11 − 2k 2(k − 1) ≤ 2 + 2 2 2 2 +b +c a +b +c a b + bc + ca

for any k > 1. Due to homogeneity, we may assume that a2 + b2 + c 2 = 3. On this hypothesis, we need to show that X

k+2 11 − 2k 2(k − 1) ≤ + . 2 (k − 1)a + 3 3 a b + bc + ca

Using the substitution m = 3/(k − 1), m > 0, the inequality can be written as m(m + 1)

X a2

1 6 ≤ 3m − 2 + . +m a b + bc + ca

By the Cauchy-Schwarz inequality, we have p p (a2 + m)[m + (m + 1 − a)2 ] ≥ [a m + m(m + 1 − a)]2 = m(m + 1)2 , and hence

m(m + 1) a2 − 1 ≤ + m + 2 − 2a, a2 + m m+1

Symmetric Rational Inequalities m(m + 1)

153 X

X 1 ≤ 3(m + 2) − 2 a. a2 + m

Thus, it suffices to show that 3(m + 2) − 2

X

a ≤ 3m − 2 +

6 ; a b + bc + ca

that is, (4 − a − b − c)(a b + bc + ca) ≤ 3. Let p = a + b + c. Since 2(a b + bc + ca) = (a + b + c)2 − (a2 + b2 + c 2 ) = p2 − 3, we get 6 − 2(4 − a − b − c)(a b + bc + ca) = 6 − (4 − p)(p2 − 3) = (p − 3)2 (p + 2) ≥ 0. This completes the proof. The equality holds for a = b = c.

P 1.110. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then p p p a b c 3 + + ≥ . b+c c+a a+b 2 (Vasile Cîrtoaje, 2006) Solution. By the Cauchy-Schwarz inequality, we have P 3/4
2 X pa a 1 €X 3/4 Š2 ≥P = a . b+c a(b + c) 6 Thus, it suffices to show that a3/4 + b3/4 + c 3/4 ≥ 3, which follows immediately by Remark 1 from the proof of the inequality in P 3.33 in Volume 1. The equality occurs for a = b = c = 1. Remark. Analogously, according to Remark 2 from the proof of P 3.33 in Volume 1, we can prove that ak bk ck 3 + + ≥ b+c c+a a+b 2 4 ln 2 4 ln 2 for all k ≥ 3 − ≈ 0.476. For k = 3 − , the equality occurs for a = b = c = 1, ln 3 ln 3 p and also for a = 0 and b = c = 3 (or any cyclic permutation).

154

Vasile Cîrtoaje

P 1.111. If a, b, c are nonnegative real numbers such that a b + bc + ca ≥ 3, then 1 1 1 1 1 1 + + ≥ + + . 2+a 2+ b 2+c 1+ b+c 1+c+a 1+a+ b (Vasile Cîrtoaje, 2014) Solution. Denote E(a, b, c) =

1 1 1 1 1 1 + + − − − . 2+a 2+ b 2+c 1+ b+c 1+c+a 1+a+ b

Consider first the case a b + bc + ca = 3. We will show that 1 1 1 1 1 1 + + ≥1≥ + + . 2+a 2+ b 2+c 1+ b+c 1+c+a 1+a+ b By direct calculation, we can show that the left inequality is equivalent to a bc ≤ 1. Indeed, applying the AM-GM inequality, we get p 3 = a b + bc + ca ≥ 3 a b · bc · ca. Also, the right inequality is equivalent to a + b + c ≥ 2 + a bc. Since a bc ≤ 1, it suffices to show that a + b + c ≥ 3. Indeed, (a + b + c)2 ≥ 3(a b + bc + ca) = 9. Consider further that a b+ bc+ca > 3. Without loss of generality, assume that a ≥ b ≥ c, a > 1. For c ≥ 1, that is, a ≥ b ≥ c ≥ 1, the desired inequality follows by summing the obvious inequalities 1 1 ≥ , 2+a 1+c+a 1 1 ≥ , 2+ b 1+a+ b 1 1 ≥ . 2+c 1+ b+c Therefore, assume now that c < 1. Consider the cases b + c ≥ 2 and b + c < 2. Case 1: b + c ≥ 2, a > 1, c < 1. We will show that E(a, b, c) ≥ E(1, b, c) ≥ 0.

Symmetric Rational Inequalities

155

We have ‹  ‹  ‹ 1 1 1 1 1 1 E(a, b, c) − E(1, b, c) = − + − + − 2+a 3 2+ b 1+a+ b 2+c 1+c+a ˜ • 1 1 −1 + + = (a − 1) 3(2 + a) (2 + b)(1 + a + b) (2 + c)(1 + c + a) ˜ • 1 −1 + > (a − 1) 3(2 + a) (2 + c)(1 + c + a) (a − 1)(1 − c)(4 + c + a) = >0 3(2 + a)(2 + c)(1 + c + a) 

and E(1, b, c) =

b+c−2 ≥ 0. 3(1 + b + c)

Case 2: b + c < 2, a > 1, c < 1. From b + c < 2, it follows that bc ≤



b+c 2

‹2

< 1.

For fixed b and c, define the function f (x) = E(x, b, c). Since 1 1 −1 1 −1 + + > + 2 2 2 2 (2 + x) (1 + c + x) (1 + x + b) (2 + x) (1 + c + x)2 (1 − c)(3 + 2x + c) = > 0, (2 + x)2 (1 + c + x)2

f 0 (x) =

f (x) is strictly increasing for x ≥ 0. Since a>

3 − bc , b+c

‹  ‹ 3 − bc 3 − bc . Therefore, it suffices to prove that f ≥ 0, which b+c b+c 3 − bc , that is, for a b + bc + ca = 3. But this was is equivalent to E(a, b, c) ≥ 0 for a = b+c proved in the first part of the proof. So, the proof is completed. The equality occurs for a = b = c = 1.

we have f (a) > f



156

Vasile Cîrtoaje

P 1.112. If a, b, c are the lengths of the sides of a triangle, then (a)

b2 − ca c2 − a b a2 − bc + + ≤ 0; 3a2 + b2 + c 2 3b2 + c 2 + a2 3c 2 + a2 + b2

(b)

a4 − b2 c 2 b4 − c 2 a2 c 4 − a2 b2 + + ≤ 0. 3a4 + b4 + c 4 3b4 + c 4 + a4 3c 4 + a4 + b4 (Nguyen Anh Tuan and Vasile Cîrtoaje, 2006)

Solution. (a) Apply the SOS method. We have 2

X

X (a − b)(a + c) + (a − c)(a + b) a2 − bc = 3a2 + b2 + c 2 3a2 + b2 + c 2 =

X (a − b)(a + c)

+

X (b − a)(b + c)

3a2 + b2 + c 2 3b2 + c 2 + a2 ‹  X b+c a+c − = (a − b) 3a2 + b2 + c 2 3b2 + c 2 + a2 = (a2 + b2 + c 2 − 2a b − 2bc − 2ca)

X

(a − b)2 . (3a2 + b2 + c 2 )(3b2 + c 2 + a2 )

Since a2 + b2 + c 2 − 2a b − 2bc − 2ca = a(a − b − c) + b(b − c − a) + c(c − a − b) ≤ 0, the conclusion follows. The equality holds for an equilateral triangle, and for a degenerate triangle with a = 0 and b = c (or any cyclic permutation). (b) Using the same way as above, we get 2

X

X a4 − b2 c 2 (a2 − b2 )2 = A , 3a4 + b4 + c 4 (3a4 + b4 + c 4 )(3b4 + c 4 + a4 )

where A = a4 + b4 + c 4 − 2a2 b2 − 2b2 c 2 − 2c 2 a2 = −(a + b + c)(a + b − c)(b + c − a)(c + a − b) ≤ 0. The equality holds for an equilateral triangle, and for a degenerate triangle with a = b+c (or any cyclic permutation).

Symmetric Rational Inequalities

157

P 1.113. If a, b, c are the lengths of the sides of a triangle, then 4a2

ca ab 1 bc + 2 + 2 ≥ . 2 2 2 2 2 2 +b +c 4b + c + a 4c + a + b 2 (Vasile Cîrtoaje and Vo Quoc Ba Can, 2010)

Solution. We apply the SOS method. Write the inequality as  X X b2 c 2 2bc ≥ 0, − 4a2 + b2 + c 2 a2 b2 + b2 c 2 + c 2 a2 X bc(2a2 − bc)(b − c)2 ≥ 0. 4a2 + b2 + c 2 Without loss of generality, assume that a ≥ b ≥ c. Then, it suffices to prove that c(2b2 − ca)(c − a)2 b(2c 2 − a b)(a − b)2 + ≥ 0. 4b2 + c 2 + a2 4c 2 + a2 + b2 Since 2b2 − ca ≥ c(b + c) − ca = c(b + c − a) ≥ 0 and (2b2 − ca) + (2c 2 − a b) = 2(b2 + c 2 ) − a(b + c) ≥ (b + c)2 − a(b + c) = (b + c)(b + c − a) ≥ 0, it is enough to show that b(a − b)2 c(a − c)2 ≥ . 4b2 + c 2 + a2 4c 2 + a2 + b2 This follows by multiplying the inequalities c 2 (a − c)2 ≥ b2 (a − b)2 and

b c ≥ 2 . 4b2 + c 2 + a2 4c + a2 + b2 These inequalities are true, since c(a − c) − b(a − b) = (b − c)(b + c − a) ≥ 0, b(4c 2 + a2 + b2 ) − c(4b2 + c 2 + a2 ) = (b − c)[(b − c)2 + a2 − bc] ≥ 0. The equality occurs for an equilateral triangle, and for a degenerate triangle with a = 0 and b = c (or any cyclic permutation).

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P 1.114. If a, b, c are the lengths of the sides of a triangle, then b2

1 1 1 9 + 2 + 2 ≤ . 2 2 2 +c c +a a +b 2(a b + bc + ca) (Vo Quoc Ba Can, 2008)

Solution. Apply the SOS method. Write the inequality as X • 3 a b + bc + ca ˜ − ≥ 0, 2 b2 + c 2 X 3(b2 + c 2 ) − 2(a b + bc + ca) b2 + c 2

≥ 0,

X 3b(b − a) + 3c(c − a) + c(a − b) + b(a − c) b2 + c 2

≥ 0,

X (a − b)(c − 3b) + (a − c)(b − 3c)

≥ 0, b2 + c 2 X (a − b)(c − 3b) X (b − a)(c − 3a) + ≥ 0, b2 + c 2 c 2 + a2 X (a2 + b2 )(a − b)2 (ca + c b + 3c 2 − 3a b) ≥ 0. Without loss of generality, assume that a ≥ b ≥ c. Since a b + ac + 3a2 − 3bc > 0, it suffices to prove that (a2 + b2 )(a − b)2 (ca + c b + 3c 2 − 3a b) + (a2 + c 2 )(a − c)2 (a b + bc + 3b2 − 3ac) ≥ 0, or, equivalently, (a2 + c 2 )(a − c)2 (a b + bc + 3b2 − 3ac) ≥ (a2 + b2 )(a − b)2 (3a b − 3c 2 − ca − c b). Since 2

a b + bc + 3b − 3ac = a



=

   bc + 3b2 bc + 3b2 + b − 3c ≥ a + b − 3c a b+c

a(b − c)(4b + 3c) ≥0 b+c

and (a b + bc + 3b2 − 3ac) − (3a b − 3c 2 − ca − c b) = 3(b2 + c 2 ) + 2bc − 2a(b + c) ≥ 3(b2 + c 2 ) + 2bc − 2(b + c)2 = (b − c)2 ≥ 0,

Symmetric Rational Inequalities

159

it suffices to show that (a2 + c 2 )(a − c)2 ≥ (a2 + b2 )(a − b)2 ). This is true, since is equivalent to (b − c)A ≥ 0, where A = 2a3 − 2a2 (b + c) + 2a(b2 + bc + c 2 ) − (b + c)(b2 + c 2 )  ‹ b + c 2 a(3b2 + 2bc + 3c 2 ) = 2a a − − (b + c)(b2 + c 2 ) + 2 2 ≥

b(3b2 + 2bc + 3c 2 ) − (b + c)(b2 + c 2 ) 2 =

(b − c)(b2 + bc + 2c 2 ) ≥ 0. 2

The equality occurs for an equilateral triangle, and for a degenerate triangle with a = 0 and b = c (or any cyclic permutation).

P 1.115. If a, b, c are the lengths of the sides of a triangle, then (a)

a + b b + c c + a a − b + b − c + c − a > 5;

(b)

2 a + b2 b2 + c 2 c 2 + a2 a2 − b2 + b2 − c 2 + c 2 − a2 ≥ 3. (Vasile Cîrtoaje, 2003)

Solution. Since the inequalities are symmetric, we consider that a > b > c. (a) Let x = a − c and y = b − c. From a > b > c and a ≤ b + c, it follows that x > y > 0 and c ≥ x − y. We have 2c + x + y 2c + y 2c + x a+b b+c c+a + + = + − a−b b−c c−a x−y y x  ‹ x+y 1 1 1 2c x + y = 2c + − + > + x−y y x x−y y x−y  ‹ 2(x − y) x + y x−y y ≥ + =2 + + 1 ≥ 5. y x−y y x−y

160

Vasile Cîrtoaje

(b) We will show that a2 + b2 b2 + c 2 c 2 + a2 + + ≥ 3; a2 − b2 b2 − c 2 c 2 − a2 that is,

b2 c2 a2 + ≥ . a2 − b2 b2 − c 2 a2 − c 2

Since

(b + c)2 a2 ≤ , a2 − c 2 a2 − c 2

it suffices to prove that b2 c2 (b + c)2 + ≥ . a2 − b2 b2 − c 2 a2 − c 2 This is equivalent to each of the following inequalities:   ‹ ‹ 1 1 1 1 2bc 2 2 b − 2 +c − 2 ≥ 2 , 2 2 2 2 2 2 a −b a −c b −c a −c a − c2 b2 (b2 − c 2 ) c 2 (a2 − b2 ) + ≥ 2bc, a2 − b2 b2 − c 2 [b(b2 − c 2 ) − c(a2 − b2 )]2 ≥ 0. This completes the proof. If a > b > c, then the equality holds for a degenerate triangle with a = b + c and b/c = x 1 , where x 1 ≈ 1.5321 is the positive root of the equation x 3 − 3x − 1 = 0.

P 1.116. If a, b, c are the lengths of the sides of a triangle, then  ‹ b+c c+a a+b a b c + + +3≥6 + + . a b c b+c c+a a+b Solution. We apply the SOS method. Write the inequality as X ‹ X b+c 2a −6≥3 −3 . a b+c Since

and

X b+c X b c ‹ X (b − c)2 −6= + −6= a c b bc X 2a X 2a − b − c X a − b X a − c −3= = + b+c b+c b+c b+c

Symmetric Rational Inequalities

=

X a−b b+c

+

161

X b − a X (a − b)2 X (b − c)2 = = , c+a (b + c)(c + a) (c + a)(a + b)

we can rewrite the inequality as X

a(b + c)Sa (b − c)2 ≥ 0,

where Sa = a(a + b + c) − 2bc. Without loss of generality, assume that a ≥ b ≥ c. Since Sa > 0, S b = b(a + b + c) − 2ca = (b − c)(a + b + c) + c(b + c − a) ≥ 0 and

X

a(b + c)Sa (b − c)2 ≥ b(c + a)S b (c − a)2 + c(a + b)Sc (a − b)2 ≥ (a − b)2 [b(c + a)S b + c(a + b)Sc ],

it suffices to prove that b(c + a)S b + c(a + b)Sc ≥ 0. This is equivalent to each of the following inequalities (a + b + c)[a(b2 + c 2 ) + bc(b + c)] ≥ 2a bc(2a + b + c), a(a + b + c)(b − c)2 + (a + b + c)[2a bc + bc(b + c)] ≥ 2a bc(2a + b + c), a(a + b + c)(b − c)2 + bc(2a + b + c)(b + c − a) ≥ 0. Since the last inequality is true, the proof is completed. The equality occurs for an equilateral triangle, and for a degenerate triangle with a/2 = b = c (or any cyclic permutation).

P 1.117. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that X 3a(b + c) − 2bc 3 ≥ . (b + c)(2a + b + c) 2 (Vasile Cîrtoaje, 2009) Solution. Use the SOS method. Write the inequality as follows ˜ X • 3a(b + c) − 2bc 1 − ≥ 0, (b + c)(2a + b + c) 2

162

Vasile Cîrtoaje X 4a(b + c) − 6bc − b2 − c 2 (b + c)(2a + b + c)

≥ 0,

X b(a − b) + c(a − c) + 3b(a − c) + 3c(a − b) ≥ 0, (b + c)(2a + b + c) X (a − b)(b + 3c) + (a − c)(c + 3b) ≥ 0, (b + c)(2a + b + c) X (a − b)(b + 3c) X (b − a)(a + 3c) + ≥ 0, (b + c)(2a + b + c) (c + a)(2b + c + a) ˜ • X a + 3c b + 3c − ≥ 0, (a − b) (b + c)(2a + b + c) (c + a)(2b + c + a) X (a − b)(b − c)(c − a) (a2 − b2 )(a + b + 2c) ≥ 0. Since

X (a2 − b2 )(a + b + 2c) = (a − b)(b − c)(c − a),

the conclusion follows. The equality holds for a = b, or b = c, or c = a.

P 1.118. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that X

a(b + c) − 2bc ≥ 0. (b + c)(3a + b + c) (Vasile Cîrtoaje, 2009)

Solution. We apply the SOS method. Since X

=

X b(a − c) + c(a − b) a(b + c) − 2bc = (b + c)(3a + b + c) (b + c)(3a + b + c)

X c(b − a) c(a − b) + (c + a)(3b + c + a) (b + c)(3a + b + c) X c(a + b − c)(a − b)2 = , (b + c)(c + a)(3a + b + c)(3b + c + a)

X

the inequality is equivalent to X c(a + b)(3c + a + b)(a + b − c)(a − b)2 ≥ 0. Without loss of generality, assume that a ≥ b ≥ c. Since a + b − c ≥ 0, it suffices to show that b(c + a)(3b + c + a)(c + a − b)(a − c)2 ≥ a(b + c)(3a + b + c)(a − b − c)(b − c)2 .

Symmetric Rational Inequalities

163

This is true since c + a − b ≥ a − b − c, b2 (a − c)2 ≥ a2 (b − c)2 , c + a ≥ b + c, a(3b + c + a) ≥ b(3a + b + c). The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.119. Let a, b, c be positive real numbers such that a2 + b2 + c 2 ≥ 3. Prove that a5 − a2 b5 − b2 c5 − c2 + + ≥ 0. a5 + b2 + c 2 b5 + c 2 + a2 c 5 + a2 + b2 (Vasile Cîrtoaje, 2005) Solution. The inequality is equivalent to a5

1 1 1 3 + 5 + 5 ≤ 2 . 2 2 2 2 2 2 +b +c b +c +a c +a +b a + b2 + c 2

Setting a = t x, b = t y and c = tz, where t > 0 and x, y, z > 0 such that x 2 + y 2 +z 2 = 3, the condition a2 + b2 + c 2 ≥ 3 implies t ≥ 1, and the inequality becomes 1 1 1 + + ≤ 1. t 3 x 5 + y 2 + z2 t 3 y 5 + z2 + x 2 t 3z5 + x 2 + y 2 We see that it suffices to prove this inequality for t = 1, when it becomes 1 1 1 + + ≤ 1. x 5 − x 2 + 3 y 5 − y 2 + 3 z5 − z2 + 3 Without loss of generality, assume that x ≥ y ≥ z. There are two cases to consider. p Case 1: z ≤ y ≤ x ≤ 2. The desired inequality follows by adding the inequalities 3 − y2 1 3 − x2 1 1 3 − z2 ≤ , ≤ , ≤ . x5 − x2 + 3 6 y5 − y2 + 3 6 z5 − z2 + 3 6 We have 1 3 − x2 (x − 1)2 (x 5 + 2x 4 − 3x 2 − 6x − 3) − = ≤ 0, x5 − x2 + 3 6 6(x 5 − x 2 + 3) since x 5 + 2x 4 − 3x 2 − 6x − 3 = x 2 (x 3 + 2x 2 − 3 −

6 3 − 2) x x

164

Vasile Cîrtoaje p p p 3 1 ≤ x 2 (2 2 + 4 − 3 − 3 2 − ) = −x 2 ( 2 + ) < 0. 2 2

Case 2: x >

p 2. From x 2 + y 2 + z 2 = 3, it follows that y 2 + z 2 < 1. Since x5

1 1 1 1 < p < < p 2 2 − x + 3 (2 2 − 1)x + 3 2(2 2 − 1) + 3 6

and y5

1 1 1 1 + 5 < + , 2 2 2 − y +3 z −z +3 3− y 3 − z2

it suffices to prove that 1 5 1 + ≤ . 2 2 3− y 3−z 6 Indeed, we have 1 1 5 9( y 2 + z 2 − 1) − 5 y 2 z 2 + − = < 0, 3 − y 2 3 − z2 6 6(3 − y 2 )(3 − z 2 ) which completes the proof. The equality occurs for a = b = c = 1. p 3 Remark. Since a bc ≥ 1 involves a2 + b2 + c 2 ≥ 3 a2 b2 c 2 ≥ 3, the original inequality is also true for a bc ≥ 1, which is a problem from IMO-2005 (by Hojoo Lee). A proof of this inequality is the following: X

=

X a5 − a2 a3 − 1 ≥ a5 + b2 + c 2 a(a2 + b2 + c 2 )

X X 1 1 1 2 (a − ) ≥ (a2 − bc) a2 + b2 + c 2 a a2 + b2 + c 2 X 1 = (a − b)2 ≥ 0. 2(a2 + b2 + c 2 )

P 1.120. Let a, b, c be positive real numbers such that a2 + b2 + c 2 = a3 + b3 + c 3 . Prove that a2 b2 c2 3 + + ≥ . b+c c+a a+b 2 (Pham Huu Duc, 2008)

Symmetric Rational Inequalities

165

First Solution. By the Cauchy-Schwarz inequality, we have P P P X a2 ( a3 )( a2 ) ( a3 )2 P P . = P ≥P b+c a4 (b + c) ( a3 )( a b) − a bc a2 Therefore, it is enough to show that X X X X X 2( a3 )( a2 ) + 3a bc a2 ≥ 3( a3 )( a b). Write this inequality as follows X X X X X 3( a3 )( a2 − a b) − ( a3 − 3a bc) a2 ≥ 0, X X X X X X X a3 )( a2 − a b) − ( a)( a2 − a b) a2 ≥ 0, 3( X X X X X ( a2 − a b)[3 a3 − ( a)( a2 )] ≥ 0. The last inequality is true, since X X X 2( a2 − a b) = (a − b)2 ≥ 0 and 3

X

X X X X X a3 − ( a)( a2 ) = (a3 + b3 ) − a b(a + b) = (a + b)(a − b)2 ≥ 0.

The equality occurs for a = b = c = 1. Second Solution. Write the inequality in the homogeneous form A ≥ B, where A=2

X X a2 − a, b+c

B=

3(a3 + b3 + c 3 ) X − a. a2 + b2 + c 2

Since A=

X a(a − b) + a(a − c)

and B=

b+c

X b(b − a) b+c c+a X (a − b)2 = (a + b + c) (b + c)(c + a) =

X a(a − b)

P 3 P (a + b3 ) − a b(a + b)

+

P (a + b)(a − b)2

= , a2 + b2 + c 2 a2 + b2 + c 2 we can write the inequality as ˜ X• a + b + c a+b − (a − b)2 ≥ 0, (b + c)(c + a) a2 + b2 + c 2 X (a − b)2 (a3 + b3 + c 3 − 2a bc) ≥ 0. (b + c)(c + a) Since a3 + b3 + c 3 ≥ 3a bc, the conclusion follows.

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Vasile Cîrtoaje

P 1.121. If a, b, c ∈ [0, 1], then (a)

b c a + + ≤ 1; bc + 2 ca + 2 a b + 2

(b)

ab bc ca + + ≤ 1. 2bc + 1 2ca + 1 2a b + 1 (Vasile Cîrtoaje, 2010)

Solution. (a) It suffices to show that b c a + + ≤ 1, a bc + 2 a bc + 2 a bc + 2 which is equivalent to a bc + 2 ≥ a + b + c. We have a bc + 2 − a − b − c = (1 − b)(1 − c) + (1 − a)(1 − bc) ≥ 0. The equality holds for a = b = c = 1, and for a = 0 and b = c = 1 (or any cyclic permutation). (b) It suffices to prove that ab bc ca + + ≤ 1; 2a bc + 1 2a bc + 1 2a bc + 1 that is, 2a bc + 1 ≥ a b + bc + ca. Since a + b + c − (a b + bc + ca) = a(1 − b) + b(1 − c) + c(1 − a) ≥ 0, we have 2a bc + 1 − a b − bc − ca ≥ 2a bc + 1 − a − b − c = (1 − b)(1 − c) + (1 − a)(1 − bc) ≥ 0. The equality holds for a = b = c = 1, and for a = 0 and b = c = 1 (or any cyclic permutation).

P 1.122. Let a, b, c be positive real numbers such that a + b + c = 2. Prove that  ‹ 1 1 1 5(1 − a b − bc − ca) + + + 9 ≥ 0. 1 − a b 1 − bc 1 − ca (Vasile Cîrtoaje, 2011)

Symmetric Rational Inequalities

167

Solution. Write the inequality as 24 −

5a(b + c) 5b(c + a) 5c(a + b) − − ≥ 0. 1 − bc 1 − ca 1 − ab

Since 4(1 − bc) ≥ 4 − (b + c)2 = (a + b + c)2 − (b + c)2 = a(a + 2b + 2c), it suffices to show that  6−5 which is equivalent to

‹ c+a a+b b+c − − ≥ 0, a + 2b + 2c b + 2c + 2a c + 2a + 2b

‹ b+c ≥ 9, a + 2b + 2c X 1 5(a + b + c) ≥ 9, a + 2b + 2c ˜ ”X — •X 1 ≥ 9. (a + 2b + 2c) a + 2b + 2c X  5 1−

The last inequality follows immediately from the AM-HM inequality. The equality holds for a = b = c = 2/3.

P 1.123. Let a, b, c be nonnegative real numbers such that a + b + c = 2. Prove that 2 − a2 2 − b2 2 − c2 + + ≤ 3. 2 − bc 2 − ca 2 − a b (Vasile Cîrtoaje, 2011) First Solution. Write the inequality as follows  X 2 − a2 1− ≥ 0, 2 − bc X a2 − bc

≥ 0, 2 − bc X (a2 − bc)(2 − ca)(2 − a b) ≥ 0, X (a2 − bc)[4 − 2a(b + c) + a2 bc] ≥ 0, X X X 4 (a2 − bc) − 2 a(b + c)(a2 − bc) + a bc a(a2 − bc) ≥ 0.

168

Vasile Cîrtoaje

By virtue of the AM-GM inequality, X a(a2 − bc) = a3 + b3 + c 3 − 3a bc ≥ 0. Then, it suffices to prove that X X 2 (a2 − bc) ≥ a(b + c)(a2 − bc). Indeed, we have X

X X a(b + c)(a2 − bc) = a3 (b + c) − a bc (b + c) X X X = a(b3 + c 3 ) − a bc (b + c) = a(b + c)(b − c)2 X X X • a + (b + c) ˜2 (b − c)2 = (b − c)2 = 2 (a2 − bc). ≤ 2 The equality holds for a = b = c = 2/3, and for a = 0 and b = c = 1 (or any cyclic permutation). Second Solution. We apply the SOS method. Write the inequality as follows X a2 − bc 2 − bc

≥ 0,

X (a − b)(a + c) + (a − c)(a + b)

≥ 0, 2 − bc X (a − b)(a + c) X (b − a)(b + c) + ≥ 0, 2 − bc 2 − ca X (a − b)2 [2 − c(a + b) − c 2 ] ≥ 0, (2 − bc)(2 − ca) X (a − b)2 (2 − a b)(1 − c) ≥ 0. Assuming that a ≥ b ≥ c, it suffices to prove that (b − c)2 (2 − bc)(1 − a) + (c − a)2 (2 − ca)(1 − b) ≥ 0. Since 2(1 − b) = a − b + c ≥ 0 and (c − a)2 ≥ (b − c)2 , it suffices to show that (2 − bc)(1 − a) + (2 − ca)(1 − b) ≥ 0. We have (2 − bc)(1 − a) + (2 − ca)(1 − b) = 4 − 2(a + b) − c(a + b) + 2a bc ˜ • (a + b) + (2 + c) 2 ≥ 4 − (a + b)(2 + c) ≥ 4 − = 0. 2

Symmetric Rational Inequalities

169

P 1.124. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 3 + 5a2 3 + 5b2 3 + 5c 2 + + ≥ 12. 3 − bc 3 − ca 3 − ab (Vasile Cîrtoaje, 2010) Solution. Use the SOS method. Write the inequality as follows  X  3 + 5a2 − 4 ≥ 0, 3 − bc X 5a2 + 4bc − 9 3 − bc

≥ 0,

X 4a2 − b2 − c 2 − 2a b + 2bc − 2ca 3 − bc

≥ 0,

X (2a2 − b2 − c 2 ) + 2(a − b)(a − c)

≥ 0, 3 − bc X [(a − b)(a + b) + (a − b)(a − c)] + [(a − c)(a + c) + (a − c)(a − b)] ≥ 0, 3 − bc X (a − b)(2a + b − c) + (a − c)(2a + c − b) ≥ 0, 3 − bc X (a − b)(2a + b − c) X (b − a)(2b + a − c) + ≥ 0, 3 − bc 3 − ca X (a − b)2 [3 − 2c(a + b) + c 2 ] ≥ 0, (3 − bc)(3 − ca) X (a − b)2 (c − 1)2 ≥ 0. (3 − bc)(3 − ca) The equality holds for a = b = c = 1, and for a = 0 and b = c = 3/2 (or any cyclic permutation).

P 1.125. Let a, b, c be nonnegative real numbers such that a + b + c = 2. If −1 7 ≤m≤ , 7 8 then

a2 + m b2 + m c2 + m 3(4 + 9m) + + ≥ . 3 − 2bc 3 − 2ca 3 − 2a b 19 (Vasile Cîrtoaje, 2010)

170

Vasile Cîrtoaje

Solution. We apply the SOS method. Write the inequality as X  a2 + m 4 + 9m  − ≥ 0, 3 − 2bc 19 X 19a2 + 2(4 + 9m)bc − 12 − 8m 3 − 2bc

≥ 0.

Since 19a2 + 2(4 + 9m)bc − 12 − 8m = 19a2 + 2(4 + 9m)bc − (3 + 2m)(a + b + c)2 = (16 − 2m)a2 − (3 + 2m)(b2 + c 2 + 2a b + 2ac) + 2(1 + 7m)bc = (3 + 2m)(2a2 − b2 − c 2 ) + 2(5 − 3m)(a2 + bc − a b − ac) + (4 − 10m)(a b + ac − 2bc) = (3 + 2m)(a2 − b2 ) + (5 − 3m)(a − b)(a − c) + (4 − 10m)c(a − b) +(3 + 2m)(a2 − c 2 ) + (5 − 3m)(a − c)(a − b) + (4 − 10m)b(a − c) = (a − b)B + (a − c)C, where B = (8 − m)a + (3 + 2m)b − (1 + 7m)c, C = (8 − m)a + (3 + 2m)c − (1 + 7m)b, the inequality can be written as B1 + C1 ≥ 0, where B1 =

X (a − b)[(8 − m)a + (3 + 2m)b − (1 + 7m)c]

, 3 − 2bc X (b − a)[(8 − m)b + (3 + 2m)a − (1 + 7m)c] C1 = . 3 − 2ca We have B1 + C1 =

X

(a − b)2 Ec , (3 − 2bc)(3 − 2ca)

where Ec =3(5 − 3m) − 2(8 − m)c(a + b) + 2(1 + 7m)c 2 =6(2m + 3)c 2 − 4(8 − m)c + 3(5 − 3m) • ˜2 8−m (1 + 7m)(7 − 8m) =6(2m + 3) c − + . 3(2m + 3) 3(2m + 3) Since Ec ≥ 0 for −1/7 ≤ m ≤ 7/8, we get B1 + C1 ≥ 0. Thus, the proof is completed. The equality holds for a = b = c = 2/3. When m = −1/7, the equality holds for

Symmetric Rational Inequalities

171

a = b = c = 2/3, and for a = 0 and b = c = 1 (or any cyclic permutation). When m = 7/8, the equality holds for a = b = c = 2/3, and for a = 1 and b = c = 1/2 (or any cyclic permutation). Remark. The inequalities in P 1.124 and P 1.125 are particular cases (k = 3 and k = 8/3, respectively) of the following more general result: • Let a, b, c be nonnegative real numbers such that a + b + c = 3. For 0 < k ≤ 3 and m1 ≤ m ≤ m2 , where  3  0 0 and m ≥ m1 , where p  36 + 4k − k2 + 4(9 + k) 3(3 + k)   , k 6= 72  72 − k m1 = , 238   , k = 72  5 then

a2 + mbc b2 + mca c 2 + ma b 3(1 + m) + + ≤ , 9 + k bc 9 + kca 9 + ka b 9+k

with equality for a = b = c = 1. When m = m1 , the equality holds also for p p 3k + 6 − 2 3(3 + k) 3(3 + k) − 3 a= , b=c= . k k

Symmetric Rational Inequalities

173

P 1.127. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 57 26 − 7a2 26 − 7b2 26 − 7c 2 + + ≤ . 1 + bc 1 + ca 1 + ab 2 (Vasile Cîrtoaje, 2011) Solution. Use the SOS method. Write the inequality as follows X  19 26 − 7a2  − ≥ 0, 2 1 + bc X 14a2 + 19bc − 33 1 + bc

≥ 0,

X 42a2 + 57bc − 11(a + b + c)2 1 + bc

≥ 0,

X 11(2a2 − b2 − c 2 ) + 9(a − b)(a − c) − 13(a b − 2bc + ca)

≥ 0, 1 + bc X 22(a − b)(a + b) + 9(a − b)(a − c) − 26c(a − b) + 1 + bc X 22(a − c)(a + c) + 9(a − c)(a − b) − 26b(a − c) ≥ 0, + 1 + bc X (a − b)(31a + 22b − 35c) X (a − c)(31a + 22c − 35b) + ≥ 0, 1 + bc 1 + bc X (a − b)(31a + 22b − 35c) X (b − a)(31b + 22a − 35c) + ≥ 0, 1 + bc 1 + ca X (a − b)2 [9 + 31c(a + b) − 35c 2 ] ≥ 0, (1 + bc)(1 + ca) X (a − b)2 (1 + a b)(1 + 11c)(3 − 2c) ≥ 0. Assume that a ≥ b ≥ c. Since 3 − 2c > 0, it suffices to show that (b − c)2 (1 + bc)(1 + 11a)(3 − 2a) + (c − a)2 (1 + a b)(1 + 11b)(3 − 2b) ≥ 0; that is, (a − c)2 (1 + a b)(1 + 11b)(3 − 2b) ≥ (b − c)2 (1 + bc)(1 + 11a)(2a − 3). Since 3 − 2b = a − b + c ≥ 0, we get this inequality by multiplying the inequalities 3 − 2b ≥ 2a − 3, a(1 + a b) ≥ b(1 + bc),

174

Vasile Cîrtoaje a(1 + 11b) ≥ b(1 + 11a), b2 (a − c)2 ≥ a2 (b − c)2 .

The equality holds for a = b = c = 1, and for a = 0 and b = c = 3/2 (or any cyclic permutation). Remark. The inequalities in P 1.127 is a particular cases (k = 9) of the following more general result: • Let a, b, c be nonnegative real numbers such that a + b + c = 3. For k > 0 and m ≤ m2 , where (3 + k)(4 + k) m2 = , 2(3 + 2k) then

a2 + mbc b2 + mca c 2 + ma b 3(1 + m) + + ≥ , 9 + k bc 9 + kca 9 + ka b 9+k

with equality for a = b = c = 1. When m = m2 , the equality holds also for a = 0 and b = c = 3/2 (or any cyclic permutation).

P 1.128. Let a, b, c be nonnegative real numbers, no all are zero. Prove that X 5a(b + c) − 6bc ≤ 3. a2 + b2 + c 2 + bc (Vasile Cîrtoaje, 2010) First Solution. Apply the SOS method. If two of a, b, c are zero, then the inequality is trivial. Consider further that a2 + b2 + c 2 = 1, a ≥ b ≥ c, b > 0, and write the inequality as follows ˜ X• 5a(b + c) − 6bc 1− 2 ≥ 0, a + b2 + c 2 + bc X a2 + b2 + c 2 − 5a(b + c) + 7bc ≥ 0, a2 + b2 + c 2 + bc X (7b + 2c − a)(c − a) − (7c + 2b − a)(a − b) ≥ 0, 1 + bc X (7c + 2a − b)(a − b) X (7c + 2b − a)(a − b) − ≥ 0, 1 + ca 1 + bc X (a − b)2 (1 + a b)(3 + ac + bc − 7c 2 ] ≥ 0.

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Since 3 + ac + bc − 7c 2 = 3a2 + 3b2 + ac + bc − 4c 2 > 0, it suffices to prove that (1 + bc)(3 + a b + ac − 7a2 )(b − c)2 + (1 + ac)(3 + a b + bc − 7b2 )(a − c)2 ≥ 0. Since 3 + a b + ac − 7b2 = 3(a2 − b2 ) + 3c 2 + b(a − b) + bc ≥ 0 and 1 + ac ≥ 1 + bc, it is enough to show that (3 + a b + ac − 7a2 )(b − c)2 + (3 + a b + bc − 7b2 )(a − c)2 ≥ 0. From b(a − c) ≥ a(b − c) ≥ 0, we get b2 (a − c)2 ≥ a2 (b − c)2 , and hence b(a − c)2 ≥ a(b − c)2 . Thus, it suffices to show that b(3 + a b + ac − 7a2 ) + a(3 + a b + bc − 7b2 ) ≥ 0. This is true if b(3 + a b − 7a2 ) + a(3 + a b − 7b2 ) ≥ 0. Indeed, b(3 + a b − 7a2 ) + a(3 + a b − 7b2 ) = 3(a + b)(1 − 2a b) ≥ 0, since 1 − 2a b = (a − b)2 + c 2 ≥ 0. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation). Second Solution. Without loss of generality, assume that a2 + b2 +c 2 = 1 and a ≤ b ≤ c. Setting p = a + b + c, q = a b + bc + ca and r = a bc, the inequality becomes X 5q − 11bc ≤ 3, 1 + bc Y X 3 (1 + bc) + (11bc − 5q)(1 + ca)(1 + a b) ≥ 0, 3(1 + q + pr + r 2 ) + 11(q + 2pr + 3r 2 ) − 5q(3 + 2q + pr) ≥ 0, 36r 2 + 5(5 − q)pr + 3 − q − 10q2 ≥ 0. Since p2 − 2q = 1, the inequality has the homogeneous form 36r 2 + 5(5p2 − 11q)pr + 3(p2 − 2q)3 − q(p2 − 2q)2 − 10q2 (p2 − 2q) ≥ 0. According to P 3.57-(a) in Volume 1, for fixed p and q, the product r = a bc is minimal when b = c or a = 0. Therefore, since 5p2 − 11q > 0, it suffices to prove the inequality for a = 0, and for b = c = 1. For a = 0, the original inequality becomes b2

−6bc 10bc + 2 ≤ 3, 2 + c + bc b + c 2

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which reduces to (b − c)2 (3b2 + 5bc + 3b2 ) ≥ 0, while for b = c = 1, we get 5−a 10a − 6 +2 2 ≤ 3, a2 + 3 a +a+2 which is equivalent to a(3a + 1)(a − 1)2 ≥ 0. Remark. Similarly, we can prove the following generalization: • Let a, b, c be nonnegative real numbers, no all are zero. If k > 0, then X (2k + 3)a(b + c) + (k + 2)(k − 3)bc a2 + b2 + c 2 + k bc

≤ 3k,

with equality for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.129. Let a, b, c be nonnegative real numbers, no two of which are zero, and let x=

a2 + b2 + c 2 . a b + bc + ca

Prove that (a) (b) (c)

a b c 1 1 + + + ≥x+ ; b+c c+a a+b 2 x  ‹ a b c 4 6 + + ≥ 5x + ; b+c c+a a+b x  ‹ a b c 3 1 1 + + − ≥ x− . b+c c+a a+b 2 3 x (Vasile Cîrtoaje, 2011)

Solution. We will prove the more general inequality 2a 2b 2c 2(1 − k) + + + 1 − 3k ≥ (2 − k)x + , b+c c+a a+b x p where 0 ≤ k ≤ (21+6 6)/25. For k = 0, k = 1/3 and k = 4/3, we get the inequalities in (a), (b) and (c), respectively. Let p = a+b+c and q = a b+bc+ca. Since x = (p2 −2q)/q, we can write the inequality as follows a b c + + ≥ f (p, q), b+c c+a a+b

Symmetric Rational Inequalities

177

 X a + 1 ≥ 3 + f (p, q), b+c p(p2 + q) ≥ 3 + f (p, q). pq − a bc According to P 3.57-(a) in Volume 1, for fixed p and q, the product a bc is minimal when b = c or a = 0. Therefore, it suffices to prove the inequality for a = 0, and for b = c = 1. For a = 0, using the substitution y = b/c + c/b, the desired inequality becomes 2 y + 1 − 3k ≥ (2 − k) y +

2(1 − k) , y

( y − 2)[k( y − 1) + 1] ≥ 0. y Since y ≥ 2, this inequality is clearly true. For b = c = 1, the desired inequality becomes a+

4 (2 − k)(a2 + 2) 2(1 − k)(2a + 1) + 1 − 3k ≥ + , a+1 2a + 1 a2 + 2

which is equivalent to a(a − 1)2 [ka2 + 3(1 − k)a + 6 − 4k] ≥ 0. p For 0 ≤ k ≤ 1, this is obvious, and for 1 < k ≤ (21 + 6 6)/25, we have Æ ka2 + 3(1 − k)a + 6 − 4k ≥ [2 k(6 − 4k) + 3(1 − k)]a ≥ 0. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation).

P 1.130. If a, b, c are real numbers, then a2

1 1 1 9 + 2 + 2 ≤ . 2 2 2 2 2 2 + 7(b + c ) b + 7(c + a ) c + 7(a + b ) 5(a + b + c)2 (Vasile Cîrtoaje, 2008)

Solution. Let p = a + b + c and q = a b + bc + ca. Write the inequality as f6 (a, b, c) ≥ 0, where Y X f6 (a, b, c) = 9 (a2 + 7b2 + 7c 2 ) − 5p2 (b2 + 7c 2 + 7a2 )(c 2 + 7a2 + 7b2 ). Since

Y Y (a2 + 7b2 + 7c 2 ) = [7(p2 − 2q) − 6a2 ],

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f6 (a, b, c) has the highest coefficient A = 9(−6)3 < 0. According to P 2.75 in Volume 1, it suffices to prove the original inequality for b = c = 1, when the inequality reduces to (a − 1)2 (a − 4)2 ≥ 0. Thus, the proof is completed. The equality holds for a = b = c, and for a/4 = b = c (or any cyclic permutation).

P 1.131. If a, b, c are real numbers, then 3a2

ca ab 3 bc + 2 + 2 ≤ . 2 2 2 2 2 2 +b +c 3b + c + a 3c + a + b 5 (Vasile Cîrtoaje and Pham Kim Hung, 2005)

Solution. Write the inequality as f6 (a, b, c) ≥ 0, where Y X f6 (a, b, c) = 3 (3a2 + b2 + c 2 ) − 5 bc(3b2 + c 2 + a2 )(3c 2 + a2 + b2 ). Let p = a + b + c and q = a b + bc + ca. From Y X f6 (a, b, c) = 3 (2a2 + p2 − 2q) − 5 bc(2b2 + p2 − 2q)(2c 2 + p2 − 2q), it follows that f6 (a, b, c) has the same highest coefficient A as X 24a2 b2 c 2 − 20 b3 c 3 ; that is, A = 24 − 60 < 0. According to P 2.75 in Volume 1, it suffices to prove the original inequality for b = c = 1, when the inequality is equivalent to (a − 1)2 (3a − 2)2 ≥ 0. Thus, the proof is completed. The equality holds for a = b = c, and for 3a/2 = b = c (or any cyclic permutation). Remark. The inequality in P 1.131 is a particular case (k = 3) of the following more general result (Vasile Cîrtoaje, 2008): • Let a, b, c be real numbers. If k > 1, then X k(k − 3)a2 + 2(k − 1)bc

3(k + 1)(k − 2) , + k+2 with equality for a = b = c, and for ka/2 = b = c (or any cyclic permutation). ka2

b2

+ c2

≤

Symmetric Rational Inequalities

179

P 1.132. If a, b, c are real numbers such that a + b + c = 3, then (a)

1 1 3 1 + + ≤ ; 2 2 2 2 2 2 2+ b +c 2+c +a 2+a + b 4

(b)

1 1 1 1 + + ≤ . 2 2 2 2 2 2 8 + 5(b + c ) 8 + 5(c + a ) 8 + 5(a + b ) 6 (Vasile Cîrtoaje, 2006, 2009)

Solution. (a) Rewrite the inequality as follows ‹ X 1 1 3 3 − ≤ − , 2 2 2+ b +c 2 4 2 3 b2 + c 2 ≥ . 2 2 2+ b +c 2 By the Cauchy-Schwarz inequality, we have X

X


2 Pp P 2 Pp b2 + c 2 a + (a2 + b2 )(a2 + c 2 ) b2 + c 2 P P ≥ = . 2 + b2 + c 2 (2 + b2 + c 2 ) 3 + a2

Thus, it suffices to show that X XÆ (a2 + b2 )(a2 + c 2 ) ≥ a2 + 9. 2 Indeed, applying again the Cauchy-Schwarz inequality, we get 2

XÆ

(a2 + b2 )(a2 + c 2 ) ≥ 2

X X €X Š2 X (a2 + bc) = a2 + a = a2 + 9.

The equality holds for a = b = c = 1. (b) Denote p = a + b + c and q = a b + bc + ca, we write the inequality in the homogeneous form 8p2

1 1 1 1 + 2 + 2 ≤ 2, 2 2 2 2 2 2 + 45(b + c ) 8p + 45(c + a ) 8p + 45(a + b ) 6p

which is equivalent to f6 (a, b, c) ≥ 0, where Y f6 (a, b, c) = (53p2 − 90q − 45a2 ) −6p2

X (53p2 − 90q − 45b2 )(53p2 − 90q − 45c 2 ).

Clearly, f6 (a, b, c) has the highest coefficient A = (−45)3 < 0.

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Vasile Cîrtoaje

By P 2.75 in Volume 1, it suffices to prove the original inequality for b = c. In this case, the inequality is equivalent to (a − 1)2 (a − 13)2 ≥ 0. The equality holds for a = b = c = 1, and for a = 13/5 and b = c = 1/5 (or any cyclic permutation).

P 1.133. If a, b, c are real numbers, then (a + b)(a + c) (b + c)(b + a) (c + a)(c + b) 4 + + ≤ . a2 + 4(b2 + c 2 ) b2 + 4(c 2 + a2 ) c 2 + 4(a2 + b2 ) 3 (Vasile Cîrtoaje, 2008) Solution. Let p = a + b + c and q = a b + bc + ca. Write the inequality as f6 (a, b, c) ≥ 0, where Y f6 (a, b, c) = 4 (a2 + 4b2 + 4c 2 ) X −3 (a + b)(a + c)(b2 + 4c 2 + 4a2 )(c 2 + 4a2 + 4b2 ) Y X =4 (4p2 − 8q − 3a2 ) − 3 (a2 + q)(4p2 − 8q − 3b2 )(4p2 − 8q − 3c 2 ). Thus, f6 (a, b, c) has the highest coefficient A = 4(−3)3 − 34 < 0. By P 2.75 in Volume 1, it suffices to prove the original inequality for b = c = 1, when the inequality is equivalent to (a − 1)2 (2a − 7)2 ≥ 0. The equality holds for a = b = c, and for 2a/7 = b = c (or any cyclic permutation).

P 1.134. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that X

1 1 ≤ . (b + c)(7a + b + c) 2(a b + bc + ca) (Vasile Cîrtoaje, 2009)

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181

First Solution. Write the inequality as ˜ X• 4(a b + bc + ca) 1− ≥ 1, (b + c)(7a + b + c) X (b − c)2 + 3a(b + c) ≥ 1. (b + c)(7a + b + c) By the Cauchy-Schwarz inequality, we have X (b − c)2 + 3a(b + c)

4(a + b + c)4 ≥P . (b + c)(7a + b + c) [(b − c)2 + 3a(b + c)](b + c)(7a + b + c)

Therefore, it suffices to show that X 4(a + b + c)4 ≥ (b2 + c 2 − 2bc + 3ca + 3a b)(b + c)(7a + b + c). Write this inequality as X X X X a4 + a bc a+3 a b(a2 + b2 ) − 8 a2 b2 ≥ 0, X X X X a4 + a bc a− a b(a2 + b2 ) + 4 a b(a − b)2 ≥ 0. P 4 P P Since a + a bc a − a b(a2 + b2 ) ≥ 0 (Schur’s inequality of degree four), the conclusion follows. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation). Second Solution. Let p = a + b + c and q = a b + bc + ca. We need to prove that f6 (a, b, c) ≥ 0, where Y f6 (a, b, c) = (b + c)(7a + b + c) X −2(a b + bc + ca) (a + b)(a + c)(7b + c + a)(7c + a + b) Y X = (p − a)(p + 6a) − 2q (p − b)(p − c)(p + 6b)(p + 6c). Clearly, f6 (a, b, c) has the highest coefficient A = (−6)3 < 0. Thus, by P 3.76-(a) in Volume 1, it suffices to prove the original inequality for b = c = 1, and for a = 0. For b = c = 1, the inequality reduces to a(a − 1)2 ≥ 0, which is obviously true. For a = 0, the inequality can be written as 1 1 1 1 + + ≤ , 2 (b + c) c(7b + c) b(7c + b) 2bc 1 b2 + c 2 + 14bc 1 + ≤ , 2 2 2 (b + c) bc[7(b + c ) + 50bc] 2bc 1 x + 14 1 + ≤ , x + 2 7x + 50 2 where x = b/c +c/b, x ≥ 2. This reduces to the obvious inequality (x −2)(5x +28) ≥ 0.

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P 1.135. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that X b2

+ c2

1 9 ≤ . + 4a(b + c) 10(a b + bc + ca) (Vasile Cîrtoaje, 2009)

Solution. Let p = a + b + c and q = a b + bc + ca. We need to prove that f6 (a, b, c) ≥ 0, where Y f6 (a, b, c) = 9 [b2 + c 2 + 4a(b + c)] X −10(a b + bc + ca) [a2 + b2 + 4c(a + b)][a2 + c 2 + 4b(a + c)] Y X =9 (p2 + 2q − a2 − 4bc) − 10q (p2 + 2q − c 2 − 4a b)(p2 + 2q − b2 − 4ca). Clearly, f6 (a, b, c) has the same highest coefficient A as f (a, b, c), where Y X X f (a, b, c) = −9 (a2 + 4bc) = −9(65a2 b2 c 2 + 16a bc a3 + 4 a3 b3 ); that is, A = −9(65 + 48 + 12) < 0. Thus, by P 3.76-(a) in Volume 1, it suffices to prove the original inequality for b = c = 1, and for a = 0. For b = c = 1, the inequality reduces to a(a − 1)2 ≥ 0, which is obviously true. For a = 0, the inequality becomes b2

1 1 1 9 + 2 + 2 ≤ , 2 +c b + 4bc c + 4bc 10bc

1 b2 + c 2 + 8bc 9 + ≤ . 2 2 2 2 2 2 b +c 4bc(b + c ) + 17b c 10bc 1 x +8 9 + ≤ , x 4x + 17 10 where x = b/c +c/b, x ≥ 2. The inequality is true, since it is equivalent to (x −2)(26x + 85) ≥ 0. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 1.136. If a, b, c are nonnegative real numbers such that a + b + c = 3, then 1 1 1 9 + + ≤ . 3 − a b 3 − bc 3 − ca 2(a b + bc + ca) (Vasile Cîrtoaje, 2011)

Symmetric Rational Inequalities

183

First Solution. We apply the SOS method. Write the inequality as X  3 a b + bc + ca ‹ − ≥ 0. 2 3 − bc X 9 − 2a(b + c) − 5bc ≥ 0, 3 − bc X a2 + b2 + c 2 − 3bc ≥ 0. 3 − bc Since 2(a2 + b2 + c 2 − 3bc) = 2(a2 − bc) + 2(b2 + c 2 − a b − ac) + 2(a b + ac − 2bc) = (a − b)(a + c) + (a − c)(a + b) − 2b(a − b) − 2c(a − c) + 2c(a − b) + 2b(a − c) = (a − b)(a − 2b + 3c) + (a − c)(a − 2c + 3b), the required inequality is equivalent to X (a − b)(a − 2b + 3c) + (a − c)(a − 2c + 3b)

≥ 0, 3 − bc X (a − b)(a − 2b + 3c) X (b − a)(b − 2a + 3c) + ≥ 0, 3 − bc 3 − ca X (a − b)2 [9 − c(a + b + 3c)] ≥ 0, (3 − bc)(3 − ca) X (a − b)2 (3 − a b)(3 + c)(3 − 2c) ≥ 0. Without loss of generality, assume that a ≥ b ≥ c. Then it suffices to prove that (b − c)2 (3 − bc)(3 + a)(3 − 2a) + (c − a)2 (3 − ca)(3 + b)(3 − 2b) ≥ 0, which is equivalent to (a − c)2 (3 − ac)(3 + b)(3 − 2b) ≥ (b − c)2 (3 − bc)(a + 3)(2a − 3). Since 3−2b = a− b+c ≥ 0, we can obtain this inequality by multiplying the inequalities b2 (a − c)2 ≥ a2 (b − c)2 , a(3 − ac) ≥ b(3 − bc) ≥ 0, a(3 + b)(3 − 2b) ≥ b(a + 3)(2a − 3) ≥ 0. We have a(3 − ac) − b(3 − bc) = (a − b)[3 − c(a + b)] = (a − b)(3 − 3c + c 2 ) ≥ 3(a − b)(1 − c) ≥ 0.

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Also, since a + b ≤ a + b + c = 3, we have a(3 + b)(3 − 2b) − b(a + 3)(2a − 3) = 9(a + b) − 6a b − 2a b(a + b) ≥ 9(a + b) − 12a b ≥ 3(a + b)2 − 12a b = 3(a − b)2 ≥ 0. The equality holds for a = b = c = 1, and for a = 0 and b = c = 3/2 (or any cyclic permutation). Second Solution. Let p = a + b + c and q = a b + bc + ca. We need to prove that f6 (a, b, c) ≥ 0, where Y X f6 (a, b, c) = 3 (p2 − 3bc) − 2q (p2 − 2ca)(p2 − 2a b). Clearly, f6 (a, b, c) has the highest coefficient A = 3(−3)3 < 0. Thus, by P 3.76-(a) in Volume 1, it suffices to prove the original inequality for b = c, and for a = 0. For b = c = 3 − a, the inequality reduces to a(9 − a)(a − 1)2 ≥ 0, which is obviously true. For a = 0, which yields b + c = 3, the inequality can be written as (9 − 4bc)(9 − bc) ≥ 0. Indeed, (9 − 4bc)(9 − bc) = (b − c)2 (b2 + c 2 + bc) ≥ 0.

P 1.137. If a, b, c are nonnegative real numbers such that a + b + c = 3, then a2

bc ca ab 3 + 2 + 2 ≤ . +a+6 b + b+6 c +c+6 8 (Vasile Cîrtoaje, 2009)

Solution. Write the inequality as X 3a2

bc 1 ≤ , 2 + ap + 2p 8

where p = a + b + c. We need to prove that f6 (a, b, c) ≥ 0, where Y X f6 (a, b, c) = (3a2 + ap + 2p2 ) − 8 bc(3b2 + bp + 2p2 )(3c 2 + c p + 2p2 ).

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185

Clearly, f6 (a, b, c) has the same highest coefficient as X 27a2 b2 c 2 − 72 b3 c 3 , that is, A = 27 − 216 < 0. Thus, by P 3.76-(a) in Volume 1, it suffices to prove that f6 (a, 1, 1) ≥ 0 and f6 (0, b, c) ≥ 0 for all a, b, c ≥ 0. Indeed, we have f6 (a, 1, 1) = 2a(a2 + 9a + 3)(a − 1)2 (6a + 1) ≥ 0 and f6 (0, b, c) = 2(b − c)2 (5b2 + 5bc + 2c 2 )(2b2 + 5bc + 5c 2 ) ≥ 0. The equality holds for a = b = c = 1, and for a = 0 and b = c = 3/2 (or any cyclic permutation).

P 1.138. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then 8a2

1 1 1 1 + 2 + 2 ≥ . − 2bc + 21 8b − 2ca + 21 8c − 2a b + 21 9 (Michael Rozenberg, 2013)

Solution. Let q = a b + bc + ca. Write the inequality in the homogeneous form f6 (a, b, c) ≥ 0, where X Y f6 (a, b, c) = 3q (8b2 − 2ca + 7q)(8c 2 − 2a b + 7q) − (8a2 − 2bc + 7q). Clearly, f6 (a, b, c) has the same highest coefficient as f (a, b, c), where Y X X f (a, b, c) = −8 (4a2 − bc) = −8(63a2 b2 c 2 − 16 a3 b3 + 4a bc a3 ); that is, A = −8(63 − 48 + 12) < 0. By P 3.76-(a) in Volume 1, it suffices to prove that f6 (a, 1, 1) ≥ 0 and f6 (0, b, c) ≥ 0 for all a, b, c ≥ 0. Indeed, we have f6 (a, 1, 1) = 0 and f6 (0, b, c) = 8b2 c 2 (b − c)2 ≥ 0.

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The equality holds when two of a, b, c are equal. Remark. The following identity holds: X

Q 8 (a − b)2 9 −1= Q . 8a2 − 2bc + 21 (a2 − 2bc + 21)

P 1.139. Let a, b, c be real numbers, no two of which are zero. Prove that (a + b + c)2 a2 + bc b2 + ca c 2 + a b + + ≥ ; b2 + c 2 c 2 + a2 a2 + b2 a2 + b2 + c 2

(a)

(b)

6(a b + bc + ca) a2 + 3bc b2 + 3ca c 2 + 3a b + 2 + 2 ≥ . b2 + c 2 c + a2 a + b2 a2 + b2 + c 2 (Vasile Cîrtoaje, 2014)

Solution. (a) Using the known inequality X

a2 3 ≥ 2 2 b +c 2

and the Cauchy-Schwarz inequality yields X a2 + bc

‹ X bc X1 a2 bc + ≥ + b2 + c 2 b2 + c 2 b2 + c 2 2 b2 + c 2 P 2 X (b + c)2 (b + c) (a + b + c)2 P = = ≥ . 2(b2 + c 2 ) a2 + b2 + c 2 2(b2 + c 2 ) =

X

The equality holds for a = b = c. (b) We have X a2 + 3bc b2 + c 2

X 3bc a2 3 X 3bc + ≥ + b2 + c 2 b2 + c 2 2 b2 + c 2 ‹ X1 X bc (b + c)2 = −3 + 3 + 2 = −3 + 3 2 b + c2 2(b2 + c 2 ) P 2 P
2 3 (b + c) 3 a 6(a b + bc + ca) ≥ −3 + P = −3 + P = . 2 2 2 a2 + b2 + c 2 2(b + c ) a =

X

The equality holds for a = b = c.

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187

P 1.140. Let a, b, c be real numbers such that a b + bc + ca ≥ 0 and no two of which are zero. Prove that a(b + c) b(c + a) c(a + b) 3 + 2 + 2 ≥ . b2 + c 2 c + a2 a + b2 10 (Vasile Cîrtoaje, 2014) Solution. Since the problem remains unchanged by replacing a, b, c by −a, −b, −c, it suffices to consider the cases a, b, c ≥ 0 and a < 0, b ≥ 0, c ≥ 0. Case 1: a, b, c ≥ 0. We have X a(b + c) b2 + c 2

≥ =

X a(b + c) X

(b + c)2 3 3 a ≥ > . b+c 2 10

Case 2: a < 0, b ≥ 0, c ≥ 0. Replacing a by −a, we need to show that b(c − a) c(b − a) a(b + c) 3 + 2 − 2 ≥ 2 2 2 2 a +c a +b b +c 10 for any nonnegative numbers a, b, c such that a≤ We show first that

where x =

bc . b+c

b(c − a) b(c − x) ≥ 2 , a2 + c 2 x + c2

bc , x ≥ a. This is equivalent to b+c b(x − a)[(c − a)x + ac + c 2 ] ≥ 0,

which is true because (c − a)x + ac + c 2 =

c 2 (a + 2b + c) ≥ 0. b+c

Similarly, we can show that c(b − a) c(b − x) ≥ 2 . 2 2 a +b x + b2 In addition,

a(b + c) x(b + c) ≤ 2 . 2 2 b +c b + c2

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Therefore, it suffices to prove that 3 b(c − x) c(b − x) x(b + c) . + 2 − 2 ≥ 2 2 2 2 x +c x +b b +c 10 Denote p= Since

and

we need to show that

Since

b , b+c

q=

p b(c − x) = , 2 2 x +c 1 + p2

c , b+c

p + q = 1.

q c(b − x) = 2 2 x +b 1 + q2

pq x(b + c) bc = 2 = , 2 2 2 b +c b +c 1 − 2pq p q pq 3 + − ≥ . 2 2 1+p 1+q 1 − 2pq 10 q 1 + pq p + = , 1 + p2 1 + q2 2 − 2pq + p2 q2

the inequality can be written as (pq + 2)2 (1 − 4pq) ≥ 0, which is true since 1 − 4pq = (p + q)2 − 4pq = (p − q)2 ≥ 0. The equality holds for −2a = b = c (or any cyclic permutation).

P 1.141. If a, b, c are positive real numbers such that a bc > 1, then 1 1 4 + ≥ . a + b + c − 3 a bc − 1 a b + bc + ca − 3 (Vasile Cîrtoaje, 2011) Solution (by Vo Quoc Ba Can). By the AM-GM inequality, we have p 3 a + b + c ≥ 3 a bc > 3, p 3 a b + bc + ca ≥ a2 b2 c 2 > 3.

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189

Without loss of generality, assume that a = min{a, b, c}. By the Cauchy-Schwarz inequality, we have 

1 1 + a + b + c − 3 a bc − 1

‹•

˜  ‹ p 1 2 a bc − 1 a+ p a(a + b + c − 3) + ≥ . a a

Therefore, it suffices to prove that (a + 1) ≥ 4a 2

Since a(a + b + c − 3) +

a bc − 1 a . a b + bc + ca − 3

a(a + b + c − 3) +

a bc − 1 (a − 1)3 = a b + bc + ca − 3 + , a a

this inequality can be written as follows (a + 1)2 (a − 1)3 −1≥ , 4a a(a b + bc + ca − 3) (a − 1)2 (a − 1)3 ≥ , 4a a(a b + bc + ca − 3) (a − 1)2 (a b + bc + ca + 1 − 4a) ≥ 0. This is true since bc ≥

Æ 3

(a bc)2 > 1,

and hence a b + bc + ca + 1 − 4a > a2 + 1 + a2 + 1 − 4a = 2(a − 1)2 ≥ 0. The equality holds for a > 1 and b = c = 1 (or any cyclic permutation). Remark. Using this inequality, we can prove P 3.84 in Volume 1, which states that ‹ 1 1 1 1 (a + b + c − 3) + + − 3 + a bc + ≥2 a b c a bc 

for any positive real numbers a, b, c. This inequality is clearly true for a bc = 1. In addition, it remains unchanged by substituting a, b, c with 1/a, 1/b, 1/c, respectively. p 3 Therefore, it suffices to consider the case a bc > 1. Since a + b + c ≥ 3 a bc > 3, we can write the required inequality as E ≥ 0, where E = a b + bc + ca − 3a bc +

(a bc − 1)2 . a+ b+c−3

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According to the inequality in P 1.141, we have  2 E ≥ a b + bc + ca − 3a bc + (a bc − 1) = (a b + bc + ca − 3) + ≥2

v t

(a b + bc + ca − 3) ·

1 4 − a b + bc + ca − 3 a bc − 1

‹

4(a bc − 1)2 − 4(a bc − 1) a b + bc + ca − 3

4(a bc − 1)2 − 4(a bc − 1) = 0. a b + bc + ca − 3

P 1.142. Let a, b, c be positive real numbers, no two of which are zero. Prove that X (4b2 − ac)(4c 2 − a b) b+c

≤

27 a bc. 2 (Vasile Cîrtoaje, 2011)

Solution. Since X (4b2 − ac)(4c 2 − a b) b+c

X bc(16bc + a2 ) X a(b3 + c 3 ) −4 b+c b+c X bc(16bc + a2 ) X = −4 a(b2 + c 2 ) + 12a bc b + c  X  a2 16bc = bc + − 4(b + c) + 12a bc b+c b+c  X  a2 (b − c)2 = bc −4 + 12a bc b+c b+c =

we can write the inequality as follows  X a a2 4(b − c)2 bc − + ≥ 0, 2 b+c b+c 8

X bc(b − c)2 X 2a − b − c ≥ a bc . b+c b+c

In addition, since X 2a − b − c b+c

= =

X (a − b) + (a − c) X

=

X a−b

+

X b−a c+a

b+c b+c X (a − b)2 (b − c)2 = , (b + c)(c + a) (c + a)(a + b)

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191

the inequality can be restated as 8

X bc(b − c)2 X (b − c)2 ≥ a bc , b+c (c + a)(a + b)

X bc(b − c)2 (8a2 + 8bc + 7a b + 7ac) ≥ 0. (a + b)(b + c)(c + a) Since the last form is obvious, the proof is completed. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 1.143. Let a, b, c be nonnegative real numbers, no two of which are zero, such that a + b + c = 3. Prove that

a b c 2 + + ≥ . 3a + bc 3b + ca 3c + a b 3

Solution. Since 3a + bc = a(a + b + c) + bc = (a + b)(a + c), we can write the inequality as follows a(b + c) + b(c + a) + c(a + b) ≥

2 (a + b)(b + c)(c + a), 3

6(a b + bc + ca) ≥ 2[(a + b + c)(a b + bc + ca) − a bc], 2a bc ≥ 0. The equality holds for a = 0, or b = 0, or c = 0.

P 1.144. Let a, b, c be positive real numbers such that  ‹ 1 1 1 (a + b + c) + + = 10. a b c Prove that

19 a b c 5 ≤ + + ≤ . 12 b+c c+a a+b 3 (Vasile Cîrtoaje, 2012)

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Vasile Cîrtoaje

First Solution. Write the hypothesis ‹ 1 1 1 + + = 10 (a + b + c) a b c 

as

b+c c+a a+b + + =7 a b c

and (a + b)(b + c)(c + a) = 9a bc. b+c c+a a+b , y = and z = , we need to show that a b c x + y + z = 7 and x yz = 9 involve

Using the substitutions x =

1 1 1 5 19 ≤ + + ≤ , 12 x y z 3 or, equivalently, 19 1 x(7 − x) 5 ≤ + ≤ . 12 x 9 3 Clearly, x, y, z ∈ (0, 7). The left inequality is equivalent to (x − 4)(2x − 3)2 ≤ 0, while the right inequality is equivalent to (x − 1)(x − 3)2 ≥ 0. These inequalities are true if 1 ≤ x ≤ 4. To show that 1 ≤ x ≤ 4, from ( y + z)2 ≥ 4 yz, we get 36 (7 − x)2 ≥ , x (x − 1)(x − 4)(x − 9) ≥ 0, 1 ≤ x ≤ 4. Thus, the proof is completed. The left inequality is an equality for 2a = b = c (or any cyclic permutation), and the right inequality is an equality for a/2 = b = c (or any cyclic permutation). Second Solution. Due to homogeneity, assume that b + c = 2; this involves bc ≤ 1. From the hypothesis  ‹ 1 1 1 (a + b + c) + + = 10, a b c we get 2a(a + 2) bc = . 9a − 2

Symmetric Rational Inequalities

193

Since bc − 1 =

(a − 2)(2a − 1) , 9a − 2

from the condition bc ≤ 1, we get 1 ≤ a ≤ 2. 2 We have b c a(b + c) + b2 + c 2 2a + 4 − 2bc + = 2 = 2 c+a a+b a + (b + c)a + bc a + 2a + bc 2 2(7a + 12a − 4) 2(7a − 2) = = , 9a2 (a + 2) 9a2 and hence

b c a 2(7a − 2) 9a3 + 28a − 8 a + + = + = . b+c c+a a+b 2 9a2 18a2

Thus, we need to show that 19 9a3 + 28a − 8 5 ≤ ≤ . 12 18a2 3 These inequalities are true, since the left inequality is equivalent to (2a − 1)(3a − 4)2 ≥ 0, and the right inequality is equivalent to (a − 2)(3a − 2)2 ≤ 0. Remark. Similarly, we can prove the following generalization. • Let a, b, c be positive real numbers such that ‹ 1 1 1 8k2 (a + b + c) + + =9+ , a b c 1 − k2 

where k ∈ (0, 1). Then, k2 a b c 3 k2 ≤ + + − ≤ . 1+k b+c c+a a+ b 2 1−k

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Vasile Cîrtoaje

P 1.145. Let a, b, c be nonnegative real numbers, no two of which are zero, such that a + b + c = 3. Prove that a b c 9 < + + ≤ 1. 10 2a + bc 2b + ca 2c + a b (Vasile Cîrtoaje, 2012) Solution. (a) Since a 1 −bc − = , 2a + bc 2 2(2a + bc) we can write the right inequality as X

bc ≥ 1. 2a + bc

According to the Cauchy-Schwarz inequality, we have P P 2 2 P X bc ( bc)2 b c + 2a bc a P ≥P = = 1. 2a + bc bc(2a + bc) 6a bc + b2 c 2 The equality holds for a = b = c = 1, and also for a = 0, or b = 0, or c = 0. (b) First Solution. For the nontrivial case a, b, c > 0, we can write the left inequality as X 1 9 . > bc 10 2+ a Using the substitutions x=

v t bc a

, y=

we need to show that X

s

v t ab ca , z= , b c

1 9 > 2 + x2 10

for all positive real numbers x, y, z satisfying x y + yz + z x = 3. By expanding, the inequality becomes X X 4 x 2 + 48 ≥ 9x 2 y 2 z 2 + 8 x 2 y 2. Since

X

X X X x2 y2 = ( x y)2 − 2x yz x = 9 − 2x yz x,

we can write the desired inequality as X X 4 x 2 + 16x yz x ≥ 9x 2 y 2 z 2 + 24,

Symmetric Rational Inequalities

195

which is equivalent to 16x yz(x + y + z) ≥ 9x 2 y 2 z 2 + 4(2x y + 2 yz + 2z x − x 2 − y 2 − z 2 ). Using Schur’s inequality 9x yz ≥ 2x y + 2 yz + 2z x − x 2 − y 2 − z 2 , x + y +z it suffices to prove that 16x yz(x + y + z) ≥ 9x 2 y 2 z 2 + This is true if 16(x + y + z) ≥ 9x yz +

36x yz . x + y +z

36 . x + y +z

Since x + y +z ≥ and 1=

Æ

3(x y + yz + z x) = 3

x y + yz + z x Æ 3 ≥ x 2 y 2z2, 3

we have 16(x + y + z) − 9x yz −

36 ≥ 48 − 9x yz − 12 = 9(4 − x yz) > 0. x + y +z

Second Solution. As it is shown at the first solution, it suffices to show that X

1 9 > 2 2+ x 10

for all positive real numbers x, y, z satisfying x y + yz + z x = 3. Rewrite this inequality as X x2 6 < . 2 2+ x 5 Let p and q be two positive real numbers such that p p + q = 3. By the Cauchy-Schwarz inequality, we have (p x + q x)2 x2 3x 2 = = 2 + x2 2(x y + yz + z x) + 3x 2 2x(x + y + z) + (x 2 + 2 yz) ≤

p2 x q2 x 2 + 2 . 2(x + y + z) x + 2 yz

196

Vasile Cîrtoaje

Therefore, X

X X q2 x 2 X p2 x p2 x2 x2 2 ≤ + = + q . 2 + x2 2(x + y + z) x 2 + 2 yz 2 x 2 + 2 yz

Thus, it suffices to prove that X p2 x2 6 + q2 < . 2 2 x + 2 yz 5 We claim that X

x2 < 2. x 2 + 2 yz

Under this assumption, we only need to show that p2 6 + 2q2 ≤ . 2 5 p p p p2 4 3 3 6 Indeed, choosing p = and q = , we have p + q = 3 and + 2q2 = . To 5 5 2 5 P x2 complete the proof, we need to prove the homogeneous inequality < 2, x 2 + 2 yz which is equivalent to X yz 1 > . x 2 + 2 yz 2 By the Cauchy-Schwarz inequality, we get P P 2 2 P X ( yz)2 y z + 2x yz x yz 1 P P ≥P > . = 2 2 2 2 x + 2 yz 2 x yz x + 2 y z yz(x + 2 yz)

P 1.146. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that b3 c3 a3 + b3 + c 3 a3 + + ≤ . 2a2 + bc 2b2 + ca 2c 2 + a b a2 + b2 + c 2 (Vasile Cîrtoaje, 2011) Solution. Write the inequality as follows  X a3 a3 − ≥ 0, a2 + b2 + c 2 2a2 + bc

Symmetric Rational Inequalities

197

X a3 (a2 + bc − b2 − c 2 ) 2a2 + bc

≥ 0,

X a3 [a2 (b + c) − b3 − c 3 ] (b + c)(2a2 + bc)

≥ 0,

X a3 b(a2 − b2 ) + a3 c(a2 − c 2 )

≥ 0, (b + c)(2a2 + bc) X a3 c(a2 − c 2 ) X a3 b(a2 − b2 ) + ≥ 0, (b + c)(2a2 + bc) (b + c)(2a2 + bc) X a3 b(a2 − b2 ) X b3 a(b2 − a2 ) + ≥ 0, (b + c)(2a2 + bc) (c + a)(2b2 + ca) X a b(a + b)(a − b)2 [2a2 b2 + c(a3 + a2 b + a b2 + b3 ) + c 2 (a2 + a b + b2 )] (b + c)(c + a)(2a2 + bc)(2b2 + ca)

≥ 0.

The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 1.147. Let a, b, c be positive real numbers, no two of which are zero. Prove that b3 c3 a+b+c a3 + + ≥ . 2 2 2 4a + bc 4b + ca 4c + a b 5 (Vasile Cîrtoaje, 2011) Solution. Assume that a ≥ b ≥ c, and write the inequality as follows  X a3 a − ≥ 0, 4a2 + bc 5 X a(a2 − bc) 4a2 + bc

≥ 0,

X a[(a − b)(a + c) + (a − c)(a + b)]

≥ 0, 4a2 + bc X a(a − b)(a + c) X a(a − c)(a + b) + ≥ 0, 4a2 + bc 4a2 + bc X a(a − b)(a + c) X b(b − a)(b + c) + ≥ 0, 4a2 + bc 4b2 + ca X c(a − b)2 [(a − b)2 + bc + ca − a b] ≥ 0. (4a2 + bc)(4b2 + ca)

198

Vasile Cîrtoaje

Clearly, it suffices to show that X c(a − b)2 (bc + ca − a b) (4a2 + bc)(4b2 + ca)

≥ 0,

we can be written as X (a − b)2 (bc + ca − a b)(4c 3 + a bc) ≥ 0. Since ca + a b − bc > 0, it is enough to prove that (c − a)2 (a b + bc − ca)(4b3 + a bc) + (a − b)2 (bc + ca − a b)(4c 3 + a bc) ≥ 0. In addition, since (c − a)2 ≥ (a − b)2 , 4b3 + a bc ≥ 4c 3 + a bc and a b + bc − ca > 0, we only need to show that (a − b)2 (a b + bc − ca)(4c 3 + a bc) + (a − b)2 (bc + ca − a b)(4c 3 + a bc) ≥ 0. This is equivalent to the obvious inequality a bc(a − b)2 (4c 2 + bc) ≥ 0. The equality holds for a = b = c.

P 1.148. If a, b, c are positive real numbers, then 1 1 3 1 + + ≥ . (2 + a)2 (2 + b)2 (2 + c)2 6 + a b + bc + ca (Vasile Cîrtoaje, 2013) Solution. By the Cauchy-Schwarz inequality, we have X

1 4(a + b + c)2 P ≥ . (2 + a)2 (2 + a)2 (b + c)2

Thus, it suffices to show that 4(a + b + c)2 (6 + a b + bc + ca) ≥

X (2 + a)2 (b + c)2 .

This inequality is equivalent to 2p2 q − 3q2 + 3pr + 12q ≥ 6(pq + 3r),

Symmetric Rational Inequalities

199

where p = a + b + c, q = a b + bc + ca,

r = a bc.

According to AM-GM inequality, 2p2 q − 3q2 + 3pr + 12q ≥ 2

Æ

12q(2p2 q − 3q2 + 3pr).

Therefore, it is enough to prove the homogeneous inequality 4q(2p2 q − 3q2 + 3pr) ≥ 3(pq + 3r)2 , which can be written as 5p2 q2 ≥ 12q3 + 6pqr + 27r 2 . Since pq ≥ 9r, we have 3(5p2 q2 − 12q3 − 6pqr − 27r 2 ) ≥ 15p2 q2 − 36q3 − 2p2 q2 − p2 q2 = 12q2 (p2 − 3q) ≥ 0. The equality holds for a = b = c = 1.

P 1.149. If a, b, c are positive real numbers, then 1 1 1 3 + + ≥ . 1 + 3a 1 + 3b 1 + 3c 3 + a bc (Vasile Cîrtoaje, 2013) Solution. Set p = a + b + c, q = a b + bc + ca,

r=

p 3

a bc,

and write the inequality as follows X (3 + r 3 ) (1 + 3b)(1 + 3c) ≥ 3(1 + 3a)(1 + 3b)(1 + 3c), (3 + r 3 )(3 + 6p + 9q) ≥ 3(1 + 3p + 9q + 27r 3 ), r 3 (2p + 3q) + 2 + 3p ≥ 26r 3 . By virtue of the AM-GM inequality, we have p ≥ 3r, q ≥ 3r 2 . Therefore, it suffices to show that r 3 (6r + 9r 2 ) + 2 + 9r ≥ 26r 3 ,

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Vasile Cîrtoaje

which is equivalent to the obvious inequality (r − 1)2 (9r 3 + 24r 2 + 13r + 2) ≥ 0. The equality holds for a = b = c = 1.

P 1.150. Let a, b, c be real numbers, no two of which are zero. If 1 ≤ k ≤ 3, then  ‹ ‹ ‹ 2a b 2bc 2ca k+ 2 k + k + ≥ (k − 1)(k2 − 1). a + b2 b2 + c 2 c 2 + a2 (Vasile Cîrtoaje and Vo Quoc Ba Can, 2011) Solution. If a, b, c are the same sign, then the inequality is obvious since  ‹ ‹ ‹ 2a b 2bc 2ca k+ 2 k+ 2 k+ 2 ≥ k3 > (k − 1)(k2 − 1). a + b2 b + c2 c + a2 Since the inequality remains unchanged by replacing a, b, c with −a, −b, −c, it suffices to consider further that a ≤ 0, b ≥ 0, c ≥ 0. Setting −a for a, we need to show that  ‹ ‹ ‹ 2a b 2bc 2ca k− 2 k+ 2 k− 2 ≥ (k − 1)(k2 − 1) a + b2 b + c2 c + a2 for a, b, c ≥ 0. Since •  ‹ ‹  ˜ 2a b (a − c) 2ca (a − b)2 k− 2 k − 1 + k − = k − 1 + a + b2 c 2 + a2 a2 + b2 c 2 + a2   (a − b)2 (a − c)2 2 ≥ (k − 1) + (k − 1) + 2 , a2 + b2 c + a2 it suffices to prove that 

(a − b)2 (a − c)2 k−1+ 2 + 2 a + b2 c + a2



k+

2bc b2 + c 2

‹

≥ k2 − 1.

According to the inequality (a) from P 2.19 in Volume 3, we have (a − b)2 (a − c)2 (b − c)2 + ≥ . a2 + b2 c 2 + a2 (b + c)2 Thus, it suffices to show that 

(b − c)2 k−1+ (b + c)2



2bc k+ 2 b + c2

‹

≥ k2 − 1,

Symmetric Rational Inequalities

201

which is equivalent to the obvious inequality (b − c)4 + 2(3 − k)bc(b − c)2 ≥ 0. The equality holds for a = b = c.

P 1.151. If a, b, c are non-zero and distinct real numbers, then • ˜  ‹ 1 1 1 1 1 1 1 1 1 + + +3 + + ≥4 + + . a2 b2 c 2 (a − b)2 (b − c)2 (c − a)2 a b bc ca Solution. Write the inequality as X ‹ X X 1 1 X 1 1 − + 3 ≥ 3 . a2 bc (b − c)2 bc In virtue of the AM-GM inequality, it suffices to prove that v  ˜ X 1 t X 1 X 1 ‹ •X 1 2 3 − ≥ 3 , a2 bc (b − c)2 bc which is true if X ‹ •X ˜ X ‹2 1 X 1 1 1 4 − ≥3 . 2 2 a bc (b − c) bc Rewrite this inequality as X X X X 4( a2 b2 − a bc a)( a2 − a b)2 ≥ 3(a + b + c)2 (a − b)2 (b − c)2 (c − a)2 . Using the notations p = a + b + c, q = a b + bc + ca, r = a bc, and the identity (a − b)2 (b − c)2 (c − a)2 = −27r 2 − 2(2p2 − 9q)pr + p2 q2 − 4q3 , we can write the inequality as 4(q2 − 3pr)(p2 − 3q)2 ≥ 3p2 [−27r 2 − 2(2p2 − 9q)pr + p2 q2 − 4q3 ], which is equivalent to (9pr + p2 q − 6q2 )2 ≥ 0.

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Vasile Cîrtoaje

P 1.152. Let a, b, c be positive real numbers, and let A=

a b + + k, b a

B=

b c + + k, c b

C=

c a + + k, a b

where −2 < k ≤ 4. Prove that 1 4 1 1 1 + + ≤ + . A B C k + 2 A + B + C − (k + 2) (Vasile Cîrtoaje, 2009) Solution. Let us denote x= We need to show that X

a , b

y=

b c , z= . c a

x 1 4 P ≤ +P x2 + kx + 1 k+2 x + x y + 2k − 2

for all positive real numbers x, y, z satisfying x yz = 1. Write this inequality as follows: ‹ X 1 x 2 4 P , − 2 ≥ −P k + 2 x + kx + 1 k+2 x + x y + 2k − 2 P X (x − 1)2 2 yz(x − 1)2 P ≥P , x2 + kx + 1 x + x y + 2k − 2 X (x − 1)2 [−x + y + z + x( y + z) − yz − 2] ≥ 0. x2 + kx + 1 Since −x + y + z + x( y + z) − yz − 2 = (x + 1)( y + z) − (x + yz + 2) = (x + 1)( y + z) − (x + 1)( yz + 1) = −(x + 1)( y − 1)(z − 1), the inequality is equivalent to −(x − 1)( y − 1)(z − 1)

X

x2 − 1 ≥ 0, x2 + kx + 1

or E ≥ 0, where E = −(x − 1)( y − 1)(z − 1) We have

X (x 2 − 1)( y 2 + k y + 1)(z 2 + kz + 1).

X (x 2 − 1)( y 2 + k y + 1)(z 2 + kz + 1) = X Š €X X Š €X = k(2 − k) xy− x + x2 y2 − x2

Symmetric Rational Inequalities

203

= −k(2 − k)(x − 1)( y − 1)(z − 1) − (x 2 − 1)( y 2 − 1)(z 2 − 1) = −(x − 1)( y − 1)(z − 1)[(x + 1)( y + 1)(z + 1) + k(2 − k)], and hence E = (x − 1)2 ( y − 1)2 (z − 1)2 [(x + 1)( y + 1)(z + 1) + k(2 − k)] ≥ 0, because p p p (x + 1)( y + 1)(z + 1) + k(2 − k) ≥ (2 x)(2 y)(2 z) + k(2 − k) = (2 + k)(4 − k) ≥ 0. The equality holds for a = b, or b = c, or c = a.

P 1.153. If a, b, c are nonnegative real numbers, no two of which are zero, then b2

1 1 1 1 1 1 + 2 + 2 ≥ 2 + 2 + 2 . 2 2 2 + bc + c c + ca + a a + ab + b 2a + bc 2b + ca 2c + a b (Vasile Cîrtoaje, 2014)

Solution. Write the inequality as follows: ‹ X 1 1 − ≥ 0, b2 + bc + c 2 2a2 + bc (a2 − b2 ) + (a2 − c 2 ) ≥ 0, (b2 + bc + c 2 )(2a2 + bc) X X a2 − b2 b2 − a2 + ≥ 0, (b2 + bc + c 2 )(2a2 + bc) (c 2 + ca + a2 )(2b2 + ca) X c(a2 − b2 )(a − b) (a2 + b2 + c 2 − a b − bc − ca) ≥ 0. (b2 + bc + c 2 )(c 2 + ca + a2 )(2a2 + bc)(2b2 + ca) X

Clearly, the last form is obvious. The equality holds for a = b = c.

P 1.154. If a, b, c are nonnegative real numbers such that a + b + c = 3, then 1 1 1 1 1 1 + + ≥ 2 + 2 + 2 . 2a b + 1 2bc + 1 2ca + 1 a +2 b +2 c +2 (Vasile Cîrtoaje, 2014)

204

Vasile Cîrtoaje

Solution. Write the inequality as X

 ‹ 3 X 1 1 1 − ≥ − , 2a b + 1 2 a2 + 2 2

X

X 1 a2 3 + ≥ . 2 2a b + 1 2(a + 2) 2

Let us denote q = a b + bc + ca,

q ≤ 3.

By the Cauchy-Schwarz inequality, we have X

and X

1 9 9 ≥P = 2a b + 1 (2a b + 1) 2q + 3

P
2 a a2 9 ≥P = . 2 2 2(a + 2) 2(a + 2) 2(15 − 2q)

Therefore, it suffices to prove that 9 3 9 + ≥ . 2q + 3 2(15 − 2q) 2 This inequality is true because it reduces to the obvious inequality (3 − q)(9 − 2q) ≥ 0. The equality holds for a = b = c = 1.

P 1.155. If a, b, c are nonnegative real numbers such that a + b + c = 4, then 1 1 1 1 1 1 + + ≥ 2 + 2 + 2 . a b + 2 bc + 2 ca + 2 a +2 b +2 c +2 (Vasile Cîrtoaje, 2014) First Solution (by Nguyen Van Quy). Rewrite the inequality as follows: ‹ X 2 1 1 − − ≥ 0, a b + 2 a2 + 2 b2 + 2 ˜ X• a(a − b) b(b − a) + ≥ 0, (a b + 2)(a2 + 2) (a b + 2)(b2 + 2)

Symmetric Rational Inequalities

205

X (2 − a b)(a − b)2 (c 2 + 2)

≥ 0. ab + 2 Without loss of generality, assume that a ≥ b ≥ c ≥ 0. Then, bc ≤ ac ≤

a(b + c) (a + b + c)2 ≤ =2 2 8

and X (2 − a b)(a − b)2 (c 2 + 2) ab + 2

(2 − a b)(a − b)2 (c 2 + 2) (2 − ac)(a − c)2 (b2 + 2) + ab + 2 ac + 2 2 2 (2 − a b)(a − b) (c + 2) (2 − ac)(a − b)2 (c 2 + 2) ≥ + ab + 2 ab + 2 (4 − a b − ac)(a − b)2 (c 2 + 2) ≥ 0. = ab + 2 ≥

The equality holds for a = b = c = 4/3, and also for a = 2 and b = c = 1 (or any cyclic permutation). Second Solution. Write the inequality as X

 ‹ 1 3 X 1 1 − ≥ − , bc + 2 2 a2 + 2 2

X

X 1 a2 3 + ≥ . bc + 2 2(a2 + 2) 2

Assume that a ≥ b ≥ c and denote s=

b+c , 2

p = bc,

0≤s≤

4 , 0 ≤ p ≤ s2 . 3

By the Cauchy-Schwarz inequality, we have b2 c2 (b + c)2 s2 + ≥ = . 2(b2 + 2) 2(c 2 + 2) 2(b2 + 2) + 2(c 2 + 2) + 4 2s2 − p + 2 In addition,

1 1 a(b + c) + 4 2as + 4 + = = 2 . ca + 2 a b + 2 (a b + 2)(ac + 2) a p + 4as + 4

Therefore, it suffices to show that E(a, b, c) ≥ 0, where E(a, b, c) =

1 2(as + 2) a2 s2 3 + 2 + + − . p + 2 a p + 4as + 4 2(a2 + 2) 2s2 − p + 2 2

We will prove that E(a, b, c) ≥ E(a, s, s) ≥ 0.

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We have ‹  ‹ 1 1 1 1 − + 2(as + 2) 2 − E(a, b, c) − E(a, s, s) = p + 2 s2 + 2 a p + 4as + 4 a2 s2 + 4as + 4  ‹ 1 1 + s2 − 2 2 2s − p + 2 s + 2 2 s −p 2a2 (as + 2)(s2 − p) = + (p + 2)(s2 + 2) (a2 p + 4as + 4)(a2 s2 + 4as + 4) s2 (s2 − p) − 2 . (s + 2)(2s2 − p + 2) 

Since s2 − p ≥ 0, it remains to show that 2a2 (as + 2) s2 1 + ≥ , (p + 2)(s2 + 2) (a2 p + 4as + 4)(a2 s2 + 4as + 4) (s2 + 2)(2s2 − p + 2) which is equivalent to p(s2 + 1) − 2 2a2 (as + 2) ≥ . (a2 p + 4as + 4)(a2 s2 + 4as + 4) (p + 2)(s2 + 2)(2s2 − p + 2) Since a2 p + 4as + 4 ≤ a2 s2 + 4as + 4 = (as + 2)2 and 2s2 − p + 2 ≥ s2 + 2, it is enough to prove that p(s2 + 1) − 2 2a2 ≥ . (as + 2)3 (p + 2)(s2 + 2)2 In addition, since as + 2 = (4 − 2s)s + 2 ≤ 4 and

p(s2 + 1) − 2 2(s2 + 2) 2(s2 + 2) = s2 + 1 − ≤ s2 + 1 − 2 = s2 − 1, p+2 p+2 s +2

it suffices to show that

a2 s2 − 1 ≥ 2 , 32 (s + 2)2

which is equivalent to (2 − s)2 (2 + s2 )2 ≥ 8(s2 − 1).

Symmetric Rational Inequalities

207

4 , we have 3  ‹ 4 2 4 2 2 2 2 (2 − s) (2 + s ) − 8(s − 1) ≥ 2 − (2 + s2 )2 − 8(s2 − 1) = (s4 − 14s2 + 22) 3 9   ‹2   4 4 88 16 2 2 = (7 − s ) − 27 ≥ − 27 = > 0. 7− 9 9 9 729

Indeed, for the non-trivial case 1 < s ≤

To end the proof, we need to show that E(a, s, s) ≥ 0. Notice that E(a, s, s) can be find from E(a, b, c) by replacing p with s2 . We get 1 2 a2 s2 3 + + + − s2 + 2 as + 2 2(a2 + 2) s2 + 2 2 (s − 1)2 (3s − 4)2 = ≥ 0. 2(s2 + 2)(1 + 2s − s2 )(2s2 − 8s + 9)

E(a, s, s) =

P 1.156. If a, b, c are nonnegative real numbers, no two of which are zero, then (a)

(b)

a b + bc + ca (a − b)2 (b − c)2 (c − a)2 + ≤ 1; a2 + b2 + c 2 (a2 + b2 )(b2 + c 2 )(c 2 + a2 ) a b + bc + ca (a − b)2 (b − c)2 (c − a)2 + ≤ 1. a2 + b2 + c 2 (a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 ) (Vasile Cîrtoaje, 2014)

Solution. (a) First Solution. Consider the non-trivial case where a, b, c are distinct and write the inequality as follows: (a − b)2 (b − c)2 (c − a)2 (a − b)2 + (b − c)2 + (c − a)2 ≤ , (a2 + b2 )(b2 + c 2 )(c 2 + a2 ) 2(a2 + b2 + c 2 ) (a2 + b2 ) + (b2 + c 2 ) + (c 2 + a2 ) (a − b)2 + (b − c)2 + (c − a)2 ≤ , (a2 + b2 )(b2 + c 2 )(c 2 + a2 ) (a − b)2 (b − c)2 (c − a)2 X X 1 1 ≤ . (b2 + c 2 )(c 2 + a2 ) (b − c)2 (c − a)2 Since a2 + b2 ≥ (a − b)2 ,

b2 + c 2 ≥ (b − c)2 ,

c 2 + a2 ≥ (c − a)2 ,

the conclusion follows. The equality holds for a = b = c.

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Second Solution. Assume that a ≥ b ≥ c. We have (a − b)2 (b − c)2 (c − a)2 a b + bc + ca (a − b)2 (a − c)2 a b + bc + ca + ≤ + a2 + b2 + c 2 (a2 + b2 )(b2 + c 2 )(c 2 + a2 ) a2 + b2 + c 2 (a2 + b2 )(a2 + c 2 ) 2a b + c 2 (a − b)2 a2 ≤ 2 + a + b2 + c 2 a2 (a2 + b2 + c 2 ) 2a b + c 2 + (a − b)2 = 1. = a2 + b2 + c 2 (b) Consider the non-trivial case where a, b, c are distinct and write the inequality as follows: (a − b)2 (b − c)2 (c − a)2 (a − b)2 + (b − c)2 + (c − a)2 ≤ , (a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 ) 2(a2 + b2 + c 2 ) 2(a2 + b2 + c 2 ) (a − b)2 + (b − c)2 + (c − a)2 ≤ , (a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 ) (a − b)2 (b − c)2 (c − a)2 X 1 2(a2 + b2 + c 2 ) ≥ . (a − b)2 (a − c)2 (a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 ) Assume that a = min{a, b, c} and use the substitution b = a + x,

c = a + y,

x, y ≥ 0.

The inequality can be written as 1 x2 y2

+

x 2 (x

1 1 + 2 ≥ 2 f (a), 2 − y) y (x − y)2

where f (a) =

3a2 + 2(x + y)a + x 2 + y 2 . (a2 + x a + x 2 )(a2 + y a + y 2 )[a2 + (x + y)a + x 2 − x y + y 2 ]

We will show that 1 x2 y2

+

x 2 (x

1 1 + 2 ≥ 2 f (0) ≥ 2 f (a). 2 − y) y (x − y)2

We have 2(x 2 + y 2 − x y) 2(x 2 + y 2 ) 1 1 1 + + − 2 f (0) = − x 2 y 2 x 2 (x − y)2 y 2 (x − y)2 x 2 y 2 (x − y)2 x 2 y 2 (x 2 − x y + y 2 ) 2 = ≥ 0. 2 2 (x − y) (x − x y + y 2 )

Symmetric Rational Inequalities

209

Also, since (a2 + x a + x 2 )(a2 + y a + y 2 ) ≥ (x 2 + y 2 )a2 + x y(x + y)a + x 2 y 2 and a2 + (x + y)a + x 2 − x y + y 2 ≥ x 2 − x y + y 2 , we get f (a) ≤ g(a), where g(a) =

3a2 + 2(x + y)a + x 2 + y 2 . [(x 2 + y 2 )a2 + x y(x + y)a + x 2 y 2 ](x 2 − x y + y 2 )

Therefore, x2 + y2 − g(a) x 2 y 2 (x 2 − x y + y 2 ) (x 4 − x 2 y 2 + y 4 )a2 + x y(x + y)(x − y)2 a = 2 2 2 ≥ 0. x y (x − x y + y 2 )[(x 2 + y 2 )a2 + x y(x + y)a + x 2 y 2 ]

f (0) − f (a) ≥

Thus, the proof is completed. The equality holds for a = b = c.

P 1.157. If a, b, c are nonnegative real numbers, no two of which are zero, then

a2

1 1 1 45 + 2 + 2 ≥ . 2 2 2 2 2 2 +b b +c c +a 8(a + b + c ) + 2(a b + bc + ca) (Vasile Cîrtoaje, 2014)

First Solution (by Nguyen Van Quy). Multiplying by a2 +b2 +c 2 , the inequality becomes X

a2 45(a2 + b2 + c 2 ) + 3 ≥ . b2 + c 2 8(a2 + b2 + c 2 ) + 2(a b + bc + ca)

Applying the Cauchy-Schwarz inequality, we have X

P 2
2 a a2 (a2 + b2 + c 2 )2 P ≥ = . b2 + c 2 a2 (b2 + c 2 ) 2(a2 b2 + b2 c 2 + c 2 a2 )

Therefore, it suffices to show that (a2 + b2 + c 2 )2 45(a2 + b2 + c 2 ) + 3 ≥ , 2(a2 b2 + b2 c 2 + c 2 a2 ) 8(a2 + b2 + c 2 ) + 2(a b + bc + ca)

210

Vasile Cîrtoaje

which is equivalent to 45(a2 + b2 + c 2 ) (a2 + b2 + c 2 )2 − 3 ≥ − 9, a2 b2 + b2 c 2 + c 2 a2 4(a2 + b2 + c 2 ) + a b + bc + ca 9(a2 + b2 + c 2 − a b − bc − ca) a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2 . ≥ a2 b2 + b2 c 2 + c 2 a2 4(a2 + b2 + c 2 ) + a b + bc + ca By Schur’s inequality of degree four, we have a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2 ≥ (a2 + b2 + c 2 − a b − bc − ca)(a b + bc + ca) ≥ 0. Therefore, it suffices to show that [4(a2 + b2 + c 2 ) + a b + bc + ca](a b + bc + ca) ≥ 9(a2 b2 + b2 c 2 + c 2 a2 ). Since (a b + bc + ca)2 ≥ a2 b2 + b2 c 2 + c 2 a2 , this inequality is true if 4(a2 + b2 + c 2 )(a b + bc + ca) ≥ 8(a2 b2 + b2 c 2 + c 2 a2 ), which is equivalent to the obvious inequality a b(a − b)2 + bc(b − c)2 + ca(c − a)2 + a bc(a + b + c) ≥ 0. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation). Second Solution. Write the inequality as f6 (a, b, c) ≥ 0, where Y  X 2 f6 (a, b, c) = 8(a2 + b2 + c 2 ) + 2(a b + bc + ca) (a + b2 )(a2 + c 2 ) − 45 (b2 + c 2 ). Clearly, f6 (a, b, c) has the same highest coefficient A as Y Y f (a, b, c) = −45 (b2 + c 2 ) = −45 (p2 − 2q − a2 ), where p = a + b + c and q = a b + bc + ca; that is, A = 45. Since A > 0, we will apply the highest coefficient cancellation method. We have f6 (a, 1, 1) = 4a(2a + 5)(a2 + 1)(a − 1)2 , f6 (0, b, c) = (b − c)2 [8(b4 + c 4 ) + 18bc(b2 + c 2 ) + 15b2 c 2 ].

Symmetric Rational Inequalities

211

Since f6 (1, 1, 1) = f6 (0, 1, 1) = 0, define the homogeneous function P(a, b, c) = a bc + B(a + b + c)3 + C(a + b + c)(a b + bc + ca) such that P(1, 1, 1) = P(0, 1, 1) = 0; that is, 1 4 P(a, b, c) = a bc + (a + b + c)3 − (a + b + c)(a b + bc + ca). 9 9 We will show that the following sharper inequality holds f6 (a, b, c) ≥ 45P 2 (a, b, c). Let us denote g6 (a, b, c) = f6 (a, b, c) − 45P 2 (a, b, c). Clearly, g6 (a, b, c) has the highest coefficient A = 0. By P 3.76-(a) in Volume 1, it suffices to prove that g6 (a, 1, 1) ≥ 0 and g6 (0, b, c) ≥ 0 for all a, b, c ≥ 0. We have P(a, 1, 1) =

a(a − 1)2 , 9

hence g6 (a, 1, 1) = f6 (a, 1, 1) − 45P 2 (a, 1, 1) = Also, we have P(0, b, c) =

a(a − 1)2 (67a3 + 190a2 + 67a + 180) ≥ 0. 9

(b + c)(b − c)2 , 9

hence g6 (0, b, c) = f6 (0, b, c) − 45P 2 (0, b, c) =

(b − c)2 [67(b4 + c 4 ) + 162bc(b2 + c 2 ) + 145b2 c 2 ] ≥ 0. 9

P 1.158. If a, b, c are real numbers, no two of which are zero, then a2 − 7bc b2 − 7ca c 2 − 7a b 9(a b + bc + ca) + 2 + 2 + ≥ 0. b2 + c 2 a + b2 a + b2 a2 + b2 + c 2 (Vasile Cîrtoaje, 2014)

212

Vasile Cîrtoaje

Solution. Let p = a + b + c,

q = a b + bc + ca,

r = a bc.

Write the inequality as f8 (a, b, c) ≥ 0, where X f8 (a, b, c) =(a2 + b2 + c 2 ) (a2 − 7bc)(a2 + b2 )(a2 + c 2 ) Y + 9(a b + bc + ca) (b2 + c 2 ) is a symmetric homogeneous polynomial of degree eight. Always, f8 (a, b, c) can be written in the form f8 (a, b, c) = A(p, q)r 2 + B(p, q)r + C(p, q), where the highest polynomial A(p, q) has the form αp2 + βq. Since X f8 (a, b, c) =(p2 − 2q) (a2 − 7bc)(p2 − 2q − c 2 )(p2 − 2q − b2 ) Y + 9q (p2 − 2q − a2 ), f8 (a, b, c) has the same highest polynomial as X g8 (a, b, c) =(p2 − 2q) (a2 − 7bc)b2 c 2 + 9q(−a2 b2 c 2 ) X € Š =(p2 − 2q) 3r 2 − 7 b3 c 3 − 9qr 2 ; that is, A(p, q) = (p2 − 2q)(3 − 21) − 9q = −9(p2 − 3q). Since A(p, q) ≤ 0 for all real a, b, c, by Lemma below, it suffices to prove that f8 (a, 1, 1) ≥ 0 for all a, b, c ≥ 0. We have f8 (a, 1, 1) = (a2 + 1)(a − 1)2 (a + 2)2 (a2 − 2a + 3) ≥ 0. The equality holds for a = b = c, and also for −a/2 = b = c (or any cyclic permutation). Lemma. Let p = a + b + c,

q = a b + bc + ca,

r = a bc,

and let f8 (a, b, c) be a symmetric homogeneous polynomial of degree eight written in the form f8 (a, b, c) = A(p, q)r 2 + B(p, q)r + C(p, q), where A(p, q) ≤ 0 for all real a, b, c. The inequality f8 (a, b, c) ≥ 0 holds for all real numbers a, b, c if and only if f8 (a, 1, 1) ≥ 0 for all real a. Proof. For fixed p and q, h8 (r) = A(p, q)r 2 + B(p, q)r + C(p, q)

Symmetric Rational Inequalities

213

is a concave quadratic function of r. Therefore, h8 (r) is minimal when r is minimal or maximal; this is, according to P 2.53 in Volume 1, when f8 (a, 1, 1) ≥ 0 and f8 (a, 0, 0) ≥ 0 for all real a. Notice that the condition " f8 (a, 0, 0) ≥ 0 for all real a" is not necessary because it follows from the condition " f8 (a, 1, 1) ≥ 0 for all real a" as follows: f8 (a, 0, 0) = lim f8 (a, t, t) = lim t 8 f8 (a/t, 1, 1) ≥ 0. t→0

t→0

Notice that A(p, q) is called the highest polynomial of f8 (a, b, c). Remark. This Lemma can be extended as follow. • The inequality f8 (a, b, c) ≥ 0 holds for all real numbers a, b, c satisfying A(p, q) ≤ 0 if and only if f8 (a, 1, 1) ≥ 0 for all real a such that A(a + 2, 2a + 1) ≤ 0.

P 1.159. If a, b, c are real numbers such that a bc 6= 0, then (b + c)2 (c + a)2 (a + b)2 10(a + b + c)2 + + ≥ 2 + . a2 b2 c2 3(a2 + b2 + c 2 ) (Vasile Cîrtoaje and Michael Rozenberg, 2014) Solution. Let p = a + b + c,

q = a b + bc + ca,

r = a bc.

Write the inequality as f8 (a, b, c) ≥ 0, where ”X — f8 (a, b, c) = 3(a2 + b2 + c 2 ) b2 c 2 (b + c)2 − 2a2 b2 c 2 − 10a2 b2 c 2 (a + b + c)2 . From X

b2 c 2 (b + c)2 − 2a2 b2 c 2 =

X

b2 c 2 (p − a)2 − 2r 2 = p2

X

b2 c 2 − 2pqr + r 2 ,

it follows that f8 (a, b, c) has the same highest polynomial as 3(a2 + b2 + c 2 )r 2 − 10r 2 (a + b + c)2 ; that is, A(p, q) = 3(p2 − 2q) − 10p2 = −7p2 − 6q. There are two cases to consider. Case 1: A(p, q) ≤ 0. According to Remark from the preceding P 1.158, it suffices to show that f8 (a, 1, 1) ≥ 0 for all real a such that A(a + 2, 2a + 1) ≤ 0. Indeed, we have f8 (a, 1, 1) = 3(a2 + 2)[4 + 2a2 (a + 1)2 − 2a2 ] − 10a2 (a + 2)2 = 2(3a6 + 6a5 + a4 − 8a3 − 14a2 + 12) = 2(a − 1)2 (3a4 + 12a3 + 22a2 + 24a + 12) = 2(a − 1)2 [3(a + 1)4 + (2a + 3)2 ] ≥ 0.

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Case 2: A(p, q) > 0. We will show that there exist two real numbers B and C such that the following sharper inequality holds: f8 (a, b, c) ≥ A(p, q)P 2 (a, b, c), where P(a, b, c) = r + Bp3 + C pq. Let us denote g8 (a, b, c) = f8 (a, b, c) − A(p, q)P 2 (a, b, c). We see that the highest polynomial of g8 (a, b, c) is zero. Thus, according to Remark from P 1.158, it suffices to prove that g(a) ≥ 0 for all real a, where g(a) = g8 (a, 1, 1). We have g(a) = f8 (a, 1, 1) − A(a + 2, 2a + 1)P 2 (a, 1, 1), where f8 (a, 1, 1) = 2(3a6 + 6a5 + a4 − 8a3 − 14a2 + 12), A(a + 2, 2a + 1) = −7(a + 2)2 − 6(2a + 1) = −7a2 − 40a − 34, P(a, 1, 1) = a + B(a + 2)3 + C(a + 2)(2a + 1). Since g(−2) = 72 − 18(−2)2 = 0, a necessary condition to have g(a) ≥ 0 in the vicinity of −2 is g 0 (−2) = 0. This condition involves C = −1. We can check that g 00 (−2) = 0 for this value of C. Thus, a necessary condition to have g(a) ≥ 0 in the vicinity of −2 is g 000 (−2) = 0 is necessary. This condition involves B = 2/3. For these values of B and C, we have • ˜2 2 2 3 g(a) = f8 (a, 1, 1) + (7a + 40a + 34) a + (a + 2) − (a + 2)(2a + 1) 3 2 4 4 3 2 = (a + 2) (14a + 52a + 117a + 154a + 113). 9 Since 14a4 + 52a3 + 117a2 + 154a + 113 = (a2 + 1)2 + 13a2 (a + 2)2 + 7(9a2 + 22a + 16) > 0, the proof is completed. The equality holds for a = b = c.

P 1.160. If a, b, c are nonnegative real numbers, no two of which are zero, then a2 − 4bc b2 − 4ca c 2 − 4a b 9(a b + bc + ca) 9 + 2 + 2 + ≥ . b2 + c 2 a + b2 a + b2 a2 + b2 + c 2 2 (Vasile Cîrtoaje, 2014)

Symmetric Rational Inequalities

215

Solution. Let p = a + b + c,

q = a b + bc + ca,

r = a bc.

Write the inequality as f8 (a, b, c) ≥ 0, where X f8 (a, b, c) =2(a2 + b2 + c 2 ) (a2 − 4bc)(a2 + b2 )(a2 + c 2 ) Y + 9(2a b + 2bc + 2ca − a2 − b2 − c 2 ) (b2 + c 2 ) is a symmetric homogeneous polynomial of degree eight. Always, f8 (a, b, c) can be written in the form f8 (a, b, c) = A(p, q)r 2 + B(p, q)r + C(p, q), where A(p, q) = αp2 + βq is called the highest polynomial of f8 (a, b, c). Since X f8 (a, b, c) =2(p2 − 2q) (a2 − 4bc)(p2 − 2q − c 2 )(p2 − 2q − b2 ) Y + 9(4q − p2 (p2 − 2q − a2 ), f8 (a, b, c) has the same highest polynomial as X g8 (a, b, c) =2(p2 − 2q) (a2 − 4bc)b2 c 2 + 9(4q − p2 )(−a2 b2 c 2 ) X € Š =2(p2 − 2q) 3r 2 − 4 b3 c 3 − 9(4q − p2 )r 2 ; that is, A(p, q) = 2(p2 − 2q)(3 − 12) − 9(4q − p2 ) = −9p2 . Since A(p, q) ≤ 0 for all a, b, c ≥ 0, by Lemma below, it suffices to prove that f8 (a, 1, 1) ≥ 0 and f8 (0, b, c) ≥ 0 for all a, b, c ≥ 0. We have f8 (a, 1, 1) = 2a(a + 4)(a2 + 1)(a − 1)4 ≥ 0 and f8 (0, b, c) = (b2 + c 2 )(2E − 9F ), where E = −4b3 c 3 + (b2 + c 2 )(b4 + c 4 ),

F = b2 c 2 (b − c)2 .

Since E ≥ −4b3 c 3 + 2bc(b4 + c 4 ) = 2bc(b2 − c 2 )2 , we have 2E − 9F ≥ 4bc(b2 − c 2 )2 − 9b2 c 2 (b − c)2 = bc(b − c)2 [4(b + c)2 − 9bc] ≥ 0. Thus, the proof is completed. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

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Lemma. Let p = a + b + c,

q = a b + bc + ca,

r = a bc,

and let f8 (a, b, c) be a symmetric homogeneous polynomial of degree eight written in the form f8 (a, b, c) = A(p, q)r 2 + B(p, q)r + C(p, q), where A(p, q) ≤ 0 for all a, b, c ≥ 0. The inequality f8 (a, b, c) ≥ 0 holds for all nonnegative real numbers a, b, c if and only if f8 (a, 1, 1) ≥ 0 and f8 (0, b, c) ≥ 0 for all a, b, c ≥ 0. Proof. For fixed p and q, h8 (r) = A(p, q)r 2 + B(p, q)r + C(p, q) is a concave quadratic function of r. Therefore, h8 (r) is minimal when r is minimal or maximal. This is, according to P 3.57 in Volume 1, when b = c or a = 0. Thus, the conclusion follows. Notice that A(p, q) is called the highest polynomial of f8 (a, b, c). Remark. This Lemma can be extended as follow. • The inequality f8 (a, b, c) ≥ 0 holds for all a, b, c ≥ 0 satisfying A(p, q) ≤ 0 if and only if f8 (a, 1, 1) ≥ 0 and f8 (0, b, c) ≥ 0 for all a, b, c ≥ 0 such that A(a + 2, 2a + 1) ≤ 0 and A(b + c, bc) ≤ 0.

P 1.161. If a, b, c are nonnegative real numbers, no two of which are zero, then 9(a − b)2 (b − c)2 (c − a)2 a2 + b2 + c 2 ≥1+ . a b + bc + ca (a + b)2 (b + c)2 (c + a)2 (Vasile Cîrtoaje, 2014) Solution. Consider the non-trivial case where a, b, c are distinct and a = min{a, b, c}. Write the inequality as follows: (a − b)2 + (b − c)2 + (c − a)2 9(a − b)2 (b − c)2 (c − a)2 ≥ , 2(a b + bc + ca) (a + b)2 (b + c)2 (c + a)2 (a − b)2 + (b − c)2 + (c − a)2 18(a b + bc + ca) ≥ , 2 2 2 (a − b) (b − c) (c − a) (a + b)2 (b + c)2 (c + a)2 X 1 18(a b + bc + ca) ≥ . 2 2 (b − a) (c − a) (a + b)2 (a + c)2 (b + c)2 Since X

1 1 1 1 2(b2 + c 2 − bc) ≥ + + = (b − a)2 (c − a)2 b2 c 2 b2 (b − c)2 c 2 (b − c)2 b2 c 2 (b − c)2

Symmetric Rational Inequalities and

217

a b + bc + ca 1 a b + bc + ca ≤ ≤ , 2 2 2 2 2 (a + b) (a + c) (b + c) (a b + bc + ca) (b + c) bc(b + c)2

it suffices to show that

b2 + c 2 − bc 9 ≥ . 2 2 2 b c (b − c) bc(b + c)2

Write this inequality as follows: (b + c)2 − 3bc 9(b + c)2 − 36bc ≥ , bc (b + c)2 (b + c)2 36bc − 12 + ≥ 0, bc (b + c)2 (b + c)4 − 12bc(b + c)2 + 36b2 c 2 ≥ 0, [(b + c)2 − 6bc]2 ≥ 0. Thus, the proof is completed. The equality holds for a = b = c, and also for a = 0 and b/c + c/b = 4 (or any cyclic permutation).

P 1.162. If a, b, c are nonnegative real numbers, no two of which are zero, then p (a − b)2 (b − c)2 (c − a)2 a2 + b2 + c 2 ≥ 1 + (1 + 2)2 2 . a b + bc + ca (a + b2 )(b2 + c 2 )(c 2 + a2 ) (Vasile Cîrtoaje, 2014) p Solution. Consider the non-trivial case where a, b, c are distinct and denote k = 1+ 2. Write the inequality as follows: (a − b)2 + (b − c)2 + (c − a)2 k2 (a − b)2 (b − c)2 (c − a)2 ≥ 2 , 2(a b + bc + ca) (a + b2 )(b2 + c 2 )(c 2 + a2 ) (a − b)2 + (b − c)2 + (c − a)2 2k2 (a b + bc + ca) ≥ , (a − b)2 (b − c)2 (c − a)2 (a2 + b2 )(b2 + c 2 )(c 2 + a2 ) X

1 2k2 (a b + bc + ca) ≥ . (b − a)2 (c − a)2 (a2 + b2 )(b2 + c 2 )(c 2 + a2 )

Assume that a = min{a, b, c}, and use the substitution b = a + x,

c = a + y,

x, y ≥ 0.

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The inequality becomes 1 x2 y2

+

x 2 (x

1 1 + 2 ≥ 2k2 f (a), 2 − y) y (x − y)2

where f (a) =

3a2 + 2(x + y)a + x y . (2a2 + 2x a + x 2 )(2a2 + 2 y a + y 2 )[2a2 + 2(x + y)a + x 2 + y 2 ]

We will show that 1 x2 y2

+

x 2 (x

1 1 + 2 ≥ 2k2 f (0) ≥ 2k2 f (a). 2 − y) y (x − y)2

We have 2(x 2 + y 2 − x y) 2k2 x y 1 1 1 2 + + − 2k f (0) = − x 2 y 2 x 2 (x − y)2 y 2 (x − y)2 x 2 y 2 (x − y)2 x 2 y 2 (x 2 + y 2 ) p 2[x 2 + y 2 − (2 + 2 )x y]2 ≥ 0. = 2 2 x y (x − y)2 (x 2 − x y + y 2 ) Also, since (2a2 + 2x a + x 2 )(2a2 + 2 y a + y 2 ) ≥ 2(x 2 + y 2 )a2 + 2x y(x + y)a + x 2 y 2 and 2a2 + 2(x + y)a + x 2 + y 2 ≥ x 2 + y 2 , we get f (a) ≤ g(a), where g(a) =

3a2 + 2(x + y)a + x y . [2(x 2 + y 2 )a2 + 2x y(x + y)a + x 2 y 2 ](x 2 + y 2 )

Therefore, 1 − g(a) x y(x 2 + y 2 ) (2x 2 + 2 y 2 − 3x y)a2 = ≥ 0. x y(x 2 + y 2 )[2(x 2 + y 2 )a2 + 2x y(x + y)a + x 2 y 2 ]

f (0) − f (a) ≥

Thus, the proof is pcompleted. The equality holds for a = b = c, and also for a = 0 and b/c + c/b = 2 + 2 (or any cyclic permutation).

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P 1.163. If a, b, c are nonnegative real numbers, no two of which are zero, then 2 2 5 5 5 2 + + ≥ + + . a+b b+c c+a 3a + b + c 3b + c + a 3c + a + b Solution. Write the inequality as follows: ‹ X 2 5 − ≥ 0, b + c 3a + b + c X 2a − b − c ≥ 0, (b + c)(3a + b + c) X X a−c a−b + ≥ 0, (b + c)(3a + b + c) (b + c)(3a + b + c) X X a−b b−a + ≥ 0, (b + c)(3a + b + c) (c + a)(3b + c + a) X (a − b)2 (a + b − c) , (b + c)(c + a)(3a + b + c)(3b + c + a) X (b − c)2 Sa ≥ 0, where Sa = (b + c − a)(b + c)(3a + b + c). Assume that a ≥ b ≥ c. Since Sc > 0, it suffices to show that (b − c)2 Sa + (a − c)2 S b ≥ 0. Since S b ≥ 0 and (a − c)2 ≥ (b − c)2 , we have (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 Sa + (b − c)2 S b = (b − c)2 (Sa + S b ). Thus, it is enough to prove that Sa + S b ≥ 0, which is equivalent to (c + a − b)(c + a)(3b + c + a) ≥ (b + c − a)(b + c)(3a + b + c). Consider the nontrivial case where b + c − a > 0. Since c + a − b ≥ b + c − a, we only need to show that (c + a)(3b + c + a) ≥ (b + c)(3a + b + c). Indeed, (c + a)(3b + c + a) − (b + c)(3a + b + c) = (a − b)(a + b − c) ≥ 0. Thus, the proof is completed. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

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P 1.164. If a, b, c are real numbers, no two of which are zero, then (a)

8b2 + 3ca 8c 2 + 3a b 8a2 + 3bc + + ≥ 11; b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2

(b)

8a2 − 5bc 8b2 − 5ca 8c 2 − 5a b + + ≥ 9. b2 − bc + c 2 c 2 − ca + a2 a2 − a b + b2 (Vasile Cîrtoaje, 2011)

Solution. Consider the more general inequality a2 + mbc b2 + mca c 2 + ma b 3(m + 1) + + ≥ . 2 2 2 2 2 2 b + k bc + c c + kca + a a + ka b + b k+2 Let p = a + b + c and q = a b + bc + ca. Write the inequality in the form f6 (a, b, c) ≥ 0, where X f6 (a, b, c) = (k + 2) (a2 + mbc)(a2 + ka b + b2 )(a2 + kac + c 2 ) Y −3(m + 1) (b2 + k bc + c 2 ). From f6 (a, b, c) = (k + 2)

X (a2 + mbc)(ka b − c 2 + p2 − 2q)(kac − b2 + p2 − 2q) Y −3(m + 1) (k bc − a2 + p2 − 2q).

it follows that f6 (a, b, c) has the same highest coefficient A as (k + 2)P2 (a, b, c) − 3(m + 1)P3 (a, b, c), where

X (a2 + mbc)(ka b − c 2 )(kac − b2 ), Y P3 (a, b, c) = (k bc − a2 ).

P2 (a, b, c) =

According to Remark 2 from the proof of P 2.75 in Volume 1, A = (k + 2)P2 (1, 1, 1) − 3(m + 1)P3 (1, 1, 1) = 3(k + 2)(m + 1)(k − 1)2 − 3(m + 1)(k − 1)3 = 9(m + 1)(k − 1)2 . Also, we have f6 (a, 1, 1) = (k + 2)(a2 + ka + 1)(a − 1)2 [a2 + (k + 2)a + 1 + 2k − 2m]. (a) For our particular case m = 3/8 and k = 1, we have A = 0. Therefore, according to P 2.75 in Volume 1, it suffices to prove that f6 (a, 1, 1) ≥ 0 for all real a. Indeed,  ‹ 3 2 f6 (a, 1, 1) = 3(a2 + a + 1)(a − 1)2 a + ≥ 0. 2

Symmetric Rational Inequalities

221

Thus, the proof is completed. The equality holds for a = b = c, and also for −2a/3 = b = c (or any cyclic permutation). (b) For m = −5/8 and k = −1, we have A = 27/2 and f6 (a, 1, 1) =

1 2 (a − a + 1)(a − 1)2 (2a + 1)2 . 4

Since A > 0, we will use the highest coefficient cancellation method. Define the homogeneous polynomial P(a, b, c) = r + Bp3 + C pq, where B and C are real constants. Since the desired inequality becomes an equality for a = b = c = 1, and also for a = −1 and b = c = 2, determine B and C such that P(1, 1, 1) = P(−1, 2, 2) = 0. We find B= when P(a, 1, 1) =

4 , 27

C=

2 (a − 1)2 (2a + 1), 27

We will show that f6 (a, b, c) ≥

−5 , 9

P(a, 0, 0) = Ba3 =

4 3 a . 27

27 2 P (a, b, c). 2

Let us denote g6 (a, b, c) = f6 (a, b, c) −

27 2 P (a, b, c). 2

Since g6 (a, b, c) has the highest coefficient A = 0, it suffices to prove that g6 (a, 1, 1) ≥ 0 for all real a (see P 2.75 in Volume 1). Indeed, g6 (a, 1, 1) = f6 (a, 1, 1) −

27 2 1 P (a, 1, 1) = (a − 1)2 (2a + 1)2 (19a2 − 11a + 19) ≥ 0. 2 108

Thus, the proof is completed. The equality holds for a = b = c, and also for −2a = b = c (or any cyclic permutation).

P 1.165. If a, b, c are real numbers, no two of which are zero, then 4a2 + bc 4b2 + ca 4c 2 + a b + + ≥ 1. 4b2 + 7bc + 4c 2 4c 2 + 7ca + 4a2 4a2 + 7a b + 4b2 (Vasile Cîrtoaje, 2011)

222

Vasile Cîrtoaje

Solution. Write the inequality as f6 (a, b, c) ≥ 0, where X Y f6 (a, b, c) = (4a2 + bc)(4a2 + 7a b + 4b2 )(4a2 + 7ac + 4c 2 ) − (4b2 + 7bc + 4c 2 ). Let p = a + b + c, q = a b + bc + ca,

r = a bc.

From f6 (a, b, c) =

X (4a2 + bc)(7a b − 4c 2 + 4p2 − 8q)(7ac − 4b2 + 4p2 − 8q) −

Y (7bc − 4a2 + 4p2 − 8q),

it follows that f6 (a, b, c) has the same highest coefficient A as P2 (a, b, c) − P3 (a, b, c), where

X (4a2 + bc)(7a b − 4c 2 )(7ac − 4b2 ), Y P3 (a, b, c) = (7bc − 4a2 ).

P2 (a, b, c) =

According to Remark 2 from the proof of P 2.75 in Volume 1, A = P2 (1, 1, 1) − P3 (1, 1, 1) = 135 − 27 = 108. Since A > 0, we will apply the highest coefficient cancellation method. Define the homogeneous polynomial P(a, b, c) = r + Bp3 + C pq, where B and C are real constants. We will show that there are two real numbers B and C such that the following sharper inequality holds f6 (a, b, c) ≥ 108P 2 (a, b, c). Let us denote g6 (a, b, c) = f6 (a, b, c) − 108P 2 (a, b, c). Clearly, g6 (a, b, c) has the highest coefficient A1 = 0. Then, by P 2.75 in Volume 1, it suffices to prove that g6 (a, 1, 1) ≥ 0 for all real a. We have g6 (a, 1, 1) = f6 (a, 1, 1) − 108P 2 (a, 1, 1), where f6 (a, 1, 1) = 4(4a2 + 7a + 4)(a − 1)2 (4a2 + 15a + 16), P(a, 1, 1) = a + B(a + 2)3 + C(a + 2)(2a + 1).

Symmetric Rational Inequalities

223

Let us denote g(a) = f6 (a, 1, 1). Since g(−2) = 0, we can have g(a) ≥ 0 in the vicinity of a = −2 only if g 0 (−2) = 0, which involves C = −5/9. On the other hand, from g(1) = 0, we get B = 4/27. For these values of B and C, we get P(a, 1, 1) = g6 (a, 1, 1) =

2(a − 1)2 (2a + 1) , 27

4 (a − 1)2 (a + 2)2 (416a2 + 728a + 431) ≥ 0. 27

The proof is completed. The equality holds for a = b = c, and for a = 0 and b + c = 0 (or any cyclic permutation).

P 1.166. If a, b, c are real numbers, no two of which are equal, then 1 1 1 27 + + ≥ . 2 2 2 2 2 2 (a − b) (b − c) (c − a) 4(a + b + c − a b − bc − ca) First Solution. Write the inequality as follows 

(a − b)2 + (b − c)2 + (a − c)2 

•

(a − b)2 (b − c)2 + +1 (a − c)2 (a − c)2

˜ 1 1 27 1 + + ≥ , (a − b)2 (b − c)2 (a − c)2 2

 (a − c)2 (a − c)2 27 + +1 ≥ , (a − b)2 (b − c)2 2  ‹ 1 1 27 (x 2 + y 2 + 1) + + 1 ≥ , x2 y2 2

where x=

a−b , a−c



y=

b−c , a−c

x + y = 1.

We have ‹ 1 1 27 (x + 1)2 (x − 2)2 (2x − 1)2 (x + y + 1) + + 1 − = ≥ 0. x2 y2 2 2x 2 (1 − x)2 2

2



The proof is completed. The equality holds for 2a = b + c (or any cyclic permutation). Second Solution. Assume that a > b > c. We have 1 1 2 8 8 + ≥ ≥ = . 2 2 2 (a − b) (b − c) (a − b)(b − c) [(a − b) + (b − c)] (a − c)2

224

Vasile Cîrtoaje

Therefore, it suffices to show that 27 9 , ≥ 2 2 2 2 (a − c) 4(a + b + c − a b − bc − ca) which is equivalent to (a − 2b + c)2 ≥ 0. Third Solution. Write the inequality as f6 (a, b, c) ≥ 0, where X f6 (a, b, c) = 4(a2 + b2 + c 2 − a b − bc − ca) (a − b)2 (a − c)2 −27(a − b)2 (b − c)2 (c − a)2 . Clearly, f6 (a, b, c) has the same highest coefficient A as −27(a − b)2 (b − c)2 (c − a)2 ; that is, A = −27(−27) = 729. Since A > 0, we will use the highest coefficient cancellation method. Define the homogeneous polynomial  ‹ 1 P(a, b, c) = a bc + B(a + b + c)3 − 3B + (a + b + c)(a b + bc + ca), 9 which satisfies the property P(1, 1, 1) = 0. We will show that there is a real value of B such that the following sharper inequality holds f6 (a, b, c) ≥ 729P 2 (a, b, c). Let us denote g6 (a, b, c) = f6 (a, b, c) − 729P 2 (a, b, c). Clearly, g6 (a, b, c) has the highest coefficient A1 = 0. Then, by P 2.75 in Volume 1, it suffices to prove that g6 (a, 1, 1) ≥ 0 for all real a. We have f6 (a, 1, 1) = 4(a − 1)6 and P(a, 1, 1) =

1 (a − 1)2 [9B(a + 2) + 2], 9

hence g6 (a, 1, 1) = f6 (a, 1, 1)−729P 2 (a, 1, 1) = (27B +2)(a −1)4 (a +2)[(2−27B)a −54B −8]. Choosing B = −2/27, we get g6 (a, 1, 1) = 0 for all real a. Remark. The inequality is equivalent to (a − 2b + c)2 (b − 2c + a)2 (c − 2a + b)2 ≥ 0.

Symmetric Rational Inequalities

225

P 1.167. If a, b, c are real numbers, no two of which are zero, then a2

1 1 1 14 + 2 + 2 ≥ . 2 2 2 2 − ab + b b − bc + c c − ca + a 3(a + b2 + c 2 ) (Vasile Cîrtoaje and BJSL, 2014)

Solution. Write the inequality as f6 (a, b, c) ≥ 0, where X f6 (a, b, c) = 3(a2 + b2 + c 2 ) (a2 − a b + b2 )(a2 − ac + c 2 ) −14(a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 ). Clearly, f6 (a, b, c) has the same highest coefficient A as −14(a2 − a b + b2 )(b2 − bc + c 2 )(c 2 − ca + a2 ), hence as f (a, b, c) = −14(−c 2 − a b)(−a2 − bc)(−b2 − ca); that is, according to Remark 2 from the proof of P 2.75 in Volume 1, A = f (1, 1, 1) = −14(−2)3 = 112. Since A > 0, we apply the highest coefficient cancellation method. Define the homogeneous polynomial P(a, b, c) = a bc + B(a + b + c)3 + C(a + b + c)(a b + bc + ca). We will show that there are two real numbers B and C such that the following sharper inequality holds f6 (a, b, c) ≥ 112P 2 (a, b, c). Let us denote g6 (a, b, c) = f6 (a, b, c) − 112P 2 (a, b, c). Clearly, g6 (a, b, c) has the highest coefficient A1 = 0. By P 2.75 in Volume 1, it suffices to prove that g6 (a, 1, 1) ≥ 0 for all real a. We have g6 (a, 1, 1) = f6 (a, 1, 1) − 112P 2 (a, 1, 1), where f6 (a, 1, 1) = (a2 − a + 1)(3a4 − 3a3 + a2 + 8a + 4), P(a, 1, 1) = 1 + B(a + 2)3 + C(a + 2)(2a + 1). Let us denote g(a) = g6 (a, 1, 1). Since g(−2) = 0, we can have g(a) ≥ 0 in the vicinity of a = −2 only if g 0 (−2) = 0, which involves C = −4/7. In addition, setting B = 9/56, we get 1 (9a3 − 10a2 + 4a + 8), P(a, 1, 1) = 56

226

Vasile Cîrtoaje 3 6 (a + 4a5 + 8a4 + 16a3 + 20a2 + 16a + 16) 28 3(a + 2)2 (a2 + 2)2 = ≥ 0. 28

g6 (a, 1, 1) =

The proof is completed. The equality holds for a = 0 and b + c = 0 (or any cyclic permutation).

P 1.168. Let a, b, c be real numbers such that a b + bc + ca ≥ 0 and no two of which are zero. Prove that (a) (b) i f a b ≤ 0, then

a b c 3 + + ≥ ; b+c c+a a+b 2 a b c + + ≥ 2. b+c c+a a+b (Vasile Cîrtoaje, 2014)

Solution. Let as show first that b + c 6= 0, c + a 6= 0 and a + b 6= 0. Indeed, if b + c = 0, then a b + bc + ca ≥ 0 yields b = c = 0, which is not possible. (a) Write the inequality as follows  9 X a +1 ≥ , b+c 2  ”X — X 1 ‹ (b + c) ≥ 9, b+c ‹ Xa + b a + c + − 2 ≥ 0, a+c a+b X (b − c)2 ≥ 0, (a + b)(a + c) X (b − c)2 ≥ 0. a2 + (a b + bc + ca) Clearly, the last inequality is true. The equality holds for a = b = c 6= 0. (b) From a b + bc + ca ≥ 0, it follows that if one of a, b, c is zero, then the others are the same sign. In this case, the desired inequality is trivial. So, due to symmetry and homogeneity, it suffices to consider that a < 0 < b ≤ c.

Symmetric Rational Inequalities

227

First Solution. We will show that F (a, b, c) > F (0, b, c) ≥ 2, where F (a, b, c) =

b c a + + . b+c c+a a+b

We have F (0, b, c) =

b c + ≥2 c b

and F (a, b, c) − F (0, b, c) = a

•

˜ 1 b c − − . b + c c(c + a) b(a + b)

Since a < 0, we need to show that c 1 b + > . c(c + a) b(a + b) b+c From a b + bc + ca ≥ 0, we get c+a≥

−ca > 0, b

a+b≥

−a b > 0, c

hence b b > 2, c(c + a) c

c c > 2. b(a + b) b

Therefore, it suffices to prove that b c 1 + 2≥ . 2 c b b+c Indeed, by virtue of the AM-GM inequality, we have b c 1 2 1 + 2− ≥ p − p > 0. 2 c b b+c bc 2 bc This completes the proof. The equality holds for a = 0 and b = c, or b = 0 and a = c. Second Solution. Since b + c > 0 and (b + c)(a + b) = b2 + (a b + bc + ca) > 0, (b + c)(c + a) = c 2 + (a b + bc + ca) > 0, we get a + b > 0 and c + a > 0. By virtue of the Cauchy-Schwarz inequality and AM-GM inequality, we have

228

Vasile Cîrtoaje

a b c a (b + c)2 + + ≥ + b+c c+a a+b b + c b(c + a) + c(a + b) a (b + c)2 > + 2a + b + c (b + c)2 + a(b + c) 2 4a 2(b + c) > + = 2. 2a + b + c 2a + b + c

P 1.169. If a, b, c are nonnegative real numbers, then a b c a b + bc + ca + + ≥ . 7a + b + c 7b + c + a 7c + a + b (a + b + c)2 (Vasile Cîrtoaje, 2014) First Solution. Write the inequality as follows: ˜ X• 2a a(b + c) − ≥ 0, 7a + b + c (a + b + c)2 X a[(a − b) + (a − c)](a − b − c)

≥ 0, 7a + b + c X a(a − b)(a − b − c) X a(a − c)(a − b − c) + ≥ 0, 7a + b + c 7a + b + c X a(a − b)(a − b − c) X b(b − a)(b − c − a) + ≥ 0, 7a + b + c 7b + c + a ˜ • X a(a − b − c) b(b − c − a) − ≥ 0, (a − b) 7a + b + c 7b + c + a X (a − b)2 (a2 + b2 − c 2 + 14a b)(a + b + 7c) ≥ 0. Since a2 + b2 − c 2 + 14a b ≥ (a + b)2 − c 2 = (a + b + c)(a + b − c), it suffices to show that X (a − b)2 (a + b − c)(a + b + 7c) ≥ 0. Assume that a ≥ b ≥ c. It suffices to show that (a − c)2 (a − b + c)(a + 7b + c) + (b − c)2 (−a + b + c)(7a + b + c) ≥ 0.

Symmetric Rational Inequalities

229

For the nontrivial case b > 0, we have (a − c)2 ≥

a2 a (b − c)2 ≥ (b − c)2 . 2 b b

Thus, it is enough to prove that a(a − b + c)(a + 7b + c) + b(−a + b + c)(7a + b + c) ≥ 0. Since a(a + 7b + c) ≥ b(7a + b + c), we have a(a − b + c)(a + 7b + c) + b(−a + b + c)(7a + b + c) ≥ ≥ b(a − b + c)(7a + b + c) + b(−a + b + c)(7a + b + c) = 2bc(7a + b + c) ≥ 0. This completes the proof. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation). Second Solution. Assume that a ≤ b ≤ c, a + b + c = 3 and use the substitution x=

2a + 1 , 3

y=

2b + 1 2c + 1 , z= , 3 3

where 1/3 ≤ x ≤ y ≤ z, x + y + z = 3. We have b + c ≥ 2, y + z ≥ 2, x ≤ 1. The inequality becomes a b c 9 − a2 − b2 − c 2 + + ≥ , 2a + 1 2b + 1 2c + 1 6 9(x 2 + y 2 + z 2 ) ≥ 4



‹ 1 1 1 + + + 17. x y z

Assume that x ≤ y ≤ z and show that E(x, y, z) ≥ E(x, t, t) ≥ 0, where t = ( y + z)/2 = (3 − x)/2 and ‹ 1 1 1 E(x, y, z) = 9(x + y + z ) − 4 + + − 17. x y z 2

2

2



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Vasile Cîrtoaje

We have 1 1 2 + − E(x, y, z) − E(x, t, t) = 9( y + z − 2t ) − 4 y z t 2 ( y − z) [9 yz( y + z) − 8] = ≥ 0, 2 yz( y + z) 2

2

2



‹

since 9 yz = (2b + 1)(2c + 1) ≥ 2(b + c) + 1 ≥ 5,

y + z ≥ 2.

Also, E(x, t, t) = 9x 2 + 2t 2 − 15 −

4 8 (x − 1)2 (3x − 1)(8 − 3x) − = ≥ 0. x t 2x(3 − x)

Third Solution. Write the inequality as f5 (a, b, c) ≥ 0, where f5 (a, b, c) is a symmetric homogeneous inequality of degree five. According to P 3.68-(a) in Volume 1, it suffices to prove the inequality for a = 0 and for b = c = 1. For a = 0, the inequality is equivalent to (b − c)2 (b2 + c 2 + 11bc) ≥ 0, while, for b = c = 1, the inequality is equivalent to a(a − 1)2 (a + 14) ≥ 0.

P 1.170. If a, b, c are the lengths of the sides of a triangle, then b2 c2 1 a2 + + ≥ . 2 2 2 4a + 5bc 4b + 5ca 4c + 5a b 3 (Vasile Cîrtoaje, 2009) Solution. Write the inequality as f6 (a, b, c) ≥ 0, where X Y f6 (a, b, c) = 3 a2 (4b2 + 5ca)(4c 2 + 5a b) − (4a2 + 5bc) X X = −45a2 b2 c 2 − 25a bc a3 + 40 a3 b3 . Since f6 (a, b, c has the highest coefficient A = −45 − 75 + 120 = 0,

Symmetric Rational Inequalities

231

according to 2.67-(b), it suffices to prove the original inequality b = c = 1 and 0 ≤ a ≤ 2, and for a = b + c. Case 1: b = c = 1, 0 ≤ a ≤ 2. The original inequality reduces to (2 − a)(a − 1)2 ≥ 0, which is true. Case 2: a = b + c. Using the Cauchy-Schwarz inequality b2 c2 (b + c)2 + ≥ , 4b2 + 5ca 4c 2 + 5a b 4(b2 + c 2 ) + 5a(b + c) it suffices to show that a2 (b + c)2 1 + ≥ , 4a2 + 5bc 4(b2 + c 2 ) + 5a(b + c) 3 which reduces to the obvious inequality (b − c)2 (3b2 + 3c 2 − 4bc) ≥ 0. The equality holds for an equilateral triangle, and for a degenerate triangle with a/2 = b = c (or any cyclic permutation).

P 1.171. If a, b, c are the lengths of the sides of a triangle, then 7a2

1 1 1 3 + 2 + 2 ≥ . 2 2 2 2 2 2 +b +c 7b + c + a 7c + a + b (a + b + c)2 (Vo Quoc Ba Can, 2010)

Solution. Let p = a + b + c and q = a b + bc + ca. Write the inequality as f6 (a, b, c) ≥ 0, where X Y f6 (a, b, c) = p2 (7b2 + c 2 + a2 )(7c 2 + a2 + b2 ) − 3 (7a2 + b2 + c 2 ) X Y = p2 (6b2 + p2 − 2q)(6c 2 + p2 − 2q) − 3 (6a2 + p2 − 2q). Since f6 (a, b, c has the highest coefficient A = −3(63 ) < 0, according to 2.67-(b), it suffices to prove the original inequality b = c = 1 and 0 ≤ a ≤ 2, and for a = b + c.

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Vasile Cîrtoaje

Case 1: b = c = 1, 0 ≤ a ≤ 2. The original inequality reduces to a(8 − a)(a − 1)2 ≥ 0, which is true. Case 2: a = b + c. Write the inequality as 4(b2

1 1 3 1 + 2 + 2 ≥ , 2 2 2 + c ) + 7bc 4b + c + bc 4c + b + bc 2(b + c)2 5x + 2 3 1 + ≥ , 2 4x + 7 4x + 5x + 10 2(x + 2)

where x = b/c + c/b, x ≥ 2. This inequality is equivalent to the obvious inequality 16x 2 + 5x − 38 ≥ 0. The equality holds for an equilateral triangle, and for a degenerate triangle with a = 0 and b = c (or any cyclic permutation).

P 1.172. Let a, b, c be the lengths of the sides of a triangle. If k > −2, then X a(b + c) + (k + 1)bc b2

+ k bc

+ c2

≤

3(k + 3) . k+2 (Vasile Cîrtoaje, 2009)

Solution. Let p = a + b + c and q = a b + bc + ca. Write the inequality as f6 (a, b, c) ≥ 0, where Y f6 (a, b, c) = 3(k + 3) (b2 + k bc + c 2 ) X −(k + 2) [a(b + c) + (k + 1)bc](c 2 + kca + a2 )(a2 + ka b + b2 ) Y = 3(k + 3) (p2 − 2q + k bc − a2 ) X −(k + 2) (q + k bc)(p2 − 2q + kca − b2 )(p2 − 2q + ka b − c 2 ). Since f6 (a, b, c has the same highest coefficient A as f (a, b, c), where Y X f (a, b, c) = 3(k + 3) (k bc − a2 ) − k(k + 2) bc(kca − b2 )(ka b − c 2 ) X X = 3(k + 3)[(k3 − 1)a2 b2 c 2 − k2 a bc a3 + k a3 b3 ] X X −k(k + 2)(3k2 a2 b2 c 2 − 2ka bc a3 + a3 b3 ),

Symmetric Rational Inequalities

233

we get A = 3(k + 3)(k3 − 1 − 3k2 + 3k) − k(k + 2)(3k2 − 6k + 3) = −9(k − 1)2 ≤ 0. According to 2.67-(b), it suffices to prove the original inequality b = c = 1 and 0 ≤ a ≤ 2, and for a = b + c. Case 1: b = c = 1, 0 ≤ a ≤ 2. The original inequality reduces to (2 − a)(a − 1)2 ≥ 0, which is true. Case 2: a = b + c. Write the inequality as follows ˜ X • a(b + c) + (k + 1)bc 3 , −1 ≤ 2 2 b + k bc + c k+2 X a b + bc + ca − b2 − c 2 b2

+ k bc

+ c2

≤

3 , k+2

3bc bc − c 2 bc − b2 3 + + ≤ . 2 2 2 2 2 2 b + k bc + c b + (k + 2)(bc + c ) c + (k + 2)(bc + b ) k+2 Since

3bc 3 ≤ , b2 + k bc + c 2 k+2

it suffices to prove that bc − b2 bc − c 2 + ≤ 0. b2 + (k + 2)(bc + c 2 ) c 2 + (k + 2)(bc + b2 ) This reduces to the obvious inequality (b − c)2 (b2 + bc + c 2 ) ≥ 0. The equality holds for an equilateral triangle, and for a degenerate triangle with a/2 = b = c (or any cyclic permutation).

P 1.173. Let a, b, c be the lengths of the sides of a triangle. If k > −2, then X 2a2 + (4k + 9)bc b2

+ k bc

+ c2

≤

3(4k + 11) . k+2 (Vasile Cîrtoaje, 2009)

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Vasile Cîrtoaje

Solution. Let p = a + b + c and q = a b + bc + ca. Write the inequality as f6 (a, b, c) ≥ 0, where Y f6 (a, b, c) = 3(4k + 11) (b2 + k bc + c 2 ) X −(k + 2) [2a2 + (4k + 9)bc](c 2 + kca + a2 )(a2 + ka b + b2 ) Y = 3(4k + 11) (p2 − 2q + k bc − a2 ) X −(k + 2) [2a2 + (4k + 9)bc](p2 − 2q + kca − b2 )(p2 − 2q + ka b − c 2 ). Since f6 (a, b, c has the same highest coefficient A as f (a, b, c), where Y f (a, b, c) = 3(4k + 11) (k bc − a2 ) X [2a2 + (4k + 9)bc](kca − b2 )(ka b − c 2 ) X X = 3(4k + 11)[(k3 − 1)a2 b2 c 2 − k2 a bc a3 + k a3 b3 ] X X −(k + 2)[3(4k3 + 9k2 + 2)a2 b2 c 2 − 6k(k + 3)a bc a3 + 9 a3 b3 ], −(k + 2)

we get A = 3(4k + 11)(k3 − 1 − 3k2 + 3k) − (k + 2)[3(4k3 + 9k2 + 2) − 18k(k + 3) + 27] = −9(4k + 11)(k − 1)2 ≤ 0. According to 2.67-(b), it suffices to prove the original inequality b = c = 1 and 0 ≤ a ≤ 2, and for a = b + c. Case 1: b = c = 1, 0 ≤ a ≤ 2. The original inequality reduces to (2 − a)(a − 1)2 ≥ 0, which is true. Case 2: a = b + c. Write the inequality as follows  X  2a2 + (4k + 9)bc 3(2k + 7) −2 ≤ , 2 2 b + k bc + c k+2 X 2(a2 − b2 − c 2 ) + (2k + 9)bc + c2

≤

3(2k + 7) , k+2

+ k bc • ˜ c b (2k + 13)bc + (2k + 5)(b + c) + b2 + k bc + c 2 b2 + (k + 2)(bc + c 2 ) c 2 + (k + 2)(bc + b2 ) b2

≤

3(2k + 7) . k+2

Symmetric Rational Inequalities

235

Using the substitution x = b/c + c/b, x ≥ 2, the inequality can be written as 2k + 13 (2k + 5)(x + 2)(x + 2k + 3) 3(2k + 7) + ≤ , 2 2 x +k (k + 2)x + (k + 2)(k + 3)x + 2k + 6k + 5 k+2 (x − 2)[4(k + 2)Ax 2 + 2(k + 2)B x + C] ≥ 0, where A = k + 4,

B = 2k2 + 13k + 22,

C = 8k3 + 51k2 + 98k + 65.

The inequality is true since A > 0, B = 2(k + 2)2 + 5(k + 2) + 4 > 0, C = 8(k + 2)3 + 2k2 + (k + 1)2 > 0. The equality holds for an equilateral triangle, and for a degenerate triangle with a/2 = b = c (or any cyclic permutation).

P 1.174. If a ≥ b ≥ c ≥ d such that a bcd = 1, then 1 1 3 1 + + ≥ . p 3 1+a 1+ b 1+c 1 + a bc (Vasile Cîrtoaje, 2008) Solution. We can get this inequality by summing the inequalities below 1 1 2 + ≥ p , 1+a 1+ b 1 + ab 2 1 3 + . ≥ p p 3 1 + c 1 + ab 1 + a bc The first inequality is true, since  ‹  ‹ 1 1 2 1 1 1 1 + − − − = + p p p 1 + a 1 + b 1 + ab 1 + a 1 + ab 1 + b 1 + ab p p p ( a − b)2 ( a b − 1) = p (1 + a)(1 + b)(1 + a b) p p and a b ≥ a bcd = 1. To prove the second inequality, we denote x = a b andpy = p 3 a bc (x ≥ y ≥ 1), which yield c = y 3 /x 2 . From a bc 2 ≥ a bcd = 1, we get a bc ≥ a b, that is, y 3 ≥ x. Since 2 3 3 x2 1 2 + = + − p − p 3 2 3 1 + c 1 + a b 1 + a bc x +y 1+ x 1+ y

236

Vasile Cîrtoaje

=



1 x2 − 2 3 x +y 1+ y =



+2



1 1 − 1+ x 1+ y

‹

(x − y)2 [( y − 2)x + 2 y 2 − y] , (1 + x)(1 + y)(x 2 + y 3 )

we still have to show that ( y − 2)x + 2 y 2 − y ≥ 0. This is clearly true for y ≥ 2, while for 1 ≤ y < 2, we have ( y − 2)x + 2 y 2 − y ≥ ( y − 2) y 3 + 2 y 2 − y = y( y − 1)( y 2 − y + 1) ≥ 0. The equality holds for a = b = c.

P 1.175. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that X

1 ≤ 1. 1 + a b + bc + ca

Solution. From p p p p 1 1 1 1 1 1 + + ≥p +p +p = d( a + b + c), a b c ca bc ab we get

p a b + bc + ca ≥

Therefore, X

a+

p p

b+

d

p

c

.

p X 1 d ≤ p p = 1, p p 1 + a b + bc + ca a+ b+ c+ d

which is just the required inequality. The equality occurs for a = b = c = d = 1.

P 1.176. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that 1 1 1 1 + + + ≥ 1. 2 2 2 (1 + a) (1 + b) (1 + c) (1 + d)2 (Vasile Cîrtoaje, 1995)

Symmetric Rational Inequalities

237

First Solution. The inequality follows by summing the following inequalities (see P 1.1): 1 1 1 + ≥ , (1 + a)2 (1 + b)2 1 + ab ab 1 1 1 = . + ≥ 2 2 (1 + c) (1 + d) 1 + cd 1 + ab The equality occurs for a = b = c = d = 1. Second Solution. Using the substitutions a = 1/x 4 , b = 1/ y 4 , c = 1/z 4 , d = 1/t 4 , where x, y, z, t are positive real numbers such that x yz t = 1, the inequality becomes as follows y6 x6 z6 t6 + + +  ‹2  ‹2  ‹2  ‹2 ≥ 1. 1 1 1 1 x3 + y3 + z3 + t3 + x y z t By the Cauchy-Schwarz inequality, we get P P X ( x 3 )2 ( x 3 )2 x6 P P .  ‹2 ≥  ‹2 = P 6 P x + 2 x 2 + x 2 y 2z2 1 1 x3 + x3 + x x Thus, it suffices to show that 2(x 3 y 3 + x 3 z 3 + x 3 t 3 + y 3 z 3 + y 3 t 3 + z 3 t 3 ) ≥ 2x yz t

X

x2 +

X

x 2 y 2z2.

This is true if 2(x 3 y 3 + x 3 z 3 + x 3 t 3 + y 3 z 3 + y 3 t 3 + z 3 t 3 ) ≥ 3x yz t and 2(x 3 y 3 + x 3 z 3 + x 3 t 3 + y 3 z 3 + y 3 t 3 + z 3 t 3 ) ≥ 3

X

X

x2

x 2 y 2z2,

Write these inequalities as X and

x 3 ( y 3 + z 3 + t 3 − 3 yz t) ≥ 0

X (x 3 y 3 + y 3 z 3 + z 3 x 3 − 3x 2 y 2 z 2 ) ≥ 0,

respectively. By the AM-GM inequality, we have y 3 + z 3 + t 3 ≥ 3 yz t and x 3 y 3 + y 3 z 3 + z 3 x 3 ≥ 3x 2 y 2 z 2 . Thus the conclusion follows. Third Solution. Using the substitutions a = yz/x 2 , b = z t/ y 2 , c = t x/z 2 , d = x y/t 2 , where x, y, z, t are positive real numbers, the inequality becomes y4 x4 z4 t4 + + + ≥ 1. (x 2 + yz)2 ( y 2 + z t)2 (z 2 + t x)2 (t 2 + x y)2

238

Vasile Cîrtoaje

Using the Cauchy-Schwarz inequality two times, we deduce z4 x4 z4 x4 + + ≥ (x 2 + yz)2 (z 2 + t x)2 (x 2 + y 2 )(x 2 + z 2 ) (z 2 + t 2 )(z 2 + x 2 )   x 2 + z2 1 x4 z4 ≥ = 2 + , x + z2 x 2 + y 2 z2 + t 2 x 2 + y 2 + z2 + t 2 and hence

x4 z4 x 2 + z2 + ≥ . (x 2 + yz)2 (z 2 + t x)2 x 2 + y 2 + z2 + t 2

Adding this to the similar inequality y2 + t2 y4 t4 + ≥ , ( y 2 + z t)2 (t 2 + x y)2 x 2 + y 2 + z2 + t 2 we get the required inequality. Fourth Solution. Using the substitutions a = x/ y, b = y/z, c = z/t, d = t/x, where x, y, z, t are positive real numbers, the inequality can be written as y2 z2 t2 x2 + + + ≥ 1. (x + y)2 ( y + z)2 (z + t)2 (t + x)2 By the Cauchy-Schwarz inequality and the AM-GM inequality, we get P X [ y( y + z)]2 y2 ≥P (x + y)2 (x + y)2 ( y + z)2 =

[(x + y)2 + ( y + z)2 + (z + t)2 + (t + x)2 ]2 ≥ 1. 4[(x + y)2 + (z + t)2 ][( y + z)2 + (t + x)2 ]

Remark. The following generalization holds true (Vasile Cîrtoaje, 2005): • Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If k ≥ then 1 1 1 n + + ··· + ≥ . 2 2 2 (1 + ka1 ) (1 + ka2 ) (1 + kan ) (1 + k)2

P 1.177. Let a, b, c, d 6=

p

n − 1,

1 be positive real numbers such that a bcd = 1. Prove that 3

1 1 1 1 + + + ≥ 1. 2 2 2 (3a − 1) (3b − 1) (3c − 1) (3d − 1)2 (Vasile Cîrtoaje, 2006)

Symmetric Rational Inequalities

239

First Solution. It suffices to show that 1 a−3 ≥ . (3a − 1)2 a−3 + b−3 + c −3 + d −3 This inequality is equivalent to 6a−2 + b−3 + c −3 + d −3 ≥ 9a−1 , which follows from the AM-GM inequality, as follows 6a−2 + b−3 + c −3 + d −3 ≥ 9

p 9

a−12 b−3 c −3 d −3 = 9a−1 .

The equality occurs for a = b = c = d = 1. Second Solution. Let a ≤ b ≤ c ≤ d. If a < 1/3, then 1 > 1, (3a − 1)2 and the desired inequality is clearly true. Otherwise, if 1/3 < a ≤ b ≤ c ≤ d, we have 4a3 − (3a − 1)2 = (a − 1)2 (4a − 1) ≥ 0. Therefore, using this result and the AM-GM inequality, we get X

v t 1 1 1X 1 4 ≥ ≥ = 1. 2 3 3 3 (3a − 1) 4 a a b c3 d 3

Third Solution. We have 1 a(a − 1)2 (a + 2)(a2 + 3) 1 − = ≥ 0. (3a − 1)2 (a3 + 1)2 (3a − 1)2 (a3 + 1)2 Therefore, X

X 1 1 ≥ , 2 3 (3a − 1) (a + 1)2

and it suffices to prove that X (a3

1 ≥ 1. + 1)2

This inequality is an immediate consequence of the inequality in P 1.176.

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Vasile Cîrtoaje

P 1.178. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that 1 1 1 1 + + + ≥ 1. 1 + a + a2 + a3 1 + b + b2 + b3 1 + c + c 2 + c 3 1 + d + d 2 + d 3 (Vasile Cîrtoaje, 1999) First Solution. We get the desired inequality by summing the inequalities 1 1 1 + ≥ , 2 3 2 3 1+a+a +a 1+ b+ b + b 1 + (a b)3/2 1 1 1 . + ≥ 2 3 2 3 1+c+c +c 1+d +d +d 1 + (cd)3/2 Thus, it suffices to show that 1 1 1 + ≥ , 1 + x2 + x4 + x6 1 + y2 + y4 + y6 1 + x3 y3 where x and y are positive real numbers. Putting p = x y and s = x 2 + x y + y 2 , this inequality becomes p3 (x 6 + y 6 ) + p2 (p − 1)(x 4 + y 4 ) − p2 (p2 − p + 1)(x 2 + y 2 ) − p6 − p4 + 2p3 − p2 + 1 ≥ 0, p3 (x 3 − y 3 )2 + p2 (p − 1)(x 2 − y 2 )2 − p2 (p2 − p + 1)(x − y)2 + p6 − p4 − p2 + 1 ≥ 0. p3 s2 (x − y)2 + p2 (p − 1)(s + p)2 (x − y)2 − p2 (p2 − p + 1)(x − y)2 + p6 − p4 − p2 + 1, p2 (s + 1)(ps − 1)(x − y)2 + (p2 − 1)(p4 − 1) ≥ 0. If ps − 1 ≥ 0, then this inequality is clearly true. Consider further that ps < 1. From ps < 1 and s ≥ 3p, we get p2 < 1/3. Write the desired inequality in the form (1 − p2 )(1 − p4 ) ≥ p2 (1 + s)(1 − ps)(x − y)2 . Since p(x − y)2 = p(s − 3p) < 1 − 3p2 < 1 − p2 , it suffices to show that 1 − p4 ≥ p(1 + s)(1 − ps). Indeed, 4p(1 + s)(1 − ps) ≤ [p(1 + s) + (1 − ps)]2 = (1 + p)2 < 2(1 + p2 ) < 4(1 − p4 ). The equality occurs for a = b = c = d = 1. Second Solution. Assume that a ≥ b ≥ c ≥ d, and write the inequality as X 1 ≥ 1. (1 + a)(1 + a2 )

Symmetric Rational Inequalities

241

Since

1 1 1 1 1 1 ≤ ≤ , ≤ ≤ , 2 2 1+a 1+ b 1+c 1+a 1+ b 1 + c2 by Chebyshev’s inequality, it suffices to prove that  ‹ ‹ 1 1 1 1 1 1 1 1 + + + + + ≥ 1. 3 1+a 1+ b 1+c 1 + a2 1 + b2 1 + c 2 (1 + d)(1 + d 2 ) In addition, from the inequality in P 1.174, we have p 3 1 1 1 3 3 d + + ≥ = p p 3 3 1+a 1+ b 1+c 1 + a bc d +1 and

p 3 3 d2 1 1 3 1 = + + ≥ . p p 3 3 1 + a2 1 + b2 1 + c 2 1 + a2 b2 c 2 d2 + 1 Thus, it suffices to prove that (1 + Putting x =

p 3

p 3

3d d)(1 +

p 3

d2

)

+

1 ≥ 1. (1 + d)(1 + d 2 )

d, this inequality becomes as follows 3x 3 1 + ≥ 1, (1 + x)(1 + x 2 ) (1 + x 3 )(1 + x 6 ) 3x 3 (1 − x + x 2 )(1 − x 2 + x 4 ) + 1 ≥ (1 + x 3 )(1 + x 6 ), x 3 (2 − 3x + 2x 3 − 3x 5 + 2x 6 ) ≥ 0, x 3 (1 − x)2 (2 + x + x 3 + 2x 4 ) ≥ 0.

Remark. The following generalization holds true (Vasile Cîrtoaje, 2004): • If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then 1 1 + a1 + · · · + a1n−1

+

1 1 + a2 + · · · + a2n−1

+ ··· +

1 ≥ 1. 1 + an + · · · + ann−1

P 1.179. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that 1 1 1 1 + + + ≥ 1. 1 + a + 2a2 1 + b + 2b2 1 + c + 2c 2 1 + d + 2d 2 (Vasile Cîrtoaje, 2006)

242

Vasile Cîrtoaje

Solution. We will show that 1 1 ≥ , 2 k 1 + a + 2a 1 + a + a2k + a3k where k = 5/6. Then, it suffices to show that X

1 ≥ 1, 1 + a k + a2k + a3k

which immediately follows from the inequality in P 1.178. Setting a = x 6 , x > 0, the claimed inequality can be written as

2x 12

1 1 ≥ , 6 5 + x + 1 1 + x + x 10 + x 15

which is equivalent to x 10 + x 5 + 1 ≥ 2x 7 + x. We can prove it by summing the AM-GM inequalities x 5 + 4 ≥ 5x and 5x 10 + 4x 5 + 1 ≥ 10x 7 . This completes the proof. The equality occurs for a = b = c = d = 1. Remark. The inequalities in P 1.176, P 1.178 and P 1.179 are particular cases of the following more general inequality (Vasile Cîrtoaje, 2009): • Let a1 , a2 , . . . , an (n ≥ 4) be positive real numbers such that a1 a2 · · · an = 1. If p, q, r are nonnegative real numbers satisfying p + q + r = n − 1, then i=n X

1

i=1

1 + pai + qai2 + r ai3

≥ 1.

P 1.180. Let a, b, c, d be positive real numbers such that a bcd = 1. Prove that 1 1 1 1 9 25 + + + + ≥ . a b c d a+b+c+d 4

Symmetric Rational Inequalities

243

Solution (by Vo Quoc Ba Can). Replacing a, b, c, d by a4 , b4 , c 4 , d 4 , respectively, the inequality becomes as follows: 1 1 1 9 25 1 + 4+ 4+ 4+ 4 ≥ , 4 4 4 4 a b c d a +b +c +d 4a bcd 1 1 1 1 4 9 9 + 4+ 4+ 4− ≥ − 4 , 4 4 a b c d a bcd 4a bcd a + b + c 4 + d 4 9(a4 + b4 + c 4 + d 4 − 4a bcd) 1 1 1 1 4 ≥ . + + + − a4 b4 c 4 d 4 a bcd 4a bcd(a4 + b4 + c 4 + d 4 ) Using the identities a4 + b4 + c 4 + d 4 − 4a bcd = (a2 − b2 )2 + (c 2 − d 2 )2 + 2(a b − cd)2 , 1 1 1 1 4 (a2 − b2 )2 (c 2 − d 2 )2 2(a b − cd)2 + + + − = + + 2 2 2 2 , a4 b4 c 4 d 4 a bcd a4 b4 c4 d 4 a b c d the inequality can be written as (a2 − b2 )2 (c 2 − d 2 )2 2(a b − cd)2 9[(a2 − b2 )2 + (c 2 − d 2 )2 + 2(a b − cd)2 ] + + ≥ , a4 b4 c4 d 4 a2 b2 c 2 d 2 4a bcd(a4 + b4 + c 4 + d 4 )     4 4 4 4 4 4 4 4 2 2 2 4cd(a + b + c + d ) 2 2 2 4a b(a + b + c + d ) (a − b ) − 9 + (c − d ) −9 a3 b3 c3 d 3   4 4 4 4 2 4(a + b + c + d ) +2(a b − cd) − 9 ≥ 0. a bcd By the AM-GM inequality, we have a4 + b4 + c 4 + d 4 ≥ 4a bcd. Therefore, it suffices to show that     4 4 4 4 4 4 4 4 2 2 2 4cd(a + b + c + d ) 2 2 2 4a b(a + b + c + d ) (a − b ) − 9 + (c − d ) − 9 ≥ 0. a3 b3 c3 d 3 Without loss of generality, assume that a ≥ c ≥ d ≥ b. Since (a2 − b2 )2 ≥ (c 2 − d 2 )2 and

4cd(a4 + b4 + c 4 + d 4 ) 4(a4 + b4 + c 4 + d 4 ) 4(a4 + 3b4 ) ≥ ≥ > 9, a3 b3 a3 b a3 b it is enough to prove that     4cd(a4 + b4 + c 4 + d 4 ) 4a b(a4 + b4 + c 4 + d 4 ) −9 + − 9 ≥ 0, a3 b3 c3 d 3

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Vasile Cîrtoaje

which is equivalent to 2(a4 + b4 + c 4 + d 4 )



cd ab + 3 3 3 3 a b c d

‹

≥ 9.

Indeed, by the AM-GM inequality,  ‹  ‹ cd ab 2 4 4 4 4 2(a + b + c + d ) 3 3 + 3 3 ≥ 8a bcd = 16 > 9. a b c d a bcd The equality occurs for a = b = c = d = 1.

P 1.181. If a, b, c, d are real numbers such that a + b + c + d = 0, then (a − 1)2 (b − 1)2 (c − 1)2 (d − 1)2 + + 2 + ≤ 4. 3a2 + 1 3b2 + 1 3c + 1 3d 2 + 1 Solution. Since 4−

3(a − 1)2 (3a + 1)2 = , 3a2 + 1 3a2 + 1

we can write the inequality as X (3a + 1)2 3a2 + 1

≥ 4.

On the other hand, since 4a2 = 3a2 + (b + c + d)2 ≤ 3a2 + 3(b2 + c 2 + d 2 ) = 3(a2 + b2 + c 2 + d 2 ), 9 2 9(a2 + b2 + c 2 + d 2 ) + 4 (a + b2 + c 2 + d 2 ) + 1 = , 4 4 P X (3a + 1)2 4 (3a + 1)2 ≥ = 4. 3a2 + 1 9(a2 + b2 + c 2 + d 2 ) + 4

3a2 + 1 ≤ we have

The equality holds for a = b = c = d = 0, and also for a = 1 and b = c = d = −1/3 (or any cyclic permutation). Remark. The following generalization is also true. • If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = 0, then (a1 − 1)2 (n − 1)a12 + 1

+

(a2 − 1)2 (n − 1)a22 + 1

+ ··· +

(an − 1)2 ≤ n, (n − 1)an2 + 1

with equality for a1 = a2 = · · · = an = 0, and also for a1 = 1 and a2 = a3 = · · · = an = −1/(n − 1 (or any cyclic permutation).

Symmetric Rational Inequalities

245

P 1.182. If a, b, c, d ≥ −5 such that a + b + c + d = 4, then 1− b 1−c 1−d 1−a + + + ≥ 0. 2 2 2 (1 + a) (1 + b) (1 + c) (1 + d)2 Solution. Assume that a ≤ b ≤ c ≤ d. We show first that x ∈ [−5, −1) ∪ (−1, ∞) involves 1− x −1 , ≥ 2 (1 + x) 8 and x ∈ [−5, −1) ∪ (−1, 1/3] involves 3 1− x ≥ . 2 (1 + x) 8 Indeed, we have

1 (x − 3)2 1− x + = ≥0 (1 + x)2 8 8(1 + x)2

and

1− x 3 (5 + x)(1 − 3x) − = ≥ 0. (1 + x)2 8 8(1 + x)2

Then, if a ≤ 1/3, then 1−a 1− b 1−c 1−d 3 1 1 1 + + + ≥ − − − = 0. 2 2 2 2 (1 + a) (1 + b) (1 + c) (1 + d) 8 8 8 8 Assume now that 1/3 ≤ a ≤ b ≤ c ≤ d. Since 1−a ≥1− b ≥1−c ≥1−d and

1 1 1 1 ≥ ≥ ≥ , (1 + a)2 (1 + b)2 (1 + c)2 (1 + d)2

by Chebyshev’s inequality, we have 1− b 1−c 1−d 1−a + + + ≥ 2 2 2 (1 + a) (1 + b) (1 + c) (1 + d)2 ˜ — •X 1 ”X 1 ≥ (1 − a) = 0. 4 (1 + a)2 The equality holds for a = b = c = d = 1, and also for a = −5 and b = c = d = 3 (or any cyclic permutation).

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Vasile Cîrtoaje

P 1.183. Let a1 , a2 , . . . , an be positive real numbers such that a1 + a2 + · · · + an = n. Prove that X 1 1 ≤ . 2 2 2 2 (n + 1)a1 + a2 + · · · + an (Vasile Cîrtoaje, 2008) First Solution. By the Cauchy-Schwarz inequality, we have n2

X

(n + 1)a12 + a22 + · · · + an2

=

(a1 + a2 + · · · + an )2

X

2a12 + (a12 + a22 ) + · · · + (a12 + an2 )   X 1 an2 a22 ≤ + + ··· + 2 2 a12 + a22 a1 + an2 =

n2 n n(n − 1) + = , 2 2 2

from which the conclusion follows. The equality holds for a1 = a2 = · · · = an = 1. Second Solution. Write the inequality as X

a12 + a22 + · · · + an2

≤

(n + 1)a12 + a22 + · · · + an2 Since

a12 + a22 + · · · + an2 (n + 1)a12 + a22 + · · · + an2

=1−

a12 + a22 + · · · + an2 2

.

na12 (n + 1)a12 + a22 + · · · + an2

,

we need to prove that a12

X

(n + 1)a12 + a22 + · · · + an2

+

a12 + a22 + · · · + an2 2n

≥ 1.

By the Cauchy-Schwarz inequality, we have X

a12

(a1 + a2 + · · · + an )2 P ≥ (n + 1)a12 + a22 + · · · + an2 [(n + 1)a12 + a22 + · · · + an2 ] n = . 2 2 2(a1 + a2 + · · · + an2 )

Then, it suffices to prove that n a12 + a22 + · · · + an2

+

a12 + a22 + · · · + an2 n

which follows immediately from the AM-GM inequality.

≥ 2,

Symmetric Rational Inequalities

247

P 1.184. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = 0. Prove that (a1 + 1)2 a12 + n − 1

+

(a2 + 1)2 a22 + n − 1

+ ··· +

(an + 1)2 n ≥ . 2 an + n − 1 n−1 (Vasile Cîrtoaje, 2010)

Solution. Without loss of generality, assume that an2 = max{a12 , a22 , · · · , an2 }. Since (an + 1)2 (n − 1 − an )2 n = , − an2 + n − 1 n − 1 (n − 1)(an2 + n − 1) we can write the inequality as n−1 X (ai + 1)2 i=1

ai2 + n − 1

≥

(n − 1 − an )2 . (n − 1)(an2 + n − 1)

From the Cauchy-Schwarz inequality ™ – n−1 – n−1 ™ – n−1 ™2 X (a + 1)2 X X i (ai2 + n − 1) ≥ (ai + 1) , 2 a + n−1 i=1 i i=1 i=1 we get n−1 X (ai + 1)2 i=1

(n − 1 − an )2 . ≥ P n−1 2 2 ai2 + n − 1 i=1 ai + (n − 1)

Thus, it suffices to show that n−1 X

ai2 + (n − 1)2 ≤ (n − 1)(an2 + n − 1),

i=1

which is clearly true. The proof is completed. The equality holds for · · · = an (or any cyclic permutation).

−a1 = a2 = a3 = n−1

P 1.185. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that 1 1 1 + + ··· + ≥ 1. 1 + (n − 1)a1 1 + (n − 1)a2 1 + (n − 1)an (Vasile Cîrtoaje, 1991)

248

Vasile Cîrtoaje

First Solution. Let k = (n − 1)/n. We can get the required inequality by summing the inequalities below for i = 1, 2, · · · , n: −k

ai 1 . ≥ −k −k 1 + (n − 1)ai a1 + a2 + · · · + an−k This inequality is equivalent to −k −k a1−k + · · · + ai−1 + ai+1 + · · · + an−k ≥ (n − 1)ai1−k ,

which follows from the AM-GM inequality. The equality holds for a1 = a2 = · · · = an = 1. Second Solution. Using the substitutions ai = 1/x i for all i, the inequality becomes xn x1 x2 + + ··· + ≥ 1, x1 + n − 1 x2 + n − 1 xn + n − 1 where x 1 , x 2 , · · · , x n are positive real numbers such that x 1 x 2 · · · x n = 1. By the CauchySchwarz inequality, we have Pp 2 X ( x1) xi ≥P . xi + n − 1 (x 1 + n − 1) Thus, we still have to prove that Xp X ( x 1 )2 ≥ x 1 + n(n − 1), which reduces to X

p

xi x j ≥

1≤i< j≤n

n(n − 1) . 2

Since x 1 x 2 · · · x n = 1, this inequality follows from the AM-GM inequality. Third Solution. For the sake of contradiction, assume that the required inequality is not true. Then, it suffices to show that the hypothesis a1 a2 · · · an = 1 does not hold. More precisely, we will prove that 1 1 1 + + ··· + 1. Let x i = ai =

1 , 0 < x i < 1, for i = 1, 2, · · · , n. Since 1 + (n − 1)ai

1 − xi for all i, we need to show that (n − 1)x i

x1 + x2 + · · · + x n < 1

Symmetric Rational Inequalities

249

implies (1 − x 1 )(1 − x 2 ) · · · (1 − x n ) > (n − 1)n x 1 x 2 · · · x n . Using the AM-GM inequality, we have 1 − xi >

X

vY u t xk. x k ≥ (n − 1) n−1 k6=i

k6=i

Multiplying the inequalities vY u t 1 − x i > (n − 1) n−1 xk. k6=i

for i = 1, 2, · · · , n, the conclusion follows. Remark. The inequality in P 1.185 is a particular case of the following more general results (Vasile Cîrtoaje, 2005): • Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If 0 < k ≤ n − 1 and p ≥ n1/k − 1, then 1 1 1 n + + ··· + ≥ . k k k (1 + pa1 ) (1 + pa2 ) (1 + pan ) (1 + p)k 1 • Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If k ≥ and n−1  n 1/k − 1, then 0 0 and k = 2 +

1 . Write the inequality as n−1 (n − 1)x k + x ≥ nx 2 .

We can get this inequality using the AM-GM inequality as follows p n (n − 1)x k + x ≥ n x (n−1)k x = nx 2 . Thus, it suffices to show that 1 1 + (n − 1)a1k

+

1 1 + (n − 1)a2k 2

+ ··· +

1 ≥ 1, 1 + (n − 1)ank

which follows immediately from the inequality in the preceding P 1.185. The equality holds for a1 = a2 = · · · = an = 1. Remark 1. Similarly, we can prove the following more general statement. • Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If p and q are p real numbers such that p + q = n − 1 and n − 1 ≤ q ≤ ( n + 1)2 , then 1 1+

pa1 + qa12

+

1 1+

pa2 + qa22

+ ··· +

1 ≥ 1. 1 + pan + qan2

Remark 2. We can extend the inequality in Remark 1 as follows (Vasile Cîrtoaje, 2009). • Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If p and q are p real numbers such that p + q = n − 1 and 0 ≤ q ≤ ( n + 1)2 , then 1 1+

pa1 + qa12

+

1 1+

pa2 + qa22

+ ··· +

1 ≥ 1. 1 + pan + qan2

P 1.187. Let a1 , a2 , . . . , an be positive real numbers such that a1 , a2 , . . . , an ≥

k(n − k − 1) , kn − k − 1

k>1

and a1 a2 · · · an = 1. Prove that

1 1 1 n + + ··· + ≤ . a1 + k a2 + k an + k 1+k (Vasile Cîrtoaje, 2005)

Symmetric Rational Inequalities

251

Solution. We use the induction on n. Let En (a1 , a2 , . . . , an ) =

1 1 n 1 + + ··· + − . a1 + k a2 + k an + k 1 + k

For n = 2, we have p p (1 − k)( a1 − a2 )2 E2 (a1 , a2 ) = ≤ 0. (1 + k)(a1 + k)(a2 + k) Assume that the inequality is true for n−1 numbers (n ≥ 3), and prove that En (a1 , a2 , . . . , an ) ≥ 0 for a1 a2 · · · an = 1 and a1 , a2 , . . . , an ≥ pn , where pn =

k(n − k − 1) . kn − k − 1

Due to symmetry, we may assume that a1 ≥ 1 and a2 ≤ 1. There are two cases to consider. Case 1: a1 a2 ≤ k2 . Since a1 a2 ≥ a2 and pn−1 < pn , from a1 , a2 , . . . , an ≥ pn it follows that a1 a2 , a3 , · · · , an > pn−1 . Then, by the inductive hypothesis, we have En−1 (a1 a2 , a2 , . . . , an ) ≤ 0, and it suffices to show that En (a1 , a2 , . . . , an ) ≤ En−1 (a1 a2 , a2 , . . . , an ). This is equivalent to 1 1 1 1 + − − ≤ 0, a1 + k a2 + k a1 a2 + k 1 + k which reduces to the obvious inequality (a1 − 1)(1 − a2 )(a1 a2 − k2 ) ≤ 0. Case 2: a1 a2 ≥ k2 . Since a1 + a2 + 2k a1 + a2 + 2k 1 1 1 + = ≤ 2 = a1 + k a2 + k a1 a2 + k(a1 + a2 ) + k2 k + k(a1 + a2 ) + k2 k and

1 1 n−2 kn − k − 1 + ··· + ≤ = , a3 + k an + k pn + k k(k + 1)

we have En (a1 , a2 , . . . , an ) ≤

1 kn − k − 1 n + − = 0. k k(k + 1) 1+k

Thus, the proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

252

Vasile Cîrtoaje

Remark. For k = n − 1, we get the inequality in P 1.185. Also, for k → ∞, we get the known inequalities 1
0 and let p r = n a1 a2 · · · an . By the Cauchy-Schwarz inequality, we have Pp Pp X ( ( a2 a3 · · · an )2 a2 a3 · · · an )2 1 =P ≥P . 1 + na1 (1 + na1 )a2 a3 · · · an a2 a3 · · · an + n2 r n Therefore, it suffices to show that X Xp a2 a3 · · · an )2 ≥ n a2 a3 · · · an + n3 r n . (n + r n )( By the AM-GM inequality, we have Xp X ( a2 a3 · · · an )2 ≥ a2 a3 · · · an + n(n − 1)r n−1 . Thus, it is enough to prove that X X (n + r n )[ a2 a3 · · · an + n(n − 1)r n−1 ] ≥ n a2 a3 · · · an + n3 r n , which is equivalent to X rn a2 a3 · · · an + n(n − 1)r 2n−1 + n2 (n − 1)r n−1 ≥ n3 r n . Also, by the AM-GM inequality, X

a2 a3 · · · an ≥ nr n−1 ,

and it suffices to show the inequality nr 2n−1 + n(n − 1)r 2n−1 + n2 (n − 1)r n−1 ≥ n3 r n , which can be rewritten as n2 r n−1 (r n − nr + n − 1) ≥ 0. Indeed, by the AM-GM inequality, we get r n − nr + n − 1 = r n + 1 + · · · + 1 − nr ≥ n The equality holds for a1 = a2 = · · · = an = 1.

p n

r n · 1 · · · 1 − nr = 0.

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Vasile Cîrtoaje

P 1.190. If a1 , a2 , . . . , an are positive real numbers, then bn an b1 b2 a1 a2 + + ··· + ≥ + + ··· + , a1 a2 an b1 b2 bn where bi = Solution. Let

1 X aj, n − 1 j6=i

i = 1, 2, · · · , n.

a1 + a2 + · · · + an , n 1 1 1 A= + + ··· + . a1 a2 an a=

Using the Cauchy-Schwarz inequality, we have (n − 1)2 1 1 1 1 ≤ + + ··· + = A− , a2 + a3 + · · · + an a2 a3 an a1 n−1 1 ≤ A− , b1 a1 Aa1 − 1 a1 ≤ , b1 n−1 n X a i=1

n

i

bi

≤

n X a i=1

Since

i

bi

A X n ai − , n − 1 i=1 n−1 ≤

naA n − . n−1 n−1

n n X bi 1 X na − ai naA n = = − , a n − 1 a n − 1 n − 1 i i i=1 i=1

the conclusion follows. The equality holds for a1 = a2 = · · · = an .

P 1.191. If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an = then

1 1 1 + + ··· + , a1 a2 an

Symmetric Rational Inequalities

255

(a)

1 1 1 + + ··· + ≥ 1; 1 + (n − 1)a1 1 + (n − 1)a2 1 + (n − 1)an

(b)

1 1 1 + + ··· + ≤ 1. n − 1 + a1 n − 1 + a2 n − 1 + an (Vasile Cîrtoaje, 1996)

Solution. (a) We use the contradiction method. So, assume that 1 1 1 + + ··· + < 1, 1 + (n − 1)a1 1 + (n − 1)a2 1 + (n − 1)an and show that a1 + a2 + · · · + an >

1 1 1 + + ··· + . a1 a2 an

Using the substitution xi =

1 , 1 + (n − 1)ai

i = 1, 2, · · · , n,

the hypothesis inequality becomes x 1 + x 2 + · · · + x n < 1. This inequality involves 1 − x i > (n − 1)bi ,

bi =

1 X x j, n − 1 j6=i

i = 1, 2, · · · , n.

Using this result and the inequality from the preceding P 1.190, we get a1 + a2 + · · · + an =

n n n X X 1 − xi bi X x i > ≥ . (n − 1)x i x b i=1 i=1 i i=1 i

Thus, it suffices to show that n X xi 1 1 1 ≥ + + ··· + . b a1 a2 an i=1 i

Indeed, we have

n n n X x i X (n − 1)x i X 1 > = . b 1 − xi a i=1 i i=1 i=1 i

The proof is completed. The equality holds for a1 = a2 = · · · = an = 1. (b) The desired inequality follows from the inequality in (a) by replacing a1 , a2 , . . . , an with 1/a1 , 1/a2 , . . . , 1/an , respectively.

256

Vasile Cîrtoaje

Chapter 2

Symmetric Nonrational Inequalities 2.1

Applications

2.1. If a, b, c are nonnegative real numbers, then Xp Æ a2 − a b + b2 ≤ 6(a2 + b2 + c 2 ) − 3(a b + bc + ca).

2.2. If a, b, c are nonnegative real numbers, then p

a2 − a b + b2 +

p

b2 − bc + c 2 +

p

c 2 − ca + a2 ≤ 3

v t a2 + b2 + c 2 2

.

2.3. If a, b, c are nonnegative real numbers, then v v v t t t p 2 2 2 a2 + b2 − a b + b2 + c 2 − bc + c 2 + a2 − ca ≥ 2 a2 + b2 + c 2 . 3 3 3

2.4. If a, b, c are nonnegative real numbers, then Xp Æ a2 + a b + b2 ≥ 4(a2 + b2 + c 2 ) + 5(a b + bc + ca).

2.5. If a, b, c are nonnegative real numbers, then Xp Æ a2 + a b + b2 ≤ 5(a2 + b2 + c 2 ) + 4(a b + bc + ca). 257

258

Vasile Cîrtoaje

2.6. If a, b, c are nonnegative real numbers, then Xp p p a2 + a b + b2 ≤ 2 a2 + b2 + c 2 + a b + bc + ca.

2.7. If a, b, c are nonnegative real numbers, then p p p p p a2 + 2bc + b2 + 2ca + c 2 + 2a b ≤ a2 + b2 + c 2 + 2 a b + bc + ca.

2.8. If a, b, c are nonnegative real numbers, then p

1 a2 + 2bc

+p

1 b2 + 2ca

+p

1 c 2 + 2a b

≥p

1 a2 + b2 + c 2

+p

2 a b + bc + ca

.

2.9. If a, b, c are positive real numbers, then p p p p p 2a2 + bc + 2b2 + ca + 2c 2 + a b ≤ 2 a2 + b2 + c 2 + a b + bc + ca.

2.10. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k = then XÆ p a(a + k b)(a + kc) ≤ 3 3.

p 3 − 1,

2.11. If a, b, c are nonnegative real numbers such that a + b + c = 3, then XÆ a(2a + b)(2a + c) ≥ 9.

2.12. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that Æ Æ Æ b2 + c 2 + a(b + c) + c 2 + a2 + b(c + a) + a2 + b2 + c(a + b) ≥ 6.

2.13. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that (a)

p

(b)

p p p p 3a2 + a bc + 3b2 + a bc + 3c 2 + a bc ≥ 3 3 + a bc.

a(3a2 + a bc) +

p

b(3b2 + a bc) +

p

c(3c 2 + a bc) ≥ 6;

Symmetric Nonrational Inequalities

259

2.14. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that Æ Æ Æ a (a + 2b)(a + 2c) + b (b + 2c)(b + 2a) + c (c + 2a)(c + 2b) ≥ 9.

2.15. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that Æ Æ Æ p a + (b − c)2 + b + (c − a)2 + c + (a − b)2 ≥ 3.

2.16. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that v v v t a(b + c) t b(c + a) t c(a + b) + + ≥ 2. a2 + bc b2 + ca c2 + a b 2.17. Let a, b, c be positive real numbers such that a bc = 1. Prove that p 3

1 a2

+ 25a + 1

+p 3

1 b2

+ 25b + 1

+p 3

1 c2

+ 25c + 1

≥ 1.

2.18. If a, b, c are nonnegative real numbers, then p

a2 + bc +

p

b2 + ca +

p

c2 + a b ≤

3 (a + b + c). 2

2.19. If a, b, c are nonnegative real numbers, then p p p p a2 + 9bc + b2 + 9ca + c 2 + 9a b ≥ 5 a b + bc + ca.

2.20. If a, b, c are nonnegative real numbers, then XÆ (a2 + 4bc)(b2 + 4ca) ≥ 5(a b + ac + bc).

2.21. If a, b, c are nonnegative real numbers, then XÆ (a2 + 9bc)(b2 + 9ca) ≥ 7(a b + ac + bc).

260

Vasile Cîrtoaje

2.22. If a, b, c are nonnegative real numbers, then XÆ (a2 + b2 )(b2 + c 2 ) ≤ (a + b + c)2 .

2.23. If a, b, c are nonnegative real numbers, then XÆ (a2 + a b + b2 )(b2 + bc + c 2 ) ≥ (a + b + c)2 .

2.24. If a, b, c are nonnegative real numbers, then XÆ (a2 + 7a b + b2 )(b2 + 7bc + c 2 ) ≥ 7(a b + ac + bc).

2.25. If a, b, c are nonnegative real numbers, then v ‹ ‹ X t 7 7 13 2 2 2 2 a + ab + b b + bc + c ≤ (a + b + c)2 . 9 9 12

2.26. If a, b, c are nonnegative real numbers, then v ‹ ‹ X t 1 1 61 2 2 2 2 a + ab + b b + bc + c ≤ (a + b + c)2 . 3 3 60

2.27. If a, b, c are nonnegative real numbers, then a b c +p +p ≥ 1. p 2 2 2 2 2 4b + bc + 4c 4c + ca + 4a 4a + a b + 4b2

2.28. If a, b, c are nonnegative real numbers, then p

a b2

+ bc

+ c2

+p

b c2

+ ca + a2

+p

c a2

+ ab +

b2

≥p

a+b+c a b + bc + ca

2.29. If a, b, c are nonnegative real numbers, then p

a a2

+ 2bc

+p

b b2

+ 2ca

+p

c c2

+ 2a b

≤p

a+b+c a b + bc + ca

.

.

Symmetric Nonrational Inequalities

261

2.30. If a, b, c are nonnegative real numbers, then p p p a3 + b3 + c 3 + 3a bc ≥ a2 a2 + 3bc + b2 b2 + 3ca + c 2 c 2 + 3a b.

2.31. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a b c +p +p ≤ 1. p 4a2 + 5bc 4b2 + 5ca 4c 2 + 5a b 2.32. Let a, b, c be nonnegative real numbers. Prove that p p p a 4a2 + 5bc + b 4b2 + 5ca + c 4c 2 + 5a b ≥ (a + b + c)2 .

2.33. Let a, b, c be nonnegative real numbers. Prove that p p p a a2 + 3bc + b b2 + 3ca + c c 2 + 3a b ≥ 2(a b + bc + ca).

2.34. Let a, b, c be nonnegative real numbers. Prove that p p p a a2 + 8bc + b b2 + 8ca + c c 2 + 8a b ≤ (a + b + c)2 .

2.35. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that p

a2 + 2bc b2 + bc + c 2

+p

b2 + 2ca c 2 + ca + a2

+p

c 2 + 2a b a2 + a b + b2

≥3

p

a b + bc + ca.

2.36. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ 1, then a k+1 b k+1 c k+1 ak + bk + c k + + ≤ . 2a2 + bc 2b2 + ca 2c 2 + a b a+b+c 2.37. If a, b, c are positive real numbers, then a2 − bc b2 − ca c2 − a b +p +p ≥ 0; p 3a2 + 2bc 3b2 + 2ca 3c 2 + 2a b

(a) (b)

a2 − bc p

8a2 + (b + c)2

+p

b2 − ca 8b2 + (c + a)2

+p

c2 − a b 8c 2 + (a + b)2

≥ 0.

262

Vasile Cîrtoaje

p 2.38. Let a, b, c be positive real numbers. If 0 ≤ k ≤ 1 + 2 2, then p

a2 − bc ka2 + b2 + c 2

+p

b2 − ca k b2 + c 2 + a2

+p

c2 − a b kc 2 + a2 + b2

≥ 0.

2.39. If a, b, c are nonnegative real numbers, then p p p (a2 − bc) b + c + (b2 − ca) c + a + (c 2 − a b) a + b ≥ 0.

2.40. If a, b, c are nonnegative real numbers, then p p p (a2 − bc) a2 + 4bc + (b2 − ca) b2 + 4ca + (c 2 − a b) c 2 + 4a b ≥ 0.

2.41. If a, b, c are nonnegative real numbers, then v t

v v t t a3 b3 c3 + + ≥ 1. a3 + (b + c)3 b3 + (c + a)3 c 3 + (a + b)3

2.42. If a, b, c are positive real numbers, then v t

1 1 1 (a + b + c) + + a b c 

‹

≥1+

v u t

1+

v t

(a2 + b2 + c 2 )



‹ 1 1 1 + + . a2 b2 c 2

2.43. If a, b, c are positive real numbers, then 5+

v t

2(a2

+

b2

+ c2)



‹  ‹ 1 1 1 1 1 1 + + − 2 ≥ (a + b + c) + + . a2 b2 c 2 a b c

2.44. If a, b, c are real numbers, then Æ 2(1 + a bc) + 2(1 + a2 )(1 + b2 )(1 + c 2 ) ≥ (1 + a)(1 + b)(1 + c).

Symmetric Nonrational Inequalities

263

2.45. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that v t a2 + bc b2

+ c2

+

v t b2 + ca c2

+ a2

+

v t c2 + a b a2

+

1 ≥2+ p . 2

b2

2.46. If a, b, c are nonnegative real numbers, then Æ

a(2a + b + c) +

Æ

b(2b + c + a) +

Æ

c(2c + a + b) ≥

Æ

12(a b + bc + ca).

2.47. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that a

Æ

(4a + 5b)(4a + 5c) + b

Æ

(4b + 5c)(4b + 5a) + c

Æ

(4c + 5a)(4c + 5b) ≥ 27.

2.48. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that a

Æ

(a + 3b)(a + 3c) + b

Æ

(b + 3c)(b + 3a) + c

Æ

(c + 3a)(c + 3b) ≥ 12.

2.49. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 3. Prove that p

2 + 7a b +

p

2 + 7bc +

p

2 + 7ca ≥ 3

Æ

3(a b + bc + ca).

2.50. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that Pp

(a)

P

(b) (c)

P

a(b + c)(a2 + bc) ≥ 6;

p p a(b + c) a2 + 2bc ≥ 6 3;

p a(b + c) (a + 2b)(a + 2c) ≥ 18.

2.51. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that a

p

p p bc + 3 + b ca + 3 + c a b + 3 ≥ 6.

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Vasile Cîrtoaje

2.52. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that (a)

P p (b + c) b2 + c 2 + 7bc ≥ 18;

(b)

p P p (b + c) b2 + c 2 + 10bc ≤ 12 3.

2.53. Let a, b, c be nonnegative real numbers such then a + b + c = 2. Prove that p p p p a + 4bc + b + 4ca + c + 4a b ≥ 4 a b + bc + ca.

2.54. If a, b, c are nonnegative real numbers, then p p p p a2 + b2 + 7a b + b2 + c 2 + 7bc + c 2 + a2 + 7ca ≥ 5 a b + bc + ca.

2.55. If a, b, c are nonnegative real numbers, then p p p Æ a2 + b2 + 5a b + b2 + c 2 + 5bc + c 2 + a2 + 5ca ≥ 21(a b + bc + ca).

2.56. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that v t2 p p p a a2 + 5 + b b2 + 5 + c c 2 + 5 ≥ (a + b + c)2 . 3

2.57. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 1. Prove that p p p a 2 + 3bc + b 2 + 3ca + c 2 + 3a b ≥ (a + b + c)2 .

2.58. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that (a)

(b)

a

a

v t 2a + bc 3

v t a(1 + b + c) 3

+b

+b

v t 2b + ca 3

+c

v t b(1 + c + a) 3

v t 2c + a b

+c

3

≥ 3;

v t c(1 + a + b) 3

≥ 3.

Symmetric Nonrational Inequalities

265

2.59. If a, b, c are nonnegative real numbers such that a + b + c = 3, then Æ

8(a2 + bc) + 9 +

Æ

8(b2 + ca) + 9 +

Æ

8(c 2 + a b) + 9 ≥ 15.

2.60. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k ≥ p

a2 + bc + k +

p

b2 + ca + k +

p

9 , then 8

p c 2 + a b + k ≥ 3 2 + k.

2.61. If a, b, c are nonnegative real numbers such that a + b + c = 3, then p

a3 + 2bc +

p

b3 + 2ca +

p

p c 3 + 2a b ≥ 3 3.

2.62. If a, b, c are positive real numbers, then p

a2 + bc + b+c

p

b2 + ca + c+a

p

p c2 + a b 3 2 ≥ . a+b 2

2.63. If a, b, c are nonnegative real numbers, no two of which are zero,then p

bc + 4a(b + c) + b+c

p

ca + 4b(c + a) + c+a

p

a b + 4c(a + b) 9 ≥ . a+b 2

2.64. If a, b, c are nonnegative real numbers, no two of which are zero,then p p p a a2 + 3bc b b2 + 3ca c c 2 + 3a b + + ≥ a + b + c. b+c c+a a+b

2.65. If a, b, c are nonnegative real numbers, no two of which are zero,then v t

v v t t 2a(b + c) 2b(c + a) 2c(a + b) + + ≥ 2. (2b + c)(b + 2c) (2c + a)(c + 2a) (2a + b)(a + 2b)

266

Vasile Cîrtoaje

2.66. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then v v v v s s t ab t bc t ab t bc ca ca + + ≤ 1 ≤ + + . 3a2 + 6 3b2 + 6 3c 2 + 6 6a2 + 3 6b2 + 3 6c 2 + 3

2.67. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. If k > 1, than a k (b + c) + b k (c + a) + c k (a + b) ≥ 6.

2.68. Let a, b, c be nonnegative real numbers such that a + b + c = 2. If 2 ≤ k ≤ 3, than a k (b + c) + b k (c + a) + c k (a + b) ≤ 2.

2.69. Let a, b, c be nonnegative real numbers, no two of which are zero. If m > n ≥ 0, than bm + c m c m + am am + bm (b + c − 2a) + (c + a − 2b) + (a + b − 2c) ≥ 0. bn + c n c n + an an + bn 2.70. Let a, b, c be positive real numbers such that a bc = 1. Prove that p p p a2 − a + 1 + a2 − a + 1 + a2 − a + 1 ≥ a + b + c.

2.71. Let a, b, c be positive real numbers such that a bc = 1. Prove that p p p 16a2 + 9 + 16b2 + 9 + 16b2 + 9 ≥ 4(a + b + c) + 3.

2.72. Let a, b, c be positive real numbers such that a bc = 1. Prove that p p p 25a2 + 144 + 25b2 + 144 + 25c 2 + 144 ≤ 5(a + b + c) + 24.

2.73. If a, b are positive real numbers such that a b + bc + ca = 3, then (a) (b)

p

a2 + 3 +

p

b2 + 3 +

p

b2 + 3 ≥ a + b + c + 3; p p p p a + b + b + c + c + a ≥ 4(a + b + c) + 6.

Symmetric Nonrational Inequalities

267

2.74. If a, b, c are nonnegative real numbers such that a + b + c = 3, then Æ Æ Æ (5a2 + 3)(5b2 + 3) + (5b2 + 3)(5c 2 + 3) + (5c 2 + 3)(5a2 + 3) ≥ 24.

2.75. If a, b, c are nonnegative real numbers such that a + b + c = 3, then v t 4(a2 + b2 + c 2 ) + 42 p p p . a2 + 1 + b2 + 1 + c 2 + 1 ≥ 3 2.76. If a, b, c are nonnegative real numbers such that a + b + c = 3, then p p p p (a) a2 + 3 + b2 + 3 + c 2 + 3 ≥ 2(a2 + b2 + c 2 ) + 30; p p p p (b) 3a2 + 1 + 3b2 + 1 + 3c 2 + 1 ≥ 2(a2 + b2 + c 2 ) + 30. 2.77. If a, b, c are nonnegative real numbers such that a + b + c = 3, then Æ Æ Æ (32a2 + 3)(32b2 + 3) + (32b2 + 3)(32c 2 + 3) + (32c 2 + 3)(32a2 + 3) ≤ 105.

2.78. If a, b, c are positive real numbers, then b+c c + a a + b ≥ 2. + − 3 − 3 + − 3 a c b

2.79. If a, b, c are real numbers such that a bc 6= 0, then b + c c + a a + b + ≥ 2. + a b c

2.80. Let a, b, c be nonnegative real numbers, no two of which are zero, and let x=

2a , b+c

y=

2b 2c , z= . c+a a+b

Prove that (a) (b)

p yz + z x ≥ 6; p p p p x + y + z ≥ 8 + x yz.

x + y +z+

p

xy+

p

268

Vasile Cîrtoaje

2.81. Let a, b, c be nonnegative real numbers, no two of which are zero, and let x=

2a , b+c

y=

2b 2c , z= . c+a a+b

Prove that p

1 + 24x +

p

1 + 24 y +

p

1 + 24z ≥ 15.

2.82. If a, b, c are positive real numbers, then v v v t t t 7a 7b 7c + + ≤ 3. a + 3b + 3c b + 3c + 3a c + 3a + 3b

2.83. If a, b, c are positive real numbers such that a + b + c = 3, then Æ Æ Æ p 3 3 3 3 a2 (b2 + c 2 ) + b2 (c 2 + a2 ) + c 2 (a2 + b2 ) ≤ 3 2.

2.84. If a, b, c are nonnegative real numbers, no two of which are zero, then 1 1 1 1 2 + + ≥ +p . a+b b+c c+a a+b+c a b + bc + ca

2.85. If a, b ≥ 1, then 1 1 1 1 + ≥p +p . p 2 3a + 1 3a b + 1 3b + 1 2.86. Let a, b, c be positive real numbers such that a ≥ 1 ≥ b ≥ c and a bc = 1. Prove that 1 1 1 3 +p ≥ . +p p 3a + 1 3c + 1 2 3b + 1 1 2.87. Let a, b, c be positive real numbers such that a + b + c = 3. If k ≥ p , then 2 (a bc)k (a2 + b2 + c 2 ) ≤ 3.

Symmetric Nonrational Inequalities

269

2.88. Let p and q be nonnegative real numbers such that p2 ≥ 3q, and let v v t 2p + w t 2p − 2w +2 , g(p, q) = 3 3 s  s 2p + 2w 2p − w  +2 ,  3 3 h(p, q) = p p p   p + p + q, where w =

p

p2 ≤ 4q p2 ≥ 4q

,

p2 − 3q. If a, b, c are nonnegative real numbers such that a + b + c = p,

a b + bc + ca = q,

then p

(a) with equality for a = (b)

a+b+

p

b+c+

p

c + a ≥ g(p, q),

p−w p + 2w and b = c = (or any cyclic permutation); 3 3 p p p a + b + b + c + c + a ≤ h(p, q),

p − 2w p+w and b = c = (or any cyclic permutation) - when 3 3 p2 ≤ 4q, and for a = 0, b+c = p and bc = q (or any cyclic permutation) - when p2 ≥ 4q.

with equality for a =

2.89. Let a, b, c, d be nonnegative real numbers such that a2 + b2 + c 2 + d 2 = 1. Prove that p p p p p p p p 1 − a + 1 − b + 1 − c + 1 − d ≥ a + b + c + d.

2.90. Let a, b, c, d be positive real numbers. Prove that p A + 2 ≥ B + 4, where

‹ 1 1 1 1 A = (a + b + c + d) + + + − 16, a b c d  ‹ 1 1 1 1 2 2 2 2 B = (a + b + c + d ) 2 + 2 + 2 + 2 − 16. a b c d 

270

Vasile Cîrtoaje

2.91. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = 1. Prove that p p p 3a1 + 1 + 3a2 + 1 + · · · + 3an + 1 ≥ n + 1. 2.92. Let 0 ≤ a < b and a1 , a2 , . . . , an ∈ [a, b]. Prove that €p p p Š2 b− a . a1 + a2 · · · + an − n n a1 a2 · · · an ≤ (n − 1)

2.93. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that 1 p

1 + (n2

− 1)a1

+p

1 1 + (n2 − 1)a2

+ ··· + p

1 1 + (n2 − 1)an

≥ 1.

2.94. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that n X i=1

1 1+

p

1 + 4n(n − 1)ai

≥

1 . 2

2.95. If f is a convex function on a real interval I and a1 , a2 , . . . , an ∈ I, then a + a + ··· + a  1 2 n f (a1 ) + f (a2 ) + · · · + f (an ) + n(n − 2) f ≥ n ≥ (n − 1)[ f (b1 ) + f (b2 ) + · · · + f (bn )], where bi =

1 X aj, n − 1 j6=i

i = 1, 2, · · · , n.

2.96. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that n X i=1

1 n−1+

p

(n − 1)2

+ 4nai

≤

1 . 2

2.97. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then v u 2 t a1 + a22 + · · · + an2 a1 + a2 + · · · + an ≥ n − 1 + . n

Symmetric Nonrational Inequalities

271

2.98. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then q Æ (n − 1)(a12 + a22 + · · · + an2 ) + n − n(n − 1) ≥ a1 + a2 + · · · + an . 2.99. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If 0 1, then X a1k ≥ 1. a1k + a2 + · · · + an 2.102. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If −2 ≤ k < 1, n−2 then X

a1k a1k + a2 + · · · + an

≤ 1.

2.103. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If k > 1, then X a1 ≤ 1. k a1 + a2 + · · · + an

272

Vasile Cîrtoaje

2.104. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If −1 − then

2 ≤ k < 1, n−2 a1

X a1k

+ a2 + · · · + an

≥ 1.

2.105. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If k ≥ 0, then X 1 ≤ 1. k a1 + a2 + · · · + an

2.106. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n. If 1 < k ≤ n + 1, then a1 a1k

+ a2 + · · · + an

+

a2 a1 + a2k

+ · · · + an

+ ··· +

an ≤ 1. a1 + a2 + · · · + ank

2.107. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≤ n. If 0 ≤ k < 1, then 1 a1k

+ a2 + · · · + an

+

1 a1 + a2k

+ · · · + an

+ ··· +

1 ≥ 1. a1 + a2 + · · · + ank

2.108. Let a1 , a2 , . . . , an be positive real numbers. If k > 1, then X ak + ak + · · · + ak 2

3

n

a2 + a3 + · · · + an

≤

n(a1k + a2k + · · · + ank ) a1 + a2 + · · · + an

.

Symmetric Nonrational Inequalities

2.2

273

Solutions

P 2.1. If a, b, c are nonnegative real numbers, then Xp Æ a2 − a b + b2 ≤ 6(a2 + b2 + c 2 ) − 3(a b + bc + ca). Solution. By squaring, the inequality becomes as follows XÆ 2(a b + bc + ca) + 2 (a2 − a b + b2 )(a2 − ac + c 2 ) ≤ 4(a2 + b2 + c 2 ), X €p

a2 − a b + b2 −

p

a2 − ac + c 2

Š2

≥ 0.

The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 2.2. If a, b, c are nonnegative real numbers, then p

a2 − a b + b2 +

p

b2 − bc + c 2 +

p

c 2 − ca + a2 ≤ 3

v t a2 + b2 + c 2 2

Solution (by Nguyen Van Quy). Assume that c = min{a, b, c}. Since b2 − bc + c 2 ≤ b2 and c 2 − ca + a2 ≤ a2 , it suffices to show that p

a2 − a b + b2 + a + b ≤ 3

v t a2 + b2 + c 2 2

.

Using the Cauchy-Schwarz inequality, we have p

a2

− ab +

=

b2

+a+b≤

v• t

(a2

− ab +

b2 ) +

˜ (a + b)2 (1 + k) k

v t (1 + k)[(1 + k)(a2 + b2 ) + (2 − k)a b] k

.

.

274

Vasile Cîrtoaje

Choosing k = 2, we get p

a2 − a b + b2 + a + b ≤ 3

v t a2 + b2 2

v t a2 + b2 + c 2 ≤3 = 3. 2

The equality holds for a = b and c = 0 (or any cyclic permutation).

P 2.3. If a, b, c are nonnegative real numbers, then v t

a2

+

b2

v v t t p 2 2 2 2 2 − a b + b + c − bc + c 2 + a2 − ca ≥ 2 a2 + b2 + c 2 . 3 3 3 (Vasile Cîrtoaje, 2012)

First Solution. By squaring, the inequality becomes XÆ 2 (3a2 + 3b2 − 2a b)(3a2 + 3c 2 − 2ac) ≥ 6(a2 + b2 + c 2 ) + 2(a b + bc + ca), 6(a2 + b2 + c 2 − a b − bc − ca) ≥

X €p

3a2 + 3b2 − 2a b −

p

3a2 + 3c 2 − 2ac

Š2

,

X X (b − c)2 (3b + 3c − 2a)2 (b − c)2 ≥
2 , p p 3a2 + 3b2 − 2a b + 3a2 + 3c 2 − 2ac   2 X (3b + 3c − 2a) (b − c)2 1 − €p Š2  . p 2 2 2 2 9a + 9b − 6a b + 9a + 9c − 6ac 3

Since Æ 9a2 + 9b2 − 6a b = (3b − a)2 + 8a2 ≥ |3b − a|, Æ p 9a2 + 9c 2 − 6ac = (3c − a)2 + 8a2 ≥ |3c − a|,

p

it suffices to show that   ‹2  X |3b + 3c − 2a| 2 (b − c) 1 − ≥ 0. |3b − a| + |3c − a| This is true since |3b + 3c − 2a| = |(3b − a) + (3c − a)| ≤ |3b − a| + |3c − a|. The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation).

Symmetric Nonrational Inequalities

275

Second Solution. Assume that a ≥ b ≥ c. Write the inequality as Æ Æ Æ (a + b)2 + 2(a − b)2 + (b + c)2 + 2(b − c)2 + (a + c)2 + 2(a − c)2 ≥ Æ ≥ 2 3(a2 + b2 + c 2 ). By Minkowski’s inequality, it suffices to show that Æ Æ [(a + b) + (b + c) + (a + c)]2 + 2[(a − b) + (b − c) + (a − c)]2 ≥ 2 3(a2 + b2 + c 2 ), which is equivalent to Æ

(a + b + c)2 + 2(a − c)2 ≥

Æ

3(a2 + b2 + c 2 ).

By squaring, the inequality turns into (a − b)(b − c) ≥ 0.

P 2.4. If a, b, c are nonnegative real numbers, then Xp Æ a2 + a b + b2 ≥ 4(a2 + b2 + c 2 ) + 5(a b + bc + ca). (Vasile Cîrtoaje, 2009) First Solution. By squaring, the inequality becomes XÆ (a2 + a b + b2 )(a2 + ac + c 2 ) ≥ (a + b + c)2 . Using the Cauchy-Schwarz inequality, we get v • ‹ ˜  XÆ Xu t b 2 3b2  c 2 3c 2 2 2 2 2 (a + a b + b )(a + ac + c ) = a+ + a+ + 2 4 2 4 ≥

X •

‹ ˜ b  c  3bc a+ a+ + = (a + b + c)2 . 2 2 4

The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation). Second Solution. Assume that a ≥ b ≥ c. By Minkowski’s inequality, we get Xp XÆ 2 a2 + a b + b2 = 3(a + b)2 + (a − b)2 ≥

Æ

3[(a + b) + (b + c) + (c + a)]2 + [(a − b) + (b − c) + (a − c)]2

276

Vasile Cîrtoaje =2

Æ

3(a + b + c)2 + (a − c)2 .

Therefore, it suffices to show that 3(a + b + c)2 + (a − c)2 ≥ 4(a2 + b2 + c 2 ) + 5(a b + bc + ca), which is equivalent to the obvious inequality (a − b)(b − c) ≥ 0. Remark. Similarly, we can prove the following generalization. • Let a, b, c be nonnegative real numbers such that a b + bc + ca = 4. If |k| ≤ 2, then Xp p a2 + ka b + b2 ≥ 2 a2 + b2 + c 2 + 3k + 2. For k = −2/3 and k = 1, we get the inequalities in P 2.3 and P 2.4, respectively. For k = −1 and k = 0, we get the inequalities Xp p a2 − a b + b2 ≥ 2 a2 + b2 + c 2 − 1, Xp

a2 + b2 ≥ 2

p

a2 + b2 + c 2 + 2.

P 2.5. If a, b, c are nonnegative real numbers, then Xp Æ a2 + a b + b2 ≤ 5(a2 + b2 + c 2 ) + 4(a b + bc + ca). (Michael Rozenberg, 2008) First Solution (by Vo Quoc Ba Can). Using the Cauchy-Schwarz inequality, we have   €X p Š2 ”X — X b2 + bc + c 2 2 2 ≤ (b + c) b + bc + c b+c X 2  X b + bc + c 2 a  2 = 2(a + b + c) =2 1+ (b + bc + c 2 ) b+c b+c = 4(a2 + b2 + c 2 ) + 2(a b + bc + ca) +

X 2a(b2 + bc + c 2 ) b+c 

bc = 4(a + b + c ) + 2(a b + bc + ca) + 2a b + c − b+c X 1 = 4(a2 + b2 + c 2 ) + 6(a b + bc + ca) − 2a bc . b+c 2

2

2

X

‹

Symmetric Nonrational Inequalities

277

Thus, it suffices to prove that 4(a2 + b2 + c 2 ) + 6(a b + bc + ca) − 2a bc

X

1 ≤ 5(a2 + b2 + c 2 ) + 4(a b + bc + ca), b+c

which is equivalent to Schur’s inequality 2(a b + bc + ca) ≤ a2 + b2 + c 2 + 2a bc

X

1 . b+c

We can prove this inequality by writing it as follows: ‹ X  bc , a a+ (a + b + c)2 ≤ 2 b+c X a (a + b + c)2 ≤ 2(a b + bc + ca) , b+c ”X —X a (a + b + c)2 ≤ a(b + c) . b+c Clearly, the last inequality follows from the Cauchy-Schwarz inequality. The equality holds for a = b = c. Second Solution. Let us denote p p A = b2 + bc + c 2 , B = c 2 + ca + a2 ,

C=

p

a2 + a b + b2 .

Without loss of generality, assume that a ≥ b ≥ c. For b = c = 0, the inequality is clearly true. Consider further b > 0. By squaring, the inequality becomes X X X 2 BC ≤ 3 a2 + 3 a b, X

a2 −

X

ab ≤

X (B − C)2 ,

X X (b − c)2 (b − c)2 ≤ 2(a + b + c)2 . (B + C)2 Since (B + C)2 ≤ 2(B 2 + C 2 ) = 2(2a2 + b2 + c 2 + ca + a b), it suffices to show that X X (b − c)2 ≤ (a + b + c)2

(b − c)2 , 2a2 + b2 + c 2 + ca + a b

which is equivalent to X (b − c)2 Sa ≥ 0,

278

Vasile Cîrtoaje

where Sa =

−a2 + a b + 2bc + ca (a + b + c)2 − 1 = , 2a2 + b2 + c 2 + ca + a b 2a2 + b2 + c 2 + ca + a b Sb =

−b2 + bc + 2ca + a b ≥ 0, 2b2 + c 2 + a2 + a b + bc

Sc =

−c 2 + ca + 2a b + bc ≥ 0. 2c 2 + a2 + b2 + bc + ca

According to X a2 (b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b b  ‹ Sb a 2 2 2 Sa ≥ (b − c) Sa + (b − c) S b = a(b − c) + , b a b it suffices to prove that Sa S b + ≥ 0, a b which is equivalent to a2 − a b − 2bc − ca −b2 + bc + 2ca + a b ≥ . b(2b2 + c 2 + a2 + a b + bc) a(2a2 + b2 + c 2 + ca + a b) Consider the non-trivial case a2 − a b − 2bc − ca ≥ 0. Since (2a2 + b2 + c 2 + ca + a b) − (2b2 + c 2 + a2 + a b + bc) = (a − b)(a + b + c) ≥ 0, it suffices to show that −b2 + bc + 2ca + a b a2 − a b − 2bc − ca ≥ . b a Indeed, a(−b2 + bc + 2ca + a b) − b(a2 − a b − 2bc − ca) = 2c(a2 + a b + b2 ) > 0.

P 2.6. If a, b, c are nonnegative real numbers, then Xp p p a2 + a b + b2 ≤ 2 a2 + b2 + c 2 + a b + bc + ca. (Vasile Cîrtoaje, 2010)

Symmetric Nonrational Inequalities

279

First Solution (by Nguyen Van Quy). Assume that a = max{a, b, c}. Since p p Æ a2 + a b + b2 + c 2 + ca + a2 ≤ 2[(a2 + a b + b2 ) + (c 2 + ca + a2 )], it suffices to show that p p p p 2 A + b2 + bc + c 2 ≤ 2 X + Y , where 1 A = a2 + (b2 + c 2 + a b + ac), 2

X = a2 + b2 + c 2 ,

Y = a b + bc + ca.

Write the desired inequality as follows p p p p 2( A − X ) ≤ Y − b2 + bc + c 2 , 2(A − X ) Y − (b2 + bc + c 2 ) , p p p ≤p A+ X Y + b2 + bc + c 2 b(a − b) + c(a − c) b(a − b) + c(a − c) ≤p . p p p A+ X Y + b2 + bc + c 2 Since b(a − b) + c(a − c) ≥ 0, we need to show that p p p p A + X ≥ Y + b2 + bc + c 2 . This inequality is true because X ≥ Y and p p A ≥ b2 + bc + c 2 . Indeed, 2(A − b2 − bc − c 2 ) = 2a2 + (b + c)a − (b + c)2 = (2a − b − c)(a + b + c) ≥ 0. The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation). Second Solution. In the first solution of P 2.5, we have shown that X €X p Š2 b2 + bc + c 2 ≤ 4(a2 + b2 + c 2 ) + 6(a b + bc + ca) − 2a bc

1 . b+c

Thus, it suffices to prove that 4(a2 + b2 +c 2 )+6(a b+ bc +ca)−2a bc

X

€ p Š2 p 1 ≤ 2 a2 + b2 + c 2 + a b + bc + ca , b+c

which is equivalent to 2a bc

X

Æ 1 + 4 (a2 + b2 + c 2 )(a b + bc + ca) ≥ 5(a b + bc + ca). b+c

280

Vasile Cîrtoaje

Since X

1 9 9 ≥P = , b+c (b + c) 2(a + b + c)

it is enough to prove that Æ 9a bc + 4 (a2 + b2 + c 2 )(a b + bc + ca) ≥ 5(a b + bc + ca), a+b+c which can be written as

Æ 9a bc + 4 q(p2 − 2q) ≥ 5q, p

where p = a + b + c, q = a b + bc + ca. p For p2 ≥ 4q, this inequality is true because 4 q(p2 − 2q) ≥ 5q. Consider further 3q ≤ p2 ≤ 4q. By Schur’s inequality of third degree, 9a bc ≥ 4q − p2 . p Therefore, it suffices to show that (4q − p2 ) + 4

Æ

q(p2 − 2q) ≥ 5q,

which is Æ 4 q(p2 − 2q) ≥ p2 + q. Indeed, 16q(p2 − 2q) − (p2 + q)2 = (p2 − 3q)(11q − p2 ) ≥ 0. Third Solution. Let us denote p p p A = b2 + bc + c 2 , B = c 2 + ca + a2 , C = a2 + a b + b2 , p p X = a2 + b2 + c 2 , Y = a b + bc + ca. By squaring, the inequality becomes X X 2 BC ≤ 2 a2 + 4X Y, X (B − C)2 ≥ 2(X − Y )2 , P 2 X (b − c)2 (b − c)2 2 2(a + b + c) ≥ . (B + C)2 (X + Y )2

Symmetric Nonrational Inequalities

281

Since B + C ≤ (c + a) + (a + b) = 2a + b + c, it suffices to show that 2

2(a + b + c)

X

(b − c)2 ≥ (2a + b + c)2

P 2 (b − c)2 (X + Y )2

.

According to the Cauchy-Schwarz inequality, we have X

P 2 (b − c)2 (b − c)2 ≥P . (2a + b + c)2 (b − c)2 (2a + b + c)2

Therefore, it is enough to prove that 2(a + b + c)2 1 P ≥ , (X + Y )2 (b − c)2 (2a + b + c)2 which is 2(a + b + c)2 (X + Y )2 ≥

X (b − c)2 (2a + b + c)2 .

We see that (a + b + c)2 (X + Y )2 ≥ =

and

€X

a2 + 2

X

ab

Š €X

a2 +

X

ab

Š

Š2 €X Š €X Š €X Š2 a2 + 3 ab a2 + 2 ab X X X ≥ a4 + 3 a b(a2 + b2 ) + 4 a2 b2

€X

X X (b − c)2 (2a + b + c)2 = (b − c)2 [4a2 + 4a(b + c) + (b + c)2 ] X X X =4 a2 (b − c)2 + 4 a(b − c)(b2 − c 2 ) + (b2 − c 2 )2 X X X ≤8 a2 b2 + 4 a(b3 + c 3 ) + 2 a4 .

Thus, it suffices to show that X X X X X X a4 + 3 a b(a2 + b2 ) + 4 a2 b2 ≥ 4 a2 b2 + 2 a(b3 + c 3 ) + a4 , which is equivalent to the obvious inequality X a b(a2 + b2 ) ≥ 0.

282

Vasile Cîrtoaje

P 2.7. If a, b, c are nonnegative real numbers, then p p p p p a2 + 2bc + b2 + 2ca + c 2 + 2a b ≤ a2 + b2 + c 2 + 2 a b + bc + ca. (Vasile Cîrtoaje and Nguyen Van Quy, 1989) Solution (by Nguyen Van Quy). Let p X = a2 + b2 + c 2 ,

Y=

p

a b + bc + ca.

Consider the nontrivial case when no two of a, b, c are zero (Y 6= 0) and write the inequality as X€ Š p X − a2 + 2bc ≥ 2(X − Y ), P X (b − c)2 (b − c)2 ≥ . p X +Y X + a2 + 2bc By the Cauchy-Schwarz inequality, we have X

P 2 (b − c)2 (b − c)2 ≥P
. p p X + a2 + 2bc (b − c)2 X + a2 + 2bc

Therefore, it suffices to show that P (b − c)2 1 ,
≥ p P 2 2 X + Y (b − c) X + a + 2bc which is equivalent to X € Š p (b − c)2 Y − a2 + 2bc ≥ 0. From €

Y−

p

a2 + 2bc

Š2

≥ 0.

we get Y− Thus,

p

a2 + 2bc ≥

Y 2 − (a2 + 2bc) (a − b)(c − a) = . 2Y 2Y

X € Š X (b − c)2 (a − b)(c − a) p (b − c)2 Y − a2 + 2bc ≥ 2Y X (a − b)(b − c)(c − a) = (b − c) = 0. 2Y The equality holds for a = b, or b = c, or c = a.

Symmetric Nonrational Inequalities

283

P 2.8. If a, b, c are nonnegative real numbers, then p

1 a2 + 2bc

+p

1 b2 + 2ca

+p

1 c 2 + 2a b

≥p

1 a2 + b2 + c 2

+p

2 a b + bc + ca

.

(Vasile Cîrtoaje, 1989) Solution . Let X=

p

a2 + b2 + c 2 ,

Y=

p

a b + bc + ca.

Consider the nontrivial case when Y > 0 and write the inequality as ‹  ‹ X 1 1 1 1 − − ≥2 , p Y X a2 + 2bc X P X (b − c)2 (b − c)2 . ≥
p p Y (X + Y ) a2 + 2bc X + a2 + 2bc By the Cauchy-Schwarz inequality, we have X

P 2 (b − c)2 (b − c)2
≥P
. p p p p a2 + 2bc X + a2 + 2bc (b − c)2 a2 + 2bc X + a2 + 2bc

Therefore, it suffices to show that P (b − c)2 1 ,
≥ p p P 2 2 2 Y (X + Y) (b − c) a + 2bc X + a + 2bc which is equivalent to X p (b − c)2 [X Y − X a2 + 2bc + (a − b)(c − a)] ≥ 0. Since

X X (b − c)2 (a − b)(c − a) = (a − b)(b − c)(c − a) (b − c) = 0,

the inequality becomes X € Š p (b − c)2 X Y − a2 + 2bc ≥ 0, X € Š p (b − c)2 Y − a2 + 2bc ≥ 0. We have proved this inequality at the preceding problem P 2.7. The equality holds for a = b, or b = c, or c = a.

284

Vasile Cîrtoaje

P 2.9. If a, b, c are positive real numbers, then p p p p p 2a2 + bc + 2b2 + ca + 2c 2 + a b ≤ 2 a2 + b2 + c 2 + a b + bc + ca. Solution. We will apply Lemma below for X = 2a2 + bc, Y = 2b2 + ca, Z = 2c 2 + a b and A = B = a2 + b2 + c 2 , C = a2 + b2 + c 2 . We have X + Y + Z = A+ B + C and A = B ≥ C. Without loss of generality, assume that a ≥ b ≥ c. By Lemma, it suffices to show that max{X , Y, Z} ≥ A,

min{X , Y, Z} ≤ C.

Indeed, we have max{X , Y, Z} − A ≥ X − A = (a2 − b2 ) + c(b − c) ≥ 0, min{X , Y, Z} − C ≤ Z − C = c(2c − a − b) ≤ 0. Equality holds for a = b = c. Lemma. If X , Y, Z and A, B, C are positive real numbers such that X + Y + Z = A + B + C, max{X , Y, Z} ≥ max{A, B, C}, then

p

X+

p

Y+

p

Z≤

min{X , Y, Z} ≤ min{A, B, C}, p p p A + B + C.

Proof. On the assumption that X ≥ Y ≥ Z and A ≥ B ≥ C, we have X ≥ A,

Z ≤ C,

and hence p p p p p p p p p p p p X + Y + Z − A− B − C = ( X − A ) + ( Y − B ) + ( Z − C )

Symmetric Nonrational Inequalities

285

X −A Y −B Z −C X −A Y −B Z −C p + p + p ≤ p + p + p 2 C 2 C 2 A 2 B 2 B 2 B  ‹ 1 C −Z Z −C 1 ≤ 0. = p + p = (C − Z) p − p 2 C 2 B 2 B 2 C

≤

Remark. This Lemma is a particular case of Karamata’s inequality.

P 2.10. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k = then XÆ p a(a + k b)(a + kc) ≤ 3 3.

p 3 − 1,

Solution. By the Cauchy-Schwarz inequality, we have XÆ

a(a + k b)(a + kc) ≤

r€X Š ”X — a (a + k b)(a + kc) .

Thus, it suffices to show that rX (a + k b)(a + kc) ≤ a + b + c, which is an identity. The equality holds for a = b = c = 1, and also for a = 3 and b = c = 0 (or any cyclic permutation).

P 2.11. If a, b, c are nonnegative real numbers such that a + b + c = 3, then XÆ a(2a + b)(2a + c) ≥ 9.

Solution. Write the inequality as follows X ”Æ — Æ a(2a + b)(2a + c) − a 3(a + b + c) ≥ 0, X (a − b)(a − c)Ea ≥ 0, where

p Ea = p

a

(2a + b)(2a + c) +

p

3a(a + b + c)

.

286

Vasile Cîrtoaje

Assume that a ≥ b ≥ c. Since (c − a)(c − b)Ec ≥ 0, it suffices to show that (a − c)Ea ≥ (b − c)E b , which is equivalent to Æ Æ Æ (a − b) 3a b(a + b + c) + (a − c) a(2b + c)(2b + a) ≥ (b − c) b(2a + b)(2a + c). This is true if Æ Æ (a − c) a(2b + c)(2b + a) ≥ (b − c) b(2a + b)(2a + c). For the non-trivial case b > c, we have p a a a−c ≥ ≥p . b−c b b Therefore, it is enough to show that a2 (2b + c)(2b + a) ≥ b2 (2a + b)(2a + c). Write this inequality as a2 (2a b + 2bc + ca) ≥ b2 (2a b + bc + 2ca). It is true if a(2a b + 2bc + ca) ≥ b(2a b + bc + 2ca). Indeed, a(2a b + 2bc + ca) − b(2a b + bc + 2ca) = (a − b)(2a b + bc + ca) ≥ 0. The equality holds for a = b = c = 1, and also for a = 0 and b = c = 3/2 (or any cyclic permutation).

P 2.12. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that Æ

b2 + c 2 + a(b + c) +

Æ

c 2 + a2 + b(c + a) +

Æ

a2 + b2 + c(a + b) ≥ 6.

Symmetric Nonrational Inequalities

287

Solution. Denote A = b2 + c 2 + a(b + c),

B = c 2 + a2 + b(c + a),

C = a2 + b2 + c(a + b).

First Solution. Write the inequality in the homogeneous form p p p A + B + C ≥ 2(a + b + c). By squaring, the inequality becomes Xp X X 2 BC ≥ 2 a2 + 6 bc, X X p p
2 (a − b)2 ≥ B− C , X (b − c)2 Sa ≥ 0, where

(b + c − a)2 Sa = 1 − p . p ( B + C)2

Since Sa ≥ 1 −

a(a + 3b + 3c) (b + c − a)2 = ≥ 0, S b ≥ 0, Sc ≥ 0, B+C B+C

the conclusion follows. The equality holds for a = b = c = 1, and also for a = 3 and b = c = 0 (or any cyclic permutation). Second Solution. Write the original inequality as follows Xp ( A − b − c) ≥ 0, X c(a − b) + b(a − c) ≥ 0, p A+ b + c X c(a − b) X c(b − a) + ≥ 0, p p A+ b + c B+c+a X c(a − b)[a − b − (pA − pB)] ≥ 0. p p ( A + b + c)( B + c + a) It suffices to show that p p (a − b)[a − b + ( B − A)] ≥ 0. Indeed,

 ‹ p p a+b−c 2 (a − b)[a − b + ( B − A)] = (a − b) 1 + p ≥ 0, p B+ A

288

Vasile Cîrtoaje

because, for the nontrivial case a + b − c < 0, we have a+b−c a+b−c > 0. 1+ p p >1+ c+c B+ A Open Generalization. Let a, b, c be nonnegative real numbers. If 0 < k ≤ XÆ

16 , then 9

(b + c)2 + k(a b − 2bc + ca) ≥ 2(a + b + c).

16 , then the equality holds for a = b = c = 1, for a = 0 and b = c (or 9 any cyclic permutation), and for b = c = 0 (or any cyclic permutation).

Notice that if k =

P 2.13. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that (a)

p

(b)

p p p p 3a2 + a bc + 3b2 + a bc + 3c 2 + a bc ≥ 3 3 + a bc.

a(3a2 + a bc) +

p

b(3b2 + a bc) +

p

c(3c 2 + a bc) ≥ 6;

(Lorian Saceanu, 2015) Solution. (a) Write the inequality in the homogeneous form X Æ 3 a (a + b)(a + c) ≥ 2(a + b + c)2 . First Solution. Write the inequality as X

a2 −

X

ab ≥

Š2 p 3 X €p a+b− a+c , a 2

X X (b − c)2 ≥ 3 p

a(b − c)2
2 , p a+b+ a+c

X (b − c)2 Sa ≥ 0, where Sa = 1 − p Since Sa ≥ 1 − p

3a
2 . p a+b+ a+c

3a p
2 > 0, S b > 0, Sc > 0, a+ a

the inequality is true. The equality holds for a = b = c = 1.

Symmetric Nonrational Inequalities

289

Second Solution. By Hölder’s inequality, we have —2 ”X Æ a (a + b)(a + c) ≥ P

P ( a)3 27 =P . a a (a + b)(a + c) (a + b)(a + c)

Therefore, it suffices to show that X

a 3 ≤ . (a + b)(a + c) 4

This inequality has the homogeneous form X

a 9 ≤ , (a + b)(a + c) 4(a + b + c)

which is equivalent to the obvious inequality X

a(b − c)2 ≥ 0.

(b) By squaring, the inequality becomes 3

X

a2 + 2

XÆ

(3b2 + a bc)(3c 2 + a bc) ≥ 27 + 6a bc.

By the Cauchy-Schwarz inequality, we have Æ

(3b2 + a bc)(3c 2 + a bc) ≥ 3bc + a bc.

Therefore, it suffices to show that X X a2 + 6 bc + 6a bc ≥ 27 + 6a bc, 3 which is an identity. The equality holds for a = b = c = 1, and also for a = 0, or b = 0, or c = 0.

P 2.14. Let a, b, c be positive real numbers such that a b + bc + ca = 3. Prove that a

Æ

(a + 2b)(a + 2c) + b

Æ

(b + 2c)(b + 2a) + c

Æ

(c + 2a)(c + 2b) ≥ 9.

290

Vasile Cîrtoaje

First Solution. Write the inequality as follows: X Æ a (a + 2b)(a + 2c) ≥ 3(a b + bc + ca), Š2 p 1 X €p a a + 2b − a + 2c , 2 X X a(b − c)2 (b − c)2 ≥ 4
2 , p p a + 2b + a + 2c X (b − c)2 Sa ≥ 0,

X

a2 −

X

ab ≥

where Sa = 1 − p Since Sa > 1 − p

4a
2 . p a + 2b + a + 2c

4a p
2 = 0, S b > 0, Sc > 0, a+ a

the inequality is true. The equality holds for a = b = c = 1. Second Solution. We use the AM-GM inequality to get X 2a(a + 2b)(a + 2c) X Æ X 2a(a + 2b)(a + 2c) ≥ a (a + 2b)(a + 2c) = p (a + 2b) + (a + 2c) 2 (a + 2b)(a + 2c) X 1 = a(a + 2b)(a + 2c). a+b+c Thus, it suffices to show that X a(a + 2b)(a + 2c) ≥ 9(a + b + c). Write this inequality in the homogeneous form X a(a + 2b)(a + 2c) ≥ 3(a + b + c)(a b + bc + ca), which is equivalent to Schur’s inequality of degree three a3 + b3 + c 3 + 3a bc ≥ a b(a + b) + bc(b + c) + ca(c + a).

P 2.15. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that Æ Æ Æ p a + (b − c)2 + b + (c − a)2 + c + (a − b)2 ≥ 3. (Phan Thanh Nam, 2007)

Symmetric Nonrational Inequalities

291

Solution. By squaring, the inequality becomes XÆ [a + (b − c)2 ][b + (a − c)2 ] ≥ 3(a b + bc + ca). Applying the Cauchy-Schwarz inequality, it suffices to show that X Xp ab + (b − c)(a − c) ≥ 3(a b + bc + ca). This is equivalent to the homogeneous inequality €X Š €X p Š X ab + a2 ≥ 4(a b + bc + ca). a Making the substitution x = €X

p p a, y = b, z = c, the inequality turns into X Š €X Š X x2 xy + x4 ≥ 4 x 2 y 2, p

which is equivalent to X X X X x4 + x y(x 2 + y 2 ) + x yz x ≥4 x 2 y 2. Since 4

X

x2 y2 ≤ 2

X

x y(x 2 + y 2 ),

it suffices to show that X

x 4 + x yz

X

x≥

X

x y(x 2 + y 2 ),

which is just Schur’s inequality of degree four. The equality holds for a = b = c = and for a = 0 and b = c =

1 (or any cyclic permutation). 2

1 , 3

P 2.16. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that v v v t a(b + c) t b(c + a) t c(a + b) + + ≥ 2. a2 + bc b2 + ca c2 + a b (Vasile Cîrtoaje, 2006) Solution. Using the AM-GM inequality gives v t a(b + c) a(b + c) 2a(b + c) 2a(b + c) =p ≥ 2 = . 2 a2 + bc (a + bc) + (a b + ac) (a + b)(a + c) (a + bc)(ab + ac)

292

Vasile Cîrtoaje

Therefore, it suffices to show that b(c + a) c(a + b) a(b + c) + + ≥ 1, (a + b)(a + c) (b + c)(b + a) (c + a)(c + b) which is equivalent to a(b + c)2 + b(c + a)2 + c(a + b)2 ≥ (a + b)(b + c)(c + a), 4a bc ≥ 0. The equality holds for a = 0 and b = c (or any cyclic permutation).

P 2.17. Let a, b, c be positive real numbers such that a bc = 1. Prove that p 3

1 a2

+ 25a + 1

+p 3

1 b2

+ 25b + 1

+p 3

1 c2

+ 25c + 1

≥ 1.

Solution. Replacing a, b, c by a3 , b3 , c 3 , respectively, we need to show that a bc = 1 yields 1 1 1 +p +p ≥ 1. p 3 3 3 a6 + 25a3 + 1 b6 + 25b3 + 1 c 6 + 25c 3 + 1 We first show that 1 1 ≥ 2 . p 3 6 3 a +a+1 a + 25a + 1 This is equivalent to (a2 + a + 1)3 ≥ a6 + 25a3 + 1, which is true since (a2 + a + 1)3 − (a6 + 25a3 + 1) = 3a(a − 1)2 (a2 + 4a + 1) ≥ 0. Therefore, it suffices to prove that 1 1 1 + + ≥ 1. a2 + a + 1 b2 + b + 1 b2 + b + 1 Putting a=

yz , x2

b=

we need to show that X

zx , y2

c=

xy , z2

x, y, z > 0

x4 ≥ 1. x 4 + x 2 yz + y 2 z 2

Symmetric Nonrational Inequalities

293

Indeed, the Cauchy-Schwarz inequality gives X

P 2
2 P 4 P x x + 2 x2 y2 x4 P P =P ≥P x 4 + x 2 yz + y 2 z 2 (x 4 + x 2 yz + y 2 z 2 ) x 4 + x yz x + x 2 y 2 P 2 2 P x y − x yz x P P =1+ P ≥ 1. x 4 + x yz x + x 2 y 2

The equality holds for a = b = c = 1.

P 2.18. If a, b, c are nonnegative real numbers, then p

a2 + bc +

p

b2 + ca +

p

c2 + a b ≤

3 (a + b + c). 2 (Pham Kim Hung, 2005)

Solution. Without loss of generality, assume that a ≥ b ≥ c. Since the equality occurs for a = b and c = 0, we use the inequalities p c a2 + bc ≤ a + 2 and

p

b2 + ca +

p

c2 + a b ≤

Æ

2(b2 + ca) + 2(c 2 + a b).

Thus, it suffices to prove that Æ

2(b2 + ca) + 2(c 2 + a b) ≤

a + 3b + 2c . 2

By squaring, this inequality becomes a2 + b2 − 4c 2 − 2a b + 12bc − 4ca ≥ 0, (a − b − 2c)2 + 8c(b − c) ≥ 0. The equality holds for a = b and c = 0 (or any cyclic permutation).

P 2.19. If a, b, c are nonnegative real numbers, then p p p p a2 + 9bc + b2 + 9ca + c 2 + 9a b ≥ 5 a b + bc + ca. (Vasile Cîrtoaje, 2012)

294

Vasile Cîrtoaje

First Solution (by Nguyen Van Quy). Assume that c = min{a, b, c}. Since the equality occurs for a = b and c = 0, we use the inequality p p c 2 + 9a b ≥ 3 a b. Also, by Minkowski’s inequality, we have p

a2 + 9bc +

p

s b2 + 9ca ≥

(a + b)2 + 9c

€p

a+

p Š2 b .

Therefore, it suffices to show that s €p p Š2 p p (a + b)2 + 9c a + b ≥ 5 a b + bc + ca − 3 a b. By squaring, this inequality becomes p Æ (a + b)2 + 18c a b + 30 a b(a b + bc + ca) ≥ 34a b + 16c(a + b). Since c(a + b) a b(a b + bc + ca) − a b + 3 •

˜2

=

c(a + b)(3a b − ac − bc) ≥ 0, 9

p it suffices to show that f (c) ≥ 0 for 0 ≤ c ≤ a b, where p f (c) = (a + b)2 + 18c a b + [30a b + 10c(a + b)] − 34a b − 16c(a + b) p = (a + b)2 − 4a b + 6c(3 a b − a − b). p Since f (c) is a linear function, we only need to prove that f (0) ≥ 0 and f ( a b) ≥ 0. We have f (0) = (a − b)2 ≥ 0, p p p f ( a b) = (a + b)2 + 14a b − 6(a + b) a b ≥ (a + b)2 + 9a b − 6(a + b) a b € p Š2 = a + b − 3 a b ≥ 0. The equality holds for a = b and c = 0 (or any cyclic permutation). Second Solution. Assume that c = min{a, b, c}. By squaring, the inequality becomes X XÆ X a2 + 2 (a2 + 9bc)(b2 + 9ca) ≥ 16 a b, X

a2 + 2

Æ

(a2 + 9bc)(b2 + 9ca) + 2

p

c 2 + 9a b

€p

a2 + 9bc +

p

The Cauchy-Schwarz inequality gives p Æ (a2 + 9bc)(b2 + 9ca) ≥ a b + 9c a b.

X Š b2 + 9ca ≥ 16 a b.

Symmetric Nonrational Inequalities

295

In addition, Minkowski’s inequality gives s €p p Š2 p p a2 + 9bc + b2 + 9ca ≥ (a + b)2 + 9c a + b ≥ a + b + 4c, hence

p

c 2 + 9a b

€p

a2 + 9bc +

p

Š p b2 + 9ca ≥ 3 a b (a + b + 4c).

Therefore, it suffices to show that f (c) ≥ 0, where X p p f (c) = a2 + b2 + 2(a b + 9c a b) + 6 a b (a + b + 4c) − 16 ab p p = a2 + b2 − 14a b + 6(a + b) a b + c[42 a b − 16(a + b)]. Since fp(c) is a linear function and 0 ≤ c ≤ and f ( a b) ≥ 0. We have f (0) = (a − b)2 + 6

p

p

a b, it is sufficient to show that f (0) ≥ 0

ab

€p

a−

p Š2 b ≥0

and p p p f ( a b) = a2 + b2 + 28a b − 10(a + b) a b ≥ (a + b)2 + 25a b − 10(a + b) a b € p Š2 = a + b − 5 a b ≥ 0.

P 2.20. If a, b, c are nonnegative real numbers, then XÆ (a2 + 4bc)(b2 + 4ca) ≥ 5(a b + ac + bc). (Vasile Cîrtoaje, 2012) First Solution (by Michael Rozenberg). Assume that a ≥ b ≥ c. For b = c = 0, the inequality is trivial. Assume further that b > 0 and write the inequality as follows: X Æ [ (b2 + 4ca)(c 2 + 4a b) − bc − 2a(b + c)] ≥ 0, X (b2 + 4ca)(c 2 + 4a b) − (bc + 2a b + 2ac)2 ≥ 0, p (b2 + 4ca)(c 2 + 4a b) + bc + 2a(b + c) X (b − c)2 Sa ≥ 0, where Sa =

Æ a(b + c − a) , A = (b2 + 4ca)(c 2 + 4a b) + bc + 2a(b + c), A

296

Vasile Cîrtoaje Sb =

b(c + a − b) , B

B=

Æ

(c 2 + 4a b)(a2 + 4bc) + ca + 2b(c + a),

Æ c(a + b − c) , C = (a2 + 4bc)(b2 + 4ac) + a b + 2c(a + b). C Since S b ≥ 0 and Sc ≥ 0, we have Sc =

X a2 (b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b b  ‹ bSa aS b a = (b − c)2 + . b a b Thus, it suffices to prove that bSa aS b + ≥ 0, a b which is equivalent to

b(b + c − a) a(c + a − b) + ≥ 0. A B

Since b(b + c − a) a(c + a − b) b(b − a) a(a − b) (a − b)(aA − bB) + ≥ + = , A B A B AB it is enough to show that aA − bB ≥ 0. Indeed, aA − bB =

p

” p — p c 2 + 4a b a b2 + 4ca − b a2 + 4bc + 2(a − b)(a b + bc + ca)

p 4c(a3 − b3 ) c 2 + 4a b = p + 2(a − b)(a b + bc + ca) ≥ 0. p a b2 + 4ca + b a2 + 4bc The equality holds for a = b = c, and also for a = b and c = 0 (or any cyclic permutation). Second Solution (by Nguyen Van Quy). Write the inequality as €p p

a2 + 4bc +

a2 + 4bc +

p

p

b2 + 4ca +

b2 + 4ca +

p

p

c 2 + 4a b

Š2

c 2 + 4a b ≥

≥ a2 + b2 + c 2 + 14(a b + bc + ca), Æ

a2 + b2 + c 2 + 14(a b + bc + ca).

Assume that c = min{a, b, c}. For t = 2c, the inequality (b) in Lemma below becomes p

a2 + 4bc +

p

b2 + 4ca ≥

Æ

(a + b)2 + 8(a + b)c.

Symmetric Nonrational Inequalities

297

Thus, it suffices to show that p Æ Æ (a + b)2 + 8(a + b)c + c 2 + 4a b ≥ a2 + b2 + c 2 + 14(a b + bc + ca). By squaring, this inequality becomes Æ [(a + b)2 + 8(a + b)c] (c 2 + 4a b) ≥ 4a b + 3(a + b)c, 2(a + b)c 3 − 2(a + b)2 c 2 + 2a b(a + b)c + a b(a + b)2 − 4a2 b2 ≥ 0, 2(a + b)(a − c)(b − c)c + a b(a − b)2 ≥ 0. Lemma. Let a,b and t be nonnegative numbers such that t ≤ 2(a + b). Then, p

(a) (b)

p

(a2 + 2bt)(b2 + 2at) ≥ a b + (a + b)t;

a2 + 2bt +

p

b2 + 2at ≥

p

(a + b)2 + 4(a + b)t.

Proof. (a) By squaring, the inequality becomes (a − b)2 t[2(a + b) − t] ≥ 0, which is clearly true. (b) By squaring, this inequality turns into the inequality in (a).

P 2.21. If a, b, c are nonnegative real numbers, then XÆ (a2 + 9bc)(b2 + 9ca) ≥ 7(a b + ac + bc). (Vasile Cîrtoaje, 2012) Solution (by Nguyen Van Quy). We see that the equality holds for a = b and c = 0. Without loss of generality, assume that c = min{a, b, c}. For t = 4c, the inequality (a) in Lemma from the preceding P 2.20 becomes Æ (a2 + 8bc)(b2 + 8ca) ≥ a b + 4(a + b)c. Thus, we have Æ

(a2 + 9bc)(b2 + 9ca) ≥ a b + 4(a + b)c,

298

Vasile Cîrtoaje

and also, p

c 2 + 9a b

€p

a2 + 9bc +

p

Š p Æ 4 b2 + 9ca ≥ 3 a b · 2 (a2 + 9bc)(b2 + 9ca)

p Æ Æ ≥ 6 a b · a b + 4(a + b)c = 3 4a2 b2 + 16a bc(a + b) Æ ≥ 3 4a2 b2 + 4a bc(a + b) + c 2 (a + b)2 = 3(2a b + bc + ca). Therefore, XÆ

(a2 + 9bc)(b2 + 9ca) ≥ (a b + 4bc + 4ca) + 3(2a b + bc + ca) = 7(a b + bc + ca).

The equality holds for a = b and c = 0 (or any cyclic permutation).

P 2.22. If a, b, c are nonnegative real numbers, then Æ Æ Æ (a2 + b2 )(b2 + c 2 ) + (b2 + c 2 )(c 2 + a2 ) + (c 2 + a2 )(a2 + b2 ) ≤ (a + b + c)2 . (Vasile Cîrtoaje, 2007) Solution. Without loss of generality, assume that a = min{a, b, c}. Let us denote y=

a + b, 2

z=

a + c. 2

Since a2 + b2 ≤ y 2 ,

b2 + c 2 ≤ y 2 + z 2 ,

c 2 + a2 ≤ z 2 ,

it suffices to prove that Æ yz + ( y + z) y 2 + z 2 ≤ ( y + z)2 . This is true since Æ y 2 + yz + z 2 − ( y + z) y 2 + z 2 =

y 2z2

≥ 0. p y 2 + yz + z 2 + ( y + z) y 2 + z 2

The equality holds for a = b = 0 (or any cyclic permutation).

P 2.23. If a, b, c are nonnegative real numbers, then XÆ (a2 + a b + b2 )(b2 + bc + c 2 ) ≥ (a + b + c)2 .

Symmetric Nonrational Inequalities

299

Solution. By the Cauchy-Schwarz inequality, we have    ‹ c 2 3c 2 b 2 3b2  2 2 2 2 + a+ + (a + a b + b )(a + ac + c ) = a + 2 4 2 4  ‹ c  3bc b a(b + c) a+ ≥ a+ + = a2 + + bc. 2 2 4 2 Then, ˜ XÆ X• a(b + c) 2 2 2 2 2 (a + a b + b )(a + ac + c ) ≥ a + + bc = (a + b + c)2 . 2 The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation).

P 2.24. If a, b, c are nonnegative real numbers, then XÆ (a2 + 7a b + b2 )(b2 + 7bc + c 2 ) ≥ 7(a b + ac + bc). (Vasile Cîrtoaje, 2012) First Solution. Without loss of generality, assume that c = min{a, b, c}. We see that the equality holds for a = b and c = 0. Since Æ (a2 + 7ac + c 2 )(b2 + 7bc + c 2 ) ≥ (a + 2c)(b + 2c) ≥ a b + 2c(a + b), it suffices to show that €p Š p p a2 + 7a b + b2 a2 + 7ac + b2 + 7bc ≥ 6a b + 5c(a + b). By Minkowski’s inequality, we have p

a2 + 7ac + ≥

p

v t

s b2 + 7bc ≥

(a + b)2 + 7c

(a + b)2 + 7c(a + b) +

€p

a+

p Š2 b

28a bc . a+b

Therefore, it suffices to show that ˜ • 28a bc 2 2 2 (a + 7a b + b ) (a + b) + 7c(a + b) + ≥ (6a b + 5bc + 5ca)2 . a+b Due to homogeneity, we may assume that a + b = 1. Let us denote d = a b, d ≤ Since c≤

2a b = 2d, a+b

1 . 4

300

Vasile Cîrtoaje

we need to show that f (c) ≥ 0 for 0 ≤ c ≤ 2d ≤

1 , where 2

f (c) = (1 + 5d)(1 + 7c + 28cd) − (6d + 5c)2 . Since f (c) is concave, it suffices to show that f (0) ≥ 0 and f (2d) ≥ 0. Indeed, f (0) = 1 + 5d − 36d 2 = (1 − 4d)(1 + 9d) ≥ 0 and f (2d) = (1 + 5d)(1 + 14d + 56d 2 ) − 256d 2 ≥ (1 + 4d)(1 + 14d + 56d 2 ) − 256d 2 = (1 − 4d)(1 + 22d − 56d 2 ) ≥ d(1 − 4d)(22 − 56d) ≥ 0. The equality holds for a = b and c = 0 (or any cyclic permutation). Second Solution. We will use the inequality Æ

x 2 + 7x y + y 2 ≥ x + y +

2x y , x+y

x, y ≥ 0,

which, by squaring, reduces to x y(x − y)2 ≥ 0. We have XÆ

(a2

+ 7a b +

≥ Since

X

b2 )(a2

a2 + 3

+ 7ac

X

ab +

+ c2)

≥

X

2a b a+b+ a+b

‹

2ac a+c+ a+c

X 2a2 b X 2a2 c X 2a bc + + . a+b a+c a+b

X 2a2 b X 2a2 c X 2a2 b X 2b2 a X + = + =2 ab a+b a+c a+b b+a

and

X 2a bc

18a bc 9a bc ≥P , = a+b a+b+c (a + b)

it suffices to show that X

a2 +

X 9a bc ≥2 a b, a+b+c

which is just Schur’s inequality of degree three.

‹

Symmetric Nonrational Inequalities

301

P 2.25. If a, b, c are nonnegative real numbers, then v ‹ ‹ X t 7 7 13 a2 + a b + b2 b2 + bc + c 2 ≤ (a + b + c)2 . 9 9 12 (Vasile Cîrtoaje, 2012) Solution (by Nguyen Van Quy). Without loss of generality, assume that c = min{a, b, c}. It is easy to see that the equality holds for a = b = 1 and c = 0. By the AM-GM inequality, the following inequality holds for any k > 0: ‚v Œ v v t t t 7 7 7 2 2 2 2 2 2 12 a + a b + b a + ac + c + b + bc + c ≤ 9 9 9 ‚v Œ2 v  ‹ t t 36 2 7 7 7 ≤ a + a b + b2 + k a2 + ac + c 2 + b2 + bc + c 2 . k 9 9 9 We can use this inequality to prove the original inequality only if ‚v Œ2 v  ‹ t t 7 7 36 2 7 2 a + ab + b = k a2 + ac + c 2 + b2 + bc + c 2 k 9 9 9 for a = b = 1 and c = 0. This necessary condition if satisfied for k = 5. Therefore, it suffices to show that v ‹ ‹ ‹  t 7 7 36 2 7 2 2 2 2 2 a + ac + c b + bc + c + a + ab + b 12 9 9 5 9 ‚v Œ2 v t t 7 7 +5 a2 + ac + c 2 + b2 + bc + c 2 ≤ 13(a + b + c)2 . 9 9 which is equivalent to v ‹ ‹ t 7 7 4(a + b)2 + 94a b 199c(a + b) 2 2 2 2 b + bc + c ≤ + 3c 2 + . 22 a + ac + c 9 9 5 9 Since 2

v t

7 a2 + ac + c 2 9

‹

v ‹ ‹ ‹ t 16 16 7 2 2 2 2 b + bc + c ≤ 2 a + ac b + bc 9 9 9 v  ‹  ‹ t 16 16 =2 a b+ c ·b a+ c 9 9  ‹  ‹ 16 16 ≤a b+ c +b a+ c 9 9 16c(a + b) = 2a b + , 9

302

Vasile Cîrtoaje

we only need to prove that ˜ • 4(a2 + b2 ) + 102a b 199c(a + b) 8c(a + b) ≤ + 3c 2 + . 22 a b + 9 5 9 This reduces to the obvious inequality 4(a − b)2 23c(a + b) + + 3c 2 ≥ 0. 5 9 Thus, the proof is completed. The equality holds for a = b and c = 0 (or any cyclic permutation).

P 2.26. If a, b, c are nonnegative real numbers, then v ‹ ‹ X t 1 1 61 2 2 2 2 a + ab + b b + bc + c ≤ (a + b + c)2 . 3 3 60 (Vasile Cîrtoaje, 2012) Solution (by Nguyen Van Quy). Without loss of generality, assume that c = min{a, b, c}. It is easy to see that the equality holds for c = 0 and 11(a2 + b2 ) = 38a b. By the AM-GM inequality, the following inequality holds for any k > 0: v v ‚v Œ t t t 1 1 1 60 a2 + a b + b2 a2 + ac + c 2 + b2 + bc + c 2 ≤ 3 3 3 v ‚v Œ2 ‹  t t 1 1 36 2 1 2 2 2 2 2 a + a b + b + 25k a + ac + c + b + bc + c . ≤ k 3 3 3 We can use this inequality to prove the original inequality only if the equality v Œ2 ‚v  ‹ t t 36 2 1 1 1 2 a + a b + b = 25k a2 + ac + c 2 + b2 + bc + c 2 k 3 3 3 holds for c = 0 and 11(a2 + b2 ) = 38a b. This necessary condition if satisfied for k = 1. Therefore, it suffices to show that v ‹ ‹  ‹ t 1 1 1 2 2 2 2 2 2 60 a + ab + b b + bc + c + 36 a + a b + b 3 3 3 +25

‚v t

v Œ2 t 1 1 a2 + ac + c 2 + b2 + bc + c 2 ≤ 61(a + b + c)2 , 3 3

Symmetric Nonrational Inequalities

303

which is equivalent to v ‹ ‹ t 1 31c(a + b) 1 2 2 2 2 b + bc + c ≤ 10a b + c 2 + . 10 a + ac + c 3 3 3 Since 2

v t

1 a2 + ac + c 2 3

‹

v ‹ ‹ ‹ t 1 4 4 b2 + bc + c 2 ≤ 2 a2 + ac b2 + bc 3 3 3 v  ‹  ‹ t 4 4 =2 a b+ c ·b a+ c 3 3  ‹  ‹ 4 4 ≤a b+ c +b a+ c 3 3 4c(a + b) = 2a b + , 3

we only need to prove that • ˜ 2c(a + b) 31c(a + b) 10 a b + ≤ 10a b + c 2 + . 3 3 This reduces to the obvious inequality 3c 2 + 11c(a + b) ≥ 0. Thus, the proof is completed. The equality holds for 11(a2 + b2 ) = 38a b and c = 0 (or any cyclic permutation).

P 2.27. If a, b, c are nonnegative real numbers, then a b c +p +p ≥ 1. p 2 2 2 2 2 4b + bc + 4c 4c + ca + 4a 4a + a b + 4b2 (Pham Kim Hung, 2006) Solution. By Hölder’s inequality, we have P
3 P 3 P ‹2 X a a + 3 a b(a + b) + 6a bc a P ≥P = . p a(4b2 + bc + 4c 2 ) 4 a b(a + b) + 3a bc 4b2 + bc + 4c 2 Thus, it suffices to show that X

a3 + 3a bc ≥

X

a b(a + b),

which is Schur’s inequality of degree three. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

304

Vasile Cîrtoaje

P 2.28. If a, b, c are nonnegative real numbers, then p

a b2

+ bc

+ c2

+p

b c2

+ ca + a2

+p

c a2

+ ab +

b2

≥p

a+b+c a b + bc + ca

.

Solution. By Hölder’s inequality, we have X

p

‹2

a b2 + bc + c 2

P
2 a = P , ≥P a(b2 + bc + c 2 ) ab P
3 a

from which the desired inequality follows. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 2.29. If a, b, c are nonnegative real numbers, then p

a a2

+ 2bc

+p

b b2

+ 2ca

+p

c c2

+ 2a b

≤p

a+b+c a b + bc + ca

.

(Ho Phu Thai, 2007) Solution. Without loss of generality, assume that a ≥ b ≥ c. First Solution. Since p

c c2

+ 2a b

≤p

c a b + bc + ca

,

it suffices to show that p

a a2

+ 2bc

+p

b b2

+ 2ca

≤p

a+b a b + bc + ca

,

which is equivalent to p p p p a( a2 + 2bc − a b + bc + ca) b( a b + bc + ca − b2 + 2ca) ≥ . p p a2 + 2bc b2 + 2ca Since

p

a2 + 2bc −

and p

p

a a2 + 2bc

a b + bc + ca ≥ 0

≥p

b b2 + 2ca

,

it suffices to show that p p p p a2 + 2bc − a b + bc + ca ≥ a b + bc + ca − b2 + 2ca,

Symmetric Nonrational Inequalities which is equivalent to p

a2 + 2bc +

305

p

b2 + 2ca ≥ 2

p

a b + bc + ca.

Using the AM-GM inequality, it suffices to show that (a2 + 2bc)(b2 + 2ca) ≥ (a b + bc + ca)2 , which is equivalent to the obvious inequality c(a − b)2 (2a + 2b − c) ≥ 0. The equality holds for a = b = c, and also for a = b and c = 0 (or any cyclic permutation). Second Solution. By the Cauchy-Schwarz inequality, we have X

p

‹2

a a2 + 2bc

≤

€X Š X a

 a . a2 + 2bc

Thus, it suffices to prove that X a2

a+b+c a ≤ . + 2bc a b + bc + ca

This is equivalent to X  a

1 1 − 2 a b + bc + ca a + 2bc X a(a − b)(a − c) a2 + 2bc

‹

≥ 0,

≥ 0.

Assuming that a ≥ b ≥ c, we get X a(a − b)(a − c) a2 + 2bc =

≥

a(a − b)(a − c) b(b − c)(b − a) + a2 + 2bc b2 + 2ca

c(a − b)2 [2a(a − c) + 2b(b − c) + 3a b] ≥ 0. (a2 + 2bc)(b2 + 2ca)

P 2.30. If a, b, c are nonnegative real numbers, then p p p a3 + b3 + c 3 + 3a bc ≥ a2 a2 + 3bc + b2 b2 + 3ca + c 2 c 2 + 3a b. (Vo Quoc Ba Can, 2008)

306

Vasile Cîrtoaje

Solution. If a = 0, then the inequality is an identity. Consider further that a, b, c > 0, and write the inequality as follows: X p a2 ( a2 + 3bc − a) ≤ 3a bc, X

3a2 bc

≤ 3a bc, a2 + 3bc + a X 1 ≤ 1. Ç + 1 1 + 3bc 2 a p

Let us denote x=Ç

1 1+

3bc a2

, +1

y=q

1 1+

3ca b2

+1

, z=Ç

1 1+

3ab c2

, +1

which implies bc 1 − 2x = , 2 a 3x 2

1 − 2y ca = , 2 b 3 y2

ab 1 − 2z 1 = , 0 < x, y, z < , 2 2 c 3z 2

and (1 − 2x)(1 − 2 y)(1 − 2z) = 27x 2 y 2 z 2 . We need to prove that x + y +z ≤1 1 such that (1 − 2x)(1 − 2 y)(1 − 2z) = 27x 2 y 2 z 2 . We will use the 2 1 contradiction method. Assume that x + y + z > 1 for 0 < x, y, z < , and show that 2 (1 − 2x)(1 − 2 y)(1 − 2z) < 27x 2 y 2 z 2 . We have

for 0 < x, y, z
0 and denote  ‹ a b c 1 x=p , y=p , z=p , x, y, z ∈ 0, . 2 4a2 + 5bc 4b2 + 5ca 4c 2 + 5a b We have

bc 1 − 4x 2 = , a2 5x 2

1 − 4 y2 ca = , b2 5 y2

ab 1 − 4z 2 = , c2 5z 2

and (1 − 4x 2 )(1 − 4 y 2 )(1 − 4z 2 ) = 125x 2 y 2 z 2 . For the sake of contradiction, assume that x + y + z > 1. Using the AM-GM inequality and the Cauchy-Schwarz inequality, we have 1 Y 1 Y (1 − 4x 2 ) < [(x + y + z)2 − 4x 2 ] 125 125 Y 1 Y = (3x + y + z) · ( y + z − x) 125  x + y + z 3 Y ≤ ( y + z − x) 3 Y 1 ≤ (x 2 + y 2 + z 2 )(x + y + z) ( y + z − x) 9 1 = (x 2 + y 2 + z 2 )[2(x 2 y 2 + y 2 z 2 + z 2 x 2 ) − x 4 − y 4 − z 4 ], 9

x 2 y 2z2 =

and hence 9x 2 y 2 z 2 < (x 2 + y 2 + z 2 )[2(x 2 y 2 + y 2 z 2 + z 2 x 2 ) − x 4 − y 4 − z 4 ], X x 6 + y 6 + z 6 + 3x 2 y 2 z 2 < x 2 y 2 (x 2 + y 2 ). The last inequality contradicts Schur’s inequality X x 6 + y 6 + z 6 + 3x 2 y 2 z 2 ≥ x 2 y 2 (x 2 + y 2 ). Thus, the proof is completed. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

308

Vasile Cîrtoaje

Second Solution. In the nontrivial case when a, b, c > 0, setting x = z=

ca bc , y = 2 and 2 a b

ab (x yz = 1), the desired inequality becomes E(x, y, z) ≤ 1, where c2 1 1 1 +p . E(x, y, z) = p +p 4 + 5x 4 + 5z 4 + 5y

Without loss of generality, we may assume that x ≥ y ≥ z, x ≥ 1, yz ≤ 1. We will prove that p p E(x, y, z) ≤ E(x, yz, yz) ≤ 1. The left inequality has the form 1

p

1 1 ≤p +p p . 4 + 5z 4 + 5y 4 + 5 yz

For the nontrivial case y 6= z, consider y > z and denote y +z p , p = yz, 2 Æ q = (4 + 5 y)(4 + 5z). s=

We have s > p, p ≤ 1 and Æ Æ q = 16 + 40s + 25p2 > 16 + 40p + 25p2 = 4 + 5p. By squaring, the desired inequality becomes in succession as follows: 1 2 4 1 + + ≤ , 4 + 5 y 4 + 5z q 4 + 5p 1 1 2 2 2 + − ≤ − , 4 + 5 y 4 + 5z 4 + 5p 4 + 5p q 2(q − 4 − 5p) 8 + 10s 2 − ≤ , 2 q 4 + 5p q(4 + 5p) (s − p)(5p − 4) 8(s − p) ≤ , 2 q (4 + 5p) q(4 + 5p)(q + 4 + 5p) 5p − 4 8 ≤ , q q + 4 + 5p 25p2 − 16 ≤ (12 − 5p)q. The last inequality is true since (12 − 5p)q − 25p2 + 16 > (12 − 5p)(4 + 5p) − 25p2 + 16 = 2(8 − 5p)(4 + 5p) > 0.

Symmetric Nonrational Inequalities

309

In order to prove the right inequality, namely 1 2 ≤ 1, +p p p 4 + 5x 4 + 5 yz let us denote

p

p 4 + 5 yz = 3t, t ∈ (2/3, 1]. Since x=

25 1 = , 2 yz (9t − 4)2

the inequality becomes 9t 2 − 4 2 + ≤ 1, p 3 36t 4 − 32t 2 + 21 3t €p Š (2 − 3t) 36t 4 − 32t 2 + 21 − 3t 2 − 2t ≤ 0. Since 2 − 3t < 0, we still have to show that p 36t 4 − 32t 2 + 21 ≥ 3t 2 + 2t. Indeed, p

3(t − 1)2 (9t 2 + 14t + 7) 36t 4 − 32t 2 + 21 − 3t 2 − 2t = p ≥ 0. 36t 4 − 32t 2 + 21 + 3t 2 + 2t

P 2.32. Let a, b, c be nonnegative real numbers. Prove that p p p a 4a2 + 5bc + b 4b2 + 5ca + c 4c 2 + 5a b ≥ (a + b + c)2 . (Vasile Cîrtoaje, 2004) First Solution. Write the inequality as X €p Š a 4a2 + 5bc − 2a ≥ 2(a b + bc + ca) − a2 − b2 − c 2 , 1 ≥ 2(a b + bc + ca) − a2 − b2 − c 2 . p 2 4a + 5bc + 2a Writing Schur’s inequality X a3 + b3 + c 3 + 3a bc ≥ a b(a2 + b2 ) 5a bc

in the form

X

9a bc ≥ 2(a b + bc + ca) − a2 − b2 − c 2 , a+b+c

310

Vasile Cîrtoaje

it suffices to prove that X

5 9 ≥ . p 2 a+b+c 4a + 5bc + 2a

Let p = a + b + c and q = a b + bc + ca. By the AM-GM inequality, we have p p 2 (16a2 + 20bc)(3b + 3c)2 (16a2 + 20bc) + (3b + 3c)2 4a2 + 5bc = ≤ 12(b + c) 12(b + c) 16a2 + 16bc + 10(b + c)2 8a2 + 5b2 + 5c 2 + 18bc = , 12(b + c) 6(b + c)

≤ hence

X 5 5 ≥ p 2 2 2 8a + 5b + 5c 2 + 18bc 4a + 5bc + 2a + 2a 6(b + c) X X 30(b + c) 30(b + c) = . = 2 2 2 2 8a + 5b + 5c + 12a b + 18bc + 12ac 5p + 2q + 3a2 + 6bc X

Thus, it suffices to show that X

30(b + c) 9 ≥ . 5p2 + 2q + 3a2 + 6bc p

By the Cauchy-Schwarz inequality, we get X

P 30[ (b + c)]2 30(b + c) ≥P 5p2 + 2q + 3a2 + 6bc (b + c)(5p2 + 2q + 3a2 + 6bc)

=

120p2 120p2 P = . 10p3 + 4pq + 9 bc(b + c) 10p3 + 13pq − 27a bc

Therefore, it is enough to show that 120p2 9 ≥ , 3 10p + 13pq − 27a bc p which is equivalent to 10p3 + 81a bc ≥ 39pq. From Schur’s inequality p3 + 9a bc ≥ 4pq and the known inequality pq ≥ 9a bc, we have 10p3 + 81a bc − 39pq = 10(p3 + 9a bc − 4pq) + pq − 9a bc ≥ 0. This completes the proof. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

Symmetric Nonrational Inequalities

311

Second Solution. By the Cauchy-Schwarz inequality, we have ‹ Š X €X p a 2 ≥ (a + b + c)2 . a 4a + 5bc p 4a2 + 5bc From this inequality and the inequality in P 2.31, namely X a ≤ 1, p 2 4a + 5bc the desired inequality follows. Remark. Using the same way as in the second solution, we can prove the following inequalities for a, b, c > 0 satisfying a bc = 1: p p p a 4a2 + 5 + b 4b2 + 5 + c 4c 2 + 5 ≥ (a + b + c)2 ; p p p 4a4 + 5 + 4b4 + 5 + 4c 4 + 5 ≥ (a + b + c)2 . The first inequality follows from the the Cauchy-Schwarz inequality ‹ €X p Š X a 2 a 4a + 5 ≥ (a + b + c)2 p 4a2 + 5 and the inequality X

a ≤ 1, p 4a2 + 5

a bc = 1,

which follows from the inequality in P 2.31 by replacing bc/a2 , ca/b2 , a b/c 2 with 1/x 2 , 1/ y 2 , 1/z 2 , respectively. The second inequality follows from the the Cauchy-Schwarz inequality   2 €X p Š X a 4a4 + 5 ≥ (a + b + c)2 p 4 4a + 5 and the inequality X

a2 ≤ 1, p 4a4 + 5

a bc = 1.

P 2.33. Let a, b, c be nonnegative real numbers. Prove that p p p a a2 + 3bc + b b2 + 3ca + c c 2 + 3a b ≥ 2(a b + bc + ca). (Vasile Cîrtoaje, 2005)

312

Vasile Cîrtoaje

First Solution (by Vo Quoc Ba Can). Using the AM-GM inequality yields X a(b + c)(a2 + 3bc) X p a a2 + 3bc = p (b + c)2 (a2 + 3bc) X 2a(b + c)(a2 + 3bc) ≥ . (b + c)2 + (a2 + 3bc) Thus, it suffices to prove the inequality X 2a(b + c)(a2 + 3bc) a2 + b2 + c 2 + 5bc

≥

X

a(b + c),

which can be written as follows X a(b + c)(a2 − b2 − c 2 + bc) a2 + b2 + c 2 + 5bc X a3 (b + c) − a(b3 + c 3 ) a2 + b2 + c 2 + 5bc

≥ 0,

≥ 0,

X a b(a2 − b2 ) − ac(c 2 − a2 )

≥ 0, a2 + b2 + c 2 + 5bc X X a b(a2 − b2 ) ba(a2 − b2 ) − ≥ 0, a2 + b2 + c 2 + 5bc b2 + c 2 + a2 + 5ca X 5a bc(a2 − b2 )(a − b) ≥ 0. (a2 + b2 + c 2 + 5bc)(a2 + b2 + c 2 + 5ac) Since the last inequality is clearly true, the proof is completed. The equality holds a = b = c, and also for a = 0 and b = c (or any cyclic permutation). Second Solution. Write the inequality as X p (a a2 + 3bc − a2 ) ≥ 2(a b + bc + ca) − a2 − b2 − c 2 . Due to homogeneity, we may assume that a + b + c = 3. By the AM-GM inequality, we have a

p

a2 + 3bc − a2 = p

3a bc a2

12a bc = p + 3bc + a 2 4(a2 + 3bc) + 4a 12a bc ≥ . 4 + a2 + 3bc + 4a

Thus, it suffices to show that 12a bc

X

1 ≥ 2(a b + bc + ca) − a2 − b2 − c 2 . 4 + a2 + 3bc + 4a

Symmetric Nonrational Inequalities

313

By the Cauchy-Schwarz inequality, we have X 9 1 ≥P 2 2 4 + a + 3bc + 4a (4 + a + 3bc + 4a) 9 P P = 2 24 + a + 3 a b 27 P P = P 2 8( a) + 3 a2 + 9 a b P 9 a 3 P P = ≥ P . 2 11( a) + 3 a b 4 a Then, it remains to show that 9a bc ≥ 2(a b + bc + ca) − a2 − b2 − c 2 , a+b+c which is equivalent to Schur’s inequality of degree three X X a3 + 3a bc ≥ a b(a + b).

P 2.34. Let a, b, c be nonnegative real numbers. Prove that p p p a a2 + 8bc + b b2 + 8ca + c c 2 + 8a b ≤ (a + b + c)2 . Solution. Multiplying by a + b + c, the inequality becomes X Æ a (a + b + c)2 (a2 + 8bc) ≤ (a + b + c)3 . Since 2

Æ

(a + b + c)2 (a2 + 8bc) ≤ (a + b + c)2 + (a2 + 8bc),

it suffices to show that X

a[(a + b + c)2 + (a2 + 8bc)] ≤ 2(a + b + c)3 ,

which can be written as a3 + b3 + c 3 + 24a bc ≤ (a + b + c)3 . This inequality is equivalent to a(b − c)2 + b(c − a)2 + c(a − b)2 ≥ 0. The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation).

314

Vasile Cîrtoaje

P 2.35. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that p

a2 + 2bc b2 + bc + c 2

+p

b2 + 2ca c 2 + ca + a2

+p

c 2 + 2a b a2 + a b + b2

≥3

p

a b + bc + ca.

(Michael Rozenberg and Marius Stanean, 2011) Solution. By the AM-GM inequality, we have X

p 2(a2 + 2bc) a b + bc + ca = p p b2 + bc + c 2 2 (b2 + bc + c 2 )(a b + bc + ca) X p 2(a2 + 2bc) ≥ ab + bc + ca (b2 + bc + c 2 ) + (a b + bc + ca) X p 2(a2 + 2bc) = ab + bc + ca . (b + c)(a + b + c) a2 + 2bc

X

Thus, it suffices to show that 3 a2 + 2bc b2 + 2ca c 2 + 2a b + + ≥ (a + b + c). b+c c+a a+b 2 This inequality is equivalent to a4 + b4 + c 4 + a bc(a + b + c) ≥

1X a b(a + b)2 . 2

We can prove this inequality by summing Schur’s inequality of fourth degree X a4 + b4 + c 4 + a bc(a + b + c) ≥ a b(a2 + b2 ) and the obvious inequality X

a b(a2 + b2 ) ≥

1X a b(a + b)2 . 2

The equality holds for a = b = c.

P 2.36. Let a, b, c be nonnegative real numbers, no two of which are zero. If k ≥ 1, then a k+1 b k+1 c k+1 ak + bk + c k + + ≤ . 2a2 + bc 2b2 + ca 2c 2 + a b a+b+c (Vasile Cîrtoaje and Vo Quoc Ba Can, 2011)

Symmetric Nonrational Inequalities

315

Solution. Write the inequality as follows X

a k+1 ak − 2 a + b + c 2a + bc

X a k (a − b)(a − c) 2a2 + bc

 ≥ 0,

≥ 0.

Assume that a ≥ b ≥ c. Since (c − a)(c − b) ≥ 0, it suffices to show that a k (a − b)(a − c) b k (b − a)(b − c) + ≥ 0. 2a2 + bc 2b2 + ca This is true if a k (a − c) b k (b − c) − ≥ 0, 2a2 + bc 2b2 + ca which is equivalent to a k (a − c)(2b2 + ca) ≥ b k (b − c)(2a2 + bc). Since a k /b k ≥ a/b, it remains to show that a(a − c)(2b2 + ca) ≥ b(b − c)(2a2 + bc), which is equivalent to the obvious inequality (a − b)c[a2 + 3a b + b2 − c(a + b)] ≥ 0. The equality holds for a = b = c, and also for a = b and c = 0 (or any cyclic permutation).

P 2.37. If a, b, c are positive real numbers, then a2 − bc b2 − ca c2 − a b +p +p ≥ 0; p 3a2 + 2bc 3b2 + 2ca 3c 2 + 2a b

(a) (b)

a2 − bc p

8a2 + (b + c)2

+p

b2 − ca 8b2 + (c + a)2

+p

c2 − a b 8c 2 + (a + b)2

≥ 0.

(Vasile Cîrtoaje, 2006)

316

Vasile Cîrtoaje

Solution. (a) Let p

A=

3a2 + 2bc,

B=

p

3b2 + 2ca,

C=

p

3c 2 + 2a b.

We have 2

X a2 − bc A

= =

X (a − b)(a + c) + (a − c)(a + b) X (a − b)(a + c)

A +

X (b − a)(b + c)

A B ‹ X a+c b+c = (a − b) − A B X a − b (a + c)2 B 2 − (b + c)2 A2 · = AB (a + c)B + (b + c)A X c(a − b)2 2(a − b)2 + c(a + b + 2c) · ≥ 0. = AB (a + c)B + (b + c)A The equality holds for a = b = c. (b) Let A=

Æ

8a2 + (b + c)2 ,

B=

Æ

8b2 + (c + a)2 ,

C=

Æ

8c 2 + (a + b)2 b.

As we have shown before, 2 hence 2

X a2 − bc A

X a2 − bc A

=

=

X a − b (a + c)2 B 2 − (b + c)2 A2 · , AB (a + c)B + (b + c)A

X (a − b)2 AB

·

C1 ≥ 0, (a + c)B + (b + c)A

since C1 = [(a + c) + (b + c)][(a + c)2 + (b + c)2 ] − 8ac(b + c) − 8bc(a + c) ≥ [(a + c) + (b + c)](4ac + 4bc) − 8ac(b + c) − 8bc(a + c) = 4c(a − b)2 ≥ 0. The equality holds for a = b = c.

p P 2.38. Let a, b, c be positive real numbers. If 0 ≤ k ≤ 1 + 2 2, then p

a2 − bc ka2 + b2 + c 2

+p

b2 − ca k b2 + c 2 + a2

+p

c2 − a b kc 2 + a2 + b2

≥ 0.

Symmetric Nonrational Inequalities

317

Solution. Let A=

p

ka2 + b2 + c 2 ,

B=

p

k b2 + c 2 + a2 ,

C=

p

kc 2 + a2 + b2 .

As we have shown at the preceding problem, 2 therefore 2

X a2 − bc A

X a2 − bc A

=

=

X a − b (a + c)2 B 2 − (b + c)2 A2 · , AB (a + c)B + (b + c)A

X (a − b)2 AB

·

C1 ≥ 0, (a + c)B + (b + c)A

where C1 = (a2 + b2 + c 2 )(a + b + 2c) − (k − 1)c(2a b + bc + ca) p ≥ (a2 + b2 + c 2 )(a + b + 2c) − 2 2 c(2a b + bc + ca). Putting a + b = 2x, we have a2 + b2 ≥ 2x 2 , a b ≤ x 2 , and hence p p C1 ≥ (2x 2 + c 2 )(2x + 2c) − 2 2 c(2x 2 + 2c x) = 2(x + c)(x 2 − c)2 ≥ 0. The equality holds for a = b = c.

P 2.39. If a, b, c are nonnegative real numbers, then p p p (a2 − bc) b + c + (b2 − ca) c + a + (c 2 − a b) a + b ≥ 0. First Solution. Let us denote v tb+c x= , 2

y=

s

v ta + b c+a , z= , 2 2

hence a = y 2 + z2 − x 2,

b = z2 + x 2 − y 2,

c = x 2 + y 2 − z2.

The inequality turns into x y(x 3 + y 3 ) + yz( y 3 + z 3 ) + z x(z 3 + x 3 ) ≥ x 2 y 2 (x + y) + y 2 z 2 ( y + z) + z 2 x 2 (z + x), which is equivalent to the obvious inequality x y(x + y)(x − y)2 + yz( y + z)( y − z)2 + z x(z + x)(z − x)2 ≥ 0.

318

Vasile Cîrtoaje

The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation). Second Solution. Write the inequality as A(a2 − bc) + B(b2 − ca) + C(c 2 − a b) ≥ 0, where A=

p

b + c,

B=

p

c + a,

C=

p

a + b.

We have 2

X

A(a2 − bc) =

X

=

X

A[(a − b)(a + c) + (a − c)(a + b)] X A(a − b)(a + c) + B(b − a)(b + c)

X (a − b)[A(a + c) − B(b + c)] X A2 (a + c)2 − B 2 (b + c)2 = (a − b) · A(a + c) + B(b + c) X (a − b)2 (a + c)(b + c) = ≥ 0. A(a + c) + B(b + c) =

P 2.40. If a, b, c are nonnegative real numbers, then p p p (a2 − bc) a2 + 4bc + (b2 − ca) b2 + 4ca + (c 2 − a b) c 2 + 4a b ≥ 0. (Vasile Cîrtoaje, 2005) Solution. If two of a, b, c are zero, then the inequality is clearly true. Otherwise, write the inequality as AX + BY + C Z ≥ 0, where

p A=

a2 + 4bc , b+c

X = (a2 − bc)(b + c),

p B=

b2 + 4ca , c+a

p C=

Y = (b2 − bc)(b + c),

c 2 + 4a b , a+b

X = (c 2 − a b)(a + b).

Without loss of generality, assume that a ≥ b ≥ c. We have X ≥ 0, Z ≤ 0 and X + Y + Z = 0. In addition, X − Y = a b(a − b) + 2(a2 − b2 )c + (a − b)c 2 ≥ 0

Symmetric Nonrational Inequalities

319

and a4 − b4 + 2(a3 − c 3 )c + (a2 − c 2 )c 2 + 4a bc(a − b) − 4(a − b)c 3 (b + c)2 (c + a)2

A2 − B 2 =

≥

4a bc(a − b) − 4(a − b)c 3 4c(a − b)(a b − c 2 ) = ≥ 0. (b + c)2 (c + a)2 (b + c)2 (c + a)2

Since 2(AX + BY + C Z) = (A − B)(X − Y ) − (A + B − 2C)Z, it suffices to show that A + B − 2C ≥ 0. This is true if AB ≥ C 2 . Using the Cauchy-Schwarz inequality gives p p a b + 4c a b a b + 2c a b + 2c 2 AB ≥ ≥ . (b + c)(c + a) (b + c)(c + a) Thus, we need to show that (a + b)2 (a b + 2c

p

a b + 2c 2 ) ≥ (b + c)(c + a)(c 2 + 4a b).

Write this inequality as a b(a − b)2 + 2c

p

a b(a + b)

€p

a−

p Š2 b + c 2 [2(a + b)2 − 5a b − c(a + b) − c 2 ] ≥ 0.

It is true since 2(a + b)2 − 5a b − c(a + b) − c 2 = a(2a − b − c) + b(b − c) + b2 − c 2 ≥ 0. The equality holds for a = b = c, and also for a = b and c = 0 (or any cyclic permutation).

P 2.41. If a, b, c are nonnegative real numbers, then v t

v v t t a3 b3 c3 + + ≥ 1. a3 + (b + c)3 b3 + (c + a)3 c 3 + (a + b)3

320

Vasile Cîrtoaje

Solution. For a = 0, the inequality reduces to the obvious inequality p p p b3 + c 3 ≥ b3 + c 3 . For a, b, c > 0, write the inequality as v Xu u u t

1 1+



b+c a

‹3 ≥ 1.

For any x ≥ 0, we have p

1 + x3 =

Æ

(1 + x)(1 − x + x 2 ≤

(1 + x) + (1 − x + x 2 ) 1 = 1 + x 2. 2 2

Therefore, we get v Xu u u t

≥

1 1+



b+c a

‹3 ≥

1

X

=

X

X

1  ‹ 1 b+c 2 1+ 2 a

a2 = 1. a2 + b2 + c 2

b +c a2 The equality holds for a = b = c, and also for b = c = 0 (or any cyclic permutation). 1+

2

2

P 2.42. If a, b, c are positive real numbers, then v v v u ‹ ‹   t t t 1 1 1 1 1 1 (a + b + c) + + ≥ 1 + 1 + (a2 + b2 + c 2 ) 2 + 2 + 2 . a b c a b c (Vasile Cîrtoaje, 2002) Solution. Using the Cauchy-Schwarz inequality, we have v X Š X 1 X 1‹ €X Š X 1 ‹ t€X a + 2 = a2 + 2 bc a a2 bc v v  ‹  t€X Š X 1 t€X Š X 1 ‹ ≥ a2 + 2 bc a2 bc v v ‹   t€X Š X 1 t€X Š X 1 ‹ 2 = a +2 a , a2 a

Symmetric Nonrational Inequalities

321

hence v v €X Š X 1 ‹ t€X Š X 1 ‹ t€X Š X 1 ‹ −2 a +1=1+ a2 a , a a a2 ™2 –v v t€X Š X 1 ‹ t€X Š X 1 ‹ a a2 −1 ≥1+ , a a2 v v v u t€X Š X 1 ‹ t€X Š X 1 ‹ t . −1≥ 1+ a a2 a a2 The equality holds if and only if €X

a2

Š X 1 ‹ X 1 ‹ €X Š = bc , bc a2

which is equivalent to (a2 − bc)(b2 − ca)(c 2 − a b) = 0. Consequently, the equality occurs for a2 = bc, or b2 = ca, or c 2 = a b.

P 2.43. If a, b, c are positive real numbers, then v  ‹  ‹ t 1 1 1 1 1 1 2 2 2 5 + 2(a + b + c ) 2 + 2 + 2 − 2 ≥ (a + b + c) + + . a b c a b c (Vasile Cîrtoaje, 2004) Solution. Let us denote x= We have

and

a b c + + , b c a

y=

b c a + + . a b c

‹ 1 1 1 (a + b + c) + + = x + y +3 a b c 

‹ 1 1 1 2(a + b + c ) 2 + 2 + 2 − 2 = a b c  2   2  2 2 a b c b c 2 a2 =2 + + +2 + + +4 b2 c 2 a2 a2 b2 c 2 2

2

2



= 2(x 2 − 2 y) + 2( y 2 − 2x) + 4 = (x + y − 2)2 + (x − y)2 ≥ (x + y − 2)2 .

322

Vasile Cîrtoaje

Therefore, v t

‹ 1 1 1 + + + −2 ≥ x + y −2 a2 b2 c 2 ‹  1 1 1 + + − 5. = (a + b + c) a b c The equality occurs for a = b, or b = c, or c = a. 2(a2

b2

+ c2)



P 2.44. If a, b, c are real numbers, then Æ 2(1 + a bc) + 2(1 + a2 )(1 + b2 )(1 + c 2 ) ≥ (1 + a)(1 + b)(1 + c). (Wolfgang Berndt, 2006) First Solution. Denoting p = a + b + c, q = a b + bc + ca,

r = a bc,

the inequality becomes Æ 2(p2 + q2 + r 2 − 2pr − 2q + 1) ≥ p + q − r − 1. It suffices to show that 2(p2 + q2 + r 2 − 2pr − 2q + 1) ≥ (p + q − r − 1)2 , which is equivalent to p2 + q2 + r 2 − 2pq + 2qr − 2pr + 2p − 2q − 2r + 1 ≥ 0, (p − q − r + 1)2 ≥ 0. The equality holds for p + 1 = q + r and q ≥ 1. The last condition follows from p + q − r − 1 ≥ 0. Second Solution. Since 2(1 + a2 ) = (1 + a)2 + (1 − a)2 and (1 + b2 )(1 + c 2 ) = (b + c)2 + (bc − 1)2 , by the Cauchy-Schwarz inequality, we get Æ 2(1 + a2 )(1 + b2 )(1 + c 2 ) ≥ (1 + a)(b + c) + (1 − a)(bc − 1) = (1 + a)(1 + b)(1 + c) − 2(1 + a bc).

Symmetric Nonrational Inequalities

323

P 2.45. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that v v v t a2 + bc t b2 + ca t c 2 + a b 1 + + ≥2+ p . 2 2 2 2 2 2 b +c c +a a +b 2 (Vo Quoc Ba Can, 2006) Solution. We may assume that a ≥ b ≥ c. Then, it suffices to show that v v v t a2 + c 2 t b2 + c 2 t a b 1 + + ≥2+ p . 2 2 2 2 2 2 b +c c +a a +b 2 Let us denote x=

v t a2 + c 2 b2

Since x2 − y2 =

+ c2

,

y=

s

a . b

(a − b)(a b − c 2 ) ≥ 0, b(b2 + c 2 )

it follows that x ≥ y ≥ 1. From

 ‹ (x − y)(x y − 1) 1 1 x+ − y+ = ≥ 0, x y xy

we have

v t a2 + c 2

v t b2 + c 2

s

v a tb + . b a

+ ≥ b2 + c 2 c 2 + a2 Therefore, it is enough to show that v v s a t b t ab 1 + + ≥2+ p . 2 2 b a a +b 2 s a , the inequality becomes Putting t = b s 1 1 t t + −2≥ p − . 2 t t +1 2 We have

(t − 1)2 (t − 1)2 s  ‹≤ 2 t +1 1 t 2(t 2 + 1) p + 2+1 t 2 2 (t − 1) 1 ≤ = t + − 2. t t The equality holds for a = b and c = 0 (or any cyclic permutation). 1 p − 2

s

t = 2 t +1

324

Vasile Cîrtoaje

P 2.46. If a, b, c are nonnegative real numbers, then Æ

a(2a + b + c) +

Æ

b(2b + c + a) +

Æ

c(2c + a + b) ≥

Æ

12(a b + bc + ca). (Vasile Cîrtoaje, 2012)

Solution. By squaring, the inequality becomes a2 + b2 + c 2 +

XÆ

bc(2b + c + a)(2c + a + b) ≥ 5(a b + bc + ca).

Using the Cauchy-Schwarz inequality yields XÆ

bc(2b + c + a)(2c + a + b) = ≥

XÆ

(2b2 + bc + a b)(2c 2 + bc + ac)

X X p p (2bc + bc + a bc) = 3(a b + bc + ca) + a bc.

Therefore, it suffices to show that X p a bc ≥ 2(a b + bc + ca).

a2 + b2 + c 2 +

We can get this inequality by summing Schur’s inequality a2 + b2 + c 2 +

X p Xp a bc ≥ a b(a + b)

and Xp

a b (a + b) ≥ 2(a b + bc + ca).

The last inequality is equivalent to the obvious inequality Xp

p p a b ( a − b )2 ≥ 0.

The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 2.47. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that a

Æ

(4a + 5b)(4a + 5c) + b

Æ

(4b + 5c)(4b + 5a) + c

Æ

(4c + 5a)(4c + 5b) ≥ 27. (Vasile Cîrtoaje, 2010)

Symmetric Nonrational Inequalities

325

Solution. Assume that a ≥ b ≥ c, consider the non-trivial case b > 0, and write the inequality in the following equivalent homogeneous forms: X Æ a (4a + 5b)(4a + 5c) ≥ 3(a + b + c)2 , X X X €p Š2 p 2( a2 − a b) = a 4a + 5b − 4a + 5c , X X 25a(b − c)2 (b − c)2 ≥ , p p ( 4a + 5b + 4a + 5c)2 X (b − c)2 Sa ≥ 0, where

25a . Sa = 1 − p p ( 4a + 5b + 4a + 5c)2

Since

and

25b 25b Sb = 1 − p ≥1− p =0 p p ( 4b + 5c + 4b + 5a)2 ( 4b + 9b)2 25c 25 25c ≥1− p =1− Sc = 1 − p > 0, p p 36 ( 9c + 9c)2 ( 4c + 5a + 4c + 5b)2

we have X a2 (b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b b  ‹ a b a = (b − c)2 Sa + S b . b a b Thus, it suffices to prove that a b Sa + S b ≥ 0. a b We have p p 25a a( 4a + 5b − 4a)2 Sa ≥ 1 − p =1− , p b2 ( 4a + 5b + 4a)2 p p 25b b( 4b + 5a − 4b)2 Sb ≥ 1 − p =1− , p a2 ( 4b + 4b + 5a)2 and hence

p p p p b a b ( 4a + 5b − 4a)2 a ( 4b + 5a − 4b)2 Sa + S b ≥ − + − a b a b b a v ‚v Œ  ‹ t 4a2 5a t 4b2 5b a b =4 + + + − 7 + − 10 b2 b a2 a b a q p = 4 4x 2 + 5x − 8 + 2 20x + 41 − 7x − 10,

326

Vasile Cîrtoaje

where x=

a b + ≥ 2. b a

To end the proof, we only need to show that x ≥ 2 yields 4

q

p 4x 2 + 5x − 8 + 2 20x + 41 ≥ 7x + 10.

By squaring, this inequality becomes 15x 2 − 60x − 228 + 32

p

20x + 41 ≥ 0.

Indeed, 15x 2 − 60x − 228 + 32

p

p 20x + 41 ≥ 15x 2 − 60x − 228 + 32 81 = 15(x − 2)2 ≥ 0.

The equality holds for a = b = c = 1, and also for c = 0 and a = b = permutation).

3 (or any cyclic 2

P 2.48. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that a

Æ

(a + 3b)(a + 3c) + b

Æ

(b + 3c)(b + 3a) + c

Æ

(c + 3a)(c + 3b) ≥ 12. (Vasile Cîrtoaje, 2010)

Solution. Assume that a ≥ b ≥ c (b > 0), and write the inequality as X Æ a (a + 3b)(a + 3c) ≥ 4(a b + bc + ca), X X X €p Š2 p 2( a2 − a b) = a a + 3b − a + 3c , X X 9a(b − c)2 (b − c)2 ≥ , p p ( a + 3b + a + 3c)2 X (b − c)2 Sa ≥ 0, where

Since

9a Sa = 1 − p . p ( a + 3b + a + 3c)2 9b 9b ≥1− p =0 Sb = 1 − p p p 2 ( b + 3c + b + 3a) ( b + 4b)2

Symmetric Nonrational Inequalities and

327

9 9c 9c =1− ≥1− p > 0, Sc = 1 − p p p 2 2 16 ( 4c + 4c) ( c + 3a + c + 3b)

we have X a2 (b − c)2 Sa ≥ (b − c)2 Sa + (a − c)2 S b ≥ (b − c)2 Sa + 2 (b − c)2 S b b  ‹ a b a = (b − c)2 Sa + S b . b a b Thus, it suffices to prove that b a Sa + S b ≥ 0. a b We have

p p a( a + 3b − a)2 =1− , Sa ≥ 1 − p p b2 ( a + 3b + a)2 p p b( b + 3a − b)2 9b =1− Sb ≥ 1 − p , p a2 ( b + b + 3a)2 9a

and hence p p p p b a b ( a + 3b − a)2 a ( b + 3a − b)2 Sa + S b ≥ − + − a b a b b a v v ‚ Œ  ‹ t a2 3a t b2 3b a b =2 + + + − + −6 b2 b a2 a b a Æ p = 2 x 2 + 3x − 2 + 2 3x + 10 − x − 6, where

a b + ≥ 2. b a To end the proof, it remains to show that Æ p 2 x 2 + 35x − 2 + 2 3x + 10 ≥ x + 6 x=

for x ≥ 2. By squaring, this inequality becomes p 3x 2 − 44 + 8 3x + 10 ≥ 0. Indeed,

p 3x 2 − 44 + 8 3x + 10 ≥ 12 − 44 + 32 = 0.

The equality holds for a = b = c = 1, and also for c = 0 and a = b = permutation).

p 3 (or any cyclic

328

Vasile Cîrtoaje

P 2.49. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 3. Prove that Æ p p p 2 + 7a b + 2 + 7bc + 2 + 7ca ≥ 3 3(a b + bc + ca). (Vasile Cîrtoaje, 2010) Solution. Consider a ≥ b ≥ c. Since the inequality is trivial for b = c = 0, we may assume that b > 0. By squaring, the desired inequality becomes XÆ a2 + b2 + c 2 + (2 + 7a b)(2 + 7ac) ≥ 10(a b + bc + ca), X €p Š2 p 6(a2 + b2 + c 2 − a b − bc − ca) = 2 + 7a b − 2 + 7ac , 3

where

X X 49a2 (b − c)2 (b − c)2 ≥ , p p ( 2 + 7a b + 2 + 7ac )2 X (b − c)2 Sa ≥ 0, 49a2 Sa = 1 − p , p ( 6 + 21a b + 6 + 21ac)2 49b2 , Sb = 1 − p p ( 6 + 21a b + 6 + 21bc)2 49c 2 . Sc = 1 − p p ( 6 + 21ac + 6 + 21bc)2

Since 6 ≥ 2(a2 + b2 ) ≥ 4a b, we have 49a2 49a2 a ≥1− p =1− , Sa ≥ 1 − p p p b ( 4a b + 21a b + 4a b + 21ac)2 (5 a b + 2 a b)2 49b2 b 49b2 ≥1− p =1− , Sb ≥ 1 − p p p a ( 4a b + 21a b + 4a b + 21bc)2 (5 a b + 2 a b)2 49c 2 49c 2 49 Sc ≥ 1 − p ≥1− =1− > 0. p 2 2 (5c + 5c) 100 ( 4a b + 21ac + 4a b + 21bc) Therefore, X (b − c)2 Sa ≥ (b − c)2 Sa + (c − a)2 S b ‹   a b 2 2 ≥ (b − c) 1 − + (c − a) 1 − b a (a − b)2 (a b − c 2 ) = ≥ 0. ab p The equality holds for a = b = c = 1, and also for c = 0 and a = b = 3 (or any cyclic permutation).

Symmetric Nonrational Inequalities

329

P 2.50. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that Pp

(a)

P

(b) P

(c)

a(b + c)(a2 + bc) ≥ 6;

p p a(b + c) a2 + 2bc ≥ 6 3;

p a(b + c) (a + 2b)(a + 2c) ≥ 18. (Vasile Cîrtoaje, 2010)

Solution. Assume that a ≥ b ≥ c (b > 0). (a) Write the inequality in the homogeneous form XÆ a(b + c)(a2 + bc) ≥ 2(a b + bc + ca). First Solution. Write the homogeneous inequality as XÆ ”p — Æ a(b + c) a2 + bc − a(b + c) ≥ 0, p X (a − b)(a − c) a(b + c) ≥ 0. p p a2 + bc + a(b + c) Since (c − a)(c − b) ≥ 0, it suffices to show that p p (a − b)(a − c) a(b + c) (b − c)(b − a) b(c + a) + p ≥ 0. p p p a2 + bc + a(b + c) b2 + ca + b(c + a) This is true if

p p (b − c) b(c + a) (a − c) a(b + c) ≥p . p p p a2 + bc + a(b + c) b2 + ca + b(c + a)

Since

Æ

a(b + c) ≥

Æ

b(c + a),

it suffices to show that p

a−c b−c ≥p . p p a2 + bc + a(b + c) b2 + ca + b(c + a)

Moreover, since p

a2 + bc ≥

Æ

a(b + c),

p

b2 + ca ≤

it is enough to show that p

a−c a2 + bc

≥p

b−c b2 + ca

.

Æ

b(c + a),

330

Vasile Cîrtoaje

Indeed, we have (a − c)2 (b2 + ca) − (b − c)2 (a2 + bc) = (a − b)(a2 + b2 + c 2 + 3a b − 3bc − 3ca) ≥ 0, because a2 + b2 + c 2 + 3a b − 3bc − 3ca = (a2 − bc) + (b − c)2 + 3a(b − c) ≥ 0. The equality holds for a = b = c = 1, and for c = 0 and a = b = permutation).

p 3 (or any cyclic

Second Solution. By squaring, the homogeneous inequality becomes X XÆ a(b + c)(a2 + bc) + 2 bc(a + b)(a + c)(b2 + ca)(c 2 + a b) ≥ 4(a b + bc + ca)2 . Since (b2 + ca)(c 2 + a b) − bc(a + b)(a + c) = a(b + c)(b − c)2 ≥ 0, it suffices to show that X X a(b + c)(a2 + bc) + 2 bc(a + b)(a + c) ≥ 4(a b + bc + ca)2 , which is equivalent to X

bc(b − c)2 ≥ 0.

(b) Write the inequality as X p p a(b + c) a2 + 2bc ≥ 2(a b + bc + ca) a b + bc + ca, X

a(b + c) X

p

”p

a2 + 2bc −

p

— a b + bc + ca ≥ 0,

a(b + c)(a − b)(a − c) ≥ 0. p a2 + 2bc + a b + bc + ca

Since (c − a)(c − b) ≥ 0, it suffices to show that p

a(b + c)(a − b)(a − c) b(c + a)(b − c)(b − a) +p ≥ 0. p p 2 a + 2bc + a b + bc + ca b2 + 2ca + a b + bc + ca

This is true if p

a(b + c)(a − c) b(c + a)(b − c) ≥p . p p a2 + 2bc + a b + bc + ca b2 + 2ca + a b + bc + ca

Since (b + c)(a − c) ≥ (c + a)(b − c),

Symmetric Nonrational Inequalities

331

it suffices to show that p

a2 + 2bc +

a p

a b + bc + ca

Moreover, since p p a2 + 2bc ≥ a b + bc + ca,

≥p

b2 + 2ca +

p

b p

b2 + 2ca ≤

a b + bc + ca

p

.

a b + bc + ca,

it is enough to show that p

a a2 + 2bc

≥p

b b2 + 2ca

.

Indeed, we have a2 (b2 + 2ca) − b2 (a2 + 2bc) = 2c(a3 − b3 ) ≥ 0. p 3 (or any cyclic

The equality holds for a = b = c = 1, and for c = 0 and a = b = permutation).

(c) Write the inequality as follows: X Æ Æ a(b + c) (a + 2b)(a + 2c) ≥ 2(a b + bc + ca) 3(a b + bc + ca), X

a(b + c) X p

Ӯ

(a + 2b)(a + 2c) −

Æ

— 3(a b + bc + ca) ≥ 0,

a(b + c)(a − b)(a − c) ≥ 0. p (a + 2b)(a + 2c) + 3(a b + bc + ca)

Since (c − a)(c − b) ≥ 0, it suffices to show that p

b(c + a)(b − c)(b − a) a(b + c)(a − b)(a − c) +p ≥ 0. p p (a + 2b)(a + 2c) + 3(a b + bc + ca) (b + 2c)(b + 2a) + 3(a b + bc + ca)

This is true if p

a(b + c)(a − c) b(c + a)(b − c) ≥p . p p (a + 2b)(a + 2c) + 3(a b + bc + ca) (b + 2c)(b + 2a) + 3(a b + bc + ca)

Since (b + c)(a − c) ≥ (c + a)(b − c), it suffices to show that a p

(a + 2b)(a + 2c) +

p

3(a b + bc + ca)

Moreover, since Æ Æ (a + 2b)(a + 2c) ≥ 3(a b + bc + ca),

≥p

b (b + 2c)(b + 2a) +

Æ

(b + 2c)(b + 2a) ≤

p

3(a b + bc + ca)

Æ

.

3(a b + bc + ca),

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Vasile Cîrtoaje

it is enough to show that a p

(a + 2b)(a + 2c)

This is true if

p p

≥p

b (b + 2c)(b + 2a) p

a

(a + 2b)(a + 2c)

≥p

b

(b + 2c)(b + 2a)

.

.

Indeed, we have a(b + 2c)(b + 2a) − b(a + 2b)(a + 2c) = (a − b)(a b + 4bc + 4ca) ≥ 0. The equality holds for a = b = c = 1, and for c = 0 and a = b = permutation).

p 3 (or any cyclic

P 2.51. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. Prove that a

p

p p bc + 3 + b ca + 3 + c a b + 3 ≥ 6. (Vasile Cîrtoaje, 2010)

First Solution. Write the inequality in the homogeneous form X p a a b + 2bc + ca ≥ 2(a b + bc + ca). If a = 0, then the inequality turns into p

p p bc( b − c)2 ≥ 0.

Consider further a, b, c > 0. By squaring, the inequality becomes X X Æ X X a2 b2 + 6a bc a. a b(a2 + b2 ) + 2 bc (bc + 2ca + a b)(ca + 2a b + bc) ≥ 4 Using the Cauchy-Schwarz inequality, we have Æ

(bc + 2ca + a b)(ca + 2a b + bc) =

Æ

(a b + bc + ca + ca)(a b + bc + ca + a b)

≥ a b + bc + ca + a

p

bc,

and hence X Æ Xp bc (bc + 2ca + a b)(ca + 2a b + bc) ≥ (a b + bc + ca)2 + a bc bc.

Symmetric Nonrational Inequalities

333

Thus, it suffices to show that X Xp X X a b(a2 + b2 ) + 2(a b + bc + ca)2 + 2a bc bc ≥ 4 a2 b2 + 6a bc a, which is equivalent to X X Xp X bc ≥ 2 a2 b2 + 2a bc a, a b(a2 + b2 ) + 2a bc X

X Xp a b(a − b)2 ≥ 2a bc( a− bc), X (a − b)2

Using the substitution x =

p

≥

Xp p ( a − b)2 .

c p p a, y = b, z = c, the last inequality becomes

X (x 2 − y 2 )2 X ≥ (x − y)2 , z2 ( y − z)2 A + (z − x)2 B + (x − y)2 C ≥ 0, where A=

y +z− x , x2

B=

z+x− y , y2

C=

x + y −z . z2

Without loss of generality, assume that x ≥ y ≥ z > 0. Since B > 0 and C > 0, we have ( y − z)2 A + (z − x)2 B + (x − y)2 C ≥ ( y − z)2 A + (z − x)2 B ≥ ( y − z)2 (A + B). Thus, we only need to show that A + B ≥ 0. Indeed, A+ B =

y +z− x z+ x − y y +z− x z+ x − y 2z + ≥ + = 2 > 0. x2 y2 x2 x2 x

The equality holds for a = b = c = 1, and for a = 0 and b = c = permutation).

p 3 (or any cyclic

Second Solution. Assume that a ≥ b ≥ c, and write the inequality as X p a a b + 2bc + ca ≥ 2(a b + bc + ca), X €p Š a a b + 2bc + ca − b − c ≥ 0, X a(a b + ac − b2 − c 2 ) ≥ 0, p a b + 2bc + ca + b + c y x z + + ≥ 0, b+c+A c+a+B a+ b+C

(*)

334

Vasile Cîrtoaje

where A=

p

a b + 2bc + ca,

x = a2 (b + c) − a(b2 + c 2 ),

B=

p

bc + 2ca + a b,

C=

p

ca + 2a b + bc,

y = b2 (c + a) − b(c 2 + a2 ), z = c 2 (a + b) − c(a2 + b2 ).

Since x + y + z = 0, we can write the inequality as  ‹  ‹ 1 1 1 1 x − +z − ≥ 0. b+c+A c+a+B a+b+C c+a+B We have x = a b(a − b) + ac(a − c) ≥ 0, z = ac(c − a) + bc(c − b) ≤ 0. Therefore, it suffices to show that 1 1 − ≥ 0, b+c+A c+a+B

1 1 − ≤ 0, a+b+C c+a+B

that is, a − b + B − A ≥ 0,

b − c + C − B ≥ 0.

It is enough to prove that A ≤ B ≤ C. Indeed, B 2 − A2 = c(a − b) ≥ 0,

C 2 − B 2 = a(b − c) ≥ 0.

Remark. We can also prove the inequality (*) as follows X a b(a − b) + ac(a − c) ≥ 0, b+c+A X a b(a − b) X ba(b − a) + ≥ 0, b+c+A c+a+B ‹  X 1 1 a b(a − b) − ≥ 0, b+c+A c+a+B X a b(a + b + C)(a − b)(a − b + B − A) ≥ 0,  X c  2 a b(a + b + C)(a − b) 1 + ≥ 0. A+ B

P 2.52. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that P p (a) (b + c) b2 + c 2 + 7bc ≥ 18; p P p (b) (b + c) b2 + c 2 + 10bc ≤ 12 3. (Vasile Cîrtoaje, 2010)

Symmetric Nonrational Inequalities

335

Solution. (a) Write the inequality in the equivalent homogeneous forms X p (b + c) b2 + c 2 + 7bc ≥ 2(a + b + c)2 , X”

— p (b + c) b2 + c 2 + 7bc − b2 − c 2 − 4bc ≥ 0,

X (b + c)2 (b2 + c 2 + 7bc) − (b2 + c 2 + 4bc)2 ≥ 0, p (b + c) b2 + c 2 + 7bc + b2 + c 2 + 4bc X bc(b − c)2 ≥ 0. p (b + c) b2 + c 2 + 7bc + b2 + c 2 + 4bc 3 (or any cyclic permuta2 tion), and for a = 3 and b = c = 0 (or any cyclic permutation).

The equality holds for a = b = c = 1, for a = 0 and b = c =

(b) Write the inequality as X Æ (b + c) 3(b2 + c 2 + 10bc) ≤ 4(a + b + c)2 , X”

— Æ 2b2 + 2c 2 + 8bc − (b + c) 3(b2 + c 2 + 10bc) ≥ 0,

X 4(b2 + c 2 + 4bc)2 − 3(b + c)2 (b2 + c 2 + 10bc) ≥ 0, p 2b2 + 2c 2 + 8bc + (b + c) 3(b2 + c 2 + 10bc) X (b − c)4 ≥ 0. p 2b2 + 2c 2 + 8bc + (b + c) 3(b2 + c 2 + 10bc) The equality holds for a = b = c = 1.

P 2.53. Let a, b, c be nonnegative real numbers such then a + b + c = 2. Prove that p p p p a + 4bc + b + 4ca + c + 4a b ≥ 4 a b + bc + ca. (Vasile Cîrtoaje, 2012) Solution. Without loss of generality, assume that c = min{a, b, c}. Using Minkowski’s inequality gives Ç p p p p p p p p p a + 4bc + b + 4ca ≥ ( a + b)2 + 4c( a + b)2 = ( a + b) 1 + 4c. Therefore, it suffices to show that p p p p p ( a + b) 1 + 4c ≥ 4 a b + bc + ca − c + 4a b.

336

Vasile Cîrtoaje

By squaring, this inequality becomes p Æ (a + b + 2 a b)(1 + 4c) + 8 (a b + bc + ca)(c + 4a b) ≥ 16(a b + bc + ca) + c + 4a b. According to Lemma below, it suffices to show that p (a + b + 2 a b)(1 + 4c) + 8(2a b + bc + ca) ≥ 16(a b + bc + ca) + c + 4a b, which is equivalent to a+ b−c+2

p

a b + 8c

p

a b ≥ 4(a b + bc + ca).

Write this inequality in the homogeneous form p p (a + b + c)(a + b − c + 2 a b) + 16c a b ≥ 8(a b + bc + ca). p 1 Due to homogeneity, we may assume that a + b = 1. Let us denote d = a b, 0 ≤ d ≤ . 2 We need to show that f (c) ≥ 0 for 0 ≤ c ≤ d, where f (c) = (1 + c)(1 − c + 2d) + 16cd − 8d 2 − 8c = (1 − 2d)(1 + 4d) + 2(9d − 4)c − c 2 . Since f (c) is concave, it suffices to show that f (0) ≥ 0 and f (d) ≥ 0. Indeed, f (0) = (1 − 2d)(1 + 4d) ≥ 0, f (d) = (3d − 1)2 ≥ 0. Thus, the proof is completed. The equality holds for a = b = 1 and c = 0 (or any cyclic permutation). Lemma (by Nguyen Van Quy). Let a, b, c be nonnegative real numbers such then c = min{a, b, c},

a + b + c = 2.

Then, Æ

(a b + bc + ca)(c + 4a b) ≥ 2a b + bc + ca.

Proof. By squaring, the inequality becomes c[a b + bc + ca − c(a + b)2 ] ≥ 0. We need to show that (a + b + c)(a b + bc + ca) − 2c(a + b)2 ≥ 0. We have (a + b + c)(a b + bc + ca) − 2c(a + b)2 ≥ (a + b)(b + c)(c + a) − 2c(a + b)2 = (a + b)(a − c)(b − c) ≥ 0.

Symmetric Nonrational Inequalities

337

P 2.54. If a, b, c are nonnegative real numbers, then p p p p a2 + b2 + 7a b + b2 + c 2 + 7bc + c 2 + a2 + 7ca ≥ 5 a b + bc + ca. (Vasile Cîrtoaje, 2012) Solution (by Nguyen Van Quy). Assume that c = min{a, b, c}. Using Minkowski’s inequality yields Ç p p p p 2 2 2 2 b + c + 7bc + a + c + 7ca ≥ (a + b)2 + 4c 2 + 7c( a + b)2 . Therefore, it suffices to show that Ç p p p p (a + b)2 + 4c 2 + 7c( a + b)2 ≥ 5 a b + bc + ca − a2 + b2 + 7a b. By squaring, this inequality becomes p Æ 2c 2 + 7c a b + 5 (a2 + b2 + 7a b)(a b + bc + ca) ≥ 15a b + 9c(a + b). Due to homogeneity, we may assume that a + b = 1. Let us denote x = a b. We need to 1 show that f (x) ≥ 0 for c 2 ≤ x ≤ , where 4 Æ p f (x) = 2c 2 + 7c x + 5 (1 + 5x)(c + x) − 15x − 9c. Since

−7c 5(5c − 1)2 0. Since E b = b(c + a)3 − b2 (a b + bc + ca) ≥ b(c + a)3 − b2 (c + a)(c + b) ≥ b(c + a)3 − b2 (c + a)2 = b(c + a)2 (c + a − b) ≥ 0, Ec = c(a + b)3 − c 2 (a b + bc + ca) ≥ c(a + b)3 − c 2 (a + b)(b + c) ≥ c(a + b)3 − c 2 (a + b)2 = c(a + b)2 (a + b − c) ≥ 0 and Ea E b (b + c)3 (c + a)3 + = + − 2(a b + bc + ca) a2 b2 a b b3 + 2b2 c a3 + 2a2 c + − 2(a b + bc + ca) ≥ a b (a2 − b2 )2 + 2c(a + b)(a − b)2 = ≥ 0, ab we get  ‹ X Eb 2 2 2 2 2 Ea (b − c) Ea ≥ (b − c) Ea + (a − c) E b ≥ a (b − c) + ≥ 0. a2 b2 The equality holds for a = b = c = 1, and also for a = b = permutation).

p 3 and c = 0 (or any cyclic

Lemma. If x ≥ 0, y ≥ 0 and d > 0, then 2

Æ

(x + d)( y + d) ≥ x + y + 2d −

1 (x − y)2 . 4d

Proof. We have 2

Æ

2(d x + d y + x y) 2(d x + d y + x y) ≥ (x+d)+( y+d) (x + d)( y + d) − 2d = p (x + d)( y + d) + d +d 2

=

4(d x + d y + x y) (x − y)2 (x − y)2 =x+y− ≥x+y− . x + y + 4d x + y + 4d 4d

342

Vasile Cîrtoaje

P 2.57. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = 1. Prove that p p p a 2 + 3bc + b 2 + 3ca + c 2 + 3a b ≥ (a + b + c)2 . (Vasile Cîrtoaje, 2010) Solution. Let q = a b + bc + ca. Write the inequality as X p a 2 + 3bc ≥ 1 + 2q. By squaring, the inequality becomes X X Æ 1 + 3a bc a+2 bc (2 + 3a b)(2 + 3ac) ≥ 4q + 4q2 . Applying Lemma from the preceding P 2.56 for x = 3a b, y = 3ac2 and d = 2, we have 9 (2 + 3a b)(2 + 3ac) ≥ 3a(b + c) + 4 − a2 (b − c)2 , 8 X Æ X X X 9 2 bc (2 + 3a b)(2 + 3ac) ≥ 3a bc a(b − c)2 (b + c) + 4 bc − a bc 8 X X 9 = 6a bc a + 4q − a bc a(b − c)2 . 8 Therefore, it suffices to show that 2

Æ

1 + 3a bc

X

a + 4q + 6a bc

X

X 9 a − a bc a(b − c)2 ≥ 4q + 4q2 , 8

which is equivalent to 1 + 9a bc

X

a − 4q2 ≥

X 9 a bc a(b − c)2 . 8

Since a4 + b4 + c 4 = 1 − 2(a2 b2 + b2 c 2 + c 2 a2 ) = 1 − 2q2 + 4a bc

X

a,

from Schur’s inequality of fourth degree a4 + b4 + c 4 + 2a bc

X

a≥

€X

a2

Š €X

Š ab ,

we get 1 ≥ 2q2 + q − 6a bc

X

a.

Thus, it is enough to prove that (2q2 + q − 6a bc

X

a) + 9a bc

X

a − 4q2 ≥

X 9 a bc a(b − c)2 ; 8

Symmetric Nonrational Inequalities

343

that is, 8(q − 2q2 + 3a bc Since

X

a) ≥ 9a bc

X

a(b − c)2 .

X X €X Š €X Š €X Š2 q − 2q2 + 3a bc a= a2 ab − 2 a b + 3a bc a X X X = bc(b2 + c 2 ) − 2 b2 c 2 = bc(b − c)2 ,

we need to show that X

bc(8 − 9a2 )(b − c)2 ≥ 0.

Also, since 8 − 9a2 = 8(b2 + c 2 ) − a2 ≥ b2 + c 2 − a2 , it suffices to prove that X

bc(b2 + c 2 − a2 )(b − c)2 ≥ 0.

Assume that a ≥ b ≥ c. It is enough to show that bc(b2 + c 2 − a2 )(b − c)2 + ca(c 2 + a2 − b2 )(c − a)2 ≥ 0. This is true if a(c 2 + a2 − b2 )(a − c)2 ≥ b(a2 − b2 − c 2 )(b − c)2 . For the non-trivial case a2 − b2 − c 2 ≥ 0, this inequality follows from a ≥ b,

c 2 + a2 − b2 ≥ a2 − b2 − c 2 , (a − c)2 ≥ (b − c)2 .

1 1 The equality holds for a = b = c = p , and for a = 0 and b = c = p (or any cyclic 3 2 permutation).

P 2.58. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that (a)

(b)

a

a

v t 2a + bc 3

v t a(1 + b + c) 3

+b

+b

v t 2b + ca 3

+c

v t b(1 + c + a) 3

v t 2c + a b

+c

3

≥ 3;

v t c(1 + a + b) 3

≥ 3.

(Vasile Cîrtoaje, 2010)

344

Vasile Cîrtoaje

Solution. (a) If two of a, b, c are zero, then the inequality is trivial. Otherwise, by Hölder’s inequality, we have ‚

v Œ P X t 2a + bc 2 ( a)3 9 ≥P =P a . a 3a 3 2a + bc 2a + bc

Therefore, it suffices to show that X Since

a ≤ 1. 2a + bc

bc 2a =1− , 2a + bc 2a + bc

we can write this inequality as X

bc ≥ 1. 2a + bc

By the Cauchy-Schwarz inequality, we have P P X bc ( bc)2 ( bc)2 P P = ≥P = 1. 2a + bc bc(2a + bc) 2a bc a + b2 c 2 The equality holds for a = b = c = 1, and for a = 0 and b = c = permutation).

3 (or any cyclic 2

(b) Write the inequality in the homogeneous form X Æ a a(a + 4b + 4c) ≥ (a + b + c)2 . By squaring, the inequality becomes X Æ X X bc bc(b + 4c + 4a)(c + 4a + 4b) ≥ 3 a2 b2 + 6a bc a. Applying the Cauchy-Schwarz inequality, we have Æ

(b + 4c + 4a)(c + 4a + 4b) =

Æ

(4a + b + c + 3c)(4a + b + c + 3b)

p ≥ 4a + b + c + 3 bc, hence p Æ bc bc(b + 4c + 4a)(c + 4a + 4b) ≥ (4a + b + c)bc bc + 3b2 c 2 , X Æ X X p bc bc(b + 4c + 4a)(c + 4a + 4b) ≥ (4a + b + c)bc bc + 3 b2 c 2 .

Symmetric Nonrational Inequalities

345

Thus, it is enough to show that X X p a. (4a + b + c)bc bc ≥ 6a bc Replacing a, b, c by a2 , b2 , c 2 , respectively, this inequality becomes X X (4a2 + b2 + c 2 )b3 c 3 ≥ 6a2 b2 c 2 a2 , X X €X Š €X Š a2 b3 c 3 + 3a2 b2 c 2 bc ≥ 6a2 b2 c 2 a2 , X Š €X Š €X Š €X a2 a3 b3 − 3a2 b2 c 2 ≥ 3a2 b2 c 2 a2 − ab . Since X

a3 b3 − 3a2 b2 c 2 =

€X

ab

Š €X

a2 b2 − a bc

and X

a2 −

X

ab =

X Š 1 €X Š X ab a2 (b − c)2 , a = 2

1X (b − c)2 , 2

we can write the inequality as X (b − c)2 Sa ≥ 0, where Sa = a 2

€X

a2

Š €X

Š a b − 3a2 b2 c 2 .

Assume that a ≥ b ≥ c. Since Sa ≥ S b ≥ 0 and €X Š €X Š S b + Sc = (b2 + c 2 ) a2 a b − 6a2 b2 c 2 €X Š €X Š ≥ 2bc a2 a b − 6a2 b2 c 2 €X Š ≥ 2bca2 a b − 6a2 b2 c 2 = 2a2 bc(a b + ac − 2bc) ≥ 0, we get X (b − c)2 Sa ≥ (c − a)2 S b + (a − b)2 Sc ≥ (a − b)2 (S b + Sc ) ≥ 0. The equality holds for a = b = c = 1, and for a = 0 and b = c = permutation).

3 (or any cyclic 2

P 2.59. If a, b, c are nonnegative real numbers such that a + b + c = 3, then Æ Æ Æ 8(a2 + bc) + 9 + 8(b2 + ca) + 9 + 8(c 2 + a b) + 9 ≥ 15. (Vasile Cîrtoaje, 2013)

346

Vasile Cîrtoaje

Solution. Let q = a b + bc + ca and A = (3a − b − c)2 + 8q,

B = (3b − c − a)2 + 8q,

C = (3c − a − b)2 + 8q.

Since 8(a2 + bc) + 9 = 8(a2 + q) + 9 − 8a(b + c) = 8(a2 + q) + 9 − 8a(3 − a) = (4a − 3)2 + 8q = (3a − b − c)2 + 8q = A, we can rewrite the inequality as follows Xp A ≥ 15, X p [ A − (3a + b + c)] ≥ 0, X 2bc − ca − a b ≥ 0, p A + 3a + b + c ˜ X• b(c − a) c(b − a) + ≥ 0, p p A + 3a + b + c A + 3a + b + c X X c(a − b) c(b − a) + ≥ 0, p p B + 3b + c + a A + 3a + b + c X p p p c(a − b)( C + 3c + a + b)[ A − B + 2(a − b)] ≥ 0, • ˜ X p 4(a + b − c) c(a − b)2 ( C + 3c + a + b) p + 1 ≥ 0. p A+ B Without loss of generality, assume that a ≥ b ≥ c. Since a + b − c > 0, it suffices to show that • ˜ p 4(c + a − b) 2 b(a − c) ( B + 3b + c + a) p +1 ≥ p A+ C • ˜ p 4(a − b − c) − 1 . a(b − c)2 ( A + 3a + b + c) p p B+ C This inequality follows from the inequalities b2 (a − c)2 ≥ a2 (b − c)2 , p p a( B + 3b + c + a) ≥ b( A + 3a + b + c), 4(a − b − c) 4(c + a − b) +1≥ p − 1. p p p A+ C B+ C Write the second inequality as a2 B − b2 A p p + (a − b)(a + b + c) ≥ 0. a B+b A

Symmetric Nonrational Inequalities

347

Since a2 B − b2 A = (a − b)(a + b + c)(a2 + b2 − 6a b + bc + ca) + 8q(a2 − b2 ) ≥ (a − b)(a + b + c)(a2 + b2 − 6a b) ≥ −4a b(a − b)(a + b + c), it suffices to show that

−4a b p p + 1 ≥ 0. a B+b A p p p p p p Indeed, since A > 8q ≥ 2 a b and B ≥ 8q ≥ 2 a b, we have p p p p p a B + b A − 4a b > 2(a + b) a b − 4a b = 2 a b(a + b − 2 a b) ≥ 0. The third inequality holds if 2(a − b − c) 1≥ p p . B+ C p p Clearly, it suffices to show that B ≥ a and C ≥ a. We have B − a2 = 8q − 2a(3b − c) + (3b − c)2 ≥ 8a b − 2a(3b − c) = 2a(b + c) ≥ 0 and C − a2 = 8q − 2a(3c − b) + (3c − b)2 ≥ 8a b − 2a(3c − b) = 2a(5b − 3c) ≥ 0. The equality holds for a = b = c = 1, and also for a = 3 and b = c = 0 (or any cyclic permutation).

P 2.60. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k ≥ p

a2 + bc + k +

p

b2 + ca + k +

p

9 , then 8

p c 2 + a b + k ≥ 3 2 + k.

Solution. We will show that XÆ XÆ Æ 8(a2 + bc + k) ≥ (3a + b + c)2 + 8k − 9 ≥ 6 2(k + 2). The right inequality is equivalent to XÆ Æ (2a + 3)2 + 8k − 9 ≥ 6 2(k + 2), and follows immediately from Jensen’s inequality applied to the convex function f : [0, ∞) → R defined by Æ f (x) = (2x + 3)2 + 8k − 9.

348

Vasile Cîrtoaje

Using the substitution A1 = 8(a2 + bc + k),

B1 = 8(b2 + ca + k),

C1 = 8(c 2 + a b + k),

A2 = (3a + b + c)2 + 8k − 9, B2 = (3b + c + a)2 + 8k − 9, C2 = (3c + a + b)2 + 8k − 9, we can write the left inequality as follows: p

B1 − B2 C1 − C2 A1 − A2 p +p p +p p ≥ 0, A1 + A2 B1 + B2 C1 + C2

2bc − ca − a b 2ca − a b − bc 2a b − bc − ca + p + p ≥ 0, p p p p A1 + A2 B1 + B2 C1 + C2  X  b(c − a) c(b − a) ≥ 0, p +p p p A1 + A2 A1 + A2 X c(a − b) c(b − a) p +p p ≥ 0, p B1 + B2 A1 + A2 X p p p p p p c(a − b)( C1 + C2 )[( A1 − B1 ) + ( A2 − B2 )] ≥ 0,   X p p 2(a + b − c) 2a + 2b + c 2 c(a − b) ( C1 + C2 ) p ≥ 0. p +p p A1 + B1 A2 + B2 Without loss of generality, assume that a ≥ b ≥ c. Clearly, the desired inequality is true for b + c ≥ a. Consider further the case b + c < a. Since a + b − c > 0, it suffices to show that   p p 2(b + c − a) 2b + 2c + a 2 + a(b − c) ( A1 + A2 ) p p +p p B1 + C1 B2 + C2   p p 2(c + a − b) 2c + 2a + b 2 +b(a − c) ( B1 + B2 ) p ≥ 0. p p +p C1 + A1 C2 + AC2 Since b2 (a − c)2 ≥ a2 (b − c)2 , it suffices to show that   p p 2(b + c − a) 2b + 2c + a + b( A1 + A2 ) p p +p p B1 + C1 B2 + C2   p p 2(c + a − b) 2c + 2a + b +a( B1 + B2 ) p ≥ 0. p +p p C1 + A1 C2 + A2 From a2 B1 − b2 A1 = 8c(a3 − b3 ) + 8k(a2 − b2 ) ≥ 0

Symmetric Nonrational Inequalities

349

and a2 B2 − b2 A2 = (a − b)(a + b + c)(a2 + b2 + 6a b + bc + ca) + (8k − 9)(a2 − b2 ) ≥ 0, we get a

p

B1 ≥ b

p

A1 and a

p

B2 ≥ b

p

A2 , hence

p p p p a( B1 + B2 ) ≥ b( A1 + A2 ). Therefore, it is enough to show that 2b + 2c + a 2c + 2a + b 2(b + c − a) 2(c + a − b) p +p p +p p p +p p ≥ 0. B1 + C1 C1 + A1 B2 + C2 C2 + A2 This is true if p and p

−2b 2b p +p p ≥0 B1 + C1 C1 + A1

−2a 2a 2a p +p p ≥ 0. p +p B1 + C1 C1 + A1 C2 + A2

The first inequality is true because A1 − B1 = 8(a − b)(a + b − c) ≥ 0. The second inequality can be written as 1 p Since

C1 +

1 p

C1 +

p

A1

p

A1

+p

+p

1 C2 +

1 C2 +

p

A2

p

A2

≥p

≥p

1 B1 +

.

p

C1

p

C2 +

4 C1 +

p

A1 +

p

A2

,

it suffices to show that 4

p

p p p p B1 + 3 C1 ≥ A1 + A2 + C2 .

Taking into account of C1 − C2 = 4(2a b − bc − ca) ≥ 0, C1 − B1 = 8(b − c)(a − b − c) ≥ 0, A2 − A1 = 4(a b − 2bc + ca) ≥ 0, we have 4

p

B1 + 3

p

C1 −

p

A1 −

p

A2 −

p

p

A1 −

p

A2 −

p

p

B1 + 2 B1 − p p = 2(3 B1 − A2 ).

p

≥4

p

B1 + 2

C1 −

C2 ≥ 4

p

p

A2

A2

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Vasile Cîrtoaje

In addition, 9B1 − A2 = 64k − 8a2 + 72b2 − 4a b + 68ac ≥ 72 − 8a2 + 72b2 − 4a b + 68ac = 8(a + b + c)2 − 8a2 + 72b2 − 4a b + 68ac = 4(20b2 + 2c 2 + 3a b + 4bc + 21ac) ≥ 0. Thus, the proof is completed. The equality holds for a = b = c = 1. If k = 9/8, then the equality holds also for a = 3 and b = c = 0 (or any cyclic permutation).

P 2.61. If a, b, c are nonnegative real numbers such that a + b + c = 3, then p p p p a3 + 2bc + b3 + 2ca + c 3 + 2a b ≥ 3 3. (Nguyen Van Quy, 2013) Solution. Since (a3 + 2bc)(a + 2bc) ≥ (a2 + 2bc)2 , it suffices to prove that X a2 + 2bc p ≥ 3 3. p a + 2bc By Hölder’s inequality, we have X 2 2 X ”X —3 a + 2bc (a2 + 2bc)(a + 2bc) ≥ (a2 + 2bc) = (a + b + c)6 . p a + 2bc Therefore, it suffices to show that (a + b + c)4 ≥ 3

X (a2 + 2bc)(a + 2bc).

which is equivalent to (a + b + c)4 ≥

X (a2 + 2bc)(a2 + 6bc + ca + a b).

Indeed, (a + b + c)4 −

X X (a2 + 2bc)(a2 + 6bc + ca + a b) = 3 a b(a − b)2 ≥ 0.

The equality holds for a = b = c = 1, and also for a = 3 and b = c = 0 (or any cyclic permutation).

Symmetric Nonrational Inequalities

351

P 2.62. If a, b, c are positive real numbers, then p

a2 + bc + b+c

p

b2 + ca + c+a

p

p 3 2 c2 + a b ≥ . a+b 2 (Vasile Cîrtoaje, 2006)

Solution. According to the well-known inequality (x + y + z)2 ≥ 3(x y + yz + z x), it suffices to show that X

p

x, y, z ≥ 0,

(b2 + ca)(c 2 + a b 3 ≥ . (c + a)(a + b) 2

Replacing a, b, c by a2 , b2 , c 2 , respectively, the inequality becomes X Æ 2 (b2 + c 2 ) (b4 + c 2 a2 )(c 4 + a2 b2 ) ≥ 3(a2 + b2 )(b2 + c 2 )(c 2 + a2 ). Multiplying the Cauchy-Schwarz inequalities Æ

(b2 + c 2 )(b4 + c 2 a2 ) ≥ b3 + ac 2 ,

Æ

(c 2 + b2 )(c 4 + a2 b2 ) ≥ c 3 + a b2 ,

we get Æ (b2 + c 2 ) (b4 + c 2 a2 )(c 4 + a2 b2 ) ≥ (b3 + ac 2 )(c 3 + a b2 ) = b3 c 3 + a(b5 + c 5 ) + a2 b2 c 2 . Therefore, it suffices to show that X X 2 b3 c 3 + 2 a(b5 + c 5 ) + 6a2 b2 c 2 ≥ 3(a2 + b2 )(b2 + c 2 )(c 2 + a2 ). This inequality is equivalent to X X X 2 b3 c 3 + 2 bc(b4 + c 4 ) ≥ 3 b2 c 2 (b2 + c 2 ), X

bc[2b2 c 2 + 2(b4 + c 4 ) − 3bc(b2 + c 2 )] ≥ 0, X bc(b − c)2 (2b2 + bc + 2c 2 ) ≥ 0.

The equality holds for a = b = c.

352

Vasile Cîrtoaje

P 2.63. If a, b, c are nonnegative real numbers, no two of which are zero,then p p p bc + 4a(b + c) ca + 4b(c + a) a b + 4c(a + b) 9 + + ≥ . b+c c+a a+b 2 (Vasile Cîrtoaje, 2006) Solution. Let us denote A = bc + 4a(b + c),

B = ca + 4b(c + a),

C = a b + 4c(a + b).

By squaring, the inequality becomes X

p X A BC 81 +2 ≥ . 2 (b + c) (c + a)(a + b) 4

Further, we use the following identity due to Sung-Yoon Kim: (b + c)2 BC − 4[a(b2 + c 2 ) + 2bc(b + c) + 3a bc]2 = a bc(b − c)2 (a + 4b + 4c), which yields 2a(b2 + c 2 ) + 4bc(b + c) + 6a bc , b+c P P p X 4 a(b2 + c 2 ) + 8 bc(b + c) + 36a bc BC 2 ≥ , (c + a)(a + b) (a + b)(b + c)c + a) P p X 12 bc(b + c) + 36a bc BC 2 ≥ . (c + a)(a + b) (a + b)(b + c)c + a) On the other hand, according to the known inequality Iran-1996, X a b + bc + ca 9 ≥ , 2 (b + c) 4 p

BC ≥

(see Remark from the proof of P 1.71), we have X a b + bc + ca X a X X a 9 A = + 3 ≥ + 3 . (b + c)2 (b + c)2 b+c 4 b+c Thus, it suffices to show that 3

X

P 12 bc(b + c) + 36a bc a + ≥ 18. b+c (a + b)(b + c)c + a)

This is equivalent to Schur’s inequality of degree three X X a3 + 3a bc ≥ bc(b + c). The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

Symmetric Nonrational Inequalities

353

P 2.64. If a, b, c are nonnegative real numbers, no two of which are zero,then p p p a a2 + 3bc b b2 + 3ca c c 2 + 3a b + + ≥ a + b + c. b+c c+a a+b (Cezar Lupu, 2006) Solution. Using the AM-GM inequality, we have p 2a(a2 + 3bc) a a2 + 3bc 2a(a2 + 3bc) 2a3 + 6a bc ≥ = p = , b+c S + 5bc 2 (b + c)2 (a2 + 3bc) (b + c)2 + (a2 + 3bc) where S = a2 + b2 + c 2 . Thus, it suffices to show that X 2a3 + 6a bc S + 5bc

≥ a + b + c.

Write this inequality as  X  2a2 + 6bc a − 1 ≥ 0, S + 5bc or, equivalently, AX + BY + X Z ≥ 0, where A=

1 , S + 5bc

X = a3 + a bc − a(b2 + c 2 ),

B=

1 , S + 5ca

C=

1 , S + 5a b

Y = b3 + a bc − b(c 2 + a2 ),

Z = c 3 + a bc − c(a2 + b2 ).

Without loss of generality, assume that a ≥ b ≥ c. We have A ≥ B ≥ C, X = a(a2 − b2 ) + ac(b − c) ≥ 0,

Z = c(c 2 − b2 ) + ac(b − a) ≤ 0

and, according to Schur’s inequality of third degree, X X X +Y +Z = a3 + 3a bc − a(b2 + c 2 ) ≥ 0. Therefore, AX + BY + C Z ≥ BX + BY + BZ = B(X + Y + Z) ≥ 0. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

354

Vasile Cîrtoaje

P 2.65. If a, b, c are nonnegative real numbers, no two of which are zero,then v v v t t t 2a(b + c) 2b(c + a) 2c(a + b) + + ≥ 2. (2b + c)(b + 2c) (2c + a)(c + 2a) (2a + b)(a + 2b) (Vasile Cîrtoaje, 2006) p

p

p

Solution. Making the substitution x = a, y = b, z = c, the inequality becomes v X t 2( y 2 + z 2 ) x ≥ 2. (2 y 2 + z 2 )( y 2 + 2z 2 ) We claim that

v t (2 y 2

2( y 2 + z 2 ) y +z ≥ 2 . 2 2 2 + z )( y + 2z ) y + yz + z 2

Indeed, be squaring and direct calculation, this inequality reduces to y 2 z 2 ( y − z)2 ≥ 0. Thus, it suffices to show that X

x( y + z) ≥ 2, + yz + z 2

y2

which is just the inequality in P 1.68. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 2.66. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then v v v v s s t bc t ab t bc t ab ca ca + + ≤ 1 ≤ + + . 3a2 + 6 3b2 + 6 3c 2 + 6 6a2 + 3 6b2 + 3 6c 2 + 3 (Vasile Cîrtoaje, 2011) Solution. By the Cauchy-Schwarz inequality, we have ‚

v Xt

bc 3a2 + 6

Œ2 ≤

X ‹ X a b + bc + ca 3a2 + 6

hence ‚

v Xt

bc 2 3a + 6

Œ2 ≤

X a2

‹ bc , a b + bc + ca

1 . +2

Therefore, to prove the original left inequality, it suffices to show that X a2

1 ≤ 1. +2

Symmetric Nonrational Inequalities

355

This inequality is equivalent to a2 ≥ 1. a2 + 2 Indeed, by the Cauchy-Schwarz inequality, we get X

X

a2 (a + b + c)2 (a + b + c)2 P P ≥ = = 1. a2 + 2 (a2 + 2) a2 + 6

The equality occurs for a = b = c = 1. By Hölder’s inequality, we have v Œ2 ‚ X t bc ”X — €X Š3 2 2 2 b c (6a + 3) ≥ bc . 6a2 + 3 To prove the original right inequality, it suffices to show that X (a b + bc + ca)3 ≥ b2 c 2 (6a2 + a b + bc + ca), which is equivalent to (a b + bc + ca)[(a b + bc + ca)2 −

X

b2 c 2 ] ≥ 18a2 b2 c 2 ,

2a bc(a b + bc + ca)(a + b + c) ≥ 18a2 b2 c 2 , X 2a bc a(b − c)2 ≥ 0. The equality occurs for a = b = c = 1, and for a = 0 and bc = 3 (or any cyclic permutation).

P 2.67. Let a, b, c be nonnegative real numbers such that a b + bc + ca = 3. If k > 1, than a k (b + c) + b k (c + a) + c k (a + b) ≥ 6. Solution. Let E = a k (b + c) + b k (c + a) + c k (a + b). We consider two cases. Case 1: k ≥ 2. Applying Jensen’s inequality to the convex function f (x) = x k−1 , x ≥ 0, we get E = (a b + ac)a k−1 + (bc + ba)b k−1 + (ca + c b)c k−1 ˜ • (a b + ac)a + (bc + ba)b + (ca + c b)c k−1 ≥ 2(a b + bc + ca) 2(a b + bc + ca)  2 k−1 a (b + c) + b2 (c + a) + c 2 (a + b) =6 . 6

356

Vasile Cîrtoaje

Thus, it suffices to show that a2 (b + c) + b2 (c + a) + c 2 (a + b) ≥ 6. Write this inequality as (a b + bc + ca)(a + b + c) − 3a bc ≥ 6, a + b + c ≥ 2 + a bc. It is true since

Æ

a+b+c ≥ and

3(a b + bc + ca) = 3

a+b+c a bc ≤ 3 

‹3

= 1.

Case 2: 1 < k ≤ 2. We have E = a k−1 (3 − bc) + b k−1 (3 − ca) + c k−1 (3 − a b)   = 3(a k−1 + b k−1 + c k−1 ) − a k−1 b k−1 c k−1 (a b)2−k + (bc)2−k + (ca)2−k . Since 0 ≤ 2 − k < 1, f (x) = x 2−k is concave for x ≥ 0. Thus, by Jensen’s inequality, we have ‹  a b + bc + ca 2−k (a b)2−k + (bc)2−k + (ca)2−k ≤ 3 = 3, 3 and hence E ≥ 3(a k−1 + b k−1 + c k−1 ) − 3a k−1 b k−1 c k−1 . Consequently, it suffices to show that a k−1 + b k−1 + c k−1 ≥ a k−1 b k−1 c k−1 + 2. Due to symmetry, we may assume that a ≥ b ≥ c. In addition, write the inequality as a k−1 + b k−1 − 2 ≥ (a k−1 b k−1 − 1)c k−1 , a

k−1

+b

k−1

− 2 ≥ (a

Let x=

p

k−1 k−1

b

3 − ab − 1) a+b

a b, 1 ≤ x ≤



p 3.

By the AM-GM inequality, we have a + b ≥ 2x,

a k−1 + b k−1 ≥ 2x k−1 .

‹k−1

.

Symmetric Nonrational Inequalities

357

Thus, it suffices to show that 2(x

k−1

− 1) ≥ (x

2k−2

3 − x2 − 1) 2x 

k−1 .

Since x ≥ 1, this is true if 2 ≥ (x

k−1

3 − x2 + 1) 2x 

k−1 ,

which can be written as  2≥

3 − x2 2

Since 1≥

k−1

3 − x2 + 2x 

k−1 .

3 − x2 3 − x2 ≥ , 2 2x

the conclusion follows. Thus, the proof is completed. The equality holds for a = b = c = 1.

P 2.68. Let a, b, c be nonnegative real numbers such that a + b + c = 2. If 2 ≤ k ≤ 3, than a k (b + c) + b k (c + a) + c k (a + b) ≤ 2. Solution. Denote by Ek (a, b, c) the left hand side of the inequality, assume that a ≤ b ≤ c and show that Ek (a, b, c) ≤ Ek (0, a + b, c) ≤ 2. The left inequality is equivalent to a b k−1 (a + b k−1 ) ≤ (a + b)k − a k − b k . c Clearly, it suffices to consider c = b, when the inequality becomes 2a k + b k−1 (a + b) ≤ (a + b)k . Since 2a k ≤ a k−1 (a + b), it remains to show that a k−1 + b k−1 ≤ (a + b)k−1 ,

358

Vasile Cîrtoaje

which is true since  ‹k−1 b a b a k−1 + b k−1  a k−1 = + ≤ + = 1. k−1 (a + b) a+b a+b a+b a+b The right inequality, Ek (0, a + b, c) ≤ 2, is equivalent to cd(c k−1 + d k−1 ) ≤ 2, where d = a + b, and hence c + d = 2. By the Power-Mean inequality (or Jensen’s k−1 inequality applied to the concave function t 2 ), we have 

c k−1 + d k−1 2 c

k−1

+d

1/(k−1)

k−1

 ≤

c2 + d 2 2

c2 + d 2 ≤2 2 

1/2 ,

(k−1)/2 .

Thus, it suffices to show that  cd which is equivalent to

c2 + d 2 2

(k−1)/2 ≤ 1,

cd(2 − cd)(k−1)/2 ≤ 1.

Since 2 − cd ≥ 1, we have cd(2 − cd)(k−1)/2 ≤ cd(2 − cd) = 1 − (1 − cd)2 ≤ 1. The equality holds for a = 0 and b = c = 1 (or any cyclic permutation).

P 2.69. Let a, b, c be nonnegative real numbers, no two of which are zero. If m > n ≥ 0, than bm + c m c m + am am + bm (b + c − 2a) + (c + a − 2b) + (a + b − 2c) ≥ 0. bn + c n c n + an an + bn (Vasile Cîrtoaje, 2006) Solution. Write the inequality as AX + BY + C Z ≥ 0,

Symmetric Nonrational Inequalities where

359

c m + am am + bm bm + c m , B = , C = , bn + c n c n + an an + bn X = b + c − 2a, Y = c + a − 2b, Z = a + b − 2c, X + Y + Z = 0. A=

Without loss of generality, assume that a ≤ b ≤ c, which involves X ≥ Y ≥ Z and X ≥ 0. Since 2(AX + BY + C Z) = (2A − B − C)X + (B + C)X + 2(BY + C Z) = (2A − B − C)X − (B + C)(Y + Z) + 2(BY + C Z) = (2A − B − C)X + (B − C)(Y − Z), it suffices to show that B ≥ C and 2A − B − C ≥ 0. The inequality B ≥ C can be written as b n c n (c m−n − b m−n ) + a n (c m − b m ) − a m (c n − b n ) ≥ 0, b n c n (c m−n − b m−n ) + a n [c m − b m − a m−n (c n − b n )] ≥ 0. This is true since c m−n ≥ b m−n and c m − b m − a m−n (c n − b n ) ≥ c m − b m − b m−n (c n − b n ) = c n (c m−n − b m−n ) ≥ 0. The inequality 2A − B − C ≥ 0 follows from 2A ≥ b m−n + c m−n ,

b m−n ≥ C,

c m−n ≥ B.

Indeed, we have 2A − b m−n − c m−n = b m−n − C =

(b n − c n )(b m−n − c m−n ) ≥ 0, bn + c n

a n (b m−n − a m−n ) ≥ 0, an + bn

a n (c m−n − a m−n ) ≥ 0. c n + an The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation). c m−n − B =

P 2.70. Let a, b, c be positive real numbers such that a bc = 1. Prove that p p p a2 − a + 1 + b2 − b + 1 + c 2 − c + 1 ≥ a + b + c. (Vasile Cîrtoaje, 2012)

360

Vasile Cîrtoaje

First Solution. Among a − 1, b − 1 and c − 1 there are two with the same sign. Let (b − 1)(c − 1) ≥ 0, that is, 1 t ≤ , t = b + c − 1. a By Minkowsky’s inequality, we have v v u u ‹2 ‹ p p t 3 t 1 1 2 3 2 2 + + + b − b+1+ c −c+1= b− c− 2 4 2 4 p ≥ t 2 + 3. Thus, it suffices to show that p p a2 − a + 1 + t 2 + 3 ≥ a + b + c, which is equivalent to p

a2 − a + 1 + f (t) ≥ a + 1,

where f (t) =

p

t 2 + 3 − t.

Clearly, f (t) is decreasing for t ≤ 0. Since f (t) = p

3

, t2 + 3 + t  ‹ 1 , and it suffices to show that f (t) is also decreasing for t ≥ 0. Then, f (t) ≥ f a  ‹ p 1 a2 − a + 1 + f ≥ a + 1, a which is equivalent to p

v t1 1 a2 − a + 1 + + 3 ≥ a + + 1. a2 a

By squaring, this inequality becomes v  ‹ t 1 2 2 2 (a − a + 1) 2 + 3 ≥ 3a + − 1. a a Indeed, by the Cauchy-Schwarz inequality, we have v  ‹ v  ‹ t t 1 1 2 2 (a − a + 1) 2 + 3 = [(2 − a)2 + 3a2 ] 2 + 3 a a 2−a 2 ≥ + 3a = 3a + − 1. a a

Symmetric Nonrational Inequalities

361

The equality holds for a = b = c. Second Solution. If the inequality p

x2

1 − x +1− x ≥ 2



3 −1 2 x + x +1

‹

holds for all x > 0, then it suffices to prove that 1 1 1 + + ≥ 1, a2 + a + 1 b2 + b + 1 c 2 + c + 1 which is just the known inequality in P 1.44. Indeed, the above inequality is equivalent to 1− x (1 − x)(2 + x) , ≥ p 2 2(x 2 + x + 1) x − x +1+ x p (x − 1)[(x + 2) x 2 − x + 1 − x 2 − 2] ≥ 0, 3x 2 (x − 1)2 ≥ 0. p (x + 2) x 2 − x + 1 + x 2 + 2

P 2.71. Let a, b, c be positive real numbers such that a bc = 1. Prove that p p p 16a2 + 9 + 16b2 + 9 + 16b2 + 9 ≥ 4(a + b + c) + 3. (MEMO, 2012) First Solution (by Vo Quoc Ba Can). Since p

16a2 + 9 − 4a = p

9 16a2

+ 9 + 4a

,

the inequality is equivalent to 1

X p

16a2 + 9 + 4a

≥

1 . 3

By the AM-GM inequality, we have 2

p

16a2 + 9 ≤

16a2 + 9 + 2a + 3, 2a + 3

p 16a2 + 9 18(2a2 + 2a + 1) 2( 16a2 + 9 + 4a) ≤ + 10a + 3 = . 2a + 3 2a + 3

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Vasile Cîrtoaje

Thus, it suffices to show that X

2a + 3 ≥ 3. 2a2 + 2a + 1

If the inequality

2a + 3 3 ≥ 8/5 2a2 + 2a + 1 a + a4/5 + 1 holds for all a > 0, then it suffices to show that X 1 ≥ 1, 8/5 a + a4/5 + 1

which follows immediately from the inequality in P 1.44. Therefore, using the substitution x = a1/5 , x > 0, we need to show that 3 2x 5 + 3 ≥ 8 , 10 5 2x + 2x + 1 x + x4 + 1 which is equivalent to 2x 4 (x 5 − 3x 2 + x + 1) + x 4 − 4x + 3 ≥ 0. This is true since, by the AM-GM inequality, we have p 3 x 5 + x + 1 ≥ 3 x 5 · x · 1 = 3x 2 and x4 + 3 = x4 + 1 + 1 + 1 ≥ 4

p 4

x 4 · 1 · 1 · 1 = 4x.

The equality holds for a = b = c = 1. Second Solution. Making the substitution p p p x = 16a2 + 9 − 4a, y = 16b2 + 9 − 4b, z = 16c 2 + 9 − 4c, which involves a=

9 − x2 , 8x

b=

9 − y2 , 8y

c=

9 − z2 , 8z

we need to show that (9 − x 2 )(9 − y 2 )(9 − z 2 ) = 512x yz yields x + y + z ≥ 3. Use the contradiction method. Assume that x + y + z < 3 and show that (9 − x 2 )(9 − y 2 )(9 − z 2 ) > 512x yz.

x, y, z > 0,

Symmetric Nonrational Inequalities

363

According to the AM-GM inequality, we get p p p 4 4 3 + x = 1 + 1 + 1 + x ≥ 4 x, 3 + y ≥ 4 4 y, 3 + z ≥ 4 z, hence

p (3 + x)(3 + y)(3 + z) ≥ 64 4 x yz.

Therefore, it suffices to prove that (3 − x)(3 − y)(3 − z) > 8(x yz)3/4 . By the AM-GM inequality, 1>

 x + y + z 3 ≥ x yz, 3

and hence (3 − x)(3 − y)(3 − z) = 9(3 − x − y − z) + 3(x y + yz + z x) − x yz > 3(x y + yz + z x) − x yz ≥ 9(x yz)2/3 − x yz > 8(x yz)2/3 > 8(x yz)3/4 .

P 2.72. Let a, b, c be positive real numbers such that a bc = 1. Prove that p p p 25a2 + 144 + 25b2 + 144 + 25c 2 + 144 ≤ 5(a + b + c) + 24. (Vasile Cîrtoaje, 2012) First Solution. Since p

144 25a2 + 144 − 5a = p , 25a2 + 144 + 5a

the inequality is equivalent to X

1 1 ≤ . p 6 25a2 + 144 + 5a

If the inequality 1 1 ≤ p p 25a2 + 144 + 5a 6 5a18/13 + 4 holds for all a > 0, then it suffices to show that 1

X p

5a18/13 + 4

≤ 1,

364

Vasile Cîrtoaje

which follows immediately from P 2.31. Therefore, using the substitution x = a1/13 , x > 0, we only need to show that p p 25x 26 + 144 + 5x 13 ≥ 6 5x 18 + 4. By squaring, the inequality becomes p 10x 13 ( 25x 26 + 144 + 5x 13 − 18x 5 ) ≥ 0. This is true if 25x 26 + 144 ≥ (18x 5 − 5x 13 )2 , which is equivalent to 5x 18 + 4 ≥ 9x 10 . By the AM-GM inequality, we have 5x 18 + 4 = x 18 + x 18 + x 18 + x 18 + x 18 + 1 + 1 + 1 + 1 p 9 ≥ 9 x 18 · x 18 · x 18 · x 18 · x 18 · 1 · 1 · 1 · 1 = 9x 10 . The equality holds for a = b = c = 1. Second Solution. Making the substitution p p p 8x = 25a2 + 144 − 5a, 8 y = 25b2 + 144 − 5b, 8z = 25c 2 + 144 − 5c, which involves 9 − 4x 2 a= , 5x

9 − 4 y2 b= , 5y

9 − 4z 2 c= , 5z

 ‹ 3 x, y, z ∈ 0, , 2

we need to show that (9 − 4x 2 )(9 − 4 y 2 )(9 − 4z 2 ) = 125x yz involves x + y + z ≤ 3. Use the contradiction method. Assume that x + y + z > 3 and show that (9 − 4x 2 )(9 − 4 y 2 )(9 − 4z 2 ) < 125x yz. Since 9 − 4x 2 < 3(x + y + z) −

3( y + z − x)( y + z + 3x) 12x 2 = , x + y +z x + y +z

it suffices to show that 27AB ≤ 125x yz(x + y + z)3 ,

Symmetric Nonrational Inequalities

365

where A = ( y + z − x)(z + x − y)(x + y − z),

B = ( y + z + 3x)(z + x + 3 y)(x + y + 3z).

Consider the nontrivial case A ≥ 0. By the AM-GM inequality, we have B≤

125 (x + y + z)3 . 27

Therefore, it suffices to show that A ≤ x yz, which is a well known inequality (equivalent to Schur’s inequality of degree three).

P 2.73. If a, b are positive real numbers such that a b + bc + ca = 3, then p

(a)

p

(b)

a2 + 3 + a+b+

p

p

b2 + 3 +

b+c+

p

p

b2 + 3 ≥ a + b + c + 3;

c+a≥

p

4(a + b + c) + 6. (Lee Sang Hoon, 2007)

Solution. (a) First Solution (by Pham Thanh Hung). By squaring, the inequality becomes XÆ (b2 + 3)(c 2 + 3) ≥ 3(1 + a + b + c). Since (b2 + 3)(c 2 + 3) = (b + c)(b + a)(c + a)(c + b) = (b + c)2 (a2 + 3) ≥

1 (b + c)2 (a + 3)2 , 4

we have XÆ X Š 1X 1€ X (b2 + 3)(c 2 + 3) ≥ (b + c)(a + 3) = 2 bc + 6 a = 3(1 + a + b + c). 2 2 The equality holds for a = b = c = 1. Second Solution. Write the inequality as follows: Æ Æ Æ (a + b)(a + c) + (b + c)(b + a) + (c + a)(c + b) ≥ a + b + c + 3, ” — X €p Š2 Æ p 2 a + b + c − 3(a b + bc + ca) ≥ a+b− a+c , 1 a+b+c+

p

3(a b + bc + ca)

X X (b − c)2 ≥ p

(b − c)2
2 , p a+b+ a+c

366

Vasile Cîrtoaje X

where €p

Sa =

a+b+

p

Sa (b − c)2
2 ≥ 0, p p a+b+ a+c a+c

Š2

−a−b−c−

Æ

(a + b)(a + c) −

Æ

3(a b + bc + ca).

The last inequality is true since Sa = 3(a + b + c) + 2 >2

Æ

Æ

a2 + (a b + bc + ca) −

Æ

3(a b + bc + ca)

3(a b + bc + ca) > 0.

Third Solution. Use the substitution p p p x = a2 + 3 − a, y = b2 + 3 − b, z = c 2 + 3 − c,

x, y, z > 0.

We need to show that x + y + z ≥ 3. We have

X

X Æ Æ [ (b + a)(b + c) − b][ (c + a)(c + b) − c] X X Æ X Æ X Æ = (b + c) (a + b)(a + c) − b (c + a)(c + b) − c (b + a)(b + c) + bc X = bc = 3. yz =

Thus, we get x + y +z ≥

Æ

3(x y + yz + z x) = 3.

(b) By squaring, we get the inequality in (a). Otherwise, using the substitution x=

p

b + c,

y=

p

c + a, z =

p

a + b,

the inequality becomes x + y +z ≥

Ç

2(x 2 + y 2 + z 2 ) +

Æ

3(2x 2 y 2 + 2 y 2 z 2 + 2z 2 x 2 − x 4 − y 4 − z 4 ).

By squaring two times, we get Æ 2(x y + yz + z x) − x 2 − y 2 − z 2 ≥ 3(2x 2 y 2 + 2 y 2 z 2 + 2z 2 x 2 − x 4 − y 4 − z 4 ), X (x − y)2 (x + y − z)2 ≥ 0.

Symmetric Nonrational Inequalities

367

P 2.74. If a, b, c are nonnegative real numbers such that a + b + c = 3, then Æ

(5a2 + 3)(5b2 + 3) +

Æ

(5b2 + 3)(5c 2 + 3) +

Æ

(5c 2 + 3)(5a2 + 3) ≥ 24. (Nguyen Van Quy, 2012)

Solution. Assume that a ≥ b ≥ c, which involves 1 ≤ a ≤ 3 and b + c ≤ 2. Denote A = 5a2 + 3,

B = 5b2 + 3,

C = 5c 2 + 3,

and write the inequality as follows: p p p p A ( B + C) + BC ≥ 24, p q p p A · A(B + C + 2 BC ) ≥ 24 − BC. This is true if

p p A(B + C + 2 BC) ≥ (24 − BC)2 ,

which is equivalent to A(A + B + C + 48) ≥ (A + 24 −

p

BC)2 .

Applying Lemma below for k = 5/3 and m = 4/15 yields p 5 BC ≥ 4(b − c)2 + 25bc + 15. Therefore, it suffices to show that 25A(A + B + C + 48) ≥ [5A + 120 − 4(b − c)2 − 25bc − 15]2 , which is equivalent to 25(5a2 + 3)[5(a2 + b2 + c 2 ) + 57] ≥ [25a2 + 120 − 4(b − c)2 − 25bc]2 . Since 5(a2 + b2 + c 2 ) + 57 = 5a2 + 5(b + c)2 − 10bc + 57 = 2(5a2 − 15a + 51 − 5bc) and 25a2 + 120 − 4(b − c)2 − 25bc = 25a2 + 120 − 4(b + c)2 − 9bc = 3(7a2 + 8a + 28 − 3bc), we need to show that 50(5a2 + 3)(5a2 − 15a + 51 − 5bc) ≥ 9(7a2 + 8a + 28 − 3bc)2 .

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Vasile Cîrtoaje

From bc ≤ (b + c)2 /4 and (a − b)(a − c) ≥ 0, we get bc ≤ (3 − a)2 /4,

bc ≥ a(b + c) − a2 = 3a − 2a2 .

Consider a fixed, a ≥ 1, and denote x = bc. So, we only need to prove that f (x) ≥ 0 for a2 − 6a + 9 , 3a − 2a2 ≤ x ≤ 4 where f (x) = 50(5a2 + 3)(5a2 − 15a + 51 − 5x) − 9(7a2 + 8a + 28 − 3x)2 .  2  a − 6a + 9 2 Since f is concave, it suffices to show that f (3a − 2a ) ≥ 0 and f ≥ 0. 4 Indeed, f (3a − 2a2 ) = 3(743a4 − 2422a3 + 2813a2 − 1332a + 198) = 3(a − 1)2 [(a − 1)(743a − 193) + 5] ≥ 0,  f

a2 − 6a + 9 4



375 (25a4 − 140a3 + 286a2 − 252a + 81) 16 375 = (a − 1)2 (5a − 9)2 ≥ 0. 16

=

Thus, the proof is completed. The equality holds for a = b = c = 1, and also for a = 9/5 and b = c = 3/5 (or any cyclic permutation). Lemma. Let b, c ≥ 0 such that b + c ≤ 2. If k > 0 and 0 ≤ m ≤ Æ

k , then 2k + 2

(k b2 + 1)(kc 2 + 1) ≥ m(b − c)2 + k bc + 1.

Proof. By squaring, the inequality becomes (b − c)2 [k − 2m − 2kmbc − m2 (b − c)2 ] ≥ 0. This is true since k − 2m − 2kmbc − m2 (b − c)2 = k − 2m − 2m(k − 2m)bc − m2 (b + c)2 m(k − 2m) (b + c)2 − m2 (b + c)2 2 km = k − 2m − (b + c)2 ≥ k − 2m − 2km ≥ 0. 2 ≥ k − 2m −

Symmetric Nonrational Inequalities

369

P 2.75. If a, b, c are nonnegative real numbers such that a + b + c = 3, then p

a2 + 1 +

p

b2 + 1 +

p

c2 + 1 ≥

v t 4(a2 + b2 + c 2 ) + 42 3

.

(Vasile Cîrtoaje, 2014) Solution. Assume that a ≥ b ≥ c, which involves a ≥ 1 and b + c ≤ 2. By squaring, the inequality becomes p p p p a2 + b2 + c 2 + 33 A ( B + C) + BC ≥ , 6 q

p p a2 + b2 + c 2 + 33 A(B + C + 2 BC ) + BC ≥ , 6

where A = a2 + 1,

B = b2 + 1,

C = c 2 + 1.

Applying Lemma from the preceding problem P 2.74 for k = 1 and m = 1/4 gives p

BC ≥

1 (b − c)2 + bc + 1. 4

Therefore, it suffices to show that v t

1 1 a2 + b2 + c 2 + 33 A[B + C + (b − c)2 + 2bc + 2] + (b − c)2 + bc + 1 ≥ , 2 4 6

which is equivalent to 6

Æ 6

2(a2 + 1)[3(b + c)2 + 8 − 4bc] ≥ 2a2 − (b + c)2 + 54 − 4bc,

Æ

2(a2 + 1)(3a2 − 18a + 35 − 4bc) ≥ a2 + 6a + 45 − 4bc.

From bc ≤ (b + c)2 /4 and (a − b)(a − c) ≥ 0, we get bc ≤ (3 − a)2 /4,

bc ≥ a(b + c) − a2 = 3a − 2a2 .

Consider a fixed, a ≥ 1, and denote x = bc. So, we only need to prove that f (x) ≥ 0 for a2 − 6a + 9 3a − 2a2 ≤ x ≤ , 4 where f (x) = 72(a2 + 1)(3a2 − 18a + 35 − 4x) − (a2 + 6a + 45 − 4x)2 .

370

Vasile Cîrtoaje

Since f is concave, it suffices to show that f (3a − 2a ) ≥ 0 and f 2



a2 − 6a + 9 4

 ≥ 0.

Indeed, f (3a − 2a2 ) = 9(79a4 − 228a3 + 274a2 − 180a + 55) = 9(a − 1)2 (79a2 − 70a + 55 ≥ 0,  f

a2 − 6a + 9 4



= 144(a4 − 6a3 + 13a2 − 12a + 4) = 144(a − 1)2 (a − 2)2 ≥ 0.

The equality holds for a = b = c = 1, and also for a = 2 and b = c = 1/2 (or any cyclic permutation).

P 2.76. If a, b, c are nonnegative real numbers such that a + b + c = 3, then p p p p (a) a2 + 3 + b2 + 3 + c 2 + 3 ≥ 2(a2 + b2 + c 2 ) + 30; p p p p (b) 3a2 + 1 + 3b2 + 1 + 3c 2 + 1 ≥ 2(a2 + b2 + c 2 ) + 30. (Vasile Cîrtoaje, 2014) Solution. Assume that a ≥ b ≥ c, which involves a ≥ 1 and b + c ≤ 2. (a) By squaring, the inequality becomes p p p p a2 + b2 + c 2 + 21 A ( B + C) + BC ≥ , 2 q

p p a2 + b2 + c 2 + 21 A(B + C + 2 BC ) + BC ≥ , 2

where A = a2 + 3,

B = b2 + 3,

C = c 2 + 3.

Applying Lemma from problem P 2.74 for k = 1/3 and m = 1/9 gives p

BC ≥

1 (b − c)2 + bc + 3. 3

Therefore, it suffices to show that v t 2 1 a2 + b2 + c 2 + 21 A[B + C + (b − c)2 + 2bc + 6] + (b − c)2 + bc + 3 ≥ , 3 3 2

Symmetric Nonrational Inequalities

371

which is equivalent to 2

Æ

3(a2 + 3)[5(b + c)2 + 36 − 8bc] ≥ 3a2 + (b + c)2 + 45 − 4bc,

Æ

3(a2 + 3)(5a2 − 30a + 81 − 8bc) ≥ 2a2 − 3a + 27 − 2bc.

From bc ≤ (b + c)2 /4 and (a − b)(a − c) ≥ 0, we get bc ≤ (3 − a)2 /4,

bc ≥ a(b + c) − a2 = 3a − 2a2 .

Consider a fixed, a ≥ 1, and denote x = bc. So, we only need to prove that f (x) ≥ 0 for a2 − 6a + 9 3a − 2a2 ≤ x ≤ , 4 where f (x) = 3(a2 + 3)(5a2 − 30a + 81 − 8x) − (2a2 − 3a + 27 − 2x)2 .  2  a − 6a + 9 2 Since f is concave, it suffices to show that f (3a − 2a ) ≥ 0 and f ≥ 0. 4 Indeed, f (3a − 2a2 ) = 27a2 (a − 1)2 ≥ 0,  f

a2 − 6a + 9 4



27 4 (a − 8a3 + 22a2 − 24a + 9) 4 27 = (a − 1)2 (a − 3)2 ≥ 0. 4 =

The equality holds for a = b = c = 1, and also for a = 3 and b = c = 0 (or any cyclic permutation). (b) By squaring, the inequality becomes p p p p 27 − a2 − b2 − c 2 A ( B + C) + BC ≥ , 2 q

p p 27 − a2 − b2 − c 2 A(B + C + 2 BC ) + BC ≥ , 2

where A = 3a2 + 1,

B = 3b2 + 1,

C = 3c 2 + 1.

Applying Lemma from problem P 2.74 for k = 3 and m = 1/3 gives p

BC ≥

1 (b − c)2 + 3bc + 1. 3

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Vasile Cîrtoaje

Therefore, it suffices to show that v t 1 27 − a2 − b2 − c 2 2 , A[B + C + (b − c)2 + 6bc + 2] + (b − c)2 + 3bc + 1 ≥ 3 3 2 which is equivalent to Æ 2 3(3a2 + 1)[11(b + c)2 + 12 − 8bc] ≥ 75 − 3a2 − 5(b + c)2 − 4bc, Æ 3(3a2 + 1)(11a2 − 66a + 111 − 8bc) ≥ 15 + 15a − 4a2 − 2bc. From bc ≤ (b + c)2 /4 and (a − b)(a − c) ≥ 0, we get bc ≤ (3 − a)2 /4,

bc ≥ a(b + c) − a2 = 3a − 2a2 .

Consider a fixed, a ≥ 1, and denote x = bc. So, we only need to prove that f (x) ≥ 0 for a2 − 6a + 9 , 3a − 2a2 ≤ x ≤ 4 where f (x) = 3(3a2 + 1)(11a2 − 66a + 111 − 8x) − (15 + 15a − 4a2 − 2x)2 .   2 a − 6a + 9 2 ≥ 0. Since f is concave, it suffices to show that f (3a − 2a ) ≥ 0 and f 4 Indeed, f (3a − 2a2 ) = 27(a − 1)2 (3a − 2)2 ≥ 0,  f

a2 − 6a + 9 4



27 (9a4 − 48a3 + 94a2 − 80a + 25) 4 27 = (a − 1)2 (3a − 5)2 ≥ 0. 4 =

The equality holds for a = b = c = 1, and also for a = 5/3 and b = c = 2/3 (or any cyclic permutation). Remark. Similarly, we can prove the following generalization. • Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k > 0, then v t 8k(a2 + b2 + c 2 ) + 3(9k2 + 10k + 9) p p p ka2 + 1 + k b2 + 1 + kc 2 + 1 ≥ , 3(k + 1) with equality for a = b = c = 1, and also for a = cyclic permutation).

3k − 1 3k + 1 and b = c = (or any 2k 4k

Symmetric Nonrational Inequalities

373

P 2.77. If a, b, c are nonnegative real numbers such that a + b + c = 3, then Æ Æ Æ (32a2 + 3)(32b2 + 3) + (32b2 + 3)(32c 2 + 3) + (32c 2 + 3)(32a2 + 3) ≤ 105. (Vasile Cîrtoaje, 2014) Solution. Assume that a ≤ b ≤ c, which involves a ≤ 1 and b + c ≥ 2. Denote A = 32a2 + 3,

B = 32b2 + 3,

C = 32c 2 + 3,

and write the inequality as follows: p p p p A ( B + C) + BC ≤ 105, p Æ p p A · B + C + 2 BC ≤ 105 − BC. By Lemma below, we have p BC ≤ 5(b + c)2 + 12bc + 3 ≤ 8(b + c)2 + 3 ≤ 8(a + b + c)2 + 3 = 75 < 105. Therefore, we can write the desired inequality as p p A(B + C + 2 BC) ≤ (105 − BC)2 , which is equivalent to A(A + B + C + 210) ≤ (A + 105 −

p

BC)2 .

According to Lemma below, it suffices to show that A(A + B + C + 210) ≤ [A + 105 − 5(b2 + c 2 ) − 22bc − 3]2 , which is equivalent to [32a2 + 105 − 5(b2 + c 2 ) − 22bc]2 ≥ (32a2 + 3)[32(a2 + b2 + c 2 ) + 219]. Since 32(a2 + b2 + c 2 ) + 219 = 32a2 + 32(b + c)2 − 64bc + 219 = 64a2 − 192a + 507 − 64bc and 32a2 +105−5(b2 +c 2 )−22bc = 32a2 +105−5(b+c)2 −12bc = 3(9a2 +10a+20−4bc), we need to show that 9(9a2 + 10a + 20 − 4bc)2 ≥ (32a2 + 3)(64a2 − 192a + 507 − 64bc).

374

Vasile Cîrtoaje

From bc ≤ (b + c)2 /4, we get

bc ≤ (3 − a)2 /4.

Consider a fixed, 0 ≤ a ≤ 1, and denote x = bc. So, we only need to prove that f (x) ≥ 0 for a2 − 6a + 9 x≤ , 4 where f (x) = 9(9a2 + 10a + 20 − 4x)2 − (32a2 + 3)(64a2 − 192a + 507 − 64x). Since f 0 (x) = 72(4x − 9a2 − 10a − 20) + 64(32a2 + 3) ≤ 72[(a2 − 6a + 9) − 9a2 − 10a − 20) + 64(32a2 + 3) = 8[184a(a − 1) + (44a − 75)] < 0,  2  a − 6a + 9 f is decreasing, hence f (x) ≥ f . Therefore, it suffices to show that 4  2  a − 6a + 9 f ≥ 0. We have 4  2  a − 6a + 9 f =9[9a2 + 10a + 20 − (a2 − 6a + 9)]2 4 − (32a2 + 3)[64a2 − 192a + 507 − 16(a2 − 6a + 9)] =9(8a2 + 16a + 11)2 − (32a2 + 3)(48a2 − 96a + 363) =192a(a − 1)2 (18 − 5a) ≥ 0. Thus, the proof is completed. The equality holds for a = b = c = 1, and also for a = 0 and b = c = 3/2 (or any cyclic permutation). Lemma. If b, c ≥ 0 such that b + c ≥ 2, then Æ (32b2 + 3)(32c 2 + 3) ≤ 5(b2 + c 2 ) + 22bc + 3. Proof. By squaring, the inequality becomes (5b2 + 5c 2 + 22bc)2 − 322 b2 c 2 ≥ 96(b2 + c 2 ) − 6(5b2 + 5c 2 + 22bc), 5(b − c)2 (5b2 + 5c 2 + 54bc) ≥ 66(b − c)2 . It suffices to show that 5(5b2 + 5c 2 + 10bc) ≥ 100, which is equivalent to the obvious inequality (b + c)2 ≥ 4.

Symmetric Nonrational Inequalities

375

P 2.78. If a, b, c are positive real numbers, then c + a b+c a + b a − 3 + b − 3 + c − 3 ≥ 2. (Vasile Cîrtoaje, 2012) Solution. Without loss of generality, assume that a ≥ b ≥ c. Case 1: a > b + c. We have b+c a + b c + a b+c b+c a − 3 + c − 3 + b − 3 ≥ a − 3 = 3 − a > 2. Case 2: a ≤ b + c. We have b+c a + b c + a c + a b+c + + + − 3 − 3 − 3 ≥ − 3 − 3 a c b a b  ‹  b+c c + a 2b b + a (a − b)(2b − a) = 3− + 3− ≥6− − =2+ ≥ 2. a b a b ab a Thus, the proof is completed. The equality holds for = b = c (or any cyclic permuta2 tion).

P 2.79. If a, b, c are real numbers such that a bc 6= 0, then b + c c + a a + b + ≥ 2. + a b c First Solution. Let |a| = max{|a|, |b|, |c|}. We have

b + c c + a a + b b + c c + a a + b + ≥ + + + a b c a a a ≥

|(−b − c) + (c + a) + (a + b)| = 2. |a|

The equality holds for a = 1, b = −1 and |c| ≤ 1 (or any permutation). Second Solution. Since the inequality remains unchanged by replacing a, b, c with −a, −b, −c, it suffices to consider two cases: a, b, c > 0, and a < 0 and b, c > 0.

376

Vasile Cîrtoaje

Case 1: a, b, c > 0. We have b + c c + a a + b  a b ‹  b c ‹  c a  + = + + + + + ≥ 6. + a b c b a c b a c Case 2: a < 0 and b, c > 0. Replacing a by −a, we need to show that b + c |a − c| |a − b| + + ≥2 a b c for all a, b, c > 0. Without loss of generality, assume that b ≥ c. For b ≥ c ≥ a, we have b + c |a − c| |a − b| b+c + + ≥ ≥ 2. a b c a For b ≥ a ≥ c, we have b + c |a − c| |a − b| b+c a−c (a − b)2 + c(b − a) + + −2≥ + −2= ≥ 0. a b c a b ab For a ≥ b ≥ c, we have b+c a−c a−b b + c |a − c| |a − b| + + = + + a b c a b c  ‹  ‹ a b a−b 1 1 (a − b)2 (a − b)(a b − c 2 ) + −2 + +c − = + ≥ 0. = b a c a b ab a bc Third Solution. Using the substitution x=

b+c , a

y=

c+a a+b , z= , b c

we need to show that x + y + z + 2 = x yz,

x, y, z ∈ R

involves |x| + | y| + |z| ≥ 2. If x yz ≤ 0, then

−x − y − z = 2 − x yz ≥ 2,

hence |x| + | y| + |z| ≥ |x + y + z| = | − x − y − z| ≥ −x − y − z ≥ 2. If x yz > 0, then either x, y, z > 0 or only one of x, y, z is positive (for instance, x > 0 and y, z < 0).

Symmetric Nonrational Inequalities

377

Case 1: x, y, z > 0. We need to show that x + y + z ≥ 2. We have x yz = x + y + z + 2 > 2 and, by the AM-GM inequality, we get p p 3 x + y + z ≥ 3 3 x yz > 3 2 > 2,

Case 2: x > 0 and y, z < 0. Replacing y, z by − y, −z, we need to prove that x − y − z + 2 = x yz involves x + y +z ≥2 for all x, y, z > 0. Since x + y + z − 2 = x + y + z − (x yz − x + y + z) = x(2 − yz), we need to show that yz ≤ 2. Indeed, we have p x + 2 = y + z + x yz ≥ 2 yz + x yz, p x(1 − yz) + 2(1 − yz ) ≥ 0, p p (1 − yz )[x(1 + yz ) + 2] ≥ 0, hence yz ≤ 1 < 2.

P 2.80. Let a, b, c be nonnegative real numbers, no two of which are zero, and let x=

2a , b+c

y=

2b 2c , z= . c+a a+b

Prove that (a) (b)

x + y +z+ p

x+

p

p

xy+

y+

p

yz +

p z x ≥ 6;

p p z ≥ 8 + x yz.

378

Vasile Cîrtoaje

Solution. (a) First Solution. Since p p p 2 bc(a + b)(c + a) 2 bc (a + bc) p yz = ≥ (a + b)(c + a) (a + b)(c + a) p 2a(b + c) bc + 2bc(b + c) 4a bc + 2bc(b + c) ≥ , = (a + b)(b + c)(c + a) (a + b)(b + c)(c + a) we have Xp

yz ≥

12a bc + 2

P

bc(b + c)

(a + b)(b + c)(c + a) 8a bc + 2 = x yz + 2. = (a + b)(b + c)(c + a)

Therefore, it suffices to show that x + y + z + x yz ≥ 4, which is equivalent to Schur’s inequality of degree three X a3 + b3 + c 3 + 3a bc ≥ a b(a + b). The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation). Second Solution. Write the inequality as X Xp p 4 (x − 1) ≥ ( y − z)2 . Since X X (a − b) + (a − c) X a − b X b − a X (a − b)2 = + = (x − 1) = b+c b+c c+a (b + c)(c + a) = and

p p ( y − z)2 =

X

(b − c)2 , (a + b)(a + c)

2(b − c)2 (a + b + c)2
2 , p p (a + b)(a + c) b2 + a b + c 2 + ac

we can write the inequality as X (b − c)2 Ea ≥ 0, where

 Ea = (b + c) 2 − p



(a + b + c)
2  . p 2 2 b + a b + c + ac 2

Symmetric Nonrational Inequalities

379

By Minkowski’s inequality, we have €p

b2 + a b +

p

c 2 + ac

Therefore,

Š2

p p ≥ (b+c)2 +a( b+ c)2 ≥ (b+c)2 +a(b+c) = (b+c)(a+b+c).

‹  a+b+c = b + c − a, Ea ≥ (b + c) 2 − b+c

and hence

X X (b − c)2 Ea ≥ (b − c)2 (b + c − a) ≥ 0.

The right inequality is just Schur’s inequality of third degree. Third Solution. Using the Cauchy-Schwarz inequality, we have a b c (a + b + c)2 (a + b + c)2 + + ≥ = . b+c c+a a+b a(b + c) + b(c + a) + c(a + b) 2(a b + bc + ca) Using Hölder’s inequality, we have ‚s

v Œ2 s t b a c (a + b + c)3 + + ≥ 2 . b+c c+a a+b a (b + c) + b2 (c + a) + c 2 (a + b)

Thus, it suffices to prove that (a + b + c)2 2(a + b + c)3 + 2 ≥ 12. a b + bc + ca a (b + c) + b2 (c + a) + c 2 (a + b) Due to homogeneity, we may assume that a + b + c = 1. Substituting q = a b + bc + ca, 3q ≤ 1, the inequality becomes 1 2 + ≥ 12. q q − 3a bc From the fourth degree Schur’s inequality 6a bc p ≥ (p2 − q)(4q − p2 ),

p = a + b + c,

we obtain 6a bc ≥ (4q − 1)(1 − q). Therefore,

1 2 1 + − 12 ≥ + q q − 3a bc q q− =

2 (4q−1)(1−q) 2

− 12

(1 − 3q)(1 − 4q)2 4 1 + 2 − 12 = ≥ 0. q 4q − 3q + 1 q(4q2 − 3q + 1)

380

Vasile Cîrtoaje (b) By squaring, the inequality becomes p p p x + y + z + 2 x y + 2 yz + 2 z x ≥ 8 + x yz.

As we have shown in the first solution of the inequality in (a), p p p x y + yz + z x ≥ x yz + 2. Thus, it suffices to show that x + y + z + 2(x yz + 2) ≥ 8 + x yz, which is equivalent to x + y + z + x yz ≥ 4. This inequality reduces to Schur’s inequality of third degree. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 2.81. Let a, b, c be nonnegative real numbers, no two of which are zero, and let x=

2a , b+c

y=

2b 2c , z= . c+a a+b

Prove that p

1 + 24x +

p

1 + 24 y +

p

1 + 24z ≥ 15. (Vasile Cîrtoaje, 2005)

Solution (by Vo Quoc Ba Can). Assume that c = min{a, b, c}, hence z = min{x, y, z}, z ≤ 1. By Hölder’s inequality ‚s

v Œ2 t b   2 a + a (b + c) + b2 (c + a) ≥ (a + b)3 , b+c c+a

we get p

x+

p
2 y ≥

2(a + b)3 2(a + b)3 ≥ c(a + b)2 + a b(a + b − 2c) c(a + b)2 + 14 (a + b)2 (a + b − 2c)

8(a + b) 8 = . a + b + 2c 1+z Using this result and Minkowski’s inequality, we have =

p

1 + 24x +

p

1 + 24 y ≥

q

(1 + 1)2

v t p 48 p 2 + 24( x + y) ≥ 2 1 + . 1+z

Symmetric Nonrational Inequalities

381

Therefore, it suffices to show that 2

v t

1+

p 48 + 1 + 24z ≥ 15. 1+z

Using the substitution p

1 + 24z = 5t,

1 ≤ t ≤ 1, 5

the inequality turns into v t t 2 + 47 2 ≥ 3 − t. 25t 2 + 23 By squaring, this inequality becomes 25t 4 − 150t 3 + 244t 2 − 138t + 19 ≤ 0, which is equivalent to the obvious inequality (t − 1)2 (5t − 1)(5t − 19) ≤ 0. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 2.82. If a, b, c are positive real numbers, then v t

v v t t 7a 7b 7c + + ≤ 3. a + 3b + 3c b + 3c + 3a c + 3a + 3b (Vasile Cîrtoaje, 2005)

First Solution. Making the substitution x= we have

v t

7a , a + 3b + 3c

y=

v t

v t 7b 7c , z= , b + 3c + 3a c + 3a + 3b

 (x 2 − 7)a + 3x 2 b + 3x 2 c = 0      3 y 2 a + ( y 2 − 7)b + 3 y 2 c = 0 ,   2 2 2    3z a + 3z b + (z − 7)c = 0

382

Vasile Cîrtoaje

which involves

x2 − 7 3 y2 3z 2

3x 2 3x 2 y2 − 7 3 y2 3z 2 z2 − 7

=0 ;

that is, F (x, y, z) = 0, where F (x, y, z) = 4x 2 y 2 z 2 + 8

X

x2 y2 + 7

X

x 2 − 49.

We need to show that F (x, y, z) = 0 involves x + y + z ≤ 3, where x, y, z > 0. To do this, we use the contradiction method. Assume that x + y + z > 3 and show that F (x, y, z) > 0. Since F (x, y, z) is strictly increasing in each of its arguments, it is enough to prove that x + y + z = 3 involves F (x, y, z) ≥ 0. Assume that x = max{x, y, z} and denote y +z t= , 0 < t ≤ 1 ≤ x. 2 We will show that F (x, y, z) ≥ F (x, t, t) ≥ 0. We have F (x, y, z) − F (x, t, t) = (8x 2 + 7)( y 2 + z 2 − 2t 2 ) − 4(x 2 + 2)(t 4 − y 2 z 2 ) 1 = (8x 2 + 7)( y − z)2 − (x 2 + 2)(t 2 + yz)( y − z)2 2 1 ≥ (8x 2 + 7)( y − z)2 − 2(x 2 + 2)( y − z)2 2 1 = (4x 2 − 1)( y − z)2 ≥ 0 2 and 3− x 3− x F (x, t, t) = F x, , 2 2 

‹

=

1 (x − 1)2 (x − 2)2 (x 2 − 6x + 23) ≥ 0. 4

a = b = c (or any cyclic permutation). 8 Second Solution. Due to homogeneity, we may assume that a + b + c = 3, when the inequality becomes v X t 7a ≤ 3. 9 − 2a

The equality holds for a = b = c, and also for

Using the substitution v t 7a x= , 9 − 2a

v v t 7b t 7c y= , z= , 9 − 2b 9 − 2c

Symmetric Nonrational Inequalities

383

we need to show that if x, y, z are positive real number such that X

x2 1 = , 2 2x + 7 3

then x + y + z ≤ 3. For the sake of contradiction, assume that x + y + z > 3 and show that F (x, y, z) > 0, where X x2 1 F (x, y, z) = − . 2 2x + 7 3 Since F (x, y, z) is strictly increasing in each of its arguments, it is enough to prove that x + y + z = 3 involves F (x, y, z) ≥ 0. This is just the inequality in P 1.30. We give here another proof. By the Cauchy-Schwarz inequality, we have P P X x2 ( x 3/2 )2 ( x 3/2 )2 = P ≥P . 2x 2 + 7 x(2x 2 + 7) 2 x 3 + 21 Therefore, it suffices to show that X X 3( x 3/2 )2 ≥ 2 x 3 + 21, which is equivalent to the homogeneous inequality X

x3 + 6

X 7 (x y)3/2 ≥ (x + y + z)3 . 9

Replacing x, y, z by x 2 , y 2 , z 2 , we need to prove that G(x, y, z) ≥ 0, where G(x, y, z) =

X

x6 + 6

X

7 x 3 y 3 − (x 2 + y 2 + z 2 )3 . 9

Assume that x = max{x, y, z} and denote v t y 2 + z2 t= , 0 < t ≤ x. 2 We will show that G(x, y, z) ≥ G(x, t, t) ≥ 0. We have G(x, y, z) − G(x, t, t) = y 6 + z 6 + 6 y 3 z 3 − 8t 4 + 6x 3 ( y 3 + z 3 − 2t 3 ). Since y 3 + z 3 − 2t 3 ≥ 0, we get 6x 3 ( y 3 + z 3 − 2t 3 ) ≥ 4x 3 ( y 3 + z 3 − 2t 3 ) ≥ ( y 3 + z 3 + 2t 3 )( y 3 + z 3 − 2t 3 ).

384

Vasile Cîrtoaje

Thus, G(x, y, z) − G(x, t, t) ≥ y 6 + z 6 + 6 y 3 z 3 − 8t 4 + [( y 3 + z 3 )2 − 4t 6 ] = 2( y 6 + z 6 + 4 y 3 z 3 − 6t 6 ) = 2[( y 2 + z 2 )3 − 3 y 2 z 2 ( y 2 + z 2 ) + 4 y 3 z 3 − 6t 6 ] = 4(t 6 − 3t 2 y 2 z 2 + 2 y 3 z 3 ) = 4(t 2 − yz)2 (t 2 + 2 yz) ≥ 0. Also,

7 G(x, t, t) = x 6 + 2t 6 + 6(t 6 + 2x 3 t 3 ) − (x 2 + 2t 2 )3 9 =

2 (x − t)2 (x − 2t)2 (x 2 + 6x t + 2t 2 ) ≥ 0. 9

P 2.83. If a, b, c are positive real numbers such that a + b + c = 3, then Æ 3

a2 (b2 + c 2 ) +

Æ 3

b2 (c 2 + a2 ) +

Æ 3

p 3 c 2 (a2 + b2 ) ≤ 3 2. (Michael Rozenberg, 2013)

Solution. By Hölder’s inequality, we have ”X Æ 3

—3 ”X —2 X b 2 + c 2 . a2 (b2 + c 2 ) ≤ a(b + c) · (b + c)2

Therefore, it suffices to show that X b2 + c 2 27 ≤ , 2 (b + c) 2(a b + bc + ca)2 which is equivalent to the homogeneous inequalities  X  b2 + c 2 p4 − 1 ≤ − 3, (b + c)2 6q2 X

p4 2bc + ≥ 3, (b + c)2 6q2

where p = a + b + c,

q = a b + bc + ca.

According to P 1.61, the following inequality holds X

p2 9 2bc + ≥ . 2 (b + c) q 2

Symmetric Nonrational Inequalities

385

Thus, it is enough to show that p4 9 p2 − + 2 ≥ 3, 2 q 6q which is equivalent to 

p2 −3 q

2 ≥ 0.

The equality holds for a = b = c = 1.

P 2.84. If a, b, c are nonnegative real numbers, no two of which are zero, then 1 1 1 2 1 + + ≥ +p . a+b b+c c+a a+b+c a b + bc + ca (Vasile Cîrtoaje, 2005) Solution. Using the notation p = a + b + c,

q = a b + bc + ca,

r = a bc,

we can write the inequality as p2 + q 1 2 ≥ +p . pq − r p q According to P 3.57-(a) in Volume 1, for fixed p and q, the product r = a bc is minimal when two of a, b, c are equal or one of a, b, c is zero. Therefore, it suffices to prove the inequality for b = c = 1 and for a = 0. For a = 0, the inequality reduces to 1 1 2 + ≥p , b c bc which is obvious. For b = c = 1, the inequality becomes as follows: 1 2 1 2 , + ≥ +p 2 a+1 a+2 2a + 1 1 1 2 2 − ≥p − , 2 a+2 2a + 1 a + 1 p a 2(a + 1 − 2a + 1) ≥ , p 2(a + 2) (a + 1) 2a + 1

386

Vasile Cîrtoaje 2a2 a . ≥ p p 2(a + 2) (a + 1) 2a + 1(a + 1 + 2a + 1)

So, we need to show that 2a 1 ≥ . p p 2(a + 2) (a + 1) 2a + 1(a + 1 + 2a + 1) Consider two cases: 0 ≤ a ≤ 1 and a > 1. Case 1: 0 ≤ a ≤ 1. Since p p p p p 2a + 1(a + 1 + 2a + 1) ≥ 2a + 1( 2a + 1 + 2a + 1) = 2(2a + 1), it suffices to prove that 1 a ≥ , 2(a + 2) (a + 1)(2a + 1) which is equivalent to 1 − a ≥ 0. Case 2: a > 1. We will show first that p (a + 1) 2a + 1 > 3a. Indeed, by squaring, we get the obvious inequality a3 + a(a − 2)2 + 1 > 0. Therefore, it suffices to show that 1 2a ≥ , 2(a + 2) (a + 1)(3a + 2a + 1) which is equivalent to (a−1)2 ≥ 0. The proof is completed. The equality holds for a = 0 and b = c.

P 2.85. If a, b ≥ 1, then 1 1 1 1 + ≥p +p . p 2 3a + 1 3a b + 1 3b + 1

Symmetric Nonrational Inequalities

387

Solution. Using the substitution 2 y=p , 3b + 1

2 , x=p 3a + 1

x, y ∈ (0, 1],

the desired inequality can be written as v t 3 ≥ x + y − 1. xy 2 2 x y − x2 − y2 + 4 Consider the non-trivial case x + y − 1 ≥ 0, and denote t = x + y − 1,

p = x y,

1 ≥ p ≥ t ≥ 0.

Since x 2 + y 2 = (x + y)2 − 2x y = (t + 1)2 − 2p, we need to prove that p

v t

3 ≥ t. p2 + 2p − t 2 − 2t + 3

By squaring, we get the inequality (p − t)[(3 − t 2 )p + t(1 − t)(3 + t)] ≥ 0, which is clearly true. The equality holds for a = b = 1.

P 2.86. Let a, b, c be positive real numbers such that a ≥ 1 ≥ b ≥ c and a bc = 1. Prove that 1 1 1 3 +p +p ≥ . p 3a + 1 3c + 1 2 3b + 1 (Vasile Cîrtoaje, 2007) Solution. Let b1 = 1/b, b1 ≥ 1. We claim that 1 1 1 +p ≥ . p 3b + 1 3b1 + 1 2 This inequality is equivalent to 1

+ p 3b + 1

v t

b 1 ≥ . b+3 2

Making the substitution 1 = t, p 3b + 1

1 ≤ t < 1, 2

388

Vasile Cîrtoaje

the inequality becomes v t 1 − t2 ≥ 1 − t. 1 + 8t 2 By squaring, we get t(1 − t)(1 − 2t)2 ≥ 0, which is clearly true. Similarly, we have 1 1 1 ≥ , +p p 3c + 1 3c1 + 1 2 where c1 = 1/c, c1 ≥ 1. Using these inequality, it suffices to show that 1 1 1 1 + ≥p +p , p 2 3a + 1 3c1 + 1 3b1 + 1 which is equivalent to 1 p

3b1 c1 + 1

+

1 1 1 ≥p . +p 2 3c1 + 1 3b1 + 1

From the inequality in the preceding P 2.85, the conclusion follows. The equality holds for a = b = c = 1.

1 P 2.87. Let a, b, c be positive real numbers such that a + b + c = 3. If k ≥ p , then 2 (a bc)k (a2 + b2 + c 2 ) ≤ 3. (Vasile Cîrtoaje, 2006) Solution. Since

a+b+c abc ≤ 3 

‹3

= 1,

p it suffices to prove the desired inequality for k = 1/ 2. Write the inequality in the homogeneous form a+b+c (a bc) (a + b + c ) ≤ 3 3 k

2

2

2



‹3k+2

.

According to P 3.57-(a) in Volume 1, for fixed a+b+c and a b+bc+ca, the product a bc is maximal when two of a, b, c are equal. Therefore, it suffices to prove the homogeneous inequality for b = c = 1; that is, f (a) ≥ 0, where f (a) = (3k + 2) ln(a + 2) − (3k + 1) ln 3 − k ln a − ln(a2 + 2).

Symmetric Nonrational Inequalities

389

From f 0 (a) =

2a 3k + 2 k − − 2 a+2 a a +2 p p 2(a − 1)(ka2 − 2a + 2k) 2(a − 1)(a − 2)2 = = , a(a + 2)(a2 + 2) a(a + 2)(a2 + 2)

it follows that f is decreasing on (0, 1] and increasing on [1, ∞); therefore, f (a) ≥ f (1) = 0. This completes the proof. The equality holds for a = b = c.

P 2.88. Let p and q be nonnegative real numbers such that p2 ≥ 3q, and let v v t 2p + w t 2p − 2w +2 , g(p, q) = 3 3 s  s 2p + 2w 2p − w  +2 ,  3 3 h(p, q) = p p p   p + p + q, where w =

p

p2 ≤ 4q p2 ≥ 4q

,

p2 − 3q. If a, b, c are nonnegative real numbers such that a + b + c = p,

a b + bc + ca = q,

then p

(a) with equality for a = (b)

a+b+

p

b+c+

p

c + a ≥ g(p, q),

p + 2w p−w and b = c = (or any cyclic permutation); 3 3 p p p a + b + b + c + c + a ≤ h(p, q),

p+w p − 2w and b = c = (or any cyclic permutation) - when 3 3 p2 ≤ 4q, and for a = 0, b + c = p and bc = q (or any cyclic permutation) - when p2 ≥ 4q.

with equality for a =

(Vasile Cîrtoaje, 2013) Solution. Consider the non-trivial case p > 0. Since b + c = p − a,

390

Vasile Cîrtoaje (a + b)(a + c) = a2 + q

and p

a+b+

p

a+c =

Ç

a+p+2

we get p

a+b+

p

b+c+

where f (a) = From

p

p−a+

Ç

p

Æ

a2 + q,

c + a = f (a),

a+p+2

Æ

a2 + q.

p a2 + q + 2a −1 f (a) = p + p , q 2 p − a 2 a 2 + q · a + p + 2p a 2 + q 0

it follows that f 0 (a) has the same sign as F (a), where p p ( a2 + q + 2a)2 −2(3a2 − 2pa + q)(a + a2 + q) −1 F (a) = + = . p p p − a (a2 + q)(a + p + 2 a2 + q) (p − a)(a2 + q)(a + p + 2 a2 + q) Therefore, f (a) is increasing on [a1 , a2 ] and decreasing on [0, a1 ] and [a2 , p], where p p p − p2 − 3q p + p2 − 3q a1 = , a2 = . 3 3 (a) From 0 ≤ (b + c)2 − 4bc = (p − a)2 − 4(a2 − pa + q) = −(3a2 − 2pa + 4q − p2 ), we get a ∈ [0, a4 ], where p+2

p

p2 − 3q . 3 Since a2 ≤ a4 , f (a) is increasing on [a1 , a2 ] and decreasing on [0, a1 ] ∪ [a2 , a4 ]; therefore, f (a) ≥ min{ f (a1 ), f (a4 )}. a4 =

We need to show that min{ f (a1 ), f (a4 )} = g(p, q). Indeed, from 2a1 + a4 = p and q Æ a12 + q = 2a1 (p − a1 ),

q 2 a42 + q = a4 + p,

we get f (a1 ) = f (a4 ) = g(p, q).

Symmetric Nonrational Inequalities

391

(b) Consider the cases 3q ≤ p2 ≤ 4q and p2 ≥ 4q. Case 1: 3q ≤ p2 ≤ 4q. From (b + c)2 − 4bc = (p − a)2 − 4(a2 − pa + q) = −(3a2 − 2pa + 4q − p2 ) ≥ 0, we get a ∈ [a3 , a4 ], where a3 =

p−2

p

p2 − 3q ≥ 0, 3

a4 =

p+2

p

p2 − 3q . 3

Since a3 ≤ a1 ≤ a2 ≤ a4 , it follows that f (a) is increasing on [a1 , a2 ] and decreasing on [a3 , a1 ] ∪ [a2 , a4 ]. Thus, f (a) ≤ max{ f (a2 ), f (a3 )}. To complete the proof, we need to show that max{ f (a2 ), f (a3 )} = h(p, q). Indeed, from 2a2 + a3 = p and q Æ a22 + q = 2a2 (p − a2 ),

q 2 a32 + q = a3 + p,

we get f (a2 ) = f (a3 ) = h(p, q). Case 2: p2 ≥ 4q. Assume that a = min{a, b, c}, a ≤ p/3. From 0 ≤ bc = q − a(b + c) = q − a(p − a) = a2 − pa + q, we get a ∈ [0, a5 ] ∪ [a6 , p], where p p − p2 − 4q a5 = , 2

a6 =

p+

p

p2 − 4q . 2

Since a6 > p/3, it follows that a ∈ [0, a5 ]. Since a1 ≤ a5 ≤ a2 , f (a) is decreasing on [0, a1 ] and increasing on [a1 , a5 ]; thus, f (a) ≤ max{ f (0), f (a5 )}. It remains to show that max{ f (0), f (a5 )} = h(p, q). Indeed, from a52 + q = pa5 and p

a5 +

p

p − a5 =

r

p+2

q

pa5 − a52 =

we get f (0) = f (a5 ) = h(p, q).

Æ

p p + 2 q,

392

Vasile Cîrtoaje

Remark. Note the following particular cases: (a) If a, b, c are nonnegative real numbers such that a + b + c = a b + bc + ca = 4, then

p p p p 2(1 + 10) p ≤ a + b + b + c + c + a ≤ 2(1 + 2), p 3

with left equality for a = 8/3 and b = c = 2/3 (or any cyclic permutation), and right equality for a = 0 and b = c = 2 (or any cyclic permutation). (b) If a, b, c are nonnegative real numbers such that a + b + c = 4,

a b + bc + ca = 5,

then p p p p p 2+2 3≤ a+ b+ b+c+ c+a ≤

p p 10 + 2 7 , p 3

with left equality for a = 2 and b = c = 1 (or any cyclic permutation), and right equality for a = 2/3 and b = c = 5/3 (or any cyclic permutation). (c) If a, b, c are nonnegative real numbers such that a + b + c = 11, then 3

p

6≤

p

a+b+

p

b+c+

a b + bc + ca = 7, p

c+a≤

Æ p p 11 + 11 + 7,

with left equality for a = 31/3 and b = c = 1/3 (or any cyclic permutation), and right equality for a = 0, b + c = 11 and bc = 7 (or any cyclic permutation).

P 2.89. Let a, b, c, d be positive real numbers such that a2 + b2 + c 2 + d 2 = 1. Prove that p p p p p p p p 1 − a + 1 − b + 1 − c + 1 − d ≥ a + b + c + d. (Vasile Cîrtoaje, 2007)

Symmetric Nonrational Inequalities

393

First Solution. We can obtain the desired inequality by summing the inequalities p p p p 1 − a + 1 − b ≥ c + d, p p p p 1 − c + 1 − d ≥ a + b. Since

p Æ p 4 1 − a + 1 − b ≥ 2 (1 − a)(1 − b)

and p

c+

p

d ≤2

v tc + d

≤2

2

v t 2 2 4 c +d 2

,

the former inequality above holds if (1 − a)(1 − b) ≥

c2 + d 2 . 2

Indeed, 2(1 − a)(1 − b) − c 2 − d 2 = 2(1 − a)(1 − b) + a2 + b2 − 1 = (a + b − 1)2 ≥ 0. Similarly, we can prove the second inequality. The equality holds for a = b = c = d = Second Solution. We can obtain the desired inequality by summing the inequalities p p 1 1 − a − a ≥ p (1 − 4a2 ), 2 2

p

p p 1 1 − c − c ≥ p (1 − 4c 2 ), 2 2

p

1− b−

p

1−d −

p

1 b ≥ p (1 − 4b2 ), 2 2

1 d ≥ p (1 − 4d 2 ). 2 2

To prove the first inequality, we write it as 1 − 2a 1 p p ≥ p (1 − 2a)(1 + 2a). 1−a+ a 2 2 1 . We need to show that 2 p p p 2 2 ≥ (1 + 2a)( 1 − a + a). p p p p Since 1 − a + a ≤ 2[(1 − a) + a] = 2, we have p p p p 2 2 − (1 + 2a)( 1 − a + a) ≥ 2(1 − 2a) ≥ 0.

Case 1: 0 < a ≤

1 . 2

394

Vasile Cîrtoaje

1 ≤ a < 1. We need to show that 2 p p p 2 2 ≤ (1 + 2a)( 1 − a + a). p Since 1 + 2a ≥ 2 2a, it suffices to prove that Æ 1 ≤ a(1 − a) + a. Case 2:

Indeed, 1−a−

Æ

p p p a(1 − a) = 1 − a ( 1 − a − a) =

p 1 − a (1 − 2a) ≤ 0. p p 1−a+ a

P 2.90. Let a, b, c, d be positive real numbers. Prove that p A + 2 ≥ B + 4, where

‹ 1 1 1 1 A = (a + b + c + d) + + + − 16, a b c d ‹  1 1 1 1 2 2 2 2 B = (a + b + c + d ) 2 + 2 + 2 + 2 − 16. a b c d 

(Vasile Cîrtoaje, 2004) Solution. By squaring, the inequality becomes A2 + 4A ≥ B. Let us denote f (x, y, z) =

y z x + + − 3, y z x

F (x, y, z) =

y 2 z2 x2 + + 2 − 3, y 2 z2 x

where x, y, z > 0. By the AM-GM inequality, it follows that f (x, y, z) ≥ 0 and F (x, y, z) ≥ 0. We can check that A = f (a, b, c) + f (b, a, d) + f (c, d, a) + f (d, c, b) = f (c, b, a) + f (d, a, b) + f (a, d, c) + f (b, c, d) and B = F (a, b, c) + F (b, a, d) + F (c, d, a) + F (d, c, b).

Symmetric Nonrational Inequalities

395

Since F (x, y, z) = [ f (x, y, z) + 3]2 − 2[ f (z, y, x) + 3] − 3 = f 2 (x, y, z) + 6 f (x, y, z) − 2 f (z, y, x), we get B = f 2 (a, b, c) + f 2 (b, a, d) + f 2 (c, d, a) + f 2 (d, c, b) + 4A. Therefore, A2 + 4A − B = [ f (a, b, c) + f (b, a, d) + f (c, d, a) + f (d, c, b)]2 − f 2 (a, b, c) − f 2 (b, a, d) − f 2 (c, d, a) − f 2 (d, c, b) ≥ 0. The equality holds for a = b = c = d.

P 2.91. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = 1. Prove that p p p 3a1 + 1 + 3a2 + 1 + · · · + 3an + 1 ≥ n + 1. First Solution. Without loss of generality, assume that a1 = max{a1 , a2 , · · · , an }. Write the inequality as follows: p p p ( 3a1 + 1 − 2) + ( 3a2 + 1 − 1) + · · · + ( 3an + 1 − 1) ≥ 0, a1 − 1 p

3a1 + 1 + 2

+p

a2 3a2 + 1 + 1

+ ··· + p

an 3an + 1 + 1

≥ 0,

a2 + · · · + an ≥p , 3an + 1 + 1 3a1 + 1 + 2 3a2 + 1 + 1     1 1 1 1 a2 p + · · · + an p ≥ 0. −p −p 3an + 1 + 1 3a1 + 1 + 2 3a1 + 1 + 2 3a2 + 1 + 1 a2

p

+ ··· + p

an

The last inequality is clearly true. The equality holds for a1 = 1 and a2 = · · · = an = 0 (or any cyclic permutation). Second Solution. We use the induction method. For n = 1, the inequality is an equality. We claim that Æ p p 3a1 + 1 + 3an + 1 ≥ 3(a1 + an ) + 1 + 1. By squaring, this becomes Æ

(3a1 + 1)(an + 1) ≥

Æ

3(a1 + an ) + 1,

396

Vasile Cîrtoaje

which is equivalent to a1 an ≥ 0. Thus, to prove the original inequality, it suffices to show that Æ p p 3(a1 + an ) + 1 + 3a2 + 1 + · · · + 3an−1 + 1 ≥ n. Using the substitution b1 = a1 + an and b2 = a2 , · · · , bn−1 = an−1 , this inequality becomes Æ Æ Æ 3b1 + 1 + 3b2 + 1 + · · · + 3bn−1 + 1 ≥ n for b1 + b2 + · · · + bn−1 = 1. Clearly, this is true by the induction hypothesis.

P 2.92. Let 0 ≤ a < b and a1 , a2 , . . . , an ∈ [a, b]. Prove that €p p p Š2 a1 + a2 · · · + an − n n a1 a2 · · · an ≤ (n − 1) b− a . (Vasile Cîrtoaje, 2005) Solution. Based on Lemma below, it suffices to consider that a1 = · · · = ak = a,

ak+1 = · · · = an = b,

where k ∈ {1, 2, · · · , n − 1}. The original inequality becomes k

(n − k − 1)a + (k − 1)b + na n b

n−k n

p ≥ (2n − 2) a b.

Clearly, this inequality follows by the weighted AM-GM inequality. For n ≥ 3, the equality holds when a = 0, one of ai is also 0 and the other ai are equal to b. Lemma. Let 0 ≤ a < b and a1 , a2 , . . . , an ∈ [a, b]. Then, the expression p a1 + a2 · · · + an − n n a1 a2 · · · an is maximal when a1 , a2 , . . . , an ∈ {a, b}. Proof. We use the contradiction method. Consider that a2 , . . . , an are fixed and define the function p f (a1 ) = a1 + a2 + · · · + an − n n a1 a2 · · · an . For the sake of contradiction, assume that there exits a1 ∈ (a, b) such that f (a1 ) > f (a) p p p n and f (a1 ) > f (b). Let us denote x i = n ai for all i, c = n a and d = b (c < x 1 < d). From f (a1 ) − f (a) = x 1n − c n − n(x 1 − c)x 2 · · · x n = (x 1 − c)(x 1n−1 + x 1n−2 c + · · · + c n−1 − nx 2 · · · x n ),

Symmetric Nonrational Inequalities

397

we get x 1n−1 + x 1n−2 c + · · · + c n−1 > nx 2 · · · x n .

(*)

Analogously, from f (a1 ) − f (b) = x 1n − d n − n(x 1 − d)x 2 · · · x n = (x 1 − d)(x 1n−1 + x 1n−2 d + · · · + d n−1 − nx 2 · · · x n ), we get nx 2 · · · x n > x 1n−1 + x 1n−2 d + · · · + d n−1 .

(**)

Summing up (*) and (**) yields x 1n−1 + x 1n−2 c + · · · + c n−1 > x 1n−1 + x 1n−2 d + · · · + d n−1 , which is clearly false.

P 2.93. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that 1 p

1 + (n2

− 1)a1

+p

1 1 + (n2 − 1)a2

+ ··· + p

1 1 + (n2 − 1)an

≥ 1.

Solution. For the sake of contradiction, assume that 1 p

1 + (n2

− 1)a1

+p

1 1 + (n2

− 1)a2

+ ··· + p

1 1 + (n2 − 1)an

It suffices to show that a1 a2 · · · an > 1. Let xi = p Since ai =

1 − x i2 (n2 − 1)x i2

1 1 + (n2 − 1)ai

, 0 < x i < 1,

i = 1, 2, · · · , n.

for all i, we need to show that x1 + x2 + · · · + x n < 1

implies (1 − x 12 )(1 − x 22 ) · · · (1 − x n2 ) > (n2 − 1)n x 12 x 22 · · · x n2 .

< 1.

398

Vasile Cîrtoaje

Using the AM-GM inequality gives i Y Y Y h€X Š2 (1 − x 12 ) > x 1 − x 12 = (x 2 + · · · + x n )(2x 1 + x 2 + · · · + x n ) ≥ (n2 − 1)n

Y€ p n−1

x2 · · · x n ·

q

n+1

Š x 12 x 2 · · · x n = (n2 − 1)n x 12 x 22 · · · x n2 .

The equality holds for a1 = a2 = · · · = an = 1.

P 2.94. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. Prove that n X i=1

1 1+

p

1 + 4n(n − 1)ai

First Solution. Write the inequality as follows n p X 1 + 4n(n − 1)ai − 1 a1

i=1

≥

1 . 2

≥ 2n(n − 1),

v n u X X 1 4n(n − 1) t1 + ≥ 2n(n − 1) + . a1 a1 a12 i=1 By squaring, the inequality becomes v ™ u – X u 1 X 4n(n − 1) 4n(n − 1) 1 1 2 2 t + + ≥ 2n (n − 1) + . 2 2 a a a a a a i j i j i j 1≤i< j≤n 1≤i< j≤n The Cauchy-Schwarz inequality gives v ™ u – u 1 4n(n − 1) 1 4n(n − 1) 1 4n(n − 1) t ≥ + p . + + 2 2 ai aj ai a j ai a j ai aj Thus, it suffices to show that X 1≤i< j≤n

p

1 n(n − 1) ≥ , ai a j 2

which follows immediately from the AM-GM inequality. The equality holds for a1 = a2 = · · · = an = 1.

Symmetric Nonrational Inequalities

399

Second Solution. For the sake of contradiction, assume that n X i=1

1 1+

p

1 + 4n(n − 1)ai


1. Using the substitution xi 1 , = p 2n 1 + 1 + 4n(n − 1)ai which yields ai =

n − xi (n − 1)x i2

i = 1, 2, · · · , n,

, 0 < x i < n,

i = 1, 2, · · · , n,

we need to show that x1 + x2 + · · · + x n < n implies (n − x 1 )(n − x 2 ) · · · (n − x n ) > (n − 1)n x 12 x 22 · · · x n2 . By the AM-GM inequality, we have x1 x2 · · · x n ≤

 x + x + · · · + x n 1 2 n (x 1 + x 2 + · · · + x n ) − x i ≥ (n − 1)

v t x1 x2 · · · x n

n−1

xi

,

i = 1, 2, · · · , n.

Therefore, we get (n − x 1 )(n − x 2 ) · · · (n − x n ) > (n − 1)n x 1 x 2 · · · x n > (n − 1)n x 12 x 22 · · · x n2 .

P 2.95. If f is a convex function on a real interval I and a1 , a2 , . . . , an ∈ I, then a + a + ··· + a  1 2 n f (a1 ) + f (a2 ) + · · · + f (an ) + n(n − 2) f ≥ n ≥ (n − 1)[ f (b1 ) + f (b2 ) + · · · + f (bn )], where bi =

1 X aj, n − 1 j6=i

i = 1, 2, · · · , n. (Tiberiu Popoviciu, 1965)

400

Vasile Cîrtoaje

Solution. Without loss of generality, we may assume that n ≥ 3 and a1 ≤ a2 ≤ · · · ≤ an . There is an integer m, 1 ≤ m ≤ n − 1, such that a1 ≤ · · · ≤ am ≤ a ≤ am+1 ≤ · · · ≤ an , b1 ≥ · · · ≥ bm ≥ a ≥ bm+1 ≥ · · · ≥ bn . We can get the desired inequality by summing the following two inequalities: f (a1 ) + f (a2 ) + · · · + f (am ) + n(n − m − 1) f (a) ≥ ≥ (n − 1)[ f (bm+1 ) + f (bm+2 ) + · · · + f (bn )],

(*)

f (am+1 ) + f (am+2 ) + · · · + f (an ) + n(m − 1) f (a) ≥ ≥ (n − 1)[ f (b1 ) + f (b2 ) + · · · + f (bm )].

(**)

In order to prove (*), we apply Jensen’s inequality to get f (a1 ) + f (a2 ) + · · · + f (am ) + (n − m − 1) f (a) ≥ (n − 1) f (b), where b=

a1 + a2 + · · · + am + (n − m − 1)a . n−1

Thus, it suffices to show that (n − m − 1) f (a) + f (b) ≥ f (bm+1 ) + f (bm+2 ) + · · · + f (bn ). Since a ≥ bm+1 ≥ bm+2 ≥ · · · ≥ bn , (n − m − 1)a + b = bm+1 + bm+2 + · · · + bn , − → − → we see that A n−m = (a, · · · , a, b) majorizes B n−m = (bm+1 , bm+2 , · · · , bn ). Therefore, the inequality is a consequence of Karamata’s inequality. Similarly, we can prove the inequality (**), by summing Jensen’s inequality f (am+1 ) + f (am+2 ) + · · · + f (an ) + (m − 1) f (a) ≥ f (c) n−1 and the inequality f (c) + (m − 1) f (a) ≥ f (b1 ) + f (b2 ) + · · · + f (bm ), where

am+1 + am+2 + · · · + an + (m − 1)a . n−1 The last inequality follows by Karamata’s inequality, because c=

c + (m − 1)a = b1 + b2 + · · · + bm , − → − → therefore C m = (c, a, · · · , a) majorizes D m = (b1 , b2 , · · · , bm ). b1 ≥ b2 ≥ · · · ≥ bm ≥ a,

Symmetric Nonrational Inequalities

401

P 2.96. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. Prove that n X 1 1 ≤ . p 2 (n − 1)2 + 4nai i=1 n − 1 + Solution. Use the contradiction method. Assume that n X i=1

1 n−1+

p

(n − 1)2

+ 4nai

>

1 , 2

and show that a1 a2 · · · an < 1. Using the substitution 1 n−1+

p

which involves ai =

(n − 1)2 + 4nai

(n − 1)2 ci , (n − ci )2

=

n − ci , 2n(n − 1)

0 < ci < n,

i = 1, 2, · · · , n,

i = 1, 2, · · · , n,

we need to show that c1 + c2 + · · · + cn < n implies  n − c 2  n − c 2 1

2

n−1 n−1 Clearly, it suffices to show that

···

 n − c 2 n

n−1

> c1 c2 · · · cn .

c1 + c2 + · · · + cn = n implies  n − c 2  n − c 2 1

2

 n − c 2 n

≥ c1 c2 · · · cn . n−1 n−1 n−1 Popoviciu’s inequality (see the preceding P 2.95) applied to the convex function f (x) = − ln x, x > 0 gives  n − c n−1  n − c n−1 1

2

n−1

n−1

···

···

 n − c n−1 n

≥ c1 c2 · · · cn

 c + c + · · · + c n(n−2) 1 2 n , n

n−1  n − c n−1  n − c n−1  n − c n−1 n 1 2 ··· ≥ c1 c2 · · · cn , n−1 n−1 n−1  n − c 2  n − c 2  n − c 2 2 n 1 2 ··· ≥ (c1 c2 · · · cn ) n−1 . n−1 n−1 n−1 Thus, it suffices to show that 2

(c1 c2 · · · cn ) n−1 ≥ c1 c2 · · · cn .

402

Vasile Cîrtoaje

For n = 3, this inequality is an equality, while for n ≥ 4, it is equivalent to c1 c2 · · · cn ≤ 1. Indeed, by the AM-GM inequality, we have c1 c2 · · · cn ≤

 c + c + · · · + c n 1 2 n = 1. n

The equality holds for a1 = a2 = · · · = an = 1.

P 2.97. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then

a1 + a2 + · · · + an ≥ n − 1 +

v u 2 t a1 + a22 + · · · + an2 n

.

Solution. Let us denote a=

a1 + a2 + · · · + an , n

b=

v P u t 2 1≤i< j≤n ai a j n(n − 1)

,

where b ≥ 1 (by the AM-GM inequality). We need to show that na − n + 1 ≥

v t n2 a2 − n(n − 1)b2 n

.

By squaring, this inequality becomes (n − 1)[n(a − 1)2 + b2 − 1] ≥ 0, which is clearly true. The equality holds for a1 = a2 = · · · = an = 1.

P 2.98. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then q

(n − 1)(a12 + a22 + · · · + an2 ) + n −

Æ

n(n − 1) ≥ a1 + a2 + · · · + an . (Vasile Cîrtoaje, 2006)

Symmetric Nonrational Inequalities

403

Solution. We use the induction method. For n = 2, the inequality is equivalent to the obvious inequality 1 ≥ 2. a1 + a1 Assume now that the inequality holds for n − 1 numbers, n ≥ 3, and prove that it holds also for n numbers. Let a1 = min{a1 , a2 , . . . , an }, and denote x= f (a1 , a2 , . . . , an ) =

q

a2 + a3 + · · · + an , n−1

y=

p

n−1

(n − 1)(a12 + a22 + · · · + an2 ) + n −

a2 a3 · · · an ,

Æ

n(n − 1) − (a1 + a2 + · · · + an ).

By the AM-GM inequality, we have x ≥ y. We will show that f (a1 , a2 , . . . , an ) ≥ f (a1 , y, · · · , y) ≥ 0.

(*)

Write the left inequality as q q p a12 + a22 + · · · + an2 − a12 + (n − 1) y 2 ≥ n − 1 (x − y). To prove this inequality, we use the induction hypothesis, written in the homogeneous form q ” — Æ (n − 2)(a22 + a32 + · · · + an2 ) + n − 1 − (n − 1)(n − 2) y ≥ (n − 1)x, which is equivalent to a22 + · · · + an2 ≥ (n − 1)A2 , where A = k x − (k − 1) y,

k=

v tn − 1 n−2

.

So, we need to prove that q q p a12 + (n − 1)A2 − a12 + (n − 1) y 2 ≥ n − 1 (x − y). Write this inequality as

q

A2 − y 2 x−y ≥p . q 2 2 n−1 a1 + (n − 1)A2 + a1 + (n − 1) y 2

Since x ≥ y and

A2 − y 2 = k(x − y)[k x − (k − 2) y],

we need to show that q

a12

k[k x − (k − 2) y] 1 ≥p . q 2 n−1 + (n − 1)A2 + a1 + (n − 1) y 2

404

Vasile Cîrtoaje

In addition, since a1 ≤ y, it suffices to show that

p We claim that Æ

k[k x − (k − 2) y] 1 ≥p . p n−1 y 2 + (n − 1)A2 + n y

p y 2 + (n − 1)A2 ≤ k n − 1 [k x − (k − 1) y] .

If this inequality is true, then it is enough to prove that p

k[k x − (k − 2) y]

k n − 1[k x − (k − 1) y] +

p

ny

≥p

1 n−1

.

Rewrite this inequality as k[k x − (k − 2) y] ≥ 1, k[k x − (k − 1) y] + m y

m=

s

n . n−1

Since m < k, we have k[k x − (k − 2) y] k[k x − (k − 2) y] > = 1. k[k x − (k − 1) y] + m y k[k x − (k − 1) y] + k y Using the relation a1 y n−1 = 1, we can write the right inequality in (*) as Æ

(n − 1)[(n − 1) y 2n + 1] ≥ (n − 1) y n − t y n−1 + 1,

where t = n−

Æ

n(n − 1).

This inequality is true if (n − 1)[(n − 1) y 2n + 1] ≥ [(n − 1) y n − t y n−1 + 1]2 , that is, 2(n − 1)t y 2n−1 − t 2 y 2n−2 − 2(n − 1) y n + 2t k y n−1 + n − 2 ≥ 0. Since t 2 = n(2t − 1), we can write this inequality in the form 2t y n−1 B + C ≥ 0, where B = (n − 1) y n − n y n−1 + 1,

C = n y 2n−2 − 2(n − 1) y n + n − 2.

Symmetric Nonrational Inequalities

405

This inequality is true if b ≥ 0 and C ≥ 0. Indeed, by the AM-GM inequality, we have Æ n (n − 1) y n + 1 ≥ n y (n−1)n · 1 = n y n−1 and n y 2n−2 + n − 2 ≥ (2n − 2)

Æ

2n−2

y n(2n−2) · 1n−2 = 2(n − 1) y n .

The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

P 2.99. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If 0p +p p 1 + pan 1 + pa1 1 + pa2 1+p and show that this implies a1 a2 · · · an < 1. Using the substitution p p p 1+p , 0 < x i < p + 1, i = 1, 2, · · · , n, 1 + pai = xi we need to show that x 1 + x 2 + · · · + x n > n yields      1+p 1+p 1+p −1 − 1 ··· − 1 < pn. x n2 x 12 x 22 Clearly, it suffices to prove that x1 + x2 + · · · + x n = n yields 

1+p x 12

 −1

1+p x 22

 1+p − 1 ··· − 1 ≤ pn. x n2 



Denoting p

1 + p = q,

1 1, then X a1k ≥ 1. a1k + a2 + · · · + an (Vasile Cîrtoaje, 2006) p First Solution. Let us denote r = n a1 a2 · · · an and bi = ai /r for i = 1, 2, · · · , n. Note that r ≥ 1 and b1 b2 · · · bn = 1. The desired inequality becomes X

b1k b1k + (b2 + · · · + bn )/r k−1

≥ 1,

408

Vasile Cîrtoaje

and we see that it suffices to prove it for r = 1; that is, for a1 a2 · · · an = 1. On this hypothesis, we will show that there exists a positive number p such that p

a1k a1k + a2 + · · · + an

≥

a1

p.

p

p

a1 + a2 + · · · + an

If this is true, by adding this inequality and the analogous inequalities for a2 , . . . , an , we get the desired inequality. Write the claimed inequality as p

a2 + · · · + anp ≥ (a2 · · · an )k−p (a2 + · · · + an ). For

(n − 1)k + 1 , p > 1, n this inequality turns into the homogeneous inequality p=

p−1

p

a2 + · · · + anp ≥ (a2 · · · an ) n−1 (a2 + · · · + an ). Based on the AM-GM inequality p−1

(a2 · · · an ) n−1 ≤

 a + · · · + a  p−1 2 n , n−1

we only need to show that p

p

a2 + · · · + a n n−1

≥

 a + · · · + a p 2 n , n−1

which is just Jensen’s inequality applied to the convex function f (x) = x p . The equality holds for a1 = a2 = · · · = an = 1. Second Solution. By the Cauchy-Schwarz inequality, we have ‹2  P k+1 P k+1 P 2 k+1 a X 1 a1 + 2 1≤i< j≤n (a1 a2 ) 2 a1k . ≥P = P k+1 P a1k + a2 + · · · + an a1 (a1k + a2 + · · · + an ) a1 + 2 1≤i< j≤n a1 a2 Thus, it suffices to show that X

(a1 a2 )

k+1 2

≥

1≤i< j≤n

X

a1 a2 .

1≤i< j≤n

Jensen’s inequality applied to the convex function f (x) = x X

(a1 a2 )

1≤i< j≤n

k+1 2

n(n − 1) ≥ 2

k+1 2

yields

‚ P Œ k+1 2 1≤i< j≤n a1 a2 2 n(n − 1)

Symmetric Nonrational Inequalities

409

On the other hand, by the AM-GM inequality, we get X 2 2 a1 a2 ≥ (a1 a2 · · · an ) n ≥ 1. n(n − 1) 1≤i< j≤n Therefore, ‚ P Œ k+1 2 1≤i< j≤n a1 a2 2 n(n − 1) hence X

(a1 a2 )

k+1 2

1≤i< j≤n

n(n − 1) ≥ 2

≥

‚ P Œ 2 1≤i< j≤n a1 a2 n(n − 1)

‚ P Œ 2 1≤i< j≤n a1 a2 n(n − 1)

=

,

X

a1 a2 .

1≤i< j≤n

P 2.102. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If −2 ≤ k < 1, n−2 then

a1k

X

≤ 1.

a1k + a2 + · · · + an

(Vasile Cîrtoaje, 2006) p Solution. Let us denote r = n a1 a2 · · · an and bi = ai /r for i = 1, 2, · · · , n. Notice that r ≥ 1 and b1 b2 · · · bn = 1. The desired inequality becomes b1k

X

b1k + (b2 + · · · + bn )r 1−k

≤ 1,

and we see that it suffices to prove it for r = 1; that is, for a1 a2 · · · an = 1. On this hypothesis, we will prove the desired inequality by summing the inequality a1k a1k + a2 + · · · + an

p

≤

a1 p

p

p

a1 + a2 + · · · + an

and the analogous inequalities for a2 , . . . , an , where p=

(n − 1)k + 1 , n

−1 ≤ p < 1. n−2

Rewrite this inequality as 1−p

p

a2 + · · · + an ≥ (a2 · · · an ) n−1 (a2 + · · · + anp ).

(*)

410

Vasile Cîrtoaje

To prove it, we use the weighted AM-GM inequality 1−p 1 + (n − 2)p n − 1 1+(n−2)p a2 + a3 + · · · + an ≥ a2 n−1 (a3 · · · an ) n−1 , 1−p 1−p

which is equivalent to 1−p 1 + (n − 2)p n−1 p a2 + a3 + · · · + an ≥ a2 (a2 a3 · · · an ) n−1 . 1−p 1−p

Adding this inequality and the analogous inequalities for a3 , · · · , an yields the inequality (*). Thus, the proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

P 2.103. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If k > 1, then X a1 ≤ 1. k a1 + a2 + · · · + an (Vasile Cîrtoaje, 2006) 1 . n−1 Case 1: 1 < k ≤ n + 1. By the AM-GM inequality, the hypothesis a1 a2 · · · an ≥ 1 yields a1 + a2 + · · · + an ≥ n. Therefore, it suffices to prove that the desired inequality holds for a1 + a2 + · · · + an ≥ n. Actually, we only need to consider that a1 + a2 + · · · + an = n. Indeed, if we denote p = (a1 + a2 + · · · + an )/n and bi = ai /p for i = 1, 2, · · · , n, then the desired inequality becomes Solution. We consider two cases: 1 < k ≤ n + 1 and k ≥ n −

b1

X p k−1 b1k

+ b2 + · · · + bn

≤ 1,

p ≥ 1.

Clearly, it suffices to consider the case p = 1; that is, a1 + a2 + · · · + an = n. On this hypothesis, we can rewrite the desired inequality as X a1 ≤ 1. k a1 − a1 + n Since k > 1, by Bernoulli’s inequality, we have a1k − a1 + n ≥ 1 + k(a1 − 1) − a1 + n = n − k + 1 + (k − 1)a1 > 0. Thus, it is enough to show that X

a1 ≤ 1, n − k + 1 + (k − 1)a1

Symmetric Nonrational Inequalities

411

which is equivalent to X

1 ≥ 1. n − k + 1 + (k − 1)a1

This inequality follows immediately from the AM-HM inequality €X Š X 1 ‹ x1 ≥ n2 , x1 for x i = n − k + 1 + (k − 1)ai , i = 1, 2, · · · , n. 1 p . Let us denote r = n a1 a2 · · · an and bi = ai /r for i = 1, 2, · · · , n. n−1 Note that r ≥ 1 and b1 b2 · · · bn = 1. The desired inequality becomes

Case 2: k ≥ n −

b1

X b1k r k−1

+ b2 + · · · + b n

≤ 1,

and we see that it suffices to prove it for r = 1; that is, for a1 a2 · · · an = 1. On this hypothesis, it suffices to show that p

(n − 1)a1 a1k + a2 + · · · + an

+

a1 p

p

p

a1 + a2 + · · · + an

≤1

for a suitable real p; then, adding this inequality and the analogous inequalities for p a2 , · · · an yields the desired inequality. Let us denote t = n−1 a2 · · · an . By the AM-GM inequality, we have a2 + · · · an ≥ (n − 1)t,

p

a2 + · · · + anp ≥ (n − 1)t p .

Thus, it suffices to show that (n − 1)a1 a1k + (n − 1)t

p

+

a1 p

a1 + (n − 1)t p

≤ 1.

Since a1 = 1/t n−1 , this inequality is equivalent to (n − 1)t n+q − (n − 1)t q − t q−np + 1 ≥ 0, where q = (n − 1)(k − 1). Choosing p=

(n − 1)(k − n − 1) , n

the inequality becomes as follows (n − 1)t n+q − (n − 1)t q − t n(n−1) + 1 ≥ 0,

412

Vasile Cîrtoaje 2

(n − 1)t q (t n − 1) − (t n − 1)(t n (t n − 1)[(t q − t n

2

−2n

−2n 2

) + (t q − t n

+ tn

−3n

2

−3n

+ · · · + 1) ≥ 0,

) + · · · + (t q − 1)] ≥ 0.

The last inequality is clearly true for q ≥ n2 − 2n; that is, for k ≥ n − proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

1 . Thus, the n−1

P 2.104. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an ≥ 1. If −1 − then

2 ≤ k < 1, n−2 a1

X a1k

+ a2 + · · · + an

≥ 1. (Vasile Cîrtoaje, 2006)

p Solution. Let us denote r = n a1 a2 · · · an and bi = ai /r for i = 1, 2, · · · , n. Note that r ≥ 1 and b1 b2 · · · bn = 1. The desired inequality becomes b1

X b1k /r 1−k

+ b2 + · · · + bn

≥ 1,

and we see that it suffices to prove it for r = 1; that is, for a1 a2 · · · an = 1. On this hypothesis, by the Cauchy-Schwarz inequality, we have P P X ( a1 )2 ( a1 )2 a1 = P ≥P P P . a1k + a2 + · · · + an a1 (a1k + a2 + · · · + an ) ( a1 )2 + a11+k − a12 Thus, we still have to show that X

a12 ≥

X

a11+k .

Case 1: −1 ≤ k < 1. Using Chebyshev’s inequality and the AM-GM inequality yields X

a12 ≥

X X 1 €X 1−k Š €X 1+k Š a1 a1 ≥ (a1 a2 · · · an )(1−k)/n a11+k = a11+k . n

2 ≤ k < −1. It is convenient to replace the numbers a1 , a2 , · · · , an n−1 (n−1)/2 (n−1)/2 by a1 , a2 , · · · , an(n−1)/2 , respectively. So, we need to show that a1 a2 · · · an = 1 involves X X q a1n−1 ≥ a1 , Case 2: −1 −

Symmetric Nonrational Inequalities where q=

413

(n − 1)(1 + k) , 2

−1 ≤ q < 0.

By the AM-GM inequality, we get X

a1n−1 =

X X 1 1 X n−1 (a2 + · · · + ann−1 ) ≥ a2 · · · an = . n−1 a1

Thus, it suffice to show that X X 1 q ≥ a1 . a1 By Chebyshev’s inequality and the AM-GM inequality, we have X 1 €X Š X 1 €X −1−q Š €X q Š q q a1 . a1 = a1 ≥ (a1 a2 · · · an )−(1+q)/n ≥ a1 a1 n Thus, the proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

P 2.105. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If k ≥ 0, then X 1 ≤ 1. k a1 + a2 + · · · + an (Vasile Cîrtoaje, 2006) Solution. Consider two cases: 0 ≤ k ≤ 1 and k ≥ 1. Case 1: 0 ≤ k ≤ 1. By the Cauchy-Schwarz inequality and the AM-GM inequality, we have a11−k + 1 + · · · + 1 1 ≤ p p p ( a1 + a2 + · · · + an )2 a1k + a2 + · · · + an =P

a11−k + n − 1 a11−k + n − 1 P ≤P , p a1 + 2 1≤i< j≤n ai a j a1 + n(n − 1)

hence X

a11−k + n(n − 1) P ≤ . a1 + n(n − 1) a1k + a2 + · · · + an P

1

Therefore, it suffices to show that X

a11−k ≤

X

a1 .

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Vasile Cîrtoaje

Indeed, by Chebyshev’s inequality and the AM-GM inequality, we have X

a1 =

X

a1k · a11−k ≥

€X Š X 1 €X k Š €X 1−k Š a1 a1 ≥ (a1 a2 · · · an )k/n a11−k = a11−k . n

Case 2: k ≥ 1. Write the inequality as X

p

n−1 a1k + a2 + · · · + an

where p=

+



a1

p

p

p

a1 + a2 + · · · + an

− 1 ≤ 0,

(n − 1)(k − 1) ≥ 0. n

To complete the proof, it suffices to show that p

n−1 a1k + a2 + · · · + an

≤1−

a1 p

p.

p

a1 + a2 + · · · + an

Let x=

p

n−1

x > 0.

a1 ,

By the AM-GM inequality, we have p n−1 n−1 a2 + · · · + an ≥ (n − 1) n−1 a2 · · · an = n−1 p = a1 x and p

a2 + · · · + anp ≥

Æ

n−1

(a2 · · · an ) p =

n−1 n−1 . q p = xp a1

n−1

Thus, it suffices to show that n−1 x (n−1)k +

n−1 x

≤1−

x (n−1)p x (n−1)p +

n−1 xp

,

which is equivalent to x x (n−1)k+1 x

(n−1)k+1

+n−1

≤

1 , x np + n − 1

− x np+1 − (n − 1)(x − 1) ≥ 0,

x q (x n−1 − 1) − (n − 1)(x − 1) ≥ 0, where q = (n − 1)(k − 1) + 1 ≥ 1.

Symmetric Nonrational Inequalities

415

The last inequality is true since x q (x n−1 − 1) − (n − 1)(x − 1) = (x q − 1)(x n−1 − 1) + (x n−1 − 1) − (n − 1)(x − 1), and (x q − 1)(x n−1 − 1) ≥ 0, (x n−1 − 1) − (n − 1)(x − 1) = (x − 1)[(x n−2 − 1) + (x n−3 − 1) + · · · + (x − 1)] ≥ 0. This completes the proof. The equality holds for a1 = a2 = · · · = an = 1.

P 2.106. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n. If 1 < k ≤ n + 1, then a1 a1k + a2 + · · · + an

+

a2 a1 + a2k + · · · + an

+ ··· +

an ≤ 1. a1 + a2 + · · · + ank (Vasile Cîrtoaje, 2006)

Solution. Using the substitutions s= and x1 =

a1 + a2 + · · · + an , n

an a1 a , x2 = 2 , · · · , x n = , s s s

the desired inequality becomes x1 s k−1 x 1k

+ ··· +

+ x2 + · · · + x n

xn ≤ 1, x 1 + x 2 + · · · + s k−1 x nk

where s ≥ 1 and x 1 + x 2 + · · · + x n = n. Clearly, if this inequality holds for s = 1, then it holds for any s ≥ 1. Therefore, we need only to consider the case s = 1, when a1 + a2 + · · · + an = n, and the desired inequality is equivalent to a1 a1k

− a1 + n

+

a2 a2k

− a2 + n

+ ··· +

an ≤ 1. ank − an + n

By Bernoulli’s inequality, we have a1k − a1 + n ≥ 1 + k(a1 − 1) − a1 + n = n − k + 1 + (k − 1)a1 ≥ 0. Consequently, it suffices to prove that n X i=1

ai ≤ 1. n − k + 1 + (k − 1)ai

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Vasile Cîrtoaje

For k = n + 1, this inequality is an equality. Otherwise, for 1 < k < n + 1, we rewrite the inequality as n X 1 ≥ 1, n − k + 1 + (k − 1)ai i=1 which follows from the AM-HM inequality as follows: n X i=1

n2 1 ≥ Pn = 1. n − k + 1 + (k − 1)ai i=1 [n − k + 1 + (k − 1)ai ]

The equality holds for a1 = a2 = · · · = an = 1.

P 2.107. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≤ n. If 0 ≤ k < 1, then 1 a1k

+ a2 + · · · + an

+

1 a1 + a2k

+ · · · + an

+ ··· +

1 ≥ 1. a1 + a2 + · · · + ank

Solution. By the AM-HM inequality X

n2 n2 P P P = ≥ a1k + a2 + · · · + an (a1k + a2 + · · · + an ) a1k + (n − 1) a1 1

and Jensen’s inequality X

a1k

 X ‹k 1 ≤n a1 , n

we get 1

X

a1k + a2 + · · · + an

n2

≥ n

P 1 n

a1


k

+ (n − 1)

P

≥ 1. a1

The equality holds for a1 = a2 = · · · = an = 1.

P 2.108. Let a1 , a2 , . . . , an be positive real numbers. If k > 1, then X ak + ak + · · · + ak 2

3

n

a2 + a3 + · · · + an

≤

n(a1k + a2k + · · · + ank ) a1 + a2 + · · · + an

.

(Wolfgang Berndt, Vasile Cîrtoaje, 2006)

Symmetric Nonrational Inequalities

417

Solution. Due to homogeneity, we may assume that a1 + a2 + ... + an = 1. Write the inequality as follows: ‹ X a1 (a2k + a3k + · · · + ank ) ≤ n(a1k + a2k + · · · + ank ); 1+ a2 + a3 + · · · + an X a1 (a k + a k + · · · + a k ) 2

3

n

≤ a1k + a2k + · · · + ank ;

a2 + a3 + · · · + an Œ ‚ k k k X a + a + · · · + a 2 3 n ≥ 0; a1 a1k−1 − a2 + a3 + · · · + an

X a1 a2 (a k−1 − a k−1 ) + a1 a3 (a k−1 − a k−1 ) + · · · + a1 an (a k−1 − a k−1 ) 1

2

1

3

1

a2 + a3 + · · · + an X

ai a j

1≤i< j≤n

aik−1 − a k−1 j 1 − ai

+

a k−1 − aik−1 j 1 − aj

X

ai a j (aik−1 − a k−1 )(ai − a j ) j

1≤i< j≤n

(1 − ai )(1 − a j )

n

≥ 0;

! ≥ 0;

≥ 0.

Since the last inequality is true for k > 1, the proof is finished. The equality holds for a1 = a2 = · · · = an .

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Vasile Cîrtoaje

Chapter 3

Symmetric Power-Exponential Inequalities 3.1

Applications

3.1. If a, b are positive real numbers such that a + b = a4 + b4 , then 3

3

aa b b ≤ 1 ≤ aa b b .

3.2. If a, b are positive real numbers, then a2a + b2b ≥ a a+b + b a+b .

3.3. If a, b are positive real numbers, then aa + b b ≥ a b + ba .

3.4. If a, b are positive real numbers, then a2a + b2b ≥ a2b + b2a .

3.5. If a, b are nonnegative real numbers such that a + b = 2, then (a)

a b + b a ≤ 1 + a b;

(b)

a2b + b2a ≤ 1 + a b. 419

420

Vasile Cîrtoaje

3.6. If a, b are nonnegative real numbers such that a2 + b2 = 2, then a2b + b2a ≤ 1 + a b.

3.7. If a, b are positive real numbers, then a a b b ≤ (a2 − a b + b2 )

3.8. If a, b ∈ (0, 1], then

a+b 2

.

a a b b ≤ 1 − a b + a2 b2 .

3.9. If a, b are positive real numbers such that a + b ≤ 2, then  a  b  b ‹a ≤ 2. + b a

3.10. If a, b are positive real numbers such that a + b = 2, then 1 2a a b b ≥ a2b + b2a + (a − b)2 . 2

3.11. If a, b ∈ (0, 1] or a, b ∈ [1, ∞), then 2a a b b ≥ a2 + b2 .

3.12. If a, b are positive real numbers, then 2a a b b ≥ a2 + b2 .

3.13. If a, b are positive real numbers, then a b

a b ≥



a2 + b2 2

 a+b 2 .

Symmetric Power-Exponential Inequalities

421

3.14. If a, b are positive real numbers such that a2 + b2 = 2, then 1 2a a b b ≥ a2b + b2a + (a − b)2 . 2

3.15. If a, b ∈ (0, 1], then 1 1 (a + b ) 2a + 2b a b 2

2



‹

≤ 4.

3.16. If a, b are positive real numbers such that a + b = 2, then a b b a + 2 ≥ 3a b.

3.17. If a, b ∈ [0, 1], then a b−a + b a−b + (a − b)2 ≤ 2.

3.18. If a, b are nonnegative real numbers such that a + b ≤ 2, then a b−a + b a−b +

7 (a − b)2 ≤ 2. 16

3.19. If a, b are nonnegative real numbers such that a + b ≤ 4, then a b−a + b a−b ≤ 2.

3.20. Let a, b be positive real numbers such that a + b = 2. If k ≥ kb

ka

a a b b ≥ 1.

3.21. If a, b are positive real numbers such that a + b = 2, then a

p

a

p

b

b

≥ 1.

1 , then 2

422

Vasile Cîrtoaje

3.22. If a, b are positive real numbers such that a + b = 2, then 1 − a a+1 b b+1 ≥

1 (1 − a b)2 . 3

3.23. If a, b are positive real numbers such that a + b = 2, then a−a + b−b ≤ 2.

3.24. If a, b are nonnegative real numbers such that a + b = 2, then a2b + b2a ≥ a b + b a ≥ a2 b2 + 1.

3.25. If a, b are positive real numbers such that a + b = 2, then a3b + b3a ≤ 2.

3.26. If a, b are nonnegative real numbers such that a + b = 2, then a3b + b3a +



a−b 2

‹4

≤ 2.

3.27. If a, b are positive real numbers such that a + b = 2, then 2

2

a a + b b ≤ 2.

3.28. If a, b are positive real numbers such that a + b = 2, then 3

3

a a + b b ≥ 2.

3.29. If a, b are positive real numbers such that a + b = 2, then 2

2

a5b + b5a ≤ 2.

Symmetric Power-Exponential Inequalities

423

3.30. If a, b are positive real numbers such that a + b = 2, then a2

p

b

+ b2

p

a

≤ 2.

3.31. If a, b are nonnegative real numbers such that a + b = 2, then a b(1 − a b)2 a b(1 − a b)2 ≤ a b+1 + b a+1 − 2 ≤ . 2 3

3.32. If a, b are nonnegative real numbers such that a + b = 1, then a2b + b2a ≤ 1.

3.33. If a, b are positive real numbers such that a + b = 1, then 2a a b b ≥ a2b + b2a .

3.34. If a, b are positive real numbers such that a + b = 1, then a−2a + b−2b ≤ 4.

424

Vasile Cîrtoaje

Symmetric Power-Exponential Inequalities

3.2

425

Solutions

P 3.1. If a, b are positive real numbers such that a + b = a4 + b4 , then 3

3

aa b b ≤ 1 ≤ aa b b . (Vasile Cîrtoaje, 2008) Solution. We will use the inequality ln x ≤ x − 1,

x > 0.

To prove this inequality, let us denote f (x) = x − 1 − ln x,

x > 0.

From

x −1 , x it follows that f (x) is decreasing on (0, 1] and increasing on [1, ∞). Therefore, f 0 (x) =

f (x) ≥ f (1) = 0. Using this inequality, we have ln a a b b = a ln a + b ln b ≤ a(a − 1) + b(b − 1) = a2 + b2 − (a + b). Therefore, the left inequality a a b b ≤ 1 is true if a2 + b2 ≤ a + b. We write this inequality in the homogeneous form (a2 + b2 )3 ≤ (a + b)2 (a4 + b4 ), which is equivalent to the obvious inequality a b(a − b)(a3 − b3 ) ≥ 0. Taking now x =

1 in the inequality ln x ≤ x − 1 yields a a ln a ≥ a − 1.

Similarly, b ln b ≥ b − 1, and hence 3

3

ln a a b b = a3 ln a + b3 ln b ≤ a2 (a − 1) + b2 (b − 1) = a3 + b3 − (a2 + b2 ).

426

Vasile Cîrtoaje 3

3

Thus, to prove the right inequality a a b b ≥ 1, it suffices to show that a3 + b3 ≥ a2 + b2 , which is equivalent to the homogeneous inequality (a + b)(a3 + b3 )3 ≥ (a4 + b4 )(a2 + b2 )3 . We can write this inequality as A − 3B ≥ 0, where A = (a + b)(a9 + b9 ) − (a4 + b4 )(a6 + b6 ), B = a2 b2 (a2 + b2 )(a4 + b4 ) − a3 b3 (a + b)(a3 + b3 ). Since A = a b(a3 − b3 )(a5 − b5 ),

B = a2 b2 (a − b)(a5 − b5 ),

we get A − 3B = a b(a − b)3 (a5 − b5 ) ≥ 0. Both inequalities become equalities for a = b = 1.

P 3.2. If a, b are positive real numbers, then a2a + b2b ≥ a a+b + b a+b . (Vasile Cîrtoaje, 2010) Solution. Assume that a ≥ b and consider the following two cases. Case 1: a ≥ 1. Write the inequality as a a+b (a a−b − 1) ≥ b2b (b a−b − 1). For b ≤ 1, we have a a+b (a a−b − 1) ≥ 0 ≥ b2b (b a−b − 1). For b ≥ 1, the inequality is also true since a a+b ≥ a2b ≥ b2b ,

a a−b − 1 ≥ b a−b − 1 ≥ 0.

Case 2: a ≤ 1. Since a2a + b2b ≥ 2a a b b , it suffices to show that 2a a b b ≥ a a+b + b a+b ,

Symmetric Power-Exponential Inequalities which can be written as

 a b b

427

 ‹a b + ≤ 2. a

By Bernoulli’s inequality, we get  a b b

+

 ‹a  ‹  ‹ b a−b b b−a a b(a − b) a(b − a) = 1+ + 1+ ≤1+ +1+ = 2. a b a b a

The equality holds for a = b. Conjecture 1. If a, b are positive real numbers, then a4a + b4b ≥ a2a+2b + b2a+2b . Conjecture 2. If a, b, c are positive real numbers, then a3a + b3b + c 3c ≥ a a+b+c + b a+b+c + c a+b+c . Conjecture 3. If a, b, c, d are positive real numbers, then a4a + b4b + c 4c + d 4d ≥ a a+b+c+d + b a+b+c+d + c a+b+c+d + d a+b+c+d .

P 3.3. If a, b are positive real numbers, then aa + b b ≥ a b + ba . (M. Laub, 1985) Solution. Assume that a ≥ b. We will show that if a ≥ 1, then the inequality is true. From a a−b ≥ b a−b , we get bb ≥

a b ba . aa

Therefore, aa + b b − a b − ba ≥ aa +

a b ba (a a − a b )(a a − b a ) b a − a − b = ≥ 0. aa aa

Consider further the case 0 < b ≤ a < 1. First Solution. Denoting c = ab,

d = bb,

k=

a , b

428

Vasile Cîrtoaje

where c ≥ d and k ≥ 1, the inequality becomes c k − d k ≥ c − d. Since the function f (x) = x k is convex for x ≥ 0, from the well-known inequality f (c) − f (d) ≥ f 0 (d)(c − d), we get c k − d k ≥ kd k−1 (c − d). Thus, it suffices to show that kd k−1 ≥ 1, which is equivalent to b1−a+b ≤ a. Indeed, since 0 < 1 − a + b ≤ 1, by Bernoulli’s inequality, we get b1−a+b = [1 + (b − 1)]1−a+b ≤ 1 + (1 − a + b)(b − 1) = a − b(a − b) ≤ a. The equality holds for a = b. Second Solution. Denoting c=

ba , a b + ba

ab , a b + ba

d=

k=

a , b

where c + d = 1 and k ≥ 1, the inequality becomes ck a + d k−b ≥ 1. By the weighted AM-GM inequality, we have ck a + d k−b ≥ k ac · k−bd = k ac−bd . Thus, it suffices to show that ac ≥ bd; that is, a1−b ≥ b1−a , which is equivalent to f (a) ≥ f (b), where f (x) =

ln x . 1− x

It is enough to prove that f (x) is an increasing function. Since f 0 (x) =

g(x) , (1 − x)2

g(x) =

1 − 1 + ln x. x

Symmetric Power-Exponential Inequalities

429

we need to show that g(x) ≥ 0 for x ∈ (0, 1). Indeed, from g 0 (x) =

x −1 < 0, x2

it follows that g(x) is strictly decreasing, hence g(x) > g(1) = 0.

P 3.4. If a, b are positive real numbers, then a2a + b2b ≥ a2b + b2a . Solution. Assume that a > b, and denote c = ab,

d = bb,

k=

a , b

where c > d and k > 1. The inequality becomes c 2k − d 2k ≥ c 2 − d 2 . We shall show that c 2k − d 2k > k(cd)k−1 (c 2 + d 2 ) > c 2 − d 2 . The left inequality follows from Lemma below for x = (c/d)2 . The right inequality is equivalent to k(cd)k−1 > 1, (a b)a−b >

b , a

1+a− b ln a > ln b. 1−a+ b For fixed a, let us define f (b) =

1+a− b ln a − ln b, 0 < b < a. 1−a+ b

If f 0 (b) < 0, then f (b) is strictly decreasing, and hence f (b) > f (a) = 0. Since f 0 (b) =

−2 1 ln a − , 2 (1 − a + b) b

we need to show that g(a) > 0, where g(a) = 2 ln a +

(1 − a + b)2 . b

430

Vasile Cîrtoaje

From

2 2(1 − a + b) 2(a − b)(a − 1) − = < 0, a b ab it follows that g(a) is strictly decreasing on [b, 1), therefore g(a) > g(1) = b > 0. This completes the proof. The equality holds for a = b. g 0 (a) =

Lemma. Let k and x be positive real numbers. If either k > 1 and x ≥ 1, or 0 < k < 1 and 0 < x ≤ 1, then k−1 x k − 1 ≥ k x 2 (x − 1). Proof. We need to show that f (x) ≥ 0, where f (x) = x k − 1 − k x We have f 0 (x) = Since

1 k−3 k x 2 g(x), 2

k−1 2

g(x) = 2x

(x − 1).

k+1 2

− (k + 1)x + k − 1.

Š € k−1 g 0 (x) = (k + 1) x 2 − 1 ≥ 0,

g(x) is increasing. If x ≥ 1, then g(x) ≥ g(1) = 0, f (x) is increasing, hence f (x) ≥ f (1) = 0. If 0 < x ≤ 1, then g(x) ≤ g(1) = 0, f (x) is decreasing, hence f (x) ≥ f (1) = 0. The equality holds for x = 1. Remark. The following more general inequality holds for 0 < k ≤ e and any positive numbers a and b (Vasile Cîrtoaje, 2006): a ka + b kb ≥ a kb + b ka . Conjecture 1. If 0 < k ≤ e and a, b ∈ (0, 4], then p 2 a ka b kb ≥ a kb + b ka . Conjecture 2. If a, b ∈ (0, 5], then 2a a b b ≥ a2b + b2a . Conjecture 3. If a, b ∈ [0, 5], then 

a2 + b2 2

 a+b 2

≥ a2b + b2a .

Note that Conjecture 1 has been proved for a, b ∈ (0, 1] by A. Coronel and F. Huancas (2014), and also by L. Matejicka (2014).

Symmetric Power-Exponential Inequalities

431

P 3.5. If a, b are nonnegative real numbers such that a + b = 2, then (a)

a b + b a ≤ 1 + a b;

(b)

a2b + b2a ≤ 1 + a b. (Vasile Cîrtoaje, 2007)

Solution. Without loss of generality, assume that a ≥ b. Since 0 ≤ b ≤ 1 and 0 ≤ a − 1 ≤ 1, by Bernoulli’s inequality, we have a b ≤ 1 + b(a − 1) = 1 + b − b2 and b a = b · b a−1 ≤ b[1 + (a − 1)(b − 1)] = b2 (2 − b). (a) We have a b + b a − 1 − a b ≤ (1 + b − b2 ) + b2 (2 − b) − 1 − (2 − b)b = −b(b − 1)2 ≤ 0. The equality holds for a = b = 1, for a = 2 and b = 0, and for a = 0 and b = 2. (b) We have a2b + b2a − 1 − a b ≤ (1 + b − b2 )2 + b4 (2 − b)2 − 1 − (2 − b)b = b3 (b − 1)2 (b − 2) = −a b3 (b − 1)2 ≤ 0. The equality holds for a = b = 1, for a = 2 and b = 0, and for a = 0 and b = 2.

P 3.6. If a, b are nonnegative real numbers such that a2 + b2 = 2, then a2b + b2a ≤ 1 + a b. (Vasile Cîrtoaje, 2007) Solution. Without loss of generality, assume that a ≥ 1 ≥ b. Applying Bernoulli’s inequality gives a b ≤ 1 + b(a − 1), hence a2b ≤ (1 + a b − b)2 . Also, since 0 ≤ b ≤ 1 and 2a ≥ 2, we have b2a ≤ b2 .

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Therefore, it suffices to show that (1 + a b − b)2 + b2 ≤ 1 + a b, which can be written as b(2 + 2a b − a − 2b − a2 b) ≥ 0. So, we need to show that 2 + 2a b − a − 2b − a2 b ≥ 0, which is equivalent to 2(1 − a)(1 − b) + a(1 − a b) ≥ 0, 4(1 − a)(1 − b) + a(a − b)2 ≥ 0. Since a ≥ 1, it suffices to show that 4(1 − a)(1 − b) + (a − b)2 ≥ 0. Indeed, 4(1 − a)(1 − b) + (a − b)2 = −4(a − 1)(1 − b) + [(a − 1) + (1 − b)]2 = [(a − 1) − (1 − b)]2 = (a + b − 2)2 ≥ 0. The equality holds for a = b = 1, for a =

p p 2 and b = 0, and for a = 0 and b = 2.

P 3.7. If a, b are positive real numbers, then a a b b ≤ (a2 − a b + b2 )

a+b 2

.

Solution. By the weighted AM-GM inequality, we have a

b

a · a + b · b ≥ (a + b)a a+b b a+b , 

a2 + b2 a+b

a+b

≥ aa b b .

Thus, it suffices to show that 2

2

a − ab + b ≥



a2 + b2 a+b

2 ,

Symmetric Power-Exponential Inequalities

433

which is equivalent to (a + b)(a3 + b3 ) ≥ (a2 + b2 )2 , a b(a − b)2 ≥ 0. The equality holds for a = b.

P 3.8. If a, b ∈ (0, 1], then

a a b b ≤ 1 − a b + a2 b2 . (Vasile Cîrtoaje, 2010)

Solution. We claim that x x ≤ x2 − x + 1 for all x ∈ (0, 1]. If this is true, then 1 − a b + a2 b2 − a a b b ≥ 1 − a b + a2 b2 − (a2 − a + 1)(b2 − b + 1) = (a + b)(1 − a)(1 − b) ≥ 0. To prove the inequality x x ≤ x 2 − x + 1, we show that f (x) ≤ 0 for x ∈ (0, 1], where f (x) = x ln x − ln(x 2 − x + 1). We have f 0 (x) = ln x + 1 − f 00 (x) =

2x − 1 , − x +1

x2

(1 − x)(1 − 2x − x 2 − x 4 ) . x(x 2 − x + 1)2

Let x 1 ∈ (0, 1) be the positive root of the equation x 4 + x 2 + 2x = 1. Then, f 00 (x) > 0 for x ∈ (0, x 1 ) and f 00 (x) < 0 for x ∈ (x 1 , 1), hence f 0 is strictly increasing on (0, x 1 ] and strictly decreasing on [x 1 , 1]. Since lim x→0 f 0 (x) = −∞ and f 0 (1) = 0, there is x 2 ∈ (0, x 1 ) such that f 0 (x 2 ) = 0, f 0 (x) < 0 for x ∈ (0, x 2 ) and f 0 (x) > 0 for x ∈ (x 2 , 1). Therefore, f is decreasing on (0, x 2 ] and increasing on [x 2 , 1]. Since lim x→0 f (x) = 0 and f (1) = 0, it follows that f (x) ≤ 0 for x ∈ (0, 1]. The proof is completed. The equality holds for a = b = 1.

P 3.9. If a, b are positive real numbers such that a + b ≤ 2, then  a  b  b ‹a + ≤ 2. b a (Vasile Cîrtoaje, 2010)

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Vasile Cîrtoaje

Solution. Using the substitution a = t c and b = t d, where c, d, t are positive real numbers such that c + d = 2 and t ≤ 1, we need to show that  c t d d

 ‹t c d + ≤ 2. c

Write this inequality as f (t) ≤ 2, where t

t

f (t) = A + B , A =

 c d d

 ‹c d , B= . c

Since f (t) is a convex function, we have f (t) ≤ max{ f (0), f (1)} = max{2, f (1)}. Therefore, it suffices to show that f (1) ≤ 2; that is, 2c c d d ≥ c 2 + d 2 . Setting c = 1 + x and d = 1 − x, where 0 ≤ x < 1, this inequality turns into (1 + x)1+x (1 − x)1−x ≥ 1 + x 2 , which is equivalent to f (x) ≥ 0, where f (x) = (1 + x) ln(1 + x) + (1 − x) ln(1 − x) − ln(1 + x 2 ). We have f 0 (x) = ln(1 + x) − ln(1 − x) − f 00 (x) =

2x , 1 + x2

1 1 2(1 − x 2 ) 8x 2 + − = . 1 + x 1 − x (1 + x 2 )2 (1 − x 2 )(1 + x 2 )2

Since f 00 (x) ≥ 0 for x ∈ [0, 1), it follows that f 0 is increasing, f 0 (x) ≥ f 0 (0) = 0, f (x) is increasing, hence f (x) ≥ f (0) = 0. The proof is completed. The equality holds for a = b.

P 3.10. If a, b are positive real numbers such that a + b = 2, then 1 2a a b b ≥ a2b + b2a + (a − b)2 . 2 (Vasile Cîrtoaje, 2010)

Symmetric Power-Exponential Inequalities

435

Solution. According to the inequalities in P 3.5-(b) and P 3.9, we have a2b + b2a ≤ 1 + a b and 2a a b b ≥ a2 + b2 . Therefore, it suffices to show that 1 a2 + b2 ≥ 1 + a b + (a − b)2 . 2 which is an identity. The equality holds for a = b = 1.

P 3.11. If a, b ∈ (0, 1] or a, b ∈ [1, ∞), then 2a a b b ≥ a2 + b2 . Solution. For a = x and b = 1, the desired inequality becomes 2x x ≥ x 2 + 1,

x > 0.

If this inequality is true, then 4a a b b − 2(a2 + b2 ) ≥ (a2 + 1)(b2 + 1) − 2(a2 + b2 ) = (a2 − 1)(b2 − 1) ≥ 0. To prove the inequality 2x x ≥ x 2 + 1, we show that f (x) ≥ 0, where f (x) = ln 2 + x ln x − ln(x 2 + 1). We have f 0 (x) = ln x + 1 − f 00 (x) =

2x , +1

x2

x 2 (x + 1)2 + (x − 1)2 . x(x 2 + 1)2

Since f 00 (x) > 0 for x > 0, f 0 is strictly increasing. Since f 0 (1) = 0, it follows that f 0 (x) < 0 for x ∈ (0, 1) and f 0 (x) > 0 for x ∈ (1, ∞). Therefore, f is decreasing on (0, 1] and increasing on [1, ∞), hence f (x) ≥ f (1) = 0 for x > 0. This completes the proof. The equality holds for a = b = 1.

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P 3.12. If a, b are positive real numbers, then 2a a b b ≥ a2 + b2 . (Vasile Cîrtoaje, 2014) Solution. By Lemma below, it suffices to show that (a4 − 2a3 + 4a2 − 2a + 3)(b4 − 2b3 + 4b2 − 2b + 3) ≥ 8(a2 + b2 ), which is equivalent to A ≥ 0, where A =a4 b4 − 2a3 b3 (a + b) + 4a2 b2 (a2 + b2 + a b) − [2a b(a3 + b3 ) + 8a2 b2 (a + b)] + [3(a4 + b4 ) + 4a b(a2 + b2 ) + 16a2 b2 ] − [6(a3 + b3 ) + 8a b(a + b)] + 4(a2 + b2 + a b) − 6(a + b) + 9. We can check that A = [a2 b2 − a b(a + b) + a2 + b2 − 1]2 + B, where B =a2 b2 (a + b)2 − 6a2 b2 (a + b) + [2(a4 + b4 ) + 4a b(a2 + b2 ) + 16a2 b2 ] − [6(a3 + b3 ) + 10a b(a + b)] + [6(a2 + b2 ) + 4a b] − 6(a + b) + 8. Also, we have B = [a b(a + b) − 3a b + 1]2 + C, where C =[2(a4 + b4 ) + 4a b(a2 + b2 ) + 7a2 b2 ] − [6(a3 + b3 ) + 12a b(a + b)] + [6(a2 + b2 ) + 10a b] − 6(a + b) + 7, and C = (a b − 1)2 + 2D, where D =[a4 + b4 + 2a b(a2 + b2 ) + 3a2 b2 ] − [3(a3 + b3 ) + 6a b(a + b)] + 3(a + b)2 − 3(a + b) + 3, It suffices to show that D ≥ 0. Indeed, D =[(a + b)4 − 2a b(a + b)2 + a2 b2 ] − 3[(a + b)3 − a b(a + b)] + 3(a + b)2 − 3(a + b) + 3 =[(a + b)2 − a b]2 − 3(a + b)[(a + b)2 − a b] + 3(a + b)2 − 3(a + b) + 3 ˜2  ‹2 • a+b 3 2 − 1 ≥ 0. = (a + b) − a b − (a + b) + 3 2 2

Symmetric Power-Exponential Inequalities

437

This completes the proof. The equality holds for a = b = 1. Lemma. If x > 0, then

1 x x ≥ x + (x − 1)2 (x 2 + 3). 4

Proof. We need to show that f (x) ≥ 0 for x > 0, where g(x) = x 4 − 2x 3 + 4x 2 − 2x + 3.

f (x) = ln 4 + x ln x − ln g(x), We have f 0 (x) = 1 + ln x − f 00 (x) =

2(2x 3 − 3x 2 + 4x − 1) , g(x)

x 8 + 6x 4 − 32x 3 + 48x 2 − 32x + 9 (x − 1)2 h(x) = , g 2 (x) g 2 (x)

where h(x) = x 6 + 2x 5 + 3x 4 + 4x 3 + 11x 2 − 14x + 9. Since h(x) > 7x 2 − 14x + 7 = 7(x − 1)2 ≥ 0, we have f 00 (x) ≥ 0, hence f 0 is strictly increasing on (0, ∞). Since f 0 (1) = 0, it follows that f 0 (x) < 0 for x ∈ (0, 1) and f 0 (x) > 0 for x ∈ (1, ∞). Therefore, f is decreasing on (0, 1] and increasing on [1, ∞), hence f (x) ≥ f (1) = 0 for x > 0.

P 3.13. If a, b are positive real numbers, then aa b b ≥



a2 + b2 2

 a+b 2 .

First Solution. Using the substitution a = b x, where x > 0, the inequality becomes as follows:  2 2  bx+b b x + b2 2 bx b (b x) b ≥ , 2 x

(b x) b ≥



b2 x 2 + b2 2

b x+1 x x ≥ b x+1



 x+1 2

x2 + 1 2

,  x+1 2 ,

438

Vasile Cîrtoaje 

x

x ≥

x2 + 1 2

 x+1 2 .

It is true if f (x) ≥ 0 for all x > 0, where f (x) = We have f 0 (x) =

x 1 x2 + 1 ln x − ln . x +1 2 2

g(x) x 1 1 − = ln x + , (x + 1)2 x + 1 x 2 + 1 (x + 1)2

where g(x) = ln x − Since g 0 (x) =

x2 − 1 . x2 + 1

(x 2 − 1)2 ≥ 0, x(x 2 + 1)2

g is strictly increasing, therefore g(x) < 0 for x ∈ (0, 1), g(1) = 0, g(x) > 0 for x ∈ (1, ∞). Thus, f is decreasing on (0, 1] and increasing on [1, ∞), hence f (x) ≥ f (1) = 0. This completes the proof. The equality holds for a = b. Second Solution. Write the inequality in the form a ln a + b ln b ≥

a+b a2 + b2 log . 2 2

Without loss of generality, consider that a + b = 2k, k > 0, and denote a = k + x,

b = k − x, 0 ≤ x < k.

We need to show that f (x) ≥ 0, where f (x) = (k + x) ln(k + x) + (k − x) ln(k − x) − k ln(x 2 + k2 ). We have f 0 (x) = ln(k + x) − ln(k − x) −

2k x , + k2

x2

1 1 2k(x 2 − k2 ) + + k+x k−x (x 2 + k2 )2 2 2 8k x = 2 . (k − x 2 )(x 2 + k2 )2

f 00 (x) =

Since f 00 (x) ≥ 0 for x ≥ 0, f 0 is increasing, hence f 0 (x) ≥ f 0 (0) = 0. Therefore, f is increasing on [0, k), hence f (x) ≥ f (0) = 0.

Symmetric Power-Exponential Inequalities

439

Remark. For a + b = 2, this inequality can be rewritten as 2a a b b ≥ a2 + b2 , 2≥

 a b b

 ‹a b + . a

Also, for a + b = 1, the inequality becomes 2a2a b2b ≥ a2 + b2 , 2≥

 a 2b b

+

 ‹2a b . a

P 3.14. If a, b are positive real numbers such that a2 + b2 = 2, then 1 2a a b b ≥ a2b + b2a + (a − b)2 . 2 (Vasile Cîrtoaje, 2010) Solution. According to the inequalities in P 3.6 and P 3.12, we have a2b + b2a ≤ 1 + a b and 2a a b b ≥ a2 + b2 . Therefore, it suffices to show that 1 a2 + b2 ≥ 1 + a b + (a − b)2 , 2 which is an identity. The equality holds for a = b = 1.

P 3.15. If a, b ∈ (0, 1], then (a2 + b2 )



1 1 + 2b 2a a b

‹

≤ 4. (Vasile Cîrtoaje, 2014)

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Vasile Cîrtoaje

Solution. For a = x and b = 1, the desired inequality becomes x 2x ≥

1 + x2 , 3 − x2

x ∈ (0, 1].

If this inequality is true, it suffices to show that   3 − a2 3 − b2 2 2 ≤ 4, + (a + b ) 1 + a2 1 + b2 which is equivalent to a2 b2 (2 + a2 + b2 ) + 2 − (a2 + b2 ) − (a2 + b2 )2 ≥ 0, (2 + a2 + b2 )(1 − a2 )(1 − b2 ) ≥ 0. To prove the inequality x 2x ≥

1 + x2 for 0 < x ≤ 1 , we show that f (x) ≥ 0, where 3 − x2

f (x) = x ln x +

1 1 ln(3 − x 2 ) − ln(1 + x 2 ). 2 2

We have f 0 (x) = 1 + ln x −

x x − , 2 3− x 1 + x2

1 3 + x2 1 − x2 − − x (3 − x 2 )2 (1 + x 2 )2 1 − x2 (1 − x)(9 + 6x − x 3 ) − . = x(3 − x)2 (1 + x 2 )2

f 00 (x) =

We will show that f 00 (x) > 0 for 0 < x < 1. This is true if 9 + 6x − x 3 1+ x − > 0. 2 x(3 − x) (1 + x 2 )2 Indeed, 9 + 6x − x 3 1+ x 9 1+ x 1 − > − = > 0. 2 2 2 2 x(3 − x) (1 + x ) 9x x(1 + x) 1+ x Since f 00 (x) > 0, f 0 is strictly increasing on (0, 1]. Since f 0 (1) = 0, it follows that f 0 (x) < 0 for x ∈ (0, 1), f is strictly decreasing on (0, 1], hence f (x) ≥ f (1) = 0. This completes the proof. The equality holds for a = b = 1.

Symmetric Power-Exponential Inequalities

441

P 3.16. If a, b are positive real numbers such that a + b = 2, then a b b a + 2 ≥ 3a b. (Vasile Cîrtoaje, 2010) Solution. Setting a = 1+ x and b = 1− x, where 0 ≤ x < 1, the inequality is equivalent to (1 + x)1−x (1 − x)1+x ≥ 1 − 3x 2 . 1 Consider further the non-trivial case 0 ≤ x < p , and write the desired inequality as 3 f (x) ≥ 0, where f (x) = (1 − x) ln(1 + x) + (1 + x) ln(1 − x) − ln(1 − 3x 2 ). We have f 0 (x) = − ln(1 + x) + ln(1 − x) +

1− x 1+ x 6x − + , 1 + x 1 − x 1 − 3x 2

1 00 −1 2(x 2 + 1) 3(3x 2 + 1) f (x) = − + . 2 1 − x 2 (1 − x 2 )2 (1 − 3x 2 )2 Making the substitution t = x 2 , 0 ≤ t
0. 2 2 2 (3t − 1) (t − 1) (t − 1)2 (3t − 1)2 Therefore, f 0 (x) is strictly increasing, f 0 (x) ≥ f 0 (0) = 0, f (x) is strictly increasing, and hence f (x) ≥ f (0) = 0. This completes the proof. The equality holds for a = b = 1.

P 3.17. If a, b ∈ [0, 1], then a b−a + b a−b + (a − b)2 ≤ 2. (Vasile Cîrtoaje, 2010) Solution (by Vo Quoc Ba Can). Without loss of generality, assume that a ≥ b. Using the substitution c = a − b, we need to show that (b + c)−c + b c + c 2 ≤ 2 for 0 ≤ b ≤ 1 − c, 0 ≤ c ≤ 1.

442

Vasile Cîrtoaje

If c = 1, then b = 0, and the inequality is an equality. Also, for c = 0, the inequality is an equality. Consider further that 0 < c < 1. We need to show that f (x) ≤ 0, where f (x) = (x + c)−c + x c + c 2 − 2,

x ∈ [0, 1 − c].

We claim that f 0 (x) > 0 for x > 0. Then, f (x) is strictly increasing on [0, 1 − c], hence f (x) ≤ f (1 − c) = (1 − c)c − (1 − c 2 ). In addition, by Bernoulli’s inequality, f (x) ≤ 1 + c(−c) − (1 − c 2 ) = 0. Since f 0 (x) =

c[(x + c)1+c − x 1−c ] , (x + c)1+c x 1−c

we have f 0 (x) > 0 for x > 0 if and only if 1−c

x + c > x 1+c . Let d =

2c 1−c .

Using the weighted AM-GM inequality, we have

  ‹  1  ‹ 1+c 1−c 1 + c 1 − c 1−c 1−c 1 − c d 1+d x 1+c . x +c =1· x +d · = ≥ (1 + d) x 2 2 2 2 Thus, it suffices to show that  ‹ 1−c 1+c 1 − c 1+c ≥ . 2 2 Indeed, using Bernoulli’s inequality, we get 

1−c 2

1−c ‹ 1+c

 ‹ 1−c 1 + c 1+c 1−c 1+c 1+c = 1− ≤1− · = . 2 1+c 2 2

The equality holds for a = b, for a = 1 and b = 0, and for a = 0 and b = 1.

P 3.18. If a, b are nonnegative real numbers such that a + b ≤ 2, then a b−a + b a−b +

7 (a − b)2 ≤ 2. 16 (Vasile Cîrtoaje, 2010)

Symmetric Power-Exponential Inequalities

443

Solution. Assume that a > b and use the substitution c = a − b. We need to show that a−c + (a − c)c +

7 2 c ≤2 16

for

c c ≤ a ≤1+ . 2 Equivalently, we need to show that f (x) ≤ 0 for h ci x ∈ c, 1 + , 0 0 for x > c. Therefore, f (x) is strictly h in the i c increasing on c, 1 + , and hence 2  c  c −c  c c 7 2 f (x) ≤ f 1 + = 1+ + 1− + c − 2. 2 2 2 16 f (x) = x −c + (x − c)c +

To end the proof we need to show that  c c 7 2 c −c  + 1− + c ≤2 1+ 2 2 16 for c ∈ [0, 2]. Indeed, by Lemma 1 and Lemma 2 below, we have  c −c 3 2 1+ + c ≤ 1, 2 16  c c 1 2 1− + c ≤ 1. 2 4 Adding these inequalities yields the desired inequality. The equality holds for a = b, for a = 2 and b = 0, and for a = 0 and b = 2. Lemma 1. If 0 ≤ x ≤ 2, then  x −x 3 2 1+ + x ≤ 1, 2 16 with equality for x = 0 and x = 2. Proof. We need to show that f (x) ≤ 0 for 0 ≤ x ≤ 2, where ‹   x 3 2 f (x) = −x ln 1 + − ln 1 − x . 2 16

444 We have

Vasile Cîrtoaje

 x(3x 2 + 6x − 4) x + , f 0 (x) = − ln 1 + 2 (2 + x)(16 − 3x 2 ) f 00 (x) =

g(x) (2 +

x)2 (16 − 3x 2 )2

,

where g(x) = −9x 5 − 18x 4 + 168x 3 + 552x 2 + 128x − 640. Since g(x 1 ) = 0 for x 1 ≈ 0, 88067, g(x) < 0 for x ∈ [0, x 1 ) and g(x) > 0 for x ∈ (x 1 , 2], f 0 is strictly decreasing on [0, x 1 ] and strictly increasing on [x 1 , 2]. Since f 0 (0) = 0 5 and f 0 (2) = − ln 2 + > 0, there is x 2 ∈ (x 1 , 2) such that f 0 (x 2 ) = 0, f 0 (x) < 0 for 2 x ∈ (0, x 2 ), and f 0 (x) > 0 for x ∈ (x 2 , 2]. Therefore, f is decreasing on [0, x 2 ] and increasing on [x 2 , 2]. Since f (0) = f (2) = 0, it follows that f (x) ≤ 0 for x ∈ [0, 2]. Lemma 2. If 0 ≤ x ≤ 2, then  x x 1 2 1− + x ≤ 1, 2 4 with equality for x = 0 and x = 2. Proof. We need to show that f (x) ≤ 0, where ‹   x 1 2 f (x) = x ln 1 − − ln 1 − x . 2 4 We have

 x x2 f 0 (x) = ln 1 − − , 2 4 − x2 −1 8x f 00 (x) = − . 2 − x (4 − x 2 )2

Since f 00 < 0 for 0 ≤ x < 2, f 0 is strictly decreasing, hence f 0 (x) ≤ f 0 (0) = 0, f is strictly decreasing, therefore f (x) ≤ f (0) = 0 for x ∈ [0, 2]. Conjecture. If a, b are nonnegative real numbers such that a + b =

1 , then 4

a2(b−a) + b2(a−b) ≤ 2.

P 3.19. If a, b are nonnegative real numbers such that a + b ≤ 4, then a b−a + b a−b ≤ 2. (Vasile Cîrtoaje, 2010)

Symmetric Power-Exponential Inequalities

445

Solution. Without loss of generality, assume that a ≥ b. Consider first that a − b ≥ 2. We have a ≥ a − b ≥ 2, and from 4 ≥ a + b = (a − b) + 2b ≥ 2 + 2b, we get b ≤ 1. Clearly, the desired inequality is true because a b−a < 1,

b a−b ≤ 1.

Since the case a − b = 0 is trivial, consider further that 0 < a − b < 2 and use the substitution c = a − b. So, we need to show that a−c + (a − c)c ≤ 2 for

c c ≤ a ≤2+ . 2 Equivalently, we need to show that f (x) ≤ 0 for h ci x ∈ c, 2 + , 0 0, hence g(x) is strictly increasing. Since g(x) → −∞ when x → c and    c c c g 2+ = (1 + c) ln 2 + + (c − 1) ln 2 − 2 2 2       c c c2 > (c − 1) ln 2 + + (c − 1) ln 2 − = (c − 1) ln 4 − > 0, 2 2 4  c there exists x 1 ∈ c, 2 + such that g(x 1 ) = 0, g(x) < 0 for x ∈ (c, x 1 ) and g(x) > 0 2 h  ci c . Thus, f (x) is decreasing on [c, x 1 ] and increasing on x 1 , 2 + . for x ∈ x 1 , 2 + 2 2  c Then, it suffices to show that f (c) ≤ 0 and f 2 + ≤ 0. We have 2 f (c) = c −c − 2 < 1 − 2 < 0.  c Also, the inequality f 2 + ≤ 0 has the form 2  c −c  c c 2+ + 2− ≤ 2, 2 2 which follows immediately from Lemma 1 below. h ci Case 3: 0 < c < 1. We claim that g(x) > 0 for x ∈ c, 2 + . From 2 c(2x − 1 − c) , x(x − c)  ˜ • ˜ 1+c 1+c c it follows that g(x) is decreasing on c, , and increasing on , 2 + . Then, 2 2 2  ‹ 1+c 1−c 1+c g(x) ≥ g = (1 + c) ln − (1 − c) ln , 2 2 2 g 0 (x) =

and it suffices to show that 

1+c 2

‹1+c

>



1−c 2

‹1−c

.

This inequality follows from Bernoulli’s inequality, as follows 

1+c 2

‹1+c

 ‹ 1 − c 1+c (1 + c)(1 − c) 1 + c 2 = 1− >1− = 2 2 2

Symmetric Power-Exponential Inequalities and 

1−c 2

‹1−c

447

 ‹ 1 + c 1−c (1 − c)(1 + c) 1 + c 2 = 1− 0 involves f 0 (x) > 0, it follows that f (x) is strictly increasing on c, 2 + , 2 and hence   c f (x) ≤ f 2 + . 2  c So, we need to show that f 2 + ≤ 0 for 0 < c < 2, which follows immediately from 2 Lemma 1. The proof is completed. The equality holds for a = b. Lemma 1. If 0 ≤ c ≤ 2, then  c c c −c  + 2− ≤ 2. 2+ 2 2 Proof. According to Lemma 2, the following inequalities hold for c ∈ [0, 2]:  c −c c3 2+ ≤ 1 − c ln 2 + , 2 9  c3 c c ≤ 1 + c ln 2 − . 2− 2 9 Summing these inequalities, the desired inequality follows. Lemma 2. If −2 ≤ x ≤ 2, then

 x x x3 2− ≤ 1 + x ln 2 − . 2 9

Proof. We have ln 2 ≈ 0.693 < 7/9. If x ∈ [0, 2], then 1 + x ln 2 −

x3 8 x3 ≥1− ≥ 1 − > 0. 9 9 9

Also, if x ∈ [−2, 0], then 1 + x ln 2 −

x3 7x x 3 8 + 7x − x 3 ≥1+ − > 9 9 9 9 2(x + 2)2 + (−x)(x + 1)2 = > 0. 9

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Vasile Cîrtoaje

So, we can write the desired inequality as f (x) ≥ 0, where    x3 x f (x) = ln 1 + d x − − x ln 2 − , d = ln 2. 9 2 We have f 0 (x) =

 9d − 3x 2 x x + − ln 2 − . 9 + 9d x − x 3 4 − x 2

Since f (0) = 0, it suffices to show that f 0 (x) ≤ 0 for x ∈ [−2, 0], and f 0 (x) ≥ 0 for x ∈ [0, 2]; that is, x f 0 (x) ≤ 0 for x ∈ [−2, 2]. According to Lemma 3, the inequality x f 0 (x) ≤ 0 is true if x g(x) ≥ 0, where ‹  9d − 3x 2 x x 1 2 g(x) = + − d− − x . 9 + 9d x − x 3 4 − x 4 32 We have   ‹ 9d − 3x 2 x x 1 2 g(x) = −d + + + x 9 + 9d x − x 3 4 − x 4 32  2  d x − 3x − 9d 2 64 − 4x − x 2 =x + , 9 + 9d x − x 3 32(4 − x) 

hence x g(x) =

x 2 g1 (x) , 32(4 − x)(9 + 9d x − x 3 )

where g1 (x) =32(4 − x)(d x 2 − 3x − 9d 2 ) + (64 − 4x − x 2 )(9 + 9d x − x 3 ) =x 5 + 4x 4 − (64 + 41d)x 3 + (87 + 92d)x 2 + 12(24d 2 + 48d − 35)x + 576(1 − 2d 2 ). Since g1 (x) ≥ 0 for x ∈ [a1 , b1 ], where a1 ≈ −12.384 and b1 =≈ 2.652, we have g1 (x) ≥ 0 for x ∈ [−2, 2]. Lemma 3. If x < 4, then xh(x) ≤ 0, where

 ‹  x x 1 2 − ln 2 − − x . h(x) = ln 2 − 2 4 32

Proof. From h0 (x) =

−x 2 ≤ 0, 16(4 − x)

Symmetric Power-Exponential Inequalities

449

it follows that h(x) is decreasing. Since h(0) = 0, we have h(x) ≥ 0 for x ≤ 0, and h(x) ≤ 0 for x ∈ [0, 4); that is, xh(x) ≤ 0 for x < 4.

P 3.20. Let a, b be positive real numbers such that a + b = 2. If k ≥ kb

1 , then 2

ka

a a b b ≥ 1. (Vasile Cîrtoaje, 2010) Solution. Setting a = 1+ x and b = 1− x, where 0 ≤ x < 1, the inequality is equivalent to (1 + x)k(1−x) ln(1 + x) + (1 − x)k(1+x) ln(1 − x) ≥ 0. Consider further the non-trivial case 0 < x < 1, and write the desired inequality as f (x) ≥ 0, where f (x) = k(1 − x) ln(1 + x) − k(1 + x) ln(1 − x) + ln ln(1 + x) − ln(− ln(1 − x)). We have 2k(1 + x 2 ) 1 1 − k ln(1 − x 2 ) + + 1 − x2 (1 + x) ln(1 + x) (1 − x) ln(1 − x) 1 1 2k + + > 2 1− x (1 + x) ln(1 + x) (1 − x) ln(1 − x) 1 1 1 ≥ + + 2 1− x (1 + x) ln(1 + x) (1 − x) ln(1 − x) g(x) = , 2 (1 − x ) ln(1 + x) ln(1 − x)

f 0 (x) =

where g(x) = ln(1 + x) ln(1 − x) + (1 + x) ln(1 + x) + (1 − x) ln(1 − x). It suffices to show that f 0 (x) > 0. Indeed, if this is true, then f (x) is strictly increasing, hence f (x) > lim f (x) = 0. x→0

Thus, we need to prove that g(x) < 0. We have g 0 (x) =

−x h(x), 1 − x2

where h(x) = (1 + x) ln(1 + x) + (1 − x) ln(1 − x).

450

Vasile Cîrtoaje

Since h0 (x) = ln

1+ x > 0, 1− x

h(x) is strictly increasing, h(x) > h(0) = 0, g 0 (x) < 0, g(x) is strictly decreasing, and hence g(x) < g(0) = 0. This completes the proof. The equality holds for a = b = 1.

P 3.21. If a, b are positive real numbers such that a + b = 2, then a

p

a

p

b

b

≥ 1. (Vasile Cîrtoaje, 2010)

Solution. For a = b = 1, the equality holds. In what follows, we will assume that a > 1 > b. Taking logarithms of both sides, the inequality becomes in succession p p a ln a + b ln b ≥ 0, p p a ln a ≥ b(− ln b), 1 1 ln a + ln ln a ≥ ln b + ln(− ln b). 2 2 Setting a = 1 + x and b = 1 − x, we need to show that f (x) ≥ 0 for 0 < x < 1, where f (x) =

1 1 ln(1 + x) − ln(1 − x) + ln ln(1 + x) − ln(− ln(1 − x)). 2 2

We have f 0 (x) =

1 1 1 + + . 2 1− x (1 + x) ln(1 + x) (1 − x) ln(1 − x)

As shown in the proof of the preceding P 3.20, we have f 0 (x) > 0. Therefore, f (x) is strictly increasing and f (x) > lim f (x) = 0. x→0

The equality holds for a = b = 1.

P 3.22. If a, b are positive real numbers such that a + b = 2, then 1 − a a+1 b b+1 ≥

1 (1 − a b)2 . 3 (Vasile Cîrtoaje, 2010)

Symmetric Power-Exponential Inequalities

451

Solution. Putting a = 1 + x and b = 1 − x, 0 ≤ x < 1, the inequality becomes (1 + x)2+x (1 − x)2−x ≤ 1 −

1 4 x . 3

Write this inequality as f (x) ≤ 0, where ‹  1 f (x) = (2 + x) ln(1 + x) + (2 − x) ln(1 − x) − ln 1 − x 4 . 3 We have f 0 (x) = ln(1 + x) − ln(1 − x) − f 00 (x) = =

2x 4x 3 + , 1 − x2 3 − x4

2 2(1 + x 2 ) 4x 2 (x 4 + 9) − + 1 − x 2 (1 − x 2 )2 (3 − x 4 )2

4x 2 (x 4 + 9) −8x 4 [x 4 + 1 + 8(1 − x 2 )] −4x 2 + = ≤ 0. (1 − x 2 )2 (3 − x 4 )2 (1 − x 2 )2 (3 − x 4 )2

Therefore, f 0 (x) is decreasing, f 0 (x) ≤ f 0 (0) = 0, f (x) is decreasing, f (x) ≤ f (0) = 0. The equality holds for a = b = 1.

P 3.23. If a, b are positive real numbers such that a + b = 2, then a−a + b−b ≤ 2. (Vasile Cîrtoaje, 2010) Solution. Consider that a ≥ b, when 0 < b ≤ 1 ≤ a < 2, and write the inequality as aa − 1 b b − 1 + ≥ 0. aa bb According to Lemma from the proof of P 3.4, we have aa − 1 ≥ a

a+1 2

(a − 1),

bb − 1 ≥ b

b+1 2

(b − 1).

Therefore, it suffices to show that a

1−a 2

(a − 1) + b

1−b 2

(b − 1) ≥ 0,

which is equivalent to a

1−a 2

≥b

1−b 2

,

452

Vasile Cîrtoaje (a b)

1−b 2

≤ 1,

a b ≤ 1. The last inequality follows immediately from the AM-GM inequality ab ≤

1 (a + b)2 = 1. 4

The equality holds for a = b = 1.

P 3.24. If a, b are nonnegative real numbers such that a + b = 2, then a2b + b2a ≥ a b + b a ≥ a2 b2 + 1. (Vasile Cîrtoaje, 2010) Solution. Since a, b ∈ [0, 2] and (1 − a)(1 − b) ≤ 0, from Lemma below, we have ab − 1 ≥

b(a b + 1)(a − 1) b(a b + 3 − a − b)(a − 1) = 2 2

and ba − 1 ≥

a(a b + 1)(b − 1) . 2

Based on these inequalities, we get a b + b a − a2 b2 − 1 = (a b − 1) + (b a − 1) + 1 − a2 b2 b(a b + 1)(a − 1) a(a b + 1)(b − 1) + + 1 − a2 b2 2 2 = (ab + 1)(a b − 1) + 1 − a2 b2 = 0

≥

and a2b + b2a − a b − b a = a b (a b − 1) + b a (b a − 1) a b b(a b + 1)(a − 1) b a a(a b + 1)(b − 1) + 2 2 a b(a b + 1)(a − b)(a b−1 − b a−1 ) = . 4 ≥

On the valid assumption a ≥ b, we only need to show that a b−1 ≥ b a−1 , which is equivalent to a

b−a 2

≥b

a−b 2

, 1 ≥ (a b)

a−b 2

, 1 ≥ a b, (a − b)2 ≥ 0.

Symmetric Power-Exponential Inequalities

453

For both inequalities, the equality holds when a = b = 1, when a = 0 and b = 2, and when a = 2 and b = 0. Lemma. If x, y ∈ [0, 2] such that (1 − x)(1 − y) ≤ 0, then xy −1≥

y(x y + 3 − x − y)(x − 1) , 2

with equality for x = 1, for y = 0, for y = 1, and for x = 0 and y = 2. Proof. For y = 0, y = 1 and y = 2, the inequality is an identity. For fixed y ∈ (0, 1) ∪ (1, 2), let us define f (x) = x y − 1 − We have 0

•

f (x) = y x

y−1

y(x y + 3 − x − y)(x − 1) . 2

˜ x y + 3 − x − y (x − 1)( y − 1) − , − 2 2

f 00 (x) = y( y − 1)(x y−2 − 1). Since f 00 (x) ≥ 0 for x ∈ (0, 2], f 0 is increasing. There are two cases to consider. Case 1: x ≥ 1 > y. We have f 0 (x) ≥ f 0 (1) = 0, f (x) is increasing, hence f (x) ≥ f (1) = 0. Case 2: y > 1 ≥ x. We have f 0 (x) ≤ f 0 (1) = 0, f (x) is decreasing, hence f (x) ≥ f (1) = 0.

P 3.25. If a, b are positive real numbers such that a + b = 2, then a3b + b3a ≤ 2. (Vasile Cîrtoaje, 2007) Solution. Without loss of generality, assume that a ≥ b. Using the substitution a = 1+ x and b = 1 − x, 0 ≤ x < 1, we can write the inequality as e3(1−x) ln(1+x) + e3(1+x) ln(1−x) ≤ 2. Applying Lemma below, it suffices to show that f (x) ≤ 2, where f (x) = e3(1−x)(x−

x2 x3 2 + 3 )

+ e−3(1+x)(x+

x2 x3 2 + 3 )

.

454

Vasile Cîrtoaje

Since f (0) = 2, it suffices to show that f 0 (x) ≤ 0 for x ∈ [0, 1). From 9x 2 5x 3 4 15 2 x − 4x 3 )e3x− 2 + 2 −x 2 9x 2 5x 3 4 15 2 −(3 + 9x + x + 4x 3 )e−3x− 2 − 2 −x , 2

f 0 (x) =(3 − 9x +

it follows that f 0 (x) ≤ 0 is equivalent to 3

e−6x−5x ≥

6 − 18x + 15x 2 − 8x 3 . 6 + 18x + 15x 2 + 8x 3

For the nontrivial case 6 − 18x + 15x 2 − 8x 3 > 0, we rewrite this inequality as g(x) ≥ 0, where g(x) = −6x − 5x 3 − ln(6 − 18x + 15x 2 − 8x 3 ) + ln(6 + 18x + 15x 2 + 8x 3 ). Since g(0) = 0, it suffices to show that g 0 (x) ≥ 0 for x ∈ [0, 1). From (6 + 8x 2 ) − 10x 1 0 (6 + 8x 2 ) + 10x g (x) = −2 − 5x 2 + + , 3 6 + 15x 2 − (18x + 8x 3 ) 6 + 15x 2 + (18x + 8x 3 ) it follows that g 0 (x) ≥ 0 is equivalent to 2(6 + 8x 2 )(6 + 15x 2 ) − 20x(18x + 8x 3 ) ≥ (2 + 5x 2 )[(6 + 15x 2 )2 − (18x + 8x 3 )2 ]. Since (6 + 15x 2 )2 − (18x + 8x 3 )2 ≤ (6 + 15x 2 )2 − 324x 2 − 288x 4 ≤ 4(9 − 36x 2 ), it suffices to show that (3 + 4x 2 )(6 + 15x 2 ) − 5x(18x + 8x 3 ) ≥ (2 + 5x 2 )(9 − 36x 2 ). This reduces to 6x 2 +200x 4 ≥ 0, which is clearly true. The equality holds for a = b = 1. Lemma. If t > −1, then ln(1 + t) ≤ t −

t2 t3 + . 2 3

Proof. We need to prove that f (t) ≥ 0, where f (t) = t −

t2 t3 + − ln(1 + t). 2 3

Since

t3 , t +1 f (t) is decreasing on (−1, 0] and increasing on [0, ∞). Therefore, f (t) ≥ f (0) = 0. f 0 (t) =

Symmetric Power-Exponential Inequalities

455

P 3.26. If a, b are nonnegative real numbers such that a + b = 2, then a3b + b3a +



a−b 2

‹4

≤ 2. (Vasile Cîrtoaje, 2007)

Solution (by M. Miyagi and Y. Nishizawa). We may assume that a = 1+ x and b = 1− x, where 0 ≤ x ≤ 1. The desired inequality is equivalent to (1 + x)3(1−x) + (1 − x)3(1+x) + x 4 ≤ 2. By Lemma below, we have (1 + x 1−x ≤

1 (1 + x)2 (2 − x 2 )(2 − 2x + x 2 ), 4

(1 − x 1+x ≤

1 (1 − x)2 (2 − x 2 )(2 + 2x + x 2 ). 4

Therefore, it suffices to show that (1 + x)6 (2 − x 2 )3 (2 − 2x + x 2 )3 + (1 − x)6 (2 − x 2 )3 (2 + 2x + x 2 )3 + 64x 2 ≤ 128, which is equivalent to x 4 (1 − x 2 )[x 6 (x 6 − 8x 4 + 31x 2 − 34) − 2(17x 6 − 38x 4 + 16x 2 + 8)] ≤ 0. Clearly, it suffices to show that t 3 − 8t 2 + 31t − 34 < 0 and 17t 3 − 38t 2 + 16t + 8 > 0 for all t ∈ [0, 1]. Indeed, we have t 3 − 8t 2 + 31t − 34 < t 3 − 8t 2 + 31t − 24 = (t − 1)(t 2 − 7t + 24) ≤ 0, 17t 3 − 38t 2 + 16t + 8 = 17t(t − 1)2 + (−4t 2 − t + 8) > 0. Lemma. If −1 ≤ t ≤ 1, then (1 + t)1−t ≤

1 (1 + t)2 (2 − t 2 )(2 − 2t + t 2 ), 4

with equality for t = −1, t = 0 and t = 1.

456

Vasile Cîrtoaje

Proof. We need to prove that f (t) ≥ 0 for −1 < t ≤ 1, where f (t) = (1 + t) ln(1 + t) + ln(2 − t 2 ) + ln(2 − 2t + t 2 ) − ln 4. We have f 0 (t) = 1 + ln(1 + t) − f 00 (t) =

2(t − 1) 2t + , 2 − t 2 2 − 2t + t 2

t 2 g(t) , (1 + t)(2 − t 2 )2 (2 − 2t + t 2 )2

where g(t) = t 6 − 8t 5 + 12t 4 + 8t 3 − 20t 2 − 16t + 16. Case 1: 0 ≤ t ≤ 1. From g 0 (t) = 6t 5 − 40t 4 + 48t 3 + 24t 2 − 40t − 16 = 6t 5 − 8t − 16 − 8t(5t 3 − 6t 2 − 3t + 4) = (6t 5 − 8t − 16) − 8t(t − 1)2 (5t + 4) < 0, it follows that g is strictly decreasing on [0, 1]. Since g(0) = 16 and g(1) = −7, there exists a number c ∈ (0, 1) such that g(c) = 0, g(t) > 0 for 0 ≤ t < c and g(t) < 0 for c < t ≤ 1. Therefore, f 0 is strictly increasing on [0, c] and strictly decreasing on [c, 1]. From f 0 (0) = 0 and f 0 (1) = ln 2 − 1 < 0, it follows that there exists a number d ∈ (0, 1) such that f 0 (d) = 0, f 0 (t) > 0 for 0 < t < d and f 0 (t) < 0 for c < t ≤ 1. As a consequence, f is strictly increasing on [0, d] and strictly decreasing on [d, 1]. Since f (0) = 0 and f (1) = 0, we have f (t) ≥ 0 for 0 ≤ t ≤ 1. Case 2: −1 < t ≤ 0. From g(t) = t 4 (t − 2)(t − 6) + 4(t + 1)(2t 2 − 7t + 3) + 4 > 0, it follows that f 0 is strictly increasing on (−1, 0]. Since f 0 (0) = 0, we have f 0 (t) < 0 for −1 < t < 0, hence f is strictly decreasing on (−1, 0]. From f (0) = 0, it follows that f (t) ≥ 0 for −1 < t ≤ 0. Conjecture. If a, b are nonnegative real numbers such that a + b = 2, then  ‹ a−b 2 a3b + b3a + ≥ 2. 2

P 3.27. If a, b are positive real numbers such that a + b = 2, then 2

2

a a + b b ≤ 2. (Vasile Cîrtoaje, 2008)

Symmetric Power-Exponential Inequalities

457

Solution. Without loss of generality, assume that 0 < a ≤ 1 ≤ b < 2, and write the inequality as 1
1 1/a 2 a

+

1
1 2/b b

≤ 2.

By Bernoulli’s inequality, we have ‹  ‹ 1 1/a 1 1 a3 − a2 + 1 ≥ 1 + − 1 = , a2 a a2 a3  ‹2/b  ‹ 1 2 1 b2 − 2b + 2 ≥1+ −1 = . b b b b2



Therefore, it suffices to show that a3 b2 + ≤ 2, a3 − a2 + 1 b2 − 2b + 2 which is equivalent to a3 (2 − b)2 ≤ , a3 − a2 + 1 b2 − 2b + 2 a3 a2 ≤ , a3 − a2 + 1 a2 − 2a + 2 a2 (a − 1)2 ≥ 0. The equality happens for a = b = 1.

P 3.28. If a, b are positive real numbers such that a + b = 2, then 3

3

a a + b b ≥ 2. (Vasile Cîrtoaje, 2008) Solution. Assume that a ≤ b; that is, 0 < a ≤ 1 ≤ b < 2. 3 3 and ≤ a ≤ 1. 5 5 3 7 Case 1: 0 < a ≤ . From a + b = 2, we get ≤ b < 2. Let 5 5 There are two cases to consider: 0 < a ≤

3

f (x) = x x , 0 < x < 2.

458

Vasile Cîrtoaje

Since

3

f 0 (x) = 3x x −2 (1 − ln x) > 0, ‹  ‹ • 7 7 , 2 , and hence f (b) ≥ f ; that is, f (x) is increasing on 5 5 3

bb ≥

 ‹15/7 7 . 5

Using Bernoulli’s inequality gives  ‹15/7  ‹  ‹ 7 7 2 8/7 7 16 51 b ≥ = 1+ > 1+ = > 2. 5 5 5 5 35 25 3 b

hence

3

3

a a + b b > 2.

Case 2:

3 ≤ a ≤ 1. By Lemma below, we have 5 3

2a a ≥ 3 − 15a + 21a2 − 7a3 and

3

2b b ≥ 3 − 15b + 21b2 − 7b3 . Summing these inequalities, we get € 3 Š 3 2 a a + b b ≥ 6 − 15(a + b) + 21(a2 + b2 ) − 7(a3 + b3 ) = 6 − 15(a + b) + 21(a + b)2 − 7(a + b)3 = 4. This completes the proof. The equality holds for a = b = 1. Lemma. If

3 ≤ x ≤ 2, then 5 3

2x x ≥ 3 − 15x + 21x 2 − 7x 3 , with equality for x = 1. Proof. We first show that h(x) > 0, where h(x) = 3 − 15x + 21x 2 − 7x 3 . From h0 (x) = 3(−5 + 14x − 7x 2 ),

Symmetric Power-Exponential Inequalities 

s

2 ,1 + 7

459   s  s 2 2 , and decreasing on 1 + ,∞ . 7 7

it follows that h(x) is increasing on 1 −  ‹ 3 Then, it suffices to show that f ≥ 0 and f (2) ≥ 0. Indeed 5  ‹ 3 6 f = , f (2) = 1. 5 125 Write now the desired inequality as f (x) ≥ 0, where f (x) = ln 2 + We have

3 ln x − ln(3 − 15x + 21x 2 − 7x 3 ), x

x2 0 f (x) = g(x), 3

g(x) = 1 − ln x +

g 0 (x) =

3 ≤ x ≤ 2. 5

x 2 (7x 2 − 14x + 5) , 3 − 15x + 21x 2 − 7x 3

g1 (x) , (3 − 15x + 21x 2 − 7x 3 )2

where g1 (x) = −49x 7 + 245x 6 − 280x 5 − 147x 4 + 471x 3 − 321x 2 + 90x − 9. In addition, g1 (x = (x − 1)2 g2 (x), g2 (x) = 11x 5 + 3g3 (x),

g2 (x) = −49x 5 + 147x 4 + 63x 3 − 168x 2 + 72x − 9, g3 (x) = −20x 5 + 49x 4 + 21x 3 − 56x 2 + 24x − 3,

g3 (x) = (4x − 1)g4 (x), g4 (x) = x 5 + g5 (x),

g4 (x) = −5x 4 + 11x 3 + 8x 2 − 12x + 3, g5 (x) = −6x 4 + 11x 3 + 8x 2 − 12x + 3,

g5 (x) = (2x − 1)g6 (x),

g6 (x) = −3x 3 + 4x 2 + 6x − 3,

g6 (x) = 1 + (2 − x)(3x 2 + 2x − 2). Therefore, we get in succession g6 (x) > 0, g5 (x) > 0, g4 (x) > 0, g3 (x) > 0, g2 (x) ‹ • > 0, 3 0 g1 (x) ≥ 0, g (x) ≥ 0, g(x) is increasing. Since g(1) = 0, we have g(x) < 0 on ,1 5 • ˜ 3 and g(x) > 0 on (1, 2]. Then, f (x) is decreasing on , 1 and increasing on [1, 2], 5 hence f (x) ≥ f (1) = 0.

460

Vasile Cîrtoaje

P 3.29. If a, b are positive real numbers such that a + b = 2, then 2

2

a5b + b5a ≤ 2. (Vasile Cîrtoaje, 2010) Solution. Assume that a ≥ b. For a = 2 and b = 0, the inequality is obvious. Otherwise, using the substitution a = 1 + x and b = 1 − x, 0 ≤ x < 1, we can write the desired inequality as 2 2 e5(1−x) ln(1+x) + e5(1+x) ln(1−x) ≤ 2. According to Lemma below, it suffices to show that f (x) ≤ 2, where f (x) = e5(u−v) + e−5(u+v) , 7 3 31 5 5 17 4 9 6 x + x , v = x2 + x + x . 3 30 2 12 20 If f 0 (x) ≤ 0, then f (x) is decreasing, hence f (x) ≤ f (0) = 2. Since u= x+

f 0 (x) = 5(u0 − v 0 )e5(u−v) − 5(u0 + v 0 )e−5(u+v) , u0 = 1 + 7x 2 +

31 4 x , 6

v 0 = 5x +

17 3 27 5 x + x , 3 10

the inequality f 0 (x) ≤ 0 becomes e−10u (u0 + v 0 ) ≥ u0 − v 0 For the nontrivial case u0 − v 0 > 0, we rewrite this inequality as g(x) ≥ 0, where g(x) = −10u + ln(u0 + v 0 ) − ln(u0 − v 0 ). If g 0 (x) ≥ 0, then g(x) is increasing, hence g(x) ≥ f (0) = 2. We have g 0 (x) = −10u0 +

u00 + v 00 u00 − v 00 − 0 , u0 + v 0 u − v0

where

62 3 27 4 x , v 00 = 5 + 17x 2 + x . 3 2 Thus, the inequality g 0 (x) ≥ 0 is equivalent to u00 = 14x +

u0 v 00 − v 0 u00 ≥ 5u0 (u0 + v 0 )(u0 − v 0 ), a1 t + a2 t 2 + a3 t 3 + a4 t 4 + a5 t 5 + a6 t 6 + a7 t 7 ≥ 0, where t = x 2 , 0 ≤ t < 1, and a1 = 2,

a2 = 321.5,

a3 ≈ 152.1,

a4 ≈ −498.2,

Symmetric Power-Exponential Inequalities a5 ≈ −168.5,

461

a6 ≈ 356.0,

a7 ≈ 188.3.

This inequality is true if 300t 2 + 150t 3 − 500t 4 − 200t 5 + 250t 6 ≥ 0. Since the last inequality is equivalent to the obvious inequality 50t 2 (1 − t)(6 + 9t − t 2 − 5t 3 ) ≥ 0, the proof is completed. The equality holds for a = b = 1. Lemma. If −1 < t < 1, then (1 − t)2 ln(1 + t) ≤ t −

9 6 5 2 7 3 17 4 31 5 t + t − t + t − t . 2 3 12 30 20

Proof. We shall show that ‹  1 1 1 1 (1 − t)2 ln(1 + t) ≤ (1 − t)2 t − t 2 + t 3 − t 4 + t 5 2 3 4 5 5 2 7 3 17 4 31 5 9 6 ≤t− t + t − t + t − t . 2 3 12 30 20 The left inequality is equivalent to f (t) ≥ 0, where f (t) = t −

1 2 1 3 1 4 1 5 t + t − t + t − ln(1 + t). 2 3 4 5

Since

t5 , 1+ t f (t) is decreasing on (−1, 0] and increasing on [0, 1); therefore, f (t) ≥ f (0) = 0. The right inequality is equivalent to t 6 (t − 1) ≤ 0, which is clearly true. f 0 (t) =

P 3.30. If a, b are positive real numbers such that a + b = 2, then a2

p

b

+ b2

p

a

≤ 2. (Vasile Cîrtoaje, 2010)

Solution. Assume that a ≥ b. For a = 2 and b = 0, the inequality is obvious. Otherwise, using the substitution a = 1 + x and b = 1 − x, 0 ≤ x < 1, we can write the desired inequality as f (x) ≤ 2, where p 1−x ln(1+x)

f (x) = e2

p

+ e2

1+x ln(1−x)

.

462

Vasile Cîrtoaje

There are two cases to consider. 13 Case 1: ≤ x < 1. If f 0 (x) ≤ 0, then f (x) is decreasing, and hence 20 f (x) ≤ f



13 33 = 20 20 ‹



q

‹

7 5

q

+



7 20

‹

33 5


0, rewrite this inequality as g(x) ≥ 0, where g(x) = −4x −

11 3 x − ln(8 − 16x + 11x 2 − 8x 3 ) + ln(8 + 16x + 11x 2 + 8x 3 ). 6

Symmetric Power-Exponential Inequalities

463

If g 0 (x) ≥ 0, then g(x) is increasing, hence g(x) ≥ g(0) = 0. From g 0 (x) = −4 −

11 2 (16 + 24x 2 ) − 22x (16 + 24x 2 ) + 22x x + + , 2 8 + 11x 2 − (16x + 8x 3 ) 8 + 11x 2 + (16x + 8x 3 )

it follows that g 0 (x) ≥ 0 is equivalent to (16 + 24x 2 )(8 + 11x 2 ) − 22x(16x + 8x 3 ) ≥

1 (8 + 11x 2 )[(8 + 11x 2 )2 − (16x + 8x 3 )2 ]. 4

Since (8 + 11x 2 )2 − (16x + 8x 3 )2 ≤ (8 + 11x 2 )2 − 256x 2 − 256x 4 ≤ 16(4 − 5x 2 ), it suffices to show that (4 + 6x 2 )(8 + 11x 2 ) − 11x(8x + 4x 3 ) ≥ (8 + 11x 2 )(4 − 5x 2 ). This reduces to 77x 4 ≥ 0. The proof is completed. The equality holds for a = b = 1. Lemma. If −1 < t ≤

13 , then 20 p 11 3 1 4 1 − t ln(1 + t) ≤ t − t 2 + t − t . 24 4

Proof. We consider two cases. Case 1: 0 ≤ t ≤ inequalities

13 . We can prove the desired inequality by multiplying the following 20

p 1 1 3 1 t , 1 − t ≤ 1 − t − t2 − 2 8 16 1 1 1 1 ln(1 + t) ≤ t − t 2 + t 3 − t 4 + t 5 , 2 3 4 5  ‹ ‹ 1 1 1 3 1 1 1 1 11 3 1 4 1 − t − t2 − t t − t2 + t3 − t4 + t5 ≤ t − t2 + t − t . 2 8 16 2 3 4 5 24 4

The first inequality is equivalent to f (t) ≥ 0, where  ‹ 1 1 2 1 3 1 t − ln(1 − t). f (t) = ln 1 − t − t − 2 8 16 2 Since f 0 (t) =

1 8 + 4t + 3t 2 5t 3 − = ≥ 0, 2(1 − t) 16 − 8t − 2t 2 − t 3 2(1 − t)(16 − 8t − 2t 2 − t 3 )

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f (t) is increasing, and hence f (t) ≥ f (0) = 0. The second inequality is equivalent to f (t) ≥ 0, where f (t) = t −

1 2 1 3 1 4 1 5 t + t − t + t − ln(1 + t). 2 3 4 5

Since f 0 (t) = 1 − t + t 2 − t 3 + t 4 −

1 t5 = ≥ 0, 1+ t 1+ t

f (t) is increasing, and hence f (t) ≥ f (0) = 0. The third inequality is equivalent to t 4 (160 − 302t + 86t 2 + 9t 3 + 12t 4 ) ≥ 0. This is true since 160 − 302t + 86t 2 + 9t 3 + 12t 4 ≥ 2(80 − 151t + 43t 2 ) > 0.

Case 2: −1 < t ≤ 0. Write the desired inequality as p 11 3 1 4 − 1 − t ln(1 + t) ≥ −t + t 2 − t + t . 24 4 This is true if

p 1 1 1 − t ≥ 1 − t − t 2, 2 8 1 1 − ln(1 + t) ≥ −t + t 2 − t 3 + t 4 , 3 4  ‹ ‹ 1 1 1 1 11 3 1 4 1 − t − t 2 −t + t 2 − t 3 + t 4 ≥ −t + t 2 − t + t . 2 8 3 4 24 4

The first inequality is equivalent to f (t) ≥ 0, where  ‹ 1 1 2 1 f (t) = ln(1 − t) − ln 1 − t − t . 2 2 8 Since f 0 (t) =

−1 2(2 + t) −3t 2 + = ≤ 0, 2(1 − t) 8 − 4t − t 2 2(1 − t)(8 − 4t − t 2 )

f (t) is decreasing, and hence f (t) ≥ f (0) = 0. The second inequality is equivalent to f (t) ≥ 0, where f (t) = t −

1 2 1 3 1 4 t + t − t − ln(1 + t). 2 3 4

Symmetric Power-Exponential Inequalities Since f 0 (t) = 1 − t + t 2 − t 3 −

465

−t 4 1 = ≤ 0, 1+ t 1+ t

f (t) is decreasing, and hence f (t) ≥ f (0) = 0. The third inequality reduces to the obvious inequality t 4 (10 − 8t − 3t 2 ) ≥ 0.

P 3.31. If a, b are nonnegative real numbers such that a + b = 2, then a b(1 − a b)2 a b(1 − a b)2 ≤ a b+1 + b a+1 − 2 ≤ . 2 3 (Vasile Cîrtoaje, 2010) Solution. Assume that a ≥ b, which yields 1 ≤ a ≤ 2 and 0 ≤ b ≤ 1. (a) To prove the left inequality we apply Lemma 1 below. For x = a and k = b, we have a b+1 ≥ 1 + (1 + b)(a − 1) + a b+1 ≥ a − b + a b +

b(1 + b)(1 − b) b(1 + b) (a − 1)2 − (a − 1)3 , 2 6 b(1 + b) b(1 + b) (a − 1)2 − (a − 1)4 . 2 6

(*)

Also, for x = b and k = a − 1, we have b a ≥ 1 + a(b − 1) +

a(a − 1) a(a − 1)(2 − a) (b − 1)2 − (b − 1)3 , 2 6

a ab b a ≥ 1 − a + a b + (a − 1)3 + (a − 1)4 , 2 6 b a+1 ≥ b − a b + a b2 +

ab a b2 (a − 1)3 + (a − 1)4 . 2 6

Summing up (*) and (**) gives a b+1 + b a+1 − 2 ≥ −b(a − 1)2 + Since −b(a − 1)2 +

b(3 − a b) b(1 + b − a b) (a − 1)2 − (a − 1)4 . 2 6

b(3 − a b) b (a − 1)2 = (a − 1)4 , 2 2

(**)

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we have a b+1 + b a+1 − 2 ≥ =

b b(1 + b − a b) (a − 1)4 − (a − 1)4 2 6

a b(1 + b) ab a b(1 − a b)2 (a − 1)4 ≥ (a − 1)4 = . 6 6 6

The equality holds for a = b = 1, for a = 2 and b = 0, and for a = 0 and b = 2. (b) To prove the right inequality we apply Lemma 2 below. For x = a and k = b, we have a b+1 ≤ 1 + (b + 1)(a − 1) + + a b+1 ≤ 1 + (b + 1)(a − 1) +

(b + 1)b (b + 1)b(b − 1) (a − 1)2 + (a − 1)3 2 6

(b + 1)b(b − 1)(b − 2) (a − 1)4 , 24 b(b + 1) b(b + 1) a b(b + 1) (a − 1)2 − (a − 1)4 + (a − 1)5 . 2 6 24

Also, for x = b and k = a, we have b a+1 ≤ 1 + (a + 1)(b − 1) +

a(a + 1) a b(a + 1) a(a + 1) (b − 1)2 − (b − 1)4 + (b − 1)5 . 2 6 24

Summing these inequalities gives a b+1 + b a+1 − 2 ≤ −2(a − 1)2 + ≤

a2 + b2 + 2 a2 + b2 + 2 ab (a − 1)2 − (a − 1)4 − (a − 1)6 2 6 12

a2 + b2 + 2 a2 + b2 + 2 a2 + b2 − 2 (a − 1)2 − (a − 1)4 = (a − 1)4 − (a − 1)4 2 6 6 =

ab a b(1 − a b)2 (a − 1)4 = . 3 3

The equality holds for a = b = 1, for a = 2 and b = 0, and for a = 0 and b = 2. Lemma 1. If x ≥ 0 and 0 ≤ k ≤ 1, then x k+1 ≥ 1 + (1 + k)(x − 1) +

k(1 + k)(1 − k) k(1 + k) (x − 1)2 − (x − 1)3 , 2 6

with equality for x = 1, for k = 0, for k = 1. Proof. For k = 0 and k = 1, the inequality is an identity. For fixed k, 0 < k < 1, let us define f (x) = x k+1 − 1 − (1 + k)(x − 1) −

k(1 + k) k(1 + k)(1 − k) (x − 1)2 + (x − 1)3 . 2 6

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467

We need to show that f (x) ≥ 0. We have 1 k(1 − k) f 0 (x) = x k − 1 − k(x − 1) + (x − 1)2 , 1+k 2 1 f 00 (x) = x k−1 − 1 + (1 − k)(x − 1), k(1 + k) 1 f 000 (x) = −x k−2 + 1. k(1 + k)(1 − k) Case 1: 0 ≤ x ≤ 1. Since f 000 ≤ 0, f 00 is decreasing, f 00 (x) ≥ f 00 (1) = 0, f 0 is increasing, f 0 (x) ≤ f 0 (1) = 0, f is decreasing, f (x) ≥ f (1) = 0. Case 2: x ≥ 1. Since f 000 ≥ 0, f 00 is increasing, f 00 (x) ≥ f 00 (1) = 0, f 0 is increasing, f 0 (x) ≥ f 0 (1) = 0, f is increasing, f (x) ≥ f (1) = 0. Lemma 2. If either x ≥ 1 and 0 ≤ k ≤ 1, or 0 ≤ x ≤ 1 and 1 ≤ k ≤ 2, then x k+1 ≤ 1 + (k + 1)(x − 1) +

(k + 1)k (k + 1)k(k − 1) (x − 1)2 + (x − 1)3 2 6

(k + 1)k(k − 1)(k − 2) (x − 1)4 , 24 with equality for x = 1, for k = 0, for k = 1, for k = 2. +

Proof. For k = 0, k = 1 and k = 2, the inequality is an identity. For fixed k, k ∈ (0, 1) ∪ (1, 2), let us define f (x) = x k+1 − 1 − (k + 1)(x − 1) − −

(k + 1)k (k + 1)k(k − 1) (x − 1)2 − (x − 1)3 2 6

(k + 1)k(k − 1)(k − 2) (x − 1)4 . 24

We need to show that f (x) ≤ 0. We have 1 k(k − 1) k(k − 1)(k − 2) f 0 (x) = x k − 1 − k(x − 1) − (x − 1)2 − (x − 1)3 , k+1 2 6 1 (k − 1)(k − 2) f 00 (x) = x k−1 − 1 − (k − 1)(x − 1) − (x − 1)2 , k(k + 1) 2 1 f 000 (x) = x k−2 − 1 − (k − 2)(x − 1), k(k + 1)(k − 1) 1 f (4) (x) = x k−3 − 1. k(k + 1)(k − 1)(k − 2)

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Case 1: x ≥ 1, 0 < k < 1. Since f (4) (x) ≤ 0, f 000 (x) is decreasing, f 000 (x) ≤ f 000 (1) = 0, f 00 is decreasing, f 00 (x) ≤ f 00 (1) = 0, f 0 is decreasing, f 0 (x) ≤ f 0 (1) = 0, f is decreasing, f (x) ≤ f (1) = 0. Case 2: 0 ≤ x ≤ 1, 1 < k < 2. Since f (4) ≤ 0, f 000 is decreasing, f 000 (x) ≥ f 000 (1) = 0, f 00 is increasing, f 00 (x) ≤ f 00 (1) = 0, f 0 is decreasing, f 0 (x) ≥ f 0 (1) = 0, f is increasing, f (x) ≤ f (1) = 0.

P 3.32. If a, b are nonnegative real numbers such that a + b = 1, then a2b + b2a ≤ 1. (Vasile Cîrtoaje, 2007) Solution. Without loss of generality, assume that 0≤ b≤

1 ≤ a ≤ 1. 2

Applying Lemma 1 below for c = 2b, 0 ≤ c ≤ 1, we get a2b ≤ (1 − 2b)2 + 4a b(1 − b) − 2a b(1 − 2b) ln a, which is equivalent to a2b ≤ 1 − 4a b2 − 2a b(a − b) ln a. Similarly, applying Lemma 2 below for d = 2a − 1, d ≥ 0, we get b2a−1 ≤ 4a(1 − a) + 2a(2a − 1) ln(2a + b − 1), which is equivalent to b2a ≤ 4a b2 + 2a b(a − b) ln a. Adding up these inequalities, the desired inequality follows. The equality holds for a = b = 1/2, for a = 0 and b = 1, and for a = 1 and b = 0. Lemma 1. If 0 < a ≤ 1 and c ≥ 0, then a c ≤ (1 − c)2 + ac(2 − c) − ac(1 − c) ln a, with equality for a = 1, for c = 0, and for c = 1. Proof. Making the substitution a = e−x , x ≥ 0, we need to prove that f (x) ≥ 0, where f (x) = (1 − c)2 e x + c(2 − c) + c(1 − c)x − e(1−c)x ,

Symmetric Power-Exponential Inequalities

469

f 0 (x) = (1 − c)[(1 − c)e x + c − e(1−c)x ]. If f 0 ≥ 0 on [0, ∞), then f is increasing, and hence f (x) ≥ f (0) = 0. In order to prove that f 0 ≥ 0, we consider two cases. Case 1: 0 ≤ c ≤ 1. By the weighted AM-GM inequality, we have (1 − c)e x + c ≥ e(1−c)x , and hence f 0 (x) ≥ 0. Case 2: c ≥ 1. By the weighted AM-GM inequality, we have (c − 1)e x + e(1−c)x ≥ c, which yields

f 0 (x) = (c − 1)[(c − 1)e x + e(1−c)x − c] ≥ 0.

Lemma 2. If 0 ≤ b ≤ 1 and d ≥ 0, then b d ≤ 1 − d 2 + d(1 + d) ln(b + d), with equality for d = 0, and for b = 0 and d = 1. Proof. Write the inequality as (1 + d)[1 − d + d ln(b + d)] ≥ b d . Excepting the equality cases, since 1 − d + d ln(b + d) ≥ 1 − d + d ln d > 0, we can rewrite the inequality in the form ln(1 + d) + ln[1 − d + d ln(b + d)] ≥ d ln b. Using the substitution b = e−x − d, where − ln(1 + d) ≤ x ≤ − ln d, we need to prove that f (x) ≥ 0, where f (x) = ln(1 + d) + ln(1 − d − d x) + d x − d ln(1 − d e x ). Since f 0 (x) =

d 2 (e x − 1 − x) ≥ 0, (1 − d − d x)(1 − d e x )

f is increasing, and hence f (x) ≥ f (− ln(1 + d)) = ln[1 − d 2 + d(1 + d) ln(1 + d)].

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To complete the proof, we only need to show that −d 2 + d(1 + d) ln(1 + d) ≥ 0; that is, (1 + d) ln(1 + d) ≥ d. This inequality follows from e x ≥ 1 + x, where x =

−d . 1+d

Conjecture. If a, b are nonnegative real numbers such that 1 ≤ a + b ≤ 15, then a2b + b2a ≤ a a+b + b a+b .

P 3.33. If a, b are positive real numbers such that a + b = 1, then 2a a b b ≥ a2b + b2a .

Solution. Taking into account the inequality a2b + b2a ≤ 1 in the preceding P 3.32, it suffices to show that 2a a b b ≥ 1. Write this inequality as 2a a b b ≥ a a+b + b a+b ,  a  b  b ‹a 2≥ + . b a Since a < 1 and b < 1, we apply Bernoulli’s inequality as follows  a b b

 ‹a  ‹  a b b + −1 +1+a − 1 = 2. ≤1+ b a b a

Thus, the proof is completed. The equality holds for a = b = 1/2.

P 3.34. If a, b are positive real numbers such that a + b = 1, then a−2a + b−2b ≤ 4.

Symmetric Power-Exponential Inequalities

471

Solution. Applying Lemma below, we have a−2a ≤ 4 − 2 ln 2 − 4(1 − ln 2)a, b−2b ≤ 4 − 2 ln 2 − 4(1 − ln 2)b. Adding these inequalities, the desired inequality follows. The equality holds for a = b = 1/2. Lemma. If x ∈ (0, 1], then x −2x ≤ 4 − 2 ln 2 − 4(1 − ln 2)x, with equality for x = 1/2. Proof. Write the inequality as 1 −2x x ≤ 1 − c − (1 − 2c)x, 4

c=

1 ln 2 ≈ 0.346. 2

This is true if f (x) ≤ 0, where f (x) = −2 ln 2 − 2x ln x − ln[1 − c − (1 − 2c)x]. We have f 0 (x) = −2 − 2 ln x + f 00 (x) = −

1 − 2c , 1 − c − (1 − 2c)x

g(x) 2 (1 − 2c)2 + = , x [1 − c − (1 − 2c)x]2 x[1 − c − (1 − 2c)x]2

where g(x) = 2(1 − 2c)2 x 2 − (1 − 2c)(5 − 6c)x + 2(1 − c)2 . Since g 0 (x) = (1 − 2c)[4(1 − 2c)x − 5 + 6c] ≤ (1 − 2c)[4(1 − 2c) − 5 + 6c] = (1 − 2c)(−1 − 2c) < 0, g is decreasing on (0, 1], hence g(x) ≥ g(1) = −2c 2 + 4c − 1 > 0, f 00 (x) > 0 for x ∈ (0, 1]. Since f 0 is increasing and f 0 (1/2) = 0, we have f 0 (x) ≤ 0 for x ∈ (0, 1/2], and f 0 (x) ≥ 0 for x ∈ [1/2, 1]. Therefore, f is decreasing on (0, 1/2] and increasing on [1/2, 1], hence f (x) ≥ f (1/2) = 0. Remark. According to the inequalities in P 3.32 and P 3.34, the following inequality holds for all positive numbers a, b such that a + b = 1: ‹
 1 1 a2b + b2a + ≤ 4. a2a b2b Actually, this inequality holds for all a, b ∈ (0, 1]. In this case, it is sharper than the inequality in P 3.15.

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Chapter 4

Bibliography

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