Cyclic and noncyclic inequalities [Volume 3]

Citation preview

Vasile Cîrtoaje C j

MATHEM MATICAL INEQUA Q ALITIES Volume 3

CYCLIC AND NONCYCLIC INEQUA ALITIES

UNIVERSITY OF PLOIESTI, ROMANIA 2015

Contents 1 Cyclic Inequalities 1 1.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2 Noncyclic Inequalities 241 2.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 2.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 3 Bibliography

387

i

ii

Vasile Cîrtoaje

Chapter 1

Cyclic Inequalities 1.1

Applications

1.1. If a, b, c are positive real numbers, then 1 1 1 3 + + ≥ . a(a + 2b) b(b + 2c) c(c + 2a) a b + bc + ca 1.2. If a, b, c are nonnegative real numbers such that a + b + c = 3, then a b2 + bc 2 + ca2 ≤ 4.

1.3. If a, b, c ≥ 1, then (a) (b)

a b2 + bc 2 + ca2 + 6 ≥ 3(a + b + c); 2(a b2 + bc 2 + ca2 ) + 3 ≥ 3(a b + bc + ca).

1.4. If a, b, c are positive real numbers such that a + b + c = 3, then b2

a b c + 2 + 2 ≥ 1. + 2c c + 2a a + 2b

1.5. If a, b, c are positive real numbers such that a + b + c ≥ 3, then a−1 b−1 c−1 + + ≥ 0. b+1 c+1 a+1 1

2

Vasile Cîrtoaje

1.6. If a, b, c are positive real numbers such that a + b + c = 3, then 1 1 1 + + ≥ 1. 2 2 2a b + 1 2bc + 1 2ca2 + 1

1.7. If a, b, c are positive real numbers such that a + b + c = 3, then bc ca 3 ab + + ≤ . 9 − 4bc 9 − 4ca 9 − 4a b 5

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

b2 c2 a2 + + ≥ 1; 2a + b2 2b + c 2 2c + a2

(b)

a2 b2 c2 + + ≥ 1. a + 2b2 b + 2c 2 c + 2a2

1.9. Let a, b, c be positive real numbers such that a + b + c = 3. Then, 1 1 1 + + ≤ 1. a + b2 + c 3 b + c 2 + a3 c + a2 + b3

1.10. If a, b, c are positive real numbers, then 1 + b2 1 + c2 1 + a2 + + ≥ 2. 1 + b + c 2 1 + c + a2 1 + a + b2

1.11. If a, b, c are nonnegative real numbers, then a b c 1 + + ≤ . 4a + 4b + c 4b + 4c + a 4c + 4a + b 3

1.12. If a, b, c are positive real numbers, then a+b b+c c+a 2 + + ≥ . a + 7b + c b + 7c + a c + 7a + b 3

Cyclic Inequalities

3

1.13. If a, b, c are positive real numbers, then 2a + b 2b + c 2c + a + + ≥ 3. 2a + c 2b + a 2c + b 1.14. If a, b, c are positive real numbers, then 5a + b 5b + c 5c + a + + ≥ 9. a+c b+a c+b 1.15. If a, b, c are positive real numbers, then a(a + b) b(b + c) c(c + a) 3(a2 + b2 + c 2 ) + + ≤ . a+c b+a c+b a+b+c 1.16. If a, b, c are real numbers, then a2 − bc b2 − ca c2 − a b + + ≥ 0. 4a2 + b2 + 4c 2 4b2 + c 2 + 4a2 4c 2 + a2 + 4b2 1.17. If a, b, c are real numbers, then (a)

a(a + b)3 + b(b + c)3 + c(c + a)3 ≥ 0;

(b)

a(a + b)5 + b(b + c)5 + c(c + a)5 ≥ 0.

1.18. If a, b, c are real numbers, then 3(a4 + b4 + c 4 ) + 4(a3 b + b3 c + c 3 a) ≥ 0.

1.19. If a, b, c are positive real numbers, then (a − b)(3a + b) (b − c)(3b + c) (c − a)(3c + a) + + ≥ 0. a2 + b2 b2 + c 2 c 2 + a2 1.20. Let a, b, c be positive real numbers such that a bc = 1. Then, 1 1 1 + + ≤ 1. 2 2 1+a+ b 1+ b+c 1 + c + a2

4

Vasile Cîrtoaje

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

1.22. If a, b, c are positive real numbers such that a b + bc + ca = 3, then (a + 2b)(b + 2c)(c + 2a) ≥ 27.

1.23. If a, b, c are positive real numbers such that a b + bc + ca = 3, then b c a + + ≤ 1. 3 3 a + a + b b + b + c c + c3 + a

1.24. If a, b, c are positive real numbers such that a ≥ b ≥ c and a b + bc + ca = 3, then 1 1 1 + + ≥ 1. a + 2b b + 2c c + 2a

1.25. If a, b, c ∈ [0, 1], then a b c 1 + 2 + 2 ≥ . + 5 4c + 5 4a + 5 3

4b2

1.26. If a, b, c ∈

•

˜ 1 , 3 , then 3 a b c 7 + + ≥ . a+b b+c c+a 5

• ˜ 1 p 1.27. If a, b, c ∈ p , 2 , then 2 3 3 3 2 2 2 + + ≥ + + . a + 2b b + 2c c + 2a a+b b+c c+a

Cyclic Inequalities

5

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

a b2 + bc 2 + ca2 + a bc ≤ 4;

(b)

b c a + + ≤ 1; 4− b 4−c 4−a

(c)

a b3 + bc 3 + ca3 + (a b + bc + ca)2 ≤ 12;

(d)

a b2 bc 2 ca2 + + ≤ 1. 1+a+ b 1+ b+c 1+c+a

1.29. If a, b, c are nonnegative real numbers, no two of which are zero, then 4a bc a2 + b2 + c 2 + ≥ 2. a b2 + bc 2 + ca2 + a bc a b + bc + ca 1.30. If a, b, c are nonnegative real numbers such that a + b + c = 3, then 1 1 1 1 + + ≥ . a b2 + 8 bc 2 + 8 ca2 + 8 3 1.31. If a, b, c are nonnegative real numbers such that a + b + c = 3, then ab bc ca 3 + + ≤ . bc + 3 ca + 3 a b + 3 4 1.32. If a, b, c are positive real numbers such that a + b + c = 3, then (a b + bc + ca)(a b2 + bc 2 + ca2 ) ≤ 9.

1.33. If a, b, c are nonnegative real numbers such that a2 + b2 + c 2 = 3, then (a)

a b2 + bc 2 + ca2 ≤ a bc + 2;

(b)

a b c + + ≤ 1. b+2 c+2 a+2

1.34. If a, b, c are nonnegative real numbers such that a2 + b2 + c 2 = 3, then a2 b3 + b2 c 3 + c 2 a3 ≤ 3.

6

Vasile Cîrtoaje

1.35. If a, b, c are nonnegative real numbers such that a + b + c = 3, then a4 b2 + b4 c 2 + c 4 a2 + 4 ≥ a3 b3 + b3 c 3 + c 3 a3 .

1.36. If a, b, c are nonnegative real numbers such that a + b + c = 3, then a b c 3 + 2 + 2 ≥ ; +3 c +3 a +3 4 b c 3 a + 3 + 3 ≥ . 3 b +1 c +1 a +1 2

(a)

b2

(b)

1.37. Let a, b, c be positive real numbers, and let x =a+

1 − 1, b

y = b+

1 1 − 1, z = c + − 1. c a

Prove that x y + yz + z x ≥ 3.

1.38. Let a, b, c be positive real numbers such that a bc = 1. Prove that 

1 p a− − 2 b

‹2

1 p + b− − 2 c 

‹2

‹ 1 p 2 + c − − 2 ≥ 6. a 

1.39. Let a, b, c be positive real numbers such that a bc = 1. Prove that 1 1 1 1 + a − + 1 + b − + 1 + c − > 2. b c a

1.40. If a, b, c are positive real numbers, no two of which are equal, then a b c + + 1 + + 1 + 1 > 2. b−c c − a a−b

1.41. Let a, b, c be positive real numbers such that a bc = 1. Prove that  ‹  ‹  ‹ 1 1 2 1 1 2 1 1 2 3 2a − − + 2b − − + 2c − − ≥ . b 2 c 2 a 2 4

Cyclic Inequalities

7

1.42. Let x =a+

1 5 − , b 4

y = b+

1 5 1 5 − , z=c+ − , c 4 a 4

where a ≥ b ≥ c > 0. Prove that x y + yz + z x ≥

27 . 16

1.43. Let a, b, c be positive real numbers, and let  ‹ ‹ ‹ 1 p 1 p 1 p E = a+ − 3 b+ − 3 c+ − 3 ; a b c 1 p F = a+ − 3 b 

‹

1 p b+ − 3 c

‹

‹ 1 p c+ − 3 . a

Prove that E ≥ F .

1.44. If a, b, c are positive real numbers such that

a b c + + = 5, then b c a

b c a 17 + + ≥ . a b c 4

1.45. If a, b, c are positive real numbers, then

(a)

(b)

v t a b c b c a 1+ + + ≥2 1+ + + ; b c a a b c  ‹ v  ‹ t a b c b c a 1+2 + + ≥ 1 + 16 + + . b c a a b c

1.46. If a, b, c are positive real numbers, then  ‹  ‹ a2 b2 c 2 b c a a b c + + + 15 + + ≥ 16 + + . b2 c 2 a2 a b c b c a

8

Vasile Cîrtoaje

1.47. If a, b, c are positive real numbers such that a bc = 1, then (a) (b) (c)

a b c + + ≥ a + b + c; b c a a b c 3 + + ≥ (a + b + c − 1); b c a 2 a b c 5 + + + 2 ≥ (a + b + c). b c a 3

1.48. If a, b, c are positive real numbers such that a2 + b2 + c 2 = 3, then (a) (b)

a b c 3 + + ≥2+ ; b c a a b + bc + ca 9 a b c + + ≥ . b c a a+b+c

1.49. If a, b, c are positive real numbers such that a2 + b2 + c 2 = 3, then ‹  a b c + + + 5(a b + bc + ca) ≥ 33. 6 b c a 1.50. If a, b, c are positive real numbers such that a + b + c = 3, then  ‹ a b c (a) 6 + + + 3 ≥ 7(a2 + b2 + c 2 ); b c a a b c (b) + + ≥ a2 + b2 + c 2 . b c a 1.51. If a, b, c are positive real numbers, then a b c 14(a2 + b2 + c 2 ) + + +2≥ . b c a (a + b + c)2 1.52. Let a, b, c be positive real numbers such that a + b + c = 3, and let x = 3a +

1 , b

1 1 y = 3b + , z = 3c + . c a

Prove that x y + yz + z x ≥ 48.

Cyclic Inequalities

9

1.53. If a, b, c are positive real numbers such that a + b + c = 3, then a+1 b+1 c+1 + + ≥ 2(a2 + b2 + c 2 ). b c a

1.54. If a, b, c are positive real numbers such that a + b + c = 3, then a2 b2 c 2 + + + 3 ≥ 2(a2 + b2 + c 2 ). b c a

1.55. If a, b, c are positive real numbers, then a3 b3 c 3 + + + 2(a b + bc + ca) ≥ 3(a2 + b2 + c 2 ). b c a

1.56. If a, b, c are positive real numbers such that a4 + b4 + c 4 = 3, then (a)

a2 b2 c 2 + + ≥ 3; b c a

(b)

a2 b2 c2 3 + + ≥ . b+c c+a a+b 2

1.57. If a, b, c are positive real numbers, then 3(a3 + b3 + c 3 ) a2 b2 c 2 + + ≥ . b c a a2 + b2 + c 2

1.58. If a, b, c are positive real numbers, then v  ‹ t a2 b2 c 2 a b c + + + a + b + c ≥ 2 (a2 + b2 + c 2 ) + + . b c a b c a

1.59. If a, b, c are positive real numbers, then  ‹ a b c a b c + + + 32 + + ≥ 51. b c a a+b b+c c+a

10

Vasile Cîrtoaje

1.60. Find the greatest positive real number K such that the inequalities below hold for any positive real numbers a, b, c: (a) (b)

 ‹ a b c a b c 3 + + −3≥ K + + − ; b c a b+c c+a a+b 2  ‹ a b c a b c + + −3+K + + − 1 ≥ 0. b c a 2a + b 2b + c 2c + a

1.61. If a, b, c ∈

•

˜ 1 , 2 , then 2  ‹  ‹ a b c b c a 8 + + ≥5 + + + 9. b c a a b c

1.62. If a, b, c are positive real numbers such that a ≤ b ≤ c, then 2a 2b 2c a b c + + ≥ + + . b c a b+c c+a a+b

1.63. Let a, b, c be positive real numbers such that a bc = 1. (a) If a ≤ b ≤ c, then a b c + + ≥ a3/2 + b3/2 + c 3/2 ; b c a

(b) If a ≤ 1 ≤ b ≤ c, then p p p a b c + + ≥ a 3 + b 3 + c 3. b c a

1.64. If k and a, b, c are positive real numbers, then 1 1 1 1 1 1 + + ≥ + + . (k + 1)a + b (k + 1)b + c (k + 1)c + a ka + b + c k b + c + a kc + a + b

Cyclic Inequalities

11

1.65. If a, b, c are positive real numbers, then (a) (b)

p a b c +p +p ≤ a + b + c; p 2c + a 2a + b 2b + c p a b c +p +p ≥ a + b + c. p c + 2a a + 2b b + 2c

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

v t a + 2b 3

+b

v t b + 2c 3

+c

v t c + 2a 3

≤ 3.

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

p

1 + b3 + b

p

1 + c3 + c

p

1 + a3 ≤ 5.

1.68. If a, b, c are positive real numbers such that a bc = 1, then v s s t b a c 3 (a) + + ≥ ; b+3 c+3 a+3 2 (b)

s 3

v s t a b c 3 3 + + 3 ≥ . b+7 c+7 a+7 2

1.69. If a, b, c are positive real numbers, then ‹  ‹  ‹  4a 2 4b 2 4c 2 1+ + 1+ + 1+ ≥ 27. a+b b+c c+a

1.70. If a, b, c are positive real numbers, then v v v t 2b t 2a t 2c + + ≤ 3. a+b b+c c+a

12

Vasile Cîrtoaje

1.71. If a, b, c are nonnegative real numbers, then v s s t a b c + + ≤ 1. 4a + 5b 4b + 5c 4c + 5a

1.72. If a, b, c are positive real numbers, then a b c +p +p ≤ 1. p 2 2 2 2 2 4a + a b + 4b 4b + bc + 4c 4c + ca + 4a2

1.73. If a, b, c are positive real numbers, then v s s t b 3 a c + + ≥ ; (a) 3b + c 3c + a 3a + b 2 v s s t b p a c 4 (b) + + ≥ 8. 2b + c 2c + a 2a + b

1.74. If a, b, c are positive real numbers, then v s s t a b c + + ≥ 1. a + b + 7c b + c + 7a c + a + 7b 1.75. If a, b, c are positive real numbers such that a b + bc + ca = 3, then (a)

1 1 3 1 + + ≥ ; (a + b)(3a + b) (b + c)(3b + c) (c + a)(3c + a) 8

(b)

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

1.76. If a, b, c are nonnegative real numbers, then a4 + b4 + c 4 + 15(a3 b + b3 c + c 3 a) ≥

47 2 2 (a b + b2 c 2 + c 2 a2 ). 4

1.77. If a, b, c are nonnegative real numbers such that a + b + c = 4, then a3 b + b3 c + c 3 a ≤ 27.

Cyclic Inequalities

13

1.78. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = Prove that a4 + b4 + c 4 ≥

10 (a b + bc + ca). 3

82 3 (a b + b3 c + c 3 a). 27

1.79. If a, b, c are positive real numbers, then a3 b3 c3 a+b+c + + ≥ . 2 2 2 2 2 2 2a + b 2b + c 2c + a 3

1.80. If a, b, c are positive real numbers, then b4 c4 a+b+c a4 + + ≥ . 3 3 3 3 3 3 a +b b +c c +a 3

1.81. If a, b, c are positive real numbers such that a bc = 1, then  ‹  a2 b2 c 2 c a b 3 + + + 4 2 + 2 + 2 ≥ 7(a2 + b2 + c 2 ); b c a a b c  3   ‹ a b3 c 3 b c a 8 + + + 5 3 + 3 + 3 ≥ 13(a3 + b3 + c 3 ). b c a a b c 

(a) (b)

1.82. If a, b, c are positive real numbers, then ab bc ca a2 + b2 + c 2 + + ≤ . b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2 a b + bc + ca

1.83. If a, b, c are positive real numbers, then b−c c−a a−b + + ≥ 0. b(2b + c) c(2c + a) a(2a + b)

14

Vasile Cîrtoaje

1.84. If a, b, c are positive real numbers, then (a)

b2 + 6ca c 2 + 6a b a2 + 6bc + + ≥ 7; a b + 2bc bc + 2ca ca + 2a b

(b)

a2 + 7bc b2 + 7ca c 2 + 7a b + + ≥ 12. a b + bc bc + ca ca + a b

1.85. If a, b, c are positive real numbers, then (a)

ab bc ca a2 + b2 + c 2 + + ≤ ; 2b + c 2c + a 2a + b a+b+c

(b)

ab bc ca 3(a2 + b2 + c 2 ) + + ≤ ; b+c c+a a+b 2(a + b + c)

(c)

ab bc ca a2 + b2 + c 2 + + ≤ . 4b + 5c 4c + 5a 4a + 5b 3(a + b + c)

1.86. If a, b, c are positive real numbers, then (a) (b)

p p p a b2 + 8c 2 + b c 2 + 8a2 + c a2 + 8b2 ≤ (a + b + c)2 ; p p p a b2 + 3c 2 + b c 2 + 3a2 + c a2 + 3b2 ≤ a2 + b2 + c 2 + a b + bc + ca.

1.87. If a, b, c are positive real numbers, then (a)

(b)

1

1

1

+ p + p ≥ a a + 2b b b + 2c c c + 2a p

1

1 1 + p + p ≥ p a a + 8b b b + 8c c c + 8a

s

3 ; a bc

s

1 . a bc

1.88. If a, b, c are positive real numbers, then a

b

c

≤ +p +p p 5c + 4a 5a + 4b 5b + 4c

v ta + b + c 3

.

Cyclic Inequalities

15

1.89. If a, b, c are positive real numbers, then (a)

(b)

p

p

a a+b

a a+b

+p

+p

b b+c

b b+c

+p

+p

p

c c+a

c c+a

≥

≥

a+

p

b+

p 2

p

c

;

v t 4 27(a b + bc + ca) 4

.

1.90. If a, b, c are nonnegative real numbers such that a + b + c = 3, then p p p 3a + b2 + 3b + c 2 + 3c + a2 ≥ 6.

1.91. If a, b, c are nonnegative real numbers such that a + b + c = 3, then p p p a2 + b2 + 2bc + b2 + c 2 + 2ca + c 2 + a2 + 2a b ≥ 2(a + b + c).

1.92. If a, b, c are nonnegative real numbers, then p p p Æ a2 + b2 + 7bc + b2 + c 2 + 7ca + c 2 + a2 + 7a b ≥ 3 3(a b + bc + ca).

1.93. If a, b, c are nonnegative real numbers, then a4 + b4 + c 4 + 5(a3 b + b3 c + c 3 a) ≥ 6(a2 b2 + b2 c 2 + c 2 a2 ).

1.94. If a, b, c are positive real numbers, then a5 + b5 + c 5 − a4 b − b4 c − c 4 a ≥ 2a bc(a2 + b2 + c 2 − a b − bc − ca).

1.95. If a, b, c are real numbers, then (a2 + b2 + c 2 )2 ≥ 3(a3 b + b3 c + c 3 a).

1.96. If a, b, c are real numbers, then a4 + b4 + c 4 + a b3 + bc 3 + ca3 ≥ 2(a3 b + b3 c + c 3 a).

16

Vasile Cîrtoaje

1.97. If a, b, c are positive real numbers, then b2 c2 a2 + + ≥ 1. a b + 2c 2 bc + 2a2 ca + 2b2 1.98. If a, b, c are positive real numbers such that a + b + c = 3, then b c 3 a + + ≥ . a b + 1 bc + 1 ca + 1 2 1.99. If a, b, c are positive real numbers such that a + b + c = 3, then a b c 3 + + ≤ . 2 2 2 3a + b 3b + c 3c + a 2 1.100. If a, b, c are positive real numbers such that a + b + c = 3, then b2

a b c 3 + 2 + 2 ≥ . +c c +a a +b 2

1.101. If a, b, c are nonnegative real numbers such that a + b + c = 3, then p p p p a a + b + b b + c + c c + a ≥ 3 2.

1.102. If a, b, c are positive real numbers such that a + b + c = 3, then a b c + 2 + 2 ≥ 1. + c 2c + a 2a + b

2b2

1.103. If a, b, c are positive real numbers such that a2 + b2 + c 2 = 3, then a3 b3 c3 3 + + ≥ . a + b5 b + c 5 c + a5 2 1.104. If a, b, c are positive real numbers such that a2 + b2 + c 2 = 3, then a b c 3 + + ≥ . 1+ b 1+c 1+a 2

Cyclic Inequalities

17

1.105. If a, b, c are real numbers such that a2 + b2 + c 2 = 3, then a2 b + b2 c + c 2 a + 9 ≥ 4(a + b + c).

1.106. If a, b, c are real numbers such that a2 + b2 + c 2 = 3, then (1 − a)(1 − a b) + (1 − b)(1 − bc) + (1 − c)(1 − ca) ≥ 0.

1.107. If a, b, c are positive real numbers such that a + b + c = a b + bc + ca, then a2

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

1.108. Let m > n > 0, and let a, b, c be positive real numbers such that a m+n + b m+n + c m+n = 3. Then,

am bm c m + n + n ≥ 3. bn c a

1.109. If a, b, c are positive real numbers, then  ‹ 1 1 1 1 1 1 1 1 1 + + + + + ≥3 + + ; (a) 4a 4b 4c a + b b + c c + a 3a + b 3b + c 3c + a  ‹ 1 1 1 1 1 1 1 1 1 (b) + + + + + ≥2 + + . 4a 4b 4c a + 3b b + 3c c + 3a 3a + b 3b + c 3c + a

1.110. If a, b, c are positive real numbers such that a6 + b6 + c 6 = 3, then a5 b5 c 5 + + ≥ 3. b c a

1.111. If a, b, c are positive real numbers, then 1 1 1 1 + + ≤ . 2 2 2 (a + 2b + 3c) (b + 2c + 3a) (c + 2a + 3b) 4(a b + bc + ca)

18

Vasile Cîrtoaje

1.112. If a, b, c ∈ [0, 1], then a(1 − b2 ) + b(1 − c 2 ) + c(1 − a2 ) ≤

5 . 4

1.113. Let a, b, c be nonnegative real numbers such that a ≤ 1 ≤ b ≤ c. (a) If a + b + c = 3, then a2 b + b2 c + c 2 a ≥ a bc + 2; (b) If a b + bc + ca = 3, then a2 b + b2 c + c 2 a ≥ 3.

1.114. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = Prove that a4 + b4 + c 4 ≥

5 (a b + bc + ca). 2

17 3 (a b + b3 c + c 3 a). 8

1.115. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 =

5 (a b + bc + ca). 2

Prove that (a)

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

(b)

11(a4 + b4 + c 4 ) ≥ 17(a3 b + b3 c + c 3 a) + 129a bc(a + b + c); p 14 + 102 2 2 3 3 3 a b+b c+c a≤ (a b + b2 c 2 + c 2 a2 ). 8

(c)

Cyclic Inequalities

19

1.116. If a, b, c are real numbers such that a3 b + b3 c + c 3 a ≤ 0, then a2 + b2 + c 2 ≥ k(a b + bc + ca), where k=

1+

p

p 21 + 8 7 . 2

1.117. If a, b, c are real numbers such that a3 b + b3 c + c 3 a ≥ 0, then a2 + b2 + c 2 + k(a b + bc + ca) ≥ 0, where k=

−1 +

p

p 21 + 8 7 . 2

1.118. If a, b, c are real numbers such that k(a2 + b2 + c 2 ) = a b + bc + ca, where k ∈



‹ −1 , 1 , then 2 αk ≤

where

a3 b + b3 c + c 3 ≤ βk , (a2 + b2 + c 2 )2

v t 7(1 − k) 27αk = 1 + 13k − 5k2 − 2(1 − k)(1 + 2k) , 1 + 2k v t 7(1 − k) 27βk = 1 + 13k − 5k2 + 2(1 − k)(1 + 2k) . 1 + 2k

1.119. If a, b, c are positive real numbers, then a2 + 3ab b2 + 3bc c 2 + 3ca + + ≥ 3. (b + c)2 (c + a)2 (a + b)2

20

Vasile Cîrtoaje

1.120. If a, b, c are positive real numbers, then b2 c + 1 c2 a + 1 a2 b + 1 + + ≥ 3. a(b + 1) b(c + 1) c(a + 1) 1.121. If a, b, c are positive real numbers such that a + b + c = 3, then p p p a3 + 3b + b3 + 3c + c 3 + 3a ≥ 6.

1.122. If a, b, c are positive real numbers such that a bc = 1, then v s s t a b c + + ≥ 1. a + 6b + 2bc b + 6c + 2ca c + 6a + 2a b 1.123. If a, b, c are positive real numbers such that a + b + c = 3, then b2 c2 3 a2 + + ≥ . 2 2 2 4a + b 4b + c 4c + a 5 1.124. If a, b, c are positive real numbers, then a2 + bc b2 + ca c 2 + a b (a + b + c)3 + + ≤ . a+b b+c c+a 3(a b + bc + ca) 1.125. If a, b, c are positive real numbers such that a + b + c = 3, then p p p p a b2 + bc 2 + bc 2 + ca2 + ca2 + a b2 ≤ 3 2.

1.126. If a, b, c are positive real numbers such that a bc = 1, then  ‹  ‹  ‹ 1 2 1 2 1 2 a+ + b+ + c+ ≥ 6(a + b + c − 1). b c a

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

12 1 ≤3+ . 2 2 b c+c a a bc

Cyclic Inequalities

21

1.128. If a, b, c are positive real numbers such that a + b + c = 3, then 1 24 + ≥ 9. a2 b + b2 c + c 2 a a bc 1.129. If a, b, c are positive real numbers, then a b c a+b+c + + ≥ . p 3 a+b b+c c+a a + b + c − a bc 1.130. If a, b, c are positive real numbers such that a + b + c = 3, then p p p p a b2 + b + 1 + b c 2 + c + 1 + c a2 + a + 1 ≤ 3 3.

1.131. If a, b, c are positive real numbers, then 1 1 1 1 + + ≤ . 2 2 2 b(a + 2b + 3c) c(b + 2c + 3a) a(c + 2a + 3b) 12a bc 1.132. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that (a)

a2 + 9b b2 + 9c c 2 + 9a + + ≥ 15; b+c c+a a+b

(b)

a2 + 3b b2 + 3c c 2 + 3a + + ≥ 6. a+b b+c c+a

1.133. If a, b, c are positive real numbers such that a5 + b5 + c 5 = 3, then a2 b2 c 2 + + ≥ 3. b c a

1.134. Let P(a, b, c) be a cyclic homogeneous polynomial of degree three. The inequality P(a, b, c) ≥ 0 holds for all a, b, c ≥ 0 if and only if the following two conditions are fulfilled: (a) P(1, 1, 1) ≥ 0; (b) P(0, b, c) ≥ 0 for all b, c ≥ 0.

22

Vasile Cîrtoaje

1.135. If a, b, c are nonnegative real numbers, then Æ p a3 + b3 + c 3 − 3a bc ≥ 9 + 6 3 (a − b)(b − c)(c − a).

1.136. If a, b, c are the lengths of the sides of a triangle, then a b c + + ≥ 1. 3a + b − c 3b + c − a 3c + a − b

1.137. If a, b, c are the lengths of the sides of a triangle, then a2 − b2 b2 − c 2 c 2 − a2 + + ≤ 0. a2 + bc b2 + ca c 2 + a b

1.138. If a, b, c are the lengths of the sides of a triangle, then a2 (a + b)(b − c) + b2 (b + c)(c − a) + c 2 (c + a)(a − b) ≥ 0.

1.139. If a, b, c are the lengths of the sides of a triangle, then  ‹ c  a  2 b a − 1 + b2 − 1 + c2 − 1 ≥ 0. c a b

1.140. If a, b, c are the lengths of the sides of a triangle, then (a)

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

(b)

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

(c)

 ‹ a3 b + b3 c + c 3 a+b+c 4 ≥ . 3 3

1.141. If a, b, c are the lengths of the sides of a triangle, then   2 b2 c 2 b2 c 2 a2 a 2 + + ≥ + + + 3. b2 c 2 a2 a2 b2 c 2

Cyclic Inequalities

23

1.142. If a, b, c are the lengths of the sides of a triangle such that a < b < c, then b2 c2 a2 + + ≤ 0. a2 − b2 b2 − c 2 c 2 − a2

1.143. If a, b, c are the lengths of the sides of a triangle, then  ‹ a b c a+b b+c c+a + + +3≥2 + + . b c a b+c c+a a+b

1.144. Let a, b, c be the lengths of the sides of a triangle. If k ≥ 2, then a k b(a − b) + b k c(b − c) + c k a(c − a) ≥ 0.

1.145. Let a, b, c be the lengths of the sides of a triangle. If k ≥ 2, then 3(a k b + b k c + c k a) ≥ (a + b + c)(a k−1 b + b k−1 c + c k−1 a).

1.146. Let a, b, c, d be positive real numbers such that a + b + c + d = 4. Prove that b c d a + + + ≥ 1. 3+ b 3+c 3+d 3+a 1.147. Let a, b, c, d be positive real numbers such that a + b + c + d = 4. Prove that a b c d + + + ≥ 2. 2 2 2 1+ b 1+c 1+d 1 + a2 1.148. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then a2 bc + b2 cd + c 2 d a + d 2 a b ≤ 4.

1.149. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then a(b + c)2 + b(c + d)2 + c(d + a)2 + d(a + b)2 ≤ 16.

24

Vasile Cîrtoaje

1.150. If a, b, c, d are positive real numbers, then a−b b−c c−d d−a + + + ≥ 0. b+c c+d d+a a+b 1.151. If a, b, c, d are positive real numbers, then (a)

b−c c−d d−a a−b + + + ≥ 0; a + 2b + c b + 2c + d c + 2d + a d + 2a + b

(b)

a b c d + + + ≤ 1. 2a + b + c 2b + c + d 2c + d + a 2d + a + b

1.152. If a, b, c, d are positive real numbers such that a bcd = 1, then 1 1 1 1 + + + ≥ 2. a(a + b) b(b + c) c(c + d) d(d + a) 1.153. If a, b, c, d are positive real numbers, then (a)

1 1 1 1 16 + + + ≥ ; p a(1 + b) b(1 + c) c(1 + d) d(1 + a) 1 + 8 a bcd

(b)

1 1 1 16 1 + + + ≥ . p a(1 + b) b(1 + a) c(1 + d) d(1 + c) 1 + 8 a bcd

1.154. If a, b, c, d are nonnegative real numbers such that a2 + b2 + c 2 + d 2 = 4, then (a) (b)

3(a + b + c + d) ≥ 2(a b + bc + cd + d a) + 4; p a + b + c + d − 4 ≥ (2 − 2)(a b + bc + cd + d a − 4).

1.155. Let a, b, c, d be positive real numbers. (a) If a, b, c, d ≥ 1, then  ‹ ‹ ‹ ‹  ‹ 1 1 1 1 1 1 1 1 a+ b+ c+ d+ ≥ (a + b + c + d) + + + ; b c d a a b c d (b) If a bcd = 1, then  ‹ ‹ ‹ ‹  ‹ 1 1 1 1 1 1 1 1 a+ b+ c+ d+ ≥ (a + b + c + d) + + + . b c d a a b c d

Cyclic Inequalities

25

1.156. If a, b, c, d are positive real numbers, then  1+

 ‹2   ‹2 b c 2 d a 2 + 1+ + 1+ + 1+ > 7. a+b b+c c+d d+a

1.157. If a, b, c, d are positive real numbers, then a2 − bd b2 − ca c2 − d b d 2 − ac + + + ≥ 0. b + 2c + d c + 2d + a d + 2a + b a + 2b + c

1.158. If a, b, c, d are positive real numbers such that a ≤ b ≤ c ≤ d, then v v v v t 2b t 2d t 2c t 2a + + + ≤ 4. a+b b+c c+d d+a

1.159. Let a, b, c, d be nonnegative real numbers, and let x=

a , b+c

y=

b c , z= , c+d d+a

t=

d . a+b

Prove that (a) (b)

p

xz +

p

y t ≤ 1;

x + y + z + t + 4(xz + y t) ≥ 4.

1.160. If a, b, c, d are nonnegative real numbers, then  ‹ ‹ ‹ ‹ 2a 2b 2c 2d 1+ 1+ 1+ 1+ ≥ 9. b+c c+d d+a a+b

1.161. Let a, b, c, d be nonnegative real numbers. If k > 0, then  ‹ ‹ ‹ ‹ ka kb kc kd 1+ 1+ 1+ 1+ ≥ (1 + k)2 . b+c c+d d+a a+b

26

Vasile Cîrtoaje

1.162. If a, b, c, d are positive real numbers such that a + b + c + d = 4, then 1 1 1 1 + + + ≥ a2 + b2 + c 2 + d 2 . a b bc cd d a

1.163. If a, b, c, d are positive real numbers, then a2 b2 c2 d2 4 + + + ≥ . 2 2 2 2 (a + b + c) (b + c + d) (c + d + a) (d + a + b) 9

1.164. If a, b, c, d are positive real numbers such that a + b + c + d = 3, then a b(b + c) + bc(c + d) + cd(d + a) + d a(a + b) ≤ 4.

1.165. If a ≥ b ≥ c ≥ d ≥ 0 and a + b + c + d = 2, then a b(b + c) + bc(c + d) + cd(d + a) + d a(a + b) ≤ 1.

1.166. Let a, b, c, d be nonnegative real numbers such that a + b + c + d = 4. If k ≥ then

37 , 27

a b(b + kc) + bc(c + kd) + cd(d + ka) + d a(a + k b) ≤ 4(1 + k).

1.167. Let a, b, c, d be positive real numbers such that a ≤ b ≤ c ≤ d. Prove that  ‹ a b c d a c b d 2 + + + ≥4+ + + + . b c d a c a d b

1.168. Let a, b, c, d be positive real numbers such that a ≤ b ≤ c ≤ d and a bcd = 1. Prove that a b c d + + + ≥ a b + bc + cd + d a. b c d a 1.169. Let a, b, c, d be positive real numbers such that a ≤ b ≤ c ≤ d and a bcd = 1. Prove that a b c d 4 + + + + ≥ 2(a + b + c + d). b c d a

Cyclic Inequalities

27

1.170. Let A = {a1 , a2 , a3 , a4 } be a set of real numbers such that a1 + a2 + a3 + a4 = 0. Prove that there exists a permutation {a, b, c, d} of A such that a2 + b2 + c 2 + d 2 + 3(a b + bc + cd + d a) ≥ 0.

1.171. Let a, b, c, d, e be positive real numbers such that a + b + c + d + e = 5. Prove that a b c d e 4 ≥ + + + + . 1+ a bcd e b c d e a

1.172. If a, b, c, d, e are real numbers such that a + b + c + d + e = 0, then p p 5+1 a b + bc + cd + d e + ea 5−1 ≤ 2 ≤ . − 2 2 2 2 4 a +b +c +d +e 4

1.173. Let a, b, c, d, e be positive real numbers such that a2 + b2 + c 2 + d 2 + e2 = 5. Prove that a2 b2 c2 d2 e2 5 + + + + ≥ . b+c+d c+d+e d+e+a e+a+b a+b+c 3

1.174. Let a, b, c, d, e be nonnegative real numbers such that a + b + c + d + e = 5. Prove that 729 (a2 + b2 )(b2 + c 2 )(c 2 + d 2 )(d 2 + e2 )(e2 + a2 ) ≤ . 2

1.175. If a, b, c, d, e ∈ [1, 5], then a−b b−c c−d d−e e−a + + + + ≥ 0. b+c c+d d+e e+a a+b

28

Vasile Cîrtoaje

1.176. If a, b, c, d, e, f ∈ [1, 3], then f −a a−b b−c c−d d−e e− f + + + + + ≥ 0. b+c c+d d+e e+ f f +a a+b

1.177. If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers, then n X i=1

ai n ≤ , ai−1 + 2ai + ai+1 4

where a0 = an and an+1 = a1 .

1.178. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. Prove that 1 1 1 + + ··· + ≤ 1. n − 2 + a1 + a2 n − 2 + a2 + a3 n − 2 + an + a1

1.179. If a1 , a2 , ..., an ≥ 1, then ‹ ‹  Y 1 1 1 1 n−2 ; a1 + + n − 2 ≥ n (a1 + a2 + ... + an ) + + ... + a2 a1 a2 an

1.180. If a1 , a2 , ..., an ≥ 1, then  ‹ ‹  ‹  ‹ ‹  ‹ an a1 a2 1 1 1 n a1 + a2 + ... an + +2 ≥2 1+ 1+ ... 1 + . a1 a2 an a2 a3 a1

1.181. Let k and n be positive integers, and let a1 , a2 , ..., an be real numbers such that a1 ≤ a2 ≤ · · · ≤ an . Consider the inequality (a1 + a2 + · · · + an )2 ≥ n(a1 ak+1 + a2 ak+2 + · · · + an an+k ), where an+i = ai for any positive integer i. Prove that this inequality holds (a) for n = 2k; (b) for n = 4k.

Cyclic Inequalities

29

1.182. If a1 , a2 , . . . , an are real numbers, then a1 (a1 + a2 ) + a2 (a2 + a3 ) · · · + an (an + a1 ) ≥

2 (a1 + a2 + · · · + an )2 . n

1.183. If f is a convex function on (0, ∞) and a1 , a2 , . . . , an are positive numbers, then  ‹ X  ‹ n n X 1 1 f ai + ≥ f ai + , ai+1 ai i=1 i=1 where an+1 = a1 .

1.184. If f is a convex function on a real interval I, a1 , a2 , . . . , an ∈ I and a=

a1 + a2 + · · · + an , n

then n ai − ai+1  n(n − 2) nX  f (a1 ) + f (a2 ) + · · · + f (an ) + f (a) ≥ f a+ , 2 2 i=1 n

where an+1 = a1 .

30

Vasile Cîrtoaje

Cyclic Inequalities

1.2

31

Solutions

P 1.1. If a, b, c are positive real numbers, then 1 1 3 1 + + ≥ . a(a + 2b) b(b + 2c) c(c + 2a) a b + bc + ca First Solution. Write the inequality as X a(b + c) + bc a(a + 2b)

≥ 3,

X X b+c bc + ≥ 3. a + 2b a(a + 2b) It suffices to show that

X b+c ≥2 a + 2b

and X

bc ≥ 1. a(a + 2b)

Indeed, by the Cauchy-Schwarz inequality, we have P P X b+c [ (b + c)]2 4( a)2 P ≥P = P =2 a + 2b (b + c)(a + 2b) 2 a2 + 4 a b and X

P P P 2 ( bc)2 ( bc)2 a (b − c)2 bc P P =1+ P ≥ = ≥ 1. a(a + 2b) a bc (a + 2b) 3a bc a 6a bc a

The equality holds for a = b = c. Second Solution. We apply the Cauchy-Schwarz inequality in the following way P P X ( c)2 ( a)2 1 P . ≥P =P a(a + 2b) c 2 a(a + 2b) a2 b2 + 2a bc a Thus, it suffices to show that P ( a)2 P

a2 b2 + 2a bc

P

3 ≥P , a ab

which is equivalent to X X X X X ( a b)( a2 + 2 a b) ≥ 3 a2 b2 + 6a bc a,

32

Vasile Cîrtoaje X

a b(a2 + b2 ) ≥

X

a2 b2 + a bc

X

a.

The last inequality follows by summing the obvious inequalities X

a b(a2 + b2 ) ≥ 2

X

a2 b2

and X

a2 b2 ≥ a bc

X

a.

P 1.2. If a, b, c are nonnegative real numbers such that a + b + c = 3, then a b2 + bc 2 + ca2 ≤ 4.

Solution. Assume that a = max{a, b, c}. Since a b2 + bc 2 + ca2 ≤ a b ·

a+b a(a + b)(b + 2c) + a bc + ca2 = , 2 2

it suffices to show that a(a + b)(b + 2c) ≤ 8. By the AM-GM inequality, we have a(a + b)(b + 2c) ≤

•

a + (a + b) + (b + 2c) 3

˜3

=8



a+b+c 3

‹3

The equality holds for a = 2, b = 0, c = 1 (or any cyclic permutation).

P 1.3. If a, b, c ≥ 1, then (a) (b)

a b2 + bc 2 + ca2 + 6 ≥ 3(a + b + c); 2(a b2 + bc 2 + ca2 ) + 3 ≥ 3(a b + bc + ca).

= 8.

Cyclic Inequalities

33

Solution. From a(b − 1)2 + b(c − 1)2 + c(a − 1)2 ≥ 0, we get a b2 + bc 2 + ca2 ≥ 2(a b + bc + ca) − (a + b + c). Using this inequality gives a b2 + bc 2 + ca2 + 6 − 3(a + b + c) ≥ 2(a b + bc + ca) − 4(a + b + c) + 6 = 2(a − 1)(b − 1) + 2(b − 1)(c − 1) + 2(c − 1)(a − 1) ≥ 0 and 2(a b2 + bc 2 + ca2 ) + 3 − 3(a b + bc + ca) ≥ a b + bc + ca − 2(a + b + c) + 3 = (a − 1)(b − 1) + (b − 1)(c − 1) + (c − 1)(a − 1) ≥ 0. The equality in both inequalities holds for a = b = c = 1.

P 1.4. If a, b, c are positive real numbers such that a + b + c = 3, then a b c + + ≥ 1. b2 + 2c c 2 + 2a a2 + 2b Solution. Using the Cauchy-Schwarz inequality, we get P P 2 P 2 X ( a)2 a − ab a P P P . ≥ = 1 + b2 + 2c a(b2 + 2c) a b2 + 2 a b Thus, it suffices to show that X

a2 −

X

a b2 ≥ 0.

Write this inequality in the homogeneous form (a + b + c)(a2 + b2 + c 2 ) ≥ 3(a b2 + bc 2 + ca2 ), which is equivalent to the obvious inequality a(a − c)2 + b(b − a)2 + c(c − b)2 ≥ 0. The equality holds for a = b = c = 1.

34

Vasile Cîrtoaje

P 1.5. If a, b, c are positive real numbers such that a + b + c ≥ 3, then a−1 b−1 c−1 + + ≥ 0. b+1 c+1 a+1 Solution. Write the inequality as (a2 − 1)(c + 1) + (b2 − 1)(a + 1) + (c 2 − 1)(b + 1) ≥ 0, a b2 + bc 2 + ca2 + a2 + b2 + c 2 ≥ a + b + c + 3. From a(b − 1)2 + b(c − 1)2 + c(a − 1)2 ≥ 0, we get a b2 + bc 2 + ca2 ≥ 2(a b + bc + ca) − (a + b + c). Using this inequality yields a b2 + bc 2 + ca2 + a2 + b2 + c 2 − a − b − c − 3 ≥ (a + b + c)2 − 2(a + b + c) − 3 = (a + b + c − 3)(a + b + c + 1) ≥ 0. The equality holds for a = b = c = 1.

P 1.6. If a, b, c are positive real numbers such that a + b + c = 3, then 1 1 1 + + ≥ 1. 2 2 2a b + 1 2bc + 1 2ca2 + 1 First Solution. By the AM-GM inequality, we have a+b+c 1= 3 

and hence 2a b2 + 1 ≤

‹3

≥ a bc,

2b 2b + c +1= . c c

Therefore, it suffices to show that c a b + + ≥ 1. 2b + c 2c + a 2a + b

Cyclic Inequalities

35

Indeed, using the Cauchy-Schwarz inequality, we get P X c ( c)2 (a + b + c)2 ≥P = = 1. 2b + c c(2b + c) (a + b + c)2 The equality holds for a = b = c = 1. Second Solution. By expanding, the inequality can be restated as a b2 + bc 2 + ca2 + 1 ≥ 4a3 b3 c 3 . According to the AM-GM inequality, we have a b2 + bc 2 + ca2 ≥ 3a bc and 1=



a+b+c 3

‹3

≥ a bc.

Then, a b2 + bc 2 + ca2 + 1 − 4a3 b3 c 3 ≥ 3a bc + 1 − 4a3 b3 c 3 = (1 − a bc)(1 + 2a bc)2 ≥ 0.

P 1.7. If a, b, c are positive real numbers such that a + b + c = 3, then ab bc ca 3 + + ≤ . 9 − 4bc 9 − 4ca 9 − 4a b 5 Solution. We have X ab

X X ab b b = = 9 − 4bc 9 − (b + c)2 3+ b+c a + 2b + 2c • ˜ 1X a + 2c 3 1 X a + 2c = 1− = − . 2 a + 2b + 2c 2 2 a + 2b + 2c Thus, it suffices to show that X a + 2c 9 ≥ . a + 2b + 2c 5 ≤

X

Using the Cauchy-Schwarz inequality, we get P X a + 2c [ (a + 2c)]2 9(a + b + c)2 9 ≥P = = . 2 a + 2b + 2c 5 (a + 2c)(a + 2b + 2c) 5(a + b + c) The equality holds for a = b = c = 1.

36

Vasile Cîrtoaje

P 1.8. If a, b, c are positive real numbers such that a + b + c = 3, then (a)

b2 c2 a2 + + ≥ 1; 2a + b2 2b + c 2 2c + a2

(b)

a2 b2 c2 + + ≥ 1. a + 2b2 b + 2c 2 c + 2a2

Solution. (a) By the Cauchy-Schwarz inequality, we have P P 4 P X a2 ( a2 )2 a + 2 a2 b2 P . ≥P = P 2a + b2 a2 (2a + b2 ) 2 a3 + a2 b2 Thus, it suffices to prove that X

a4 +

X

a2 b2 ≥ 2

X

a3 ,

which is equivalent to the homogeneous inequalities X X X X 3 a4 + 3 a2 b2 ≥ 2( a)( a3 ), X X X a4 + 3 a2 b2 − 2 a b(a2 + b2 ) ≥ 0, X (a − b)4 ≥ 0. The equality holds for a = b = c = 1. (b) By the Cauchy-Schwarz inequality, we get P P 4 P X a2 ( a2 )2 a + 2 a2 b2 P ≥P =P . a + 2b2 a2 (a + 2b2 ) a3 + 2 a2 b2 Thus, it suffices to prove that X

a4 ≥

X

a3 .

We have X

a4 −

X

a3 =

X X (a4 − a3 − a + 1) = (a − 1)(a3 − 1) ≥ 0.

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

P 1.9. Let a, b, c be positive real numbers such that a + b + c = 3. Then, 1 1 1 + + ≤ 1. 2 3 2 3 a+b +c b+c +a c + a2 + b3 (Vasile Cîrtoaje, 2009)

Cyclic Inequalities

37

Solution (by Vo Quoc Ba Can). By the Cauchy-Schwarz inequality, we have P 3 P 2 X X a3 + b2 + c a + a +3 1 ≤ = . 2 3 2 2 2 2 a+b +c (a + b + c ) (a2 + b2 + c 2 )2 Therefore, it suffices to show that (a2 + b2 + c 2 )2 ≥ a3 + b3 + c 3 + (a2 + b2 + c 2 ) + 3, or, equivalently, (a2 + b2 + c 2 )2 +

X

a2 (3 − a) ≥ 4(a2 + b2 + c 2 ) + 3.

Let us denote t = a2 + b2 + c 2 . Since P X [ a(3 − a)]2 (9 − a2 − b2 − c 2 )2 2 = , a (3 − a) ≥ P 6 (3 − a) it is enough to show that t2 +

(9 − t)2 ≥ 4t + 3. 6

This inequality reduces to (t − 3)2 ≥ 0. The equality occurs for a = b = c = 1.

P 1.10. If a, b, c are positive real numbers, then 1 + a2 1 + b2 1 + c2 + + ≥ 2. 1 + b + c 2 1 + c + a2 1 + a + b2 Solution. From 1 + b + c2 ≤ 1 +

1 + b2 + c2, 2

we get 1 + a2 2(1 + a2 ) ≥ . 1 + b + c2 1 + b2 + 2(1 + c 2 ) Thus, it suffices to show that y x z + + ≥ 1, y + 2z z + 2x x + 2 y where x = 1 + a2 ,

y = 1 + b2 , z = 1 + c 2 .

38

Vasile Cîrtoaje

Using the Cauchy-Schwarz inequality gives y (x + y + z)2 z x + + ≥ y + 2z z + 2x x + 2 y x( y + 2z) + y(z + 2x) + z(x + 2 y) (x + y + z)2 = ≥ 1. 3(x y + yz + z x) The equality occurs for a = b = c = 1.

P 1.11. If a, b, c are nonnegative real numbers, then a b c 1 + + ≤ . 4a + 4b + c 4b + 4c + a 4c + 4a + b 3 (Pham Kim Hung, 2007) Solution. If two of a, b, c are zero, then the inequality is trivial. Otherwise, multiplying by 4(a + b + c), the inequality becomes as follows X 4a(a + b + c) 4a + 4b + c

≤

4 (a + b + c), 3

X • 4a(a + b + c)

˜ 1 − a ≤ (a + b + c), 4a + 4b + c 3 X ca 1 ≤ (a + b + c). 4a + 4b + c 9

Since

we have

9 9 1 2 = ≤ + , 4a + 4b + c (2b + c) + 2(2a + b) 2b + c 2a + b  ‹ ca 1X 1 2 ≤ ca + 4a + 4b + c 9 2b + c 2a + b  X 2a b ‹ 1 X 1 X ca = + = a, 9 2b + c 2b + c 9

X

as desired. The equality occurs for a = b = c, and also for a = 2b and c = 0 (or any cyclic permutation).

Cyclic Inequalities

39

P 1.12. If a, b, c are positive real numbers, then b+c c+a 2 a+b + + ≥ . a + 7b + c b + 7c + a c + 7a + b 3 (Vasile Cîrtoaje, 2007) Solution. Without loss of generality, assume that a = max{a, b, c}. Write the inequality as ‹ X a + b 1 2 3 − ≥ − , k > 0, a + 7b + c k 3 k X (k − 1)a + (k − 7)b − c 2k − 9 ≥ . a + 7b + c 3 Consider that all fractions in the left hand side are nonnegative and apply the CauchySchwarz inequality, as follows P P P X (k − 1)a + (k − 7)b − c [(k − 1) a + (k − 7) b − c]2 ≥P a + 7b + c (a + 7b + c)[(k − 1)a + (k − 7)b − c] P (2k − 9)2 ( a)2 P P . = (8k − 51) a2 + 2(5k − 15) a b We choose k = 12 to have 8k − 51 = 5k − 15, hence X X (8k − 51) a2 + 2(5k − 15) a b = 45 (suma)2 . For this value of k, the desired inequality X (k − 1)a + (k − 7)b − c a + 7b + c can be restated as

X 11a + 5b − c a + 7b + c

≥

2k − 9 3

≥ 5.

Consider further two cases. Case 1: 11b + 5c − a ≥ 0. By the Cauchy-Schwarz inequality, we have P
2 a ≥P = P
2 = 5. a + 7b + c (a + 7b + c)(11a + 5b − c) 45 a

X 11a + 5b − c

[

P (11a + 5b − c)]2

225

Case 2: 11b + 5c − a < 0. We have X a+b a+b 2 a − 11b − 2c 2 > = + > . a + 7b + c a + 7b + c 3 3(a + 7b + c) 3 Thus, the proof is completed. The equality holds for a = b = c.

40

Vasile Cîrtoaje

P 1.13. If a, b, c are positive real numbers, then 2a + b 2b + c 2c + a + + ≥ 3. 2a + c 2b + a 2c + b (Pham Kim Hung, 2007) Solution. Without loss of generality, assume that a = max{a, b, c}. There are two cases to consider. Case 1: 2b + 2c ≥ a. Write the inequality as X  2a + b 2a + c

−

‹ 1 3 ≥ , 2 2

X 2a + 2b − c 2a + c

≥ 3.

Since 2a + 2b − c > 0, 2b + 2c − a ≥ 0, 2c + 2a − b > 0, we may apply the Cauchy-Schwarz inequality to get X 2a + 2b − c 2a + c

P 2 (2a + 2b − c)

P 9( a)2 ≥P = 3. = P (2a + 2b − c)(2a + c) 3( a)2

Case 2: a > 2b + 2c. Since 2a + c − (2b + a) = (a − 2b − 2c) + 3c > 0, we have

2a + b 2b + c 2a + b 2b + c 3b + > + =1+ > 1. 2a + c 2b + a 2a + c 2a + c 2a + c

Therefore, it suffices to show that 2c + a ≥ 2. 2c + b Indeed,

2c + a 2c + 2b + 2c > = 2. 2c + b 2c + b

Thus, the proof is completed. The equality holds for a = b = c.

Cyclic Inequalities

41

P 1.14. If a, b, c are positive real numbers, then 5a + b 5b + c 5c + a + + ≥ 9. a+c b+a c+b (Vasile Cîrtoaje, 2007) Solution. Write the inequality in succession as follows: X (5a + b)(b + a)(c + b) ≥ 9(a + c)(b + a)(c + b), 2 2

X

a b2 +

X

a3 ≥ 3

X

a2 b,

X 4X 3 1X 3 a − b ≥3 a2 b, a b2 + 3 3 X (6a b2 + 4a3 − b3 − 9a2 b) ≥ 0,

X

(a − b)2 (4a − b) + (b − c)2 (4b − c) + (c − a)2 (4c − a) ≥ 0. Assume that a = min{a, b, c}, and use the substitution b = a + p,

c = a + q,

p, q ≥ 0.

The inequality becomes p2 (3a − p) + (p − q)2 (3a + 4p − q) + q2 (3a + 4q) ≥ 0, 2Aa + B ≥ 0, where A = p2 − pq + q2 ,

B = p3 − 3p2 q + 2pq2 + q3 .

Since A ≥ 0, we only need to show that B ≥ 0. For q = 0, we have B = p3 ≥ 0, while for q > 0, the inequality B ≥ 0 is equivalent to 1 ≥ x(x − 1)(2 − x), where x = p/q > 0. For the non-trivial case x ∈ [1, 2], we get this inequality by multiplying the obvious inequalities 1≥ x −1 and 1 ≥ x(2 − x). The proof is completed. The equality holds for a = b = c.

42

Vasile Cîrtoaje

P 1.15. If a, b, c are positive real numbers, then a(a + b) b(b + c) c(c + a) 3(a2 + b2 + c 2 ) + + ≤ . a+c b+a c+b a+b+c (Pham Huu Duc, 2007) Solution. Write the inequality as X a(a + b)(a + b + c) a+c

≤ 3(a2 + b2 + c 2 ),

X a b(a + b) + a(a + b)(a + c) a+c

≤ 3(a2 + b2 + c 2 ),

X a b(a + b)

≤ 2(a2 + b2 + c 2 ) − (a b + bc + ca). a+c Let (x, y, z) be a permutation of (a, b, c) such that x ≥ y ≥ z. Since x + y ≥z+ x ≥ y +z and x y(x + y) ≥ z x(z + x) ≥ yz( y + z), by the rearrangement inequality, we have X a b(a + b) a+c

≤

x y(x + y) z x(z + x) yz( y + z) + + . y +z z+x x+y

Consequently, it suffices to show that x y(x + y) yz( y + z) + ≤ 2(x 2 + y 2 + z 2 ) − x y − yz − 2z x. y +z x+y Write this inequality as follows:  ‹  ‹ x+y y +z xy − 1 + yz − 1 ≤ 2(x 2 + y 2 + z 2 − x y − yz − z x), y +z x+y x y(x − z) yz(z − x) + ≤ (x − y)2 + ( y − z)2 + (z − x)2 , y +z x+y y(x + y + z)(z − x)2 ≤ (x − y)2 + ( y − z)2 + (z − x)2 . (x + y)( y + z) Since y(x + y + z) < (x + y)( y + z), the last inequality is clearly true. The equality holds for a = b = c.

Cyclic Inequalities

43

P 1.16. If a, b, c are real numbers, then b2 − ca c2 − a b a2 − bc + + ≥ 0. 4a2 + b2 + 4c 2 4b2 + c 2 + 4a2 4c 2 + a2 + 4b2 (Vasile Cîrtoaje, 2006) Solution. Since

4(a2 − bc) (b + 2c)2 = 1 − , 4a2 + b2 + 4c 2 4a2 + b2 + 4c 2 we may rewrite the inequality as (b + 2c)2 (c + 2a)2 (a + 2b)2 + + ≤ 3. 4a2 + b2 + 4c 2 4b2 + c 2 + 4a2 4c 2 + a2 + 4b2 Using the Cauchy-Schwarz inequality gives (b + 2c)2 (b + 2c)2 b2 2c 2 = ≤ + . 4a2 + b2 + 4c 2 (2a2 + b2 ) + 2(2c 2 + a2 ) 2a2 + b2 2c 2 + a2 Therefore, X

X X 2c 2 X X 2a2 b2 b2 (b + 2c)2 ≤ + = + = 3. 4a2 + b2 + 4c 2 2a2 + b2 2c 2 + a2 2a2 + b2 2a2 + b2

The equality occurs when a(2b2 + c 2 ) = b(2c 2 + a2 ) = c(2a2 + b2 ); that is, when a = b = c, and also when a = 2b = 4c (or any cyclic permutation). Remark. Similarly, we can prove the following generalization. • Let a, b, c be real numbers. If k > 0, then a2 − bc b2 − ca c2 − a b + + ≥ 0, 2ka2 + b2 + k2 c 2 2k b2 + c 2 + k2 a2 2kc 2 + a2 + k2 b2 with equality for a = b = c, and also for a = k b = k2 c (or any cyclic permutation).

P 1.17. If a, b, c are real numbers, then (a)

a(a + b)3 + b(b + c)3 + c(c + a)3 ≥ 0;

(b)

a(a + b)5 + b(b + c)5 + c(c + a)5 ≥ 0. (Vasile Cîrtoaje, 1989)

44

Vasile Cîrtoaje

Solution. (a) Using the substitution b + c = 2x,

c + a = 2 y,

a + b = 2z,

which are equivalent to a = y + z − x,

b = z + x − y,

c = x + y − z,

the inequality becomes in succession x 4 + y 4 + z 4 + x y 3 + yz 3 + z x 3 ≥ x 3 y + y 3 z + z 3 x, X (x 4 + 2x y 3 − 2x 3 y + y 4 ) ≥ 0, X X (x 2 − x y − y 2 )2 + x 2 y 2 ≥ 0, the last being clearly true. The equality occurs for a = b = c = 0. (b) Using the same substitution, the inequality transforms into x 6 + y 6 + z 6 + x y 5 + yz 5 + z x 5 ≥ x 5 y + y 5 z + z 5 x, which is equivalent to X [x 6 + y 6 − 2x y(x 4 − y 4 )] ≥ 0, X [(x 2 + y 2 )(x 4 − x 2 y 2 + y 4 ) − 2x y(x 2 + y 2 )(x 2 − y 2 )] ≥ 0, X (x 2 + y 2 )(x 2 − x y − y 2 )2 ≥ 0. The equality occurs for a = b = c = 0.

P 1.18. If a, b, c are real numbers, then 3(a4 + b4 + c 4 ) + 4(a3 b + b3 c + c 3 a) ≥ 0. (Vasile Cîrtoaje, 2005) Solution. If a, b, c are nonnegative, then the inequality is trivial. Since the inequality remains unchanged by replacing a, b, c with −a, −b, −c, respectively, it suffices to consider the case when only one of a, b, c is negative; let c < 0. Replacing now c with −c, the inequality can be restated as 3(a4 + b4 + c 4 ) + 4a3 b ≥ 4(b3 c + c 3 a),

Cyclic Inequalities

45

where a, b, c ≥ 0. It is enough to prove that 3(a4 + b4 + c 4 + a3 b) ≥ 4(b3 c + c 3 a). Case 1: a ≤ b. Since 3a3 b ≥ 3a4 , it suffices to show that 6a4 + 3b4 + 3c 4 ≥ 4(b3 c + ac 3 ). Using the AM-GM inequality yields v t 4 12 1 1 1 4 4 a c = p ac 3 ≥ 2ac 3 3a4 + c 4 = 3a4 + c 4 + c 4 + c 4 ≥ 4 3 3 3 9 3 and 3b4 + c 4 ≥ 4

p 4

b12 c 4 = 4b3 c.

Thus, 6a4 + 3b4 + 3c 4 = 2(3a4 + c 4 ) + (3b4 + c 4 ) ≥ 4ac 3 + 4b3 c. Case 2: a ≥ b. Since 3a3 b ≥ 3b4 , it suffices to show that 3a4 + 6b4 + 3c 4 ≥ 4(b3 c + ac 3 ). Since

p c4 c4 4 = 2b4 + 2b4 + 2b4 + ≥ 4 b12 c 4 = 4b3 c, 8 8 we still have to show that 23 4 3a4 + c ≥ 4ac 3 . 8 We will prove the sharper inequality 6b4 +

5 3a4 + c 4 ≥ 4ac 3 . 2 Indeed, v t 4 12 5 4 5 4 5 4 5 4 4 125a c 4 3a + c = 3a + c + c + c ≥ 4 ≥ 4ac 3 . 2 6 6 6 72 4

The equality occurs for a = b = c = 0.

P 1.19. If a, b, c are positive real numbers, then (a − b)(3a + b) (b − c)(3b + c) (c − a)(3c + a) + + ≥ 0. a2 + b2 b2 + c 2 c 2 + a2 (Vasile Cîrtoaje, 2006)

46

Vasile Cîrtoaje

Solution. Since (a − b)(3a + b) = (a − b)2 + 2(a2 − b2 ), we can write the inequality as X (a − b)2 a2 + b2

+2

X a2 − b2 a2 + b2

≥ 0.

Using the identity X a2 − b2 a2 + b2

=

Y a2 − b2 a2 + b2

,

the inequality becomes X (a − b)2 a2 + b2

+2

Y a2 − b2 a2 + b2

≥ 0.

By the AM-GM inequality, we have X (a − b)2 a2 + b2

≥3

v tY (a − b)2 3 a2 + b2

.

Thus, it suffices to show that v tY Y a2 − b2 (a − b)2 3 3 + 2 ≥ 0. a2 + b2 a2 + b2 If two of a, b, c are equal, then the equality holds. Otherwise, the inequality is equivalent to Y Y 27 (a2 + b2 )2 + 8 (a + b)2 (a2 − b2 ) ≥ 0. Assume that a = max{a, b, c}. For the nontrivial case a > b > c, we can get this inequality by multiplying the inequalities 3(a2 + b2 )2 ≥ 2(a + b)2 (a2 − b2 ), 3(b2 + c 2 )2 ≥ 2(b + c)2 (b2 − c 2 ), 3(c 2 + a2 )2 ≥ 2(a + c)2 (a2 − c 2 ). These inequalities are true because 3(a2 + b2 )2 − 2(a + b)2 (a2 − b2 ) = a2 (a − 2b)2 + b2 (2a2 + 4a b + 5b2 ) > 0. The equality holds for a = b = c.

Cyclic Inequalities

47

P 1.20. Let a, b, c be positive real numbers such that a bc = 1. Then, 1 1 1 + + ≤ 1. 2 2 1+a+ b 1+ b+c 1 + c + a2 (Vasile Cîrtoaje, 2005) Solution. Using the substitution a = x 3,

b = y 3,

c = z3,

we have to show that x yz = 1 involves 1 1+

x3

+

y6

+

1 1 + ≤ 1. 3 6 1+ y +z 1 + z3 + x 6

By the Cauchy-Schwarz inequality, we have X

X z 4 + x + y −2 ≤ = 1 + x3 + y6 (z 2 + x 2 + y 2 )2 1

P 4 (z + x 2 yz + x 2 z 2 ) (x 2 + y 2 + z 2 )2

So, it remains to show that (x 2 + y 2 + z 2 )2 ≥

X

x 4 + x yz

X

x+

X

x 2 y 2,

which is equivalent to the known inequality X X x 2 y 2 ≥ x yz x. The equality occurs for a = b = c = 1. Remark. Actually, the following generalization holds. • Let a, b, c be positive real numbers such that a bc = 1. If k > 0, then 1 1 1 + + ≤ 1. k k 1+a+ b 1+ b+c 1 + c + ak

P 1.21. Let a, b, c be positive real numbers such that a bc = 1. Then, a b c 1 + + ≥ . (a + 1)(b + 2) (b + 1)(c + 2) (c + 1)(a + 2) 2

.

48

Vasile Cîrtoaje

Solution. Using the substitution a=

x , y

b=

y , z

c=

z , x

where x, y, z are positive real numbers, the inequality can be restated as xy yz zx 1 + + ≥ . (x + y)( y + 2z) ( y + z)(z + 2x) (z + x)(x + 2 y) 2 By the Cauchy-Schwarz inequality, we have X

P ( z x)2 zx 1 = . ≥P (x + y)( y + 2z) z x(x + y)( y + 2z) 2

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

P 1.22. If a, b, c are positive real numbers such that a b + bc + ca = 3, then (a + 2b)(b + 2c)(c + 2a) ≥ 27. (Michael Rozenberg, 2007) Solution. Write the inequality as A + B ≥ 0, where A = (a + 2b)(b + 2c)(c + 2a) − 3(a + b + c)(a b + bc + ca) = −(a − b)(b − c)(c − a), and B = 3(a b + bc + ca)[a + b + c − Since B= ≥

Æ

3(a b + bc + ca)].

3(a b + bc + ca)[(a − b)2 + (b − c)2 + (c − a)2 ] p 2(a + b + c + 3(a b + bc + ca)]

3(a b + bc + ca)[(a − b)2 + (b − c)2 + (c − a)2 ] , 4(a + b + c)

it suffices to show that −4(a + b + c)(a − b)(b − c)(c − a) + 3(a b + bc + ca)[(a − b)2 + (b − c)2 + (c − a)2 ] ≥ 0. Consider c = min{a, b, c}, and use the substitution a = c + x and b = c + y, where x, y ≥ 0. The inequality becomes 4x y(x − y)(3c + x + y) + 6(x 2 − x y + y 2 )[3c 2 + 2(x + y)c + x y] ≥ 0,

Cyclic Inequalities

49

which is equivalent to 9(x 2 − x y + y 2 )c 2 + 6Ac + B ≥ 0, where A = x 3 + x 2 y − x y 2 + y 3 ≥ y(x 2 − x y + y 2 ), p B = x y(5x 2 + y 2 − 3x y) ≥ (2 5 − 3)x 2 y 2 . Since A ≥ 0 and B ≥ 0, the proof is completed. The equality holds for a = b = c = 1.

P 1.23. If a, b, c are positive real numbers such that a b + bc + ca = 3, then a b c + + ≤ 1. 3 3 a + a + b b + b + c c + c3 + a (Andrei Ciupan, 2005) Solution. Write the inequality as 1 1 1 + + ≤ 1. 2 2 + b/a 1 + b + c/b 1 + c + a/c

1 + a2

By the Cauchy-Schwarz inequality, we have X

X c2 + 1 + a b 1 ≤ = 1. 1 + a2 + b/a (c + a + b)2

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

P 1.24. If a, b, c are positive real numbers such that a ≥ b ≥ c and a b + bc + ca = 3, then 1 1 1 + + ≥ 1. a + 2b b + 2c c + 2a Solution. According to the well known inequality Æ x + y + z ≥ 3(x y + yz + z x), where x, y, z are positive real numbers, it suffices to prove that 1 1 1 1 + + ≥ . (a + 2b)(b + 2c) (b + 2c)(c + 2a) (c + 2a)(a + 2b) 3

50

Vasile Cîrtoaje

This is equivalent to the following inequalities 9(a + b + c) ≥ (a + 2b)(b + 2c)(c + 2a), 3(a + b + c)(a b + bc + ca) ≥ (a + 2b)(b + 2c)(c + 2a), a2 b + b2 c + c 2 a ≥ a b2 + bc 2 + ca2 , (a − b)(b − c)(a − c) ≥ 0. The last inequality is clearly true under the hypothesis a ≥ b ≥ c. The equality occurs for a = b = c = 1.

P 1.25. If a, b, c ∈ [0, 1], then a b c 1 + + ≤ . 4b2 + 5 4c 2 + 5 4a2 + 5 3 Solution. Let E(a, b, c) =

a b c + 2 + 2 . + 5 4c + 5 4a + 5

4b2

We have  ‹ a−1 1 1 E(a, b, c) − E(1, b, c) = +c − 4b2 + 5 4a2 + 5 9 ˜ • 1 4c(1 + a) − = (1 − a) 9(4a2 + 5) 4b2 + 5 • ˜ 4(1 + a) 1 ≤ (1 − a) − 9(4a2 + 5) 9 −(1 − a)(1 − 2a)2 = ≤ 0, 9(4a2 + 5) and, similarly, E(a, b, c) − E(a, 1, c) ≤ 0,

E(a, b, c) − E(a, b, 1) ≤ 0.

Therefore, E(a, b, c) ≤ E(1, b, c) ≤ E(1, 1, c) ≤ E(1, 1, 1) = The equality occurs for a = b = c = 1, and also for a = permutation).

1 . 3

1 and b = c = 1 (or any cyclic 2

Cyclic Inequalities

51

˜ 1 P 1.26. If a, b, c ∈ , 3 , then 3 •

a b c 7 + + ≥ . a+b b+c c+a 5 Solution. Assume that a = max{a, b, c} and show that E(a, b, c) ≥ E(a, b, where E(a, b, c) =

p

ab ) ≥

7 , 5

a b c + + . a+b b+c c+a

We have p b c 2 b E(a, b, c) − E(a, b, a b ) = + −p p b+c c+a a+ b p p p ( a − b)( a b − c)2 = p ≥ 0. p (b + c)(c + a)( a + b) • ˜ • ˜ s 1 1 a , the hypothesis a, b, c ∈ , 3 involves x ∈ , 3 . Then, Substituting x = b 3 3 p p 7 7 a 2 b E(a, b, a b) − = +p p − 5 a+b a+ b 5 p

2 7 x2 + − 2 x +1 x +1 5 3 − 7x + 8x 2 − 2x 3 = 5(x + 1)(x 2 + 1) (3 − x)[x 2 + (1 − x)2 ] = ≥ 0. 5(x + 1)(x 2 + 1) =

The equality holds for a = 3, b =

1 and c = 1 (or any cyclic permutation). 3

˜ 1 p P 1.27. If a, b, c ∈ p , 2 , then 2 •

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

52

Vasile Cîrtoaje

Solution. Write the inequality as ‹ X 3 2 1 1 − + − ≥ 0, a + 2b a + b 6a 6b X (a − b)2 (2b − a) ≥ 0. 6a b(a + 2b)(a + b) Since

p 2 2b − a ≥ p − 2 = 0, 2 the conclusion follows. The equality holds for a = b = c.

P 1.28. If a, b, c are nonnegative real numbers such that a + b + c = 3, then (a)

a b2 + bc 2 + ca2 + a bc ≤ 4;

(b)

b c a + + ≤ 1; 4− b 4−c 4−a

(c)

a b3 + bc 3 + ca3 + (a b + bc + ca)2 ≤ 12;

(d)

a b2 bc 2 ca2 + + ≤ 1. 1+a+ b 1+ b+c 1+c+a

Solution. (a) First Solution. Let (x, y, z) be a permutation of (a, b, c) such that x ≥ y ≥ z; we have also x y ≥ z x ≥ yz Then, by the rearrangement inequality, we get a b2 + bc 2 + ca2 = b · a b + c · bc + a · ca ≤ x · x y + y · z x + z · yz = y(x 2 + xz + z 2 ), hence a b2 + bc 2 + ca2 + a bc ≤ y(x + z)2 . Therefore, it suffices to show that x + y + z = 3 involves y(x + z)2 ≤ 4. Indeed, according to the AM-GM inequality, we have 

1 x +z x +z  y(x + z)2 = y · · ≤ 4 2 2

y+

x + z x + z 3 + 2 2  = 1. 3

The equality holds for a = b = c = 1, and also for a = 0, b = 1, c = 2 (or any cyclic permutation).

Cyclic Inequalities

53

Second Solution. Without loss of generality, assume that b is between a and c; that is, (b − a)(b − c) ≤ 0,

b2 + ca ≤ b(c + a).

Therefore, a b2 + bc 2 + ca2 + a bc = a(b2 + ca) + bc 2 + a bc ≤ a b(c + a) + bc 2 + a bc • ˜ 1 1 2b + (a + c) + (a + c) 3 2 = b(a + c) = · 2b(a + c)(a + c) ≤ = 4. 2 2 3 Third Solution. Write the inequality in the homogeneous form 4(a + b + c)3 ≥ 27(a b2 + bc 2 + ca2 + a bc). Without loss of generality, suppose that a = min{a, b, c}. Putting b = a+x and c = a+ y, where x, y ≥ 0, the inequality can be restated as 9(x 2 − x y + y 2 )a + (2x − y)2 (x + 4 y) ≥ 0, which is obviously true. (b) First Solution. Write the inequality in the homogeneous form X 1 a ≤ . 4a + b + 4c 3 Multiplying by a + b + c, the inequality becomes as follows: X a2 + a b + ac

a+b+c , 4a + b + 4c 3 X  a2 + a b + ac a  a + b + c − ≤ , 4a + b + 4c 4 12 X 9a b ≤ a + b + c. 4a + b + 4c ≤

Since 9 9 1 1 1 = ≤ + + 4a + b + 4c (2a + c) + (2a + c) + (2c + b) 2a + c 2a + c 2c + b 2 1 = + , 2a + c 2c + b we have X

X 2a b X ab X 2a b X bc 9a b ≤ + = + 4a + b + 4c 2a + c 2c + b 2a + c 2a + c X 2a b + bc X = = b = a + b + c. 2a + c

54

Vasile Cîrtoaje

The equality holds for a = b = c = 1, and also for a = 0, b = 1, c = 2 (or any cyclic permutation). Second Solution. Write the inequality as follows X a(4 − a)(4 − c) ≤ (4 − a)(4 − b)(4 − c), 32 +

X

X Š €X a b2 + a bc ≤ 4 a2 + 2 ab , X €X Š2 32 + a b2 + a bc ≤ 4 a , a b2 + bc 2 + ca2 + a bc ≤ 4.

The last inequality is just the inequality in (a). (c) Using the inequality in (a), we get (a + b + c)(a b2 + bc 2 + ca2 + a bc) ≤ 12, which is equivalent to the desired inequality a b3 + bc 3 + ca3 + (a b + bc + ca)2 ≤ 12.

X

(d) Let q = a b + bc + ca. Since X X a b2 (1 + b + c)(1 + c + a) = a b2 (4 + q + c + c 2 ) = (4 + q) a b2 + (3 + q)a bc

and Y X X Y (1 + a + b) = 1 + (a + b) + (b + c)(c + a) + (a + b) X = 7 + 3q + c 2 + (3q − a bc) = 16 + 4q − a bc, the inequality is equivalent to X (4 + q) a b2 + (3 + q)a bc ≤ 16 + 4q − a bc, (4 + q)

€X

Š a b2 + a bc − 4 ≤ 0.

According to (a), the desired inequality is clearly true.

P 1.29. If a, b, c are nonnegative real numbers, no two of which are zero, then 4a bc a2 + b2 + c 2 + ≥ 2. a b2 + bc 2 + ca2 + a bc a b + bc + ca (Vo Quoc Ba Can, 2009)

Cyclic Inequalities

55

Solution. Without loss of generality, assume that b is between a and c; that is, b2 + ca ≤ b(c + a). Then, a b2 + bc 2 + ca2 + a bc = a(b2 + ca) + bc 2 + a bc ≤ a b(c + a) + bc 2 + a bc = b(a + c)2 , and it suffices to prove that a2 + b2 + c 2 4ac + ≥ 2. (a + c)2 a b + bc + ca This inequality is equivalent to [a2 + c 2 − b(a + c)]2 ≥ 0. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 1.30. If a, b, c are nonnegative real numbers such that a + b + c = 3, then 1 1 1 1 + 2 + 2 ≥ . + 8 bc + 8 ca + 8 3

a b2

(Vasile Cîrtoaje, 2007) Solution. By expanding, we can write the inequality as 64 ≥ r 3 + 16A + 5r B, 64 ≥ r 3 + (16 − 5r)A + 5r(A + B), where r = a bc,

A = a b2 + bc 2 + ca2 ,

B = a2 b + b2 c + c 2 a.

By the AM-GM inequality, we have a+b+c r≤ 3 

‹3

= 1.

On the other hand, by the inequality (a) in P 1.28, we have A ≤ 4 − r,

56

Vasile Cîrtoaje

and by Schur’s inequality, we have (a + b + c)3 + 9a bc ≥ 4(a + b + c)(a b + bc + ca), which is equivalent to A+ B ≤

27 − 3r . 4

Therefore, it suffices to prove that 64 ≥ r 3 + (16 − 5r)(4 − r) +

5r(27 − 3r) . 4

We can write this inequality in the obvious form r(1 − r)(9 + 4r) ≥ 0. The equality holds for a = b = c = 1, and also for a = 0, b = 1, c = 2 (or any cyclic permutation).

P 1.31. If a, b, c are nonnegative real numbers such that a + b + c = 3, then ab bc ca 3 + + ≤ . bc + 3 ca + 3 a b + 3 4 (Vasile Cîrtoaje, 2008) Solution. Using the inequality (a) in P 1.28, namely a2 b + b2 c + c 2 a ≤ 4 − a bc, we have X

a b(ca + 3)(a b + 3) = a bc

X

a2 b + 9a bc + 3

≤ 13a bc − a2 b2 c 2 + 3

X

a2 b2 + 9

X

X

a2 b2 + 9

X

a b.

On the other hand, (a b + 3)(bc + 3)(ca + 3) = a2 b2 c 2 + 9a bc + 9

X

a b + 27.

X

a b,

Therefore, it suffices to prove that 7a2 b2 c 2 + 81 ≥ 25a bc + 12

X

a2 b2 + 9

ab

Cyclic Inequalities

57

which is equivalent to 7a2 b2 c 2 + 47a bc ≥ 3(q + 3)(4q − 9), where q = a b + bc + ca, q ≤ 3. Since abc ≤



a+b+c 3

‹3

= 1,

we have 7a2 b2 c 2 + 47a bc ≥ 9a2 b2 c 2 + 45a bc. Consequently, it suffices to show that 3a2 b2 c 2 + 15a bc ≥ (q + 3)(4q − 9). For the non-trivial case

9 < q ≤ 3, we apply the fourth degree Schur’s inequality 4

a bc ≥

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

Thus, it remains to show that (9 − q)2 (4q − 9)2 5(9 − q)(4q − 9) + ≥ (q + 3)(4q − 9), 108 6 which is equivalent to (4q − 9)(3 − q)(69q − 4q2 − 81) ≥ 0. This is true since 69q − 4q2 − 81 = (3 − q)(4q − 9) + 6(8q − 9) > 0. The equality holds for a = b = c = 1, and also for a = 0 and b = c = permutation).

P 1.32. If a, b, c are positive real numbers such that a + b + c = 3, then (a b + bc + ca)(a b2 + bc 2 + ca2 ) ≤ 9.

3 (or any cyclic 2

58

Vasile Cîrtoaje

Solution. Let (x, y, z) be a permutation of (a, b, c) such that x ≥ y ≥ z. Since x y ≥ z x ≥ yz, by the rearrangement inequality, we have a b2 + bc 2 + ca2 = a · ca + b · a b + c · bc ≤ x · x y + y · z x + z · yz = y(x 2 + xz + z 2 ). Consequently, it suffices to show that y(x y + yz + z x)(x 2 + xz + z 2 ) ≤ 9. By the AM-GM inequality, we get 4(x y + yz + z x)(x 2 + xz + z 2 ) ≤ (x y + yz + z x + x 2 + xz + z 2 )2 = (x + z)2 (x + y + z)2 = 9(x + z)2 . Thus, we still have to show that y(x + z)2 ≤ 4. This follows from the AM-GM inequality, as follows 2 y + (x + z) + (x + z) 2 y(x + z) ≤ 3 2

•

˜3

= 8.

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

P 1.33. If a, b, c are nonnegative real numbers such that a2 + b2 + c 2 = 3, then (a)

a b2 + bc 2 + ca2 ≤ a bc + 2;

(b)

b c a + + ≤ 1. b+2 c+2 a+2

(Vasile Cîrtoaje, 2005)

Solution. (a) Without loss of generality, assume that b is between a and c; that is, (b − a)(b − c) ≤ 0,

b2 + ac ≤ b(a + c).

Since a b2 + bc 2 + ca2 = a(b2 + ac) + bc 2 ≤ a b(a + c) + bc 2 = b(a2 + c 2 ) + a bc, it suffices to show that b(a2 + c 2 ) ≤ 2.

Cyclic Inequalities

59

Indeed, we have 2 − b(a2 + c 2 ) = 2 − b(3 − b2 ) = (b − 1)2 (b + 2) ≥ 0. The equality holds for a = b = c = 1, and also for a = 0, b = 1, c = permutation).

p 2 (or any cyclic

(b) Write the inequality as follows X a(a + 2)(c + 2) ≤ (a + 2)(b + 2)(c + 2), a b2 + bc 2 + ca2 + 2(a2 + b2 + c 2 ) ≤ a bc + 8, a b2 + bc 2 + ca2 ≤ a bc + 2. The last inequality is just the inequality in (a).

P 1.34. If a, b, c are nonnegative real numbers such that a2 + b2 + c 2 = 3, then a2 b3 + b2 c 3 + c 2 a3 ≤ 3. (Vasile Cîrtoaje, 2005) Solution. Let (x, y, z) be a permutation of (a, b, c) such that x ≥ y ≥ z. Since x 2 y 2 ≥ z 2 x 2 ≥ y 2 z 2 , the rearrangement inequality yields a2 b3 + b2 c 3 + c 2 a3 = b · a2 b2 + c · b2 c 2 + a · c 2 a2 ≤ x · x 2 y 2 + y · z 2 x 2 + z · y 2 z 2   2 2 2 2 3 2 2 3 2 x + y 2 2 2 y +z = y(x y + z x + yz ) ≤ y x · +z x +z · 2 2 y(x 2 + z 2 )(x 2 + y 2 + z 2 ) 3 y(x 2 + z 2 ) = . 2 2 Thus, it suffices to show that y(x 2 + z 2 ) ≤ 2 =

for x 2 + y 2 + z 2 = 3. We can prove this using the AM-GM inequality, as follows Æ 3 6 = 2 y 2 + (x 2 + z 2 ) + (x 2 + z 2 ) ≥ 3 2 y 2 (x 2 + z 2 )2 . The equality holds for a = b = c.

60

Vasile Cîrtoaje

P 1.35. If a, b, c are nonnegative real numbers such that a + b + c = 3, then a4 b2 + b4 c 2 + c 4 a2 + 4 ≥ a3 b3 + b3 c 3 + c 3 a3 .

Solution. Write the inequality as a2 (a2 b2 + c 4 − a b3 − ac 3 ) + 4 ≥ b2 c 2 (bc − b2 ). Since 2

X X (a2 b2 + c 4 − a b3 − ac 3 ) = [a4 + b4 + 2a2 b2 − 2a b(a2 + b2 )] X = (a2 + b2 )(a − b)2 ≥ 0,

we may assume that a2 b2 + c 4 − a b3 − ac 3 ≥ 0. Thus, it suffices to show that 4 ≥ b2 c 2 (bc − b2 ). Since bc − b2 ≤

c2 , 4

it is enough to prove that 16 ≥ b2 c 4 . Indeed, from v  c 2 t c c 3 , 3=a+b+c ≥ b+ + ≥3 b 2 2 2 the conclusion follows. The equality holds for a = 0, b = 1, c = 2 (or any cyclic permutation).

P 1.36. If a, b, c are nonnegative real numbers such that a + b + c = 3, then (a) (b)

b2

a b c 3 + 2 + 2 ≥ ; +3 c +3 a +3 4

b3

b c 3 a + 3 + 3 ≥ . +1 c +1 a +1 2 (Vasile Cîrtoaje and Bin Zhao, 2005)

Cyclic Inequalities

61

Solution. (a) By the AM-GM inequality, we have p p 4 b2 + 3 = b2 + 1 + 1 + 1 ≥ 4 b2 · 13 = 4 b. Therefore,

1 p 3a a b2 a b2 = a − = a − ≥ a − a b b. p b2 + 3 b2 + 3 4 4 b Taking account of this inequality and the similar ones, it suffices to prove that p p p a b b + bc c + ca a ≤ 3.

This inequality follows immediately from the inequality in P 1.34. The equality holds for a = b = c = 1. (b) Using the AM-GM Inequality gives a a b3 a b3 1 p = a − ≥ a − = a − a b b, p b3 + 1 b3 + 1 2 2b b and, similarly,

b 1 p c 1 p ≥ b − bc c, ≥ c − ca a. 3 +1 2 a +1 2 Thus, it suffices to show that p p p a b b + bc c + ca a ≤ 3, c3

which follows from the inequality in P 1.34. The equality holds for a = b = c = 1. Conjecture. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If p 0 < k ≤ 3 + 2 3, then b2

a b c 3 + 2 + 2 ≥ . +k c +k a +k 1+k

p For k =p3 + 2 3, the p equality occurs when a = b = c = 1, and again when a = 0, b = 3 − 3 and c = 3 (or any cyclic permutation thereof ).

P 1.37. Let a, b, c be positive real numbers, and let x =a+

1 − 1, b

y = b+

1 1 − 1, z = c + − 1. c a

Prove that x y + yz + z x ≥ 3. (Vasile Cîrtoaje, 1991)

62

Vasile Cîrtoaje

First Solution. Among x, y, z, there are two numbers either less than or equal to 1, or greater than or equal to 1. Let y and z be these numbers; that is, ( y − 1)(z − 1) ≥ 0. Since x y + yz + z x − 3 = ( y − 1)(z − 1) + (x + 1)( y + z) − 4, it suffices to show that (x + 1)( y + z) ≥ 4. This inequality is true since y +z = b+

1 1 1 +c+ −2≥ b+ , a c a

and hence ‹ 1 1 − 4 = ab + − 2 ≥ 0. (x + 1)( y + z) − 4 ≥ (x + 1) b + a ab 

The equality holds for a = b = c = 1. Second Solution. Without loss of generality, assume that x = max{x, y, z}. Then, • ‹  ‹  ‹ ˜ 1 1 1 1 1 x ≥ (x + y + z) = a+ + b+ + c+ −3 3 3 a b c 1 ≥ (2 + 2 + 2 − 3) = 1. 3 On the other hand, (x + 1)( y + 1)(z + 1) = a bc + ≥2+a+ b+c+

1 1 1 1 +a+b+c+ + + a bc a b c

1 1 1 + + = 5 + x + y + z, a b c

and hence x yz + x y + yz + z x ≥ 4. Since

1 (c − 1)2 +b+ > 0, a c two cases are possible: yz ≤ 0 and y, z > 0. y +z =

Case 1: yz ≤ 0. We have x yz ≤ 0, and from x yz + x y + yz + z x ≥ 4 it follows that x y + yz + z x ≥ 4 > 3.

Cyclic Inequalities

63

Case 2: y, z > 0. Let d=

s

x y + yz + z x . 3

We have to show that d ≥ 1. By the AM-GM inequality, we have x yz ≤ d 3 . Thus, from x yz + x y + yz + z x ≥ 4, we get d 3 + 3d 2 ≥ 4, (d − 1)(d + 2)2 ≥ 0, and hence d ≥ 1.

P 1.38. Let a, b, c be positive real numbers such that a bc = 1. Prove that 

1 p a− − 2 b

‹2

1 p + b− − 2 c 

‹2

‹ 1 p 2 + c − − 2 ≥ 6. a 

Solution (by Nguyen Van Quy). Using the substitution a = x, y, z > 0, the inequality becomes as follows y −z x

y x z , b = , c = , where x z y

 ‹ p 2 z − x p 2  x − y p 2 − 2 + − 2 + − 2 ≥ 6, y z

‹  ‹ p y −z z− x x − y z − x 2  x − y 2 + −2 2 + + + ≥ 0, x y z x y z  y − z 2  z − x ‹2  x − y 2 2p2( y − z)(z − x)(x − y) + + ≥ 0. + x y z x yz

 y − z 2



Assume that x = max{x, y, z}. For x ≥ z ≥ y, the inequality is clearly true. Consider further that x ≥ y ≥ z and write the desired inequality as p u2 + v 2 + w2 ≥ 2 2 uvw, where u=

y −z ≥ 0, x

v=

x −z ≥ 0, y

w=

x−y ≥ 0. z

In addition, we have  ‹ z z  uv = 1 − 1− < 1 · 1 = 1. y x

64

Vasile Cîrtoaje

According to the AM-GM inequality, we get p u2 + v 2 + w2 ≥ 2uv + w2 ≥ 2u2 v 2 + w2 ≥ 2 2 uvw. This completes the proof. The equality holds for a = b = c.

P 1.39. Let a, b, c be positive real numbers such that a bc = 1. Prove that 1 1 + a − + 1 + b − b

1 + 1+c− c

1 > 2. a

Solution. Using the substitution a=

y , x

b=

x , z

c=

z , y

x, y, z > 0,

the inequality can be restated as x − y y −z z − x > 2. 1 + + 1 + + 1 + x z y Without loss of generality, assume that x = max{x, y, z}. We have y −z x− 1 + + 1 + x z =

y −z y x− z − x − 2 ≥ + 1 + 1 + + 1 + y x z

y −2

x + y −z z+ x − y y −z x − y y −z x − y x −z + −2= + ≥ + = ≥ 0. x z x z x x x

P 1.40. If a, b, c are positive real numbers, no two of which are equal, then 1 +

a b c + 1 + + + 1 > 2. b−c c − a a−b (Vasile Cîrtoaje, 2012)

Cyclic Inequalities

65

Solution. Without loss of generality, assume that a = max{a, b, c}. It suffices to show that c a + + + 1 > 2, 1 b−c a−b which is equivalent to a+b−c a−b+c + > 2. |b − c| a−b For b > c, this inequality is true since a+b−c a = + 1 > 1 + 1 = 2. |b − c| b−c Also, for b < c, we have a+b−c a−b+c a+b−c a−b+c + −2= + −2 |b − c| a−b c−b a−b s c a c−b a a + −2> + −2≥2 − 2 > 0. = c−b a−b c−b a−b a−b

P 1.41. Let a, b, c be positive real numbers such that a bc = 1. Prove that  ‹  ‹  ‹ 1 1 2 1 1 2 1 1 2 3 2a − − + 2b − − + 2c − − ≥ . b 2 c 2 a 2 4 (Vasile Cîrtoaje, 2012) Solution. Using the substitution 1 x = 2b − , c

1 1 y = 2c − , z = 2a − , a b

we can write the inequality as x 2 + y 2 + z 2 ≥ x + y + z. Since x + y +z =2

X

a−

and x yz = 7 − 4

X

a+2

X1 a X1 a

it follows that 2(x + y + z) + x yz = 7.

,

66

Vasile Cîrtoaje

From 2(|x| + | y| + |z|) +



|x| + | y| + |z| 3

‹3

≥ 2(|x| + | y| + |z|) + |x yz|

≥ 2(x + y + z) + x yz = 7, we get |x| + | y| + |z| ≥ 3. Therefore, we have x 2 + y 2 + z2 ≥

1 (|x| + | y| + |z|)2 ≥ |x| + | y| + |z| ≥ x + y + z. 3

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

P 1.42. Let

1 5 − , b 4 where a ≥ b ≥ c > 0. Prove that x =a+

y = b+

1 5 1 5 − , z=c+ − , c 4 a 4

x y + yz + z x ≥

27 . 16 (Vasile Cîrtoaje, 2011)

Solution. Write the inequality as ‹ X  ‹ X 1 b 5X 1 ab + + − a+ + 6 ≥ 0. ab a 2 a Since

we have

Xb Xa (a − b)(b − c)(a − c) − = ≥ 0, a b a bc Xb Xb Xa X X 1 ‹ 2 ≥ + =( a) − 3, a a b a

and it suffices to prove the symmetric inequality ‹ X X ‹ ‹ X X 1 1 1 2 +( a) −5 + 9 ≥ 0. ab + a+ ab a a Setting p = a + b + c, q = a b + bc + ca,

r = a bc,

we need to show that (2q − 5p + 9)r + pq − 5q + 2p ≥ 0

Cyclic Inequalities

67

for any nonnegative a, b, c. For fixed p and q, the linear function f (r) = (2q − 5p + 9)r + pq − 5q + 2p is minimal when r is either minimal or maximal. Thus, according to P 3.57 in Volume 1, it suffices to prove that f (r) ≥ 0 for a = 0 and for b = c. For a = 0, we need to show that (b + c)bc − 5bc + 2(b + c) ≥ 0. Indeed, putting x =

p

bc, we have

(b + c)bc − 5bc + 2(b + c) ≥ 2x 3 − 5x 2 + 4x > 0. For b = c, since p = a + 2b, q = 2a b + b2 and r = a b2 , we need to show that (4a b + 2b2 − 5a − 10b + 9)a b2 + (a + 2b)(2a b + b2 ) − 10a b − 5b2 + 2a + 4b ≥ 0; that is, Aa2 + 2Ba + C ≥ 0, where A = b(4b2 − 5b + 2),

B = b4 − 5b3 + 7b2 − 5b + 1,

Let x = b+

C = b(2b2 − 5b + 4).

1 , x ≥ 2. b

The inequality B ≥ 0 is equivalent to  ‹ 1 1 b + 2 −5 b+ + 7 ≥ 0, b b 2

x 2 − 5x + 5 ≥ 0, p x ≥ (5 + 5)/2. Consider two cases. p Case 1: x ≥ (5 + 5)/2. Since A ≥ 0, B ≥ 0, C ≥ 0, we have Aa2 + 2Ba + C > 0. p Case 2: 2 ≤ x < (5 + 5)/2. Since p Aa2 + 2Ba + C = (Aa2 + C) + 2Ba ≥ 2a( AC + B), we need to show that AC ≥ B 2 , which is equivalent to  ‹  ‹ •  ‹ ˜2 1 1 1 1 8 b2 + 2 − 30 b + + 45 ≥ b2 + 2 − 5 b + +7 , b b b b

68

Vasile Cîrtoaje 8x 2 − 30x + 45 ≥ (x 2 − 5x + 5)2 ,

(x − 2)2 (x 2 − 6x − 1) ≤ 0. p p This inequality is true for x ≤ 3 + 10, and hence for x < (5 + 5)/2. Thus, the proof is completed. The equality holds for a = b = c = 1.

P 1.43. Let a, b, c be positive real numbers, and let  ‹ ‹ ‹ 1 p 1 p 1 p E = a+ − 3 b+ − 3 c+ − 3 ; a b c ‹ ‹ ‹  1 p 1 p 1 p F = a+ − 3 b+ − 3 c+ − 3 . b c a Prove that E ≥ F . (Vasile Cîrtoaje, 2011) Solution. By expanding, the inequality becomes X X p X (a2 − bc) + bc(bc − a2 ) ≥ 3 a b(b − c). Since X X X X X X (a2 − bc) = a2 − a b ≥ 0, bc(bc − a2 ) = a2 b2 − a bc a ≥ 0, by the AM-GM inequality, we have r”X X X — ”X — 2 2 (a2 − bc) bc(bc − a2 ) . (a − bc) + bc(bc − a ) ≥ 2 Thus, it suffices to show that r”X — ”X — p X 2 (a2 − bc) bc(bc − a2 ) ≥ 3 a b(b − c), which is equivalent to v  ‹˜ ‹ — •X  1 t”X p a b c 1 2 (a2 − bc) − ≥ 3 + + − 3 , a2 bc b c a v   ‹2  ‹2  u t 1 1 1 1 2 [(a + c − 2b)2 + 3(c − a)2 ] 3 − + − − ≥ b c a b c  ‹ p a b c ≥2 3 + + −3 . b c a

Cyclic Inequalities

69

Applying the Cauchy-Schwarz inequality, it suffices to show that ‹  ‹  ‹  2 1 1 a b c 1 1 − + (c − a) − − ≥2 + + −3 , (a + c − 2b) b c a b c b c a which is an identity. The equality holds when the following two equations are satisfied: a2 + b2 + c 2 − a b − bc − ca = a2 b2 + b2 c 2 + c 2 a2 − a bc(a + b + c) and 3+

 ‹ a b c b c a + + =2 + + . b c a a b c

P 1.44. If a, b, c are positive real numbers such that

a b c + + = 5, then b c a

b c a 17 + + ≥ . a b c 4 (Vasile Cîrtoaje, 2007) Solution. Making the substitution x=

a , b

y=

b c , z= , c a

we need to show that if x, y, z are positive real numbers satisfying x yz = 1 and x + y + z = 5, then 1 1 1 17 + + ≥ . x y z 4 Since

y + z 17 1 1 1 17 1 1 17 + + − = + − = + x(5 − x) − x y z 4 x yz 4 x 4 =

4 − 17x + 20x 2 − 4x 3 (4 − x)(1 − 2x)2 = , 4x 4x

we need to show that x ≤ 4. From ( y + z)2 ≥ 4 yz, we get (5 − x)2 ≥

4 . x

Using this result, by the AM-GM inequality, we obtain s x x (5 − x) + (5 − x) + ≥ 3 3 (5 − x)2 ≥ 3. 4 4

70

Vasile Cîrtoaje

hence (5 − x) + (5 − x) +

x ≥ 3, 4

which is equivalent to x ≤ 4. The equality holds when one of x, y, z is 4 and the others 1 are ; that is, when 2 c b a= = 2 4 (or any cyclic permutation).

P 1.45. If a, b, c are positive real numbers, then (a)

v t b c a a b c 1+ + + ≥2 1+ + + ; b c a a b c

(b)

‹ v  ‹ t a b c b c a 1+2 + + ≥ 1 + 16 + + . b c a a b c 

(Vasile Cîrtoaje, 2007) Solution. Let x=

a , b

y=

b c , z= c a

and p = x + y + z, q = x y + yz + z x. By the AM-GM inequality, we have p p ≥ 3 3 x yz = 3. (a) We need to show that x yz = 1 involves p 1 + x + y + z ≥ 2 1 + x y + yz + z x, which is equivalent to (1 + p)2 ≥ 4 + 4q or p+3≥2

p

p + q + 3.

First Solution. By Schur’s inequality of degree three, we have p3 + 9 ≥ 4pq.

Cyclic Inequalities

71

Thus, ‹ (p − 3)(2p + 3) 9 = ≥ 0. (1 + p) − 4 − 4q ≥ 1 + p) − 4 − p + p p 2



2

2

The equality holds for a = b = c. Second Solution. Without loss of generality, assume that b is between a and c. By the AM-GM inequality, we have 2

p

p+q+3=2

v t

(a + b + c)



‹  ‹ 1 1 1 a+b+c 1 1 1 + + ≤ +b + + . a b c b a b c

Therefore, p+3−2

p

 ‹ a b c a+b+c 1 1 1 p+q+3≥ + + +3− −b + + b c a b a b c (a − b)(b − c) = ≥ 0. ab

(b) We have to show that x yz = 1 involves 1 + 2(x + y + z) ≥

Æ

1 + 16(x y + yz + z x),

which is equivalent to p2 + p ≥ 4q. By Schur’s inequality of degree three, we have p3 + 9 ≥ 4pq. Thus, ‹ (p − 3)(p + 3) 9 = ≥ 0. p + p − 4q ≥ p + p − p + p 9 2

2



2

The equality holds for a = b = c.

P 1.46. If a, b, c are positive real numbers, then  ‹  ‹ a2 b2 c 2 b c a a b c + + + 15 + + ≥ 16 + + . b2 c 2 a2 a b c b c a

72

Vasile Cîrtoaje

Solution. Making the substitution x=

a , b

y=

b c , z= , c a

we have to show that x yz = 1 involves x 2 + y 2 + z 2 + 15(x y + yz + z x) ≥ 16(x + y + z), which is equivalent to (x + y + z)2 − 16(x + y + z) + 13(x y + yz + z x) ≥ 0. According to P 3.83 in Volume 1, for fixed x + y + z and x yz = 1, the expression x y + yz + z x is minimal when two of x, y, z are equal. Therefore, due to symmetry, it suffices to consider that x = y. We need to show that (2x + z)2 − 16(2x + z) + 13(x 2 + 2xz) ≥ 0 for x 2 z = 1. Write this inequality as 17x 6 − 32x 5 + 30x 3 − 16x 2 + 1 ≥ 0, or (x − 1)2 g(x) ≥ 0,

g(x) = 17x 4 + 2x 3 − 13x 2 + 2x + 1.

Since g(x) = (2x − 1)4 + x(x 3 + 34x 2 − 37x + 10), it suffices to show that x 3 + 34x 2 − 37x + 10 ≥ 0. There are two cases to consider.  ˜ • ‹ 1 10 Case 1: x ∈ 0, , ∞ . We have ⊂ 2 17 x 3 + 34x 2 − 37x + 10 > 34x 2 − 37x + 10 = (2x − 1)(17x − 10) ≥ 0.  ‹ 1 10 Case 2: x ∈ , . We have 2 17 ‹  1 2 2 3 2 2(x + 34x − 37x + 10) > 2 x + 34x − 37x + 10 = 69x 2 − 74x + 20 > 0. 2 Thus, the proof is completed. The equality holds for a = b = c.

Cyclic Inequalities

73

P 1.47. If a, b, c are positive real numbers such that a bc = 1, then (a) (b) (c)

a b c + + ≥ a + b + c; b c a 3 a b c + + ≥ (a + b + c − 1); b c a 2 a b c 5 + + + 2 ≥ (a + b + c). b c a 3

Solution. (a) We write the inequality as ‹  ‹   b c c a a b + 2 + + 2 + ≥ 3(a + b + c). 2 + b c c a a b In virtue of the AM-GM inequality, we get v v v  ‹  ‹  t t t 2 2 2 a b b c c a 3 c 3 a 3 b 2 + + 2 + + 2 + ≥3 +3 +3 = 3(a + b + c). b c c a a b bc ca ab The equality holds for a = b = c = 1. (b) Using the substitution a=

y , x

b=

z , y

c=

x , z

where x, y, z > 0, the inequality can be restated as 2(x 3 + y 3 + z 3 ) + 3x yz ≥ 3(x 2 y + y 2 z + z 2 x). First Solution. We get the desired inequality by summing Schur’s inequality x 3 + y 3 + z 3 + 3x yz ≥ (x 2 y + y 2 z + z 2 x) + (x y 2 + yz 2 + z x 2 ) and x 3 + y 3 + z 3 + x y 2 + yz 2 + z x 2 ≥ 2(x 2 y + y 2 z + z 2 x). The last inequality is equivalent to x(x − y)2 + y( y − z)2 + z(z − x)2 ≥ 0. The equality holds for a = b = c = 1. Second Solution. Multiplying by x + y + z, the desired inequality in x, y, z turns into X X X X 2 x4 − x3 y − 3 x2 y2 + 2 x y 3 ≥ 0.

74

Vasile Cîrtoaje

Write this inequality as X [(1 + k)x 4 − x 3 y − 3x 2 y 2 + 2x y 3 + (1 − k) y 4 ] ≥ 0, X (x − y)[x 3 − 3x y 2 − y 3 + k(x 3 + x 2 y + x y 2 + y 3 )] ≥ 0. Choosing k =

3 , we get the obvious inequality 4 X (x − y)2 (7x 2 + 10x y + y 2 ) ≥ 0.

(c) Making the substitution a=

y , x

b=

z , y

c=

x , z

x, y, z > 0,

we need to show that 3(x 3 + y 3 + z 3 ) + 6x yz ≥ 5(x 2 y + y 2 z + z 2 x). Assuming that x = min{x, y, z} and substituting y = x + p, z = x + q,

p, q ≥ 0,

the inequality turns into (p2 − pq + q2 )x + 3p3 + 3q3 − 5p2 q ≥ 0. This is true since, by the AM-GM inequality, we get 6p3 + 6q3 = 3p3 + 3p3 + 6q3 ≥ 3

Æ 3

p 3 3p3 · 3p3 · 6q3 = 9 2 p2 q ≥ 10p2 q.

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

P 1.48. If a, b, c are positive real numbers such that a2 + b2 + c 2 = 3, then (a)

a b c 3 + + ≥2+ ; b c a a b + bc + ca

(b)

a b c 9 + + ≥ . b c a a+b+c

Cyclic Inequalities

75

Solution. (a) By the Cauchy-Schwarz inequality, we have a b c (a + b + c)2 3 + + ≥ =2+ . b c a a b + bc + ca a b + bc + ca The equality holds for a = b = c = 1. (b) Using the inequality in (a), it suffices to show that 2+ Let t =

9 3 ≥ . a b + bc + ca a+b+c

a+b+c , t ≤ 1. Since 3 2(a b + bc + ca) = (a + b + c)2 − (a2 + b2 + c 2 ) = 9t 2 − 3,

the inequality becomes 2+

2 3 ≥ , −1 t

3t 2

or (t − 1)2 (2t + 1) ≥ 0. The equality holds for a = b = c = 1.

P 1.49. If a, b, c are positive real numbers such that a2 + b2 + c 2 = 3, then  ‹ a b c 6 + + + 5(a b + bc + ca) ≥ 33. b c a Solution. Write the inequality in the homogeneous form  ‹ a b c 5 a b + bc + ca + + −3≥ 1− 2 . b c a 2 a + b2 + c 2 We will prove the sharper inequality  ‹ a b + bc + ca a b c + + −3≥ m 1− 2 , b c a a + b2 + c 2 where

p 5 m=4 2−3> . 2

Write this inequality as follows X X €X Š €X Š a2 a b2 + ma bc a b − (m + 3)a bc a2 ≥ 0,

76

Vasile Cîrtoaje X

X X X a b4 + a3 b2 + (m + 1)a bc a b − (m + 3)a bc a2 ≥ 0, X X X X p p a b4 + a3 b2 + 2(2 2 − 1)a bc a b − 4 2 a bc a2 ≥ 0, On the other hand, from we get X

a b4 +

Choosing k =

X

X

a3 b2 + (k2 − 2)

a(a − b)2 (b − kc)2 ≥ 0,

X

a2 b3 + k(4 − k)a bc

X

a b − 4ka bc

X

a2 ≥ 0.

p 2, we get the desired inequality. The equality holds for a = b = c = 1.

P 1.50. If a, b, c are positive real numbers such that a + b + c = 3, then 6

(a)



‹ a b c + + + 3 ≥ 7(a2 + b2 + c 2 ); b c a a b c + + ≥ a2 + b2 + c 2 . b c a

(b)

Solution. (a) Write the inequality in the homogeneous form 2

X Š €X Š2 €X Š2 €X a ≥ 21a bc a2 , a a b2 + a bc

which is equivalent to X X X X X a b4 + a3 b2 + 2 a2 b3 + 4a bc a b − 8a bc a2 ≥ 0. On the other hand, from we get X

a b4 +

X

X

a3 b2 + (k2 − 2)

a(a − b)2 (b − kc)2 ≥ 0,

X

a2 b3 + k(4 − k)a bc

X

a b − 4ka bc

X

a2 ≥ 0.

Choosing k = 2, we get the desired inequality. The equality holds for a = b = c = 1. (b) We get the desired inequality by adding the inequality in (a) and the obvious inequality a2 + b2 + c 2 ≥ 3. The equality holds for a = b = c = 1.

Cyclic Inequalities

77

P 1.51. If a, b, c are positive real numbers, then a b c 14(a2 + b2 + c 2 ) + + +2≥ . b c a (a + b + c)2 (Vo Quoc Ba Can, 2010) Solution. By expanding, the inequality becomes as follows: X a  €X X Š X X a2 + 2 ab + 4 a b ≥ 12 a2 , b X a3

+

b

where A=

X a3 b

+

Since A= and B=

X a2 b

+2

X a b2

+7

X

a b ≥ 10 c c X X A + B ≥ 10 a2 − 10 a b,

X a2 b c

−2

X a b2 c

B=4

,

X  b3

a2 b 2a b2 + − c c c

X  4ca2 b





=

− 12ca + 9bc =

X

X a b2 c

a2 ,

−3

X

a b.

X b(a − b)2 c

X c(2a − 3b)2 b

,

we get X  b(a − b)2

 c(2a − 3b)2 A+ B = + c b X X X ≥2 (a − b)(2a − 3b) = 10 a2 − 10 a b. Thus, the proof is completed. The equality holds for a b c =p . p π =p 4π 2π 7 − tan 7 − tan 7 − tan 7 7 7 Remark. Based on this inequality, we can prove the following weaker inequality a b c 7(a b + bc + ca) 17 + + + ≥ , b c a a2 + b2 + c 2 2 where the equality occurs in the same conditions (which involve a2 + b2 + c 2 = 2a b + 2bc + 2ca).

78

Vasile Cîrtoaje

P 1.52. Let a, b, c be positive real numbers such that a + b + c = 3, and let x = 3a +

1 , b

1 1 y = 3b + , z = 3c + . c a

Prove that x y + yz + z x ≥ 48. (Vasile Cîrtoaje, 2007) Solution. Write the inequality as follows  ‹ 1 b c a 3(a b + bc + ca) + + + + ≥ 13. a bc a b c We get this inequality by adding the inequality P 1.50-(a), namely  ‹ b c a 6 + + + 3 ≥ 7(a2 + b2 + c 2 ), a b c and the inequality 18(a b + bc + ca) +

6 + 7(a2 + b2 + c 2 ) ≥ 81. a bc

Since a2 + b2 + c 2 = 9 − 2(a b + bc + ca), the last inequality is equivalent to 2(a b + bc + ca) +

3 ≥ 9. a bc

By the known inequality (a b + bc + ca)2 ≥ 3a bc(a + b + c), we get 1 9 ≥ . a bc (a b + bc + ca)2 Thus, it suffices to show that 2q +

27 ≥ 9, q2

where q = a b + bc + ca. Indeed, by the AM-GM inequality, we have v t 27 27 27 2q + 2 = q + q + 2 ≥ 3 3 q · q · 2 = 9. q q q The equality holds for a = b = c = 1.

Cyclic Inequalities

79

P 1.53. If a, b, c are positive real numbers such that a + b + c = 3, then a+1 b+1 c+1 + + ≥ 2(a2 + b2 + c 2 ). b c a Solution. We get the desired inequality by summing the inequality in P 1.50-(a), namely,  ‹ a b c 6 + + + 3 ≥ 7(a2 + b2 + c 2 ), b c a and the inequality ‹ 1 1 1 + + 6 ≥ 5(a2 + b2 + c 2 ) + 3. a b c 

Write the last inequality as F (a, b, c) ≥ 0, where  ‹ 1 1 1 F (a, b, c) = 6 + + − 5(a2 + b2 + c 2 ) − 3, a b c assume that a = max{a, b, c} and show that  ‹ b+c b+c F (a, b, c) ≥ F a, , ≥ 0. 2 2 Indeed, we have b + c ≤ 2, hence  ‹  ‹ • ˜ b+c b+c 1 b+c 4 F (a, b, c) − F a, , =6 − − 5 b2 + c 2 − (b + c)2 2 2 bc b+c 2 • • ˜ ˜ 24 6 5 5 2 2 − ≥ (b − c) − ≥ 0. = (b − c) bc(b + c) 2 (b + c)3 2 Also,  ‹  ‹ b+c b+c 3−a 3−a 3(a − 1)2 (12 − 15a + 5a2 ) F a, , = F a, , = ≥ 0. 2 2 2 2 2a(3 − a) The equality holds for a = b = c = 1.

P 1.54. If a, b, c are positive real numbers such that a + b + c = 3, then a2 b2 c 2 + + + 3 ≥ 2(a2 + b2 + c 2 ). b c a (Pham Huu Duc, 2007)

80

Vasile Cîrtoaje

First Solution. We homogenize the inequality and write it in succession as follows: 6(a2 + b2 + c 2 ) a2 b2 c 2 + + +a+b+c ≥ , b c a a+b+c   2  X  b2 a + b2 + c 2 a + b + c − 2b + c ≥ 6 − , c a+b+c 3 X (b − c)2 c

≥

X 2 (b − c)2 , a+b+c

2

(b − c) A + (c − a)2 B + (a − b)2 C ≥ 0, where A=

a+b − 1, c

B=

b+c − 1, a

C=

c+a − 1. b

Assume that a = max{a, b, c}. Since A > 0 and C > 0, by the Cauchy-Schwarz inequality, we have (a − c)2 AC (b − c)2 A + (a − b)2 C ≥ = (a − c)2 . 1 1 A+ C + A C Therefore, it suffices to show that AC + B ≥ 0. A+ C Indeed, by the third degree Schur’s inequality, we get AB + BC + CA = 3 +

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

The equality holds for a = b = c = 1. Second Solution (by Michael Rozenberg). Write the inequality in the homogeneous form X X X X ( a)( a b3 ) + a bc( a)2 ≥ 6a bc a2 . By expanding, we get X (a b4 + a2 b3 + 2a b2 c 2 − 4a3 bc) ≥ 0, which is equivalent to X

a(b2 − 2bc + ac)2 ≥ 0.

Cyclic Inequalities

81

P 1.55. If a, b, c are positive real numbers, then a3 b3 c 3 + + + 2(a b + bc + ca) ≥ 3(a2 + b2 + c 2 ). b c a (Michael Rozenberg, 2010) Solution. Write the inequality as  X  a3 2 + a b − 2a ≥ a2 + b2 + c 2 − a b − bc − ca, b a(a − b)2 b(b − c)2 c(c − a)2 + + ≥ a2 + b2 + c 2 − a b − bc − ca. b c a Assume that a = max{a, b, c}. Case 1: a ≥ b ≥ c. By the Cauchy-Schwarz inequality, we have a(a − b)2 b(b − c)2 [(a − b) + (b − c)]2 a b(a − c)2 + ≥ = . b b c b2 + ac +c a

b

On the other hand, a2 + b2 + c 2 − a b − bc − ca = (a − c)2 + (b − a)(b − c) ≤ (a − c)2 . Therefore, it suffice to show that a b(a − c)2 c(a − c)2 + ≥ (a − c)2 , b2 + ac a which is equivalent to (a − c)2 (a2 b + b2 c + c 2 a − a b2 − ca2 ) ≥ 0, (a − c)2 [bc 2 − (a − b)(b − c)(c − a)] ≥ 0. Case 2: a ≥ c ≥ b. By the Cauchy-Schwarz inequality, we have b(b − c)2 c(c − a)2 bc(a − b)2 [(b − c) + (c − a)]2 + ≥ = . c a c a c2 + a b b + c On the other hand, a2 + b2 + c 2 − a b − bc − ca = (a − b)2 + (c − a)(c − b) ≤ (a − b)2 . Therefore, it suffice to show that a(a − b)2 bc(a − b)2 + 2 ≥ (a − b)2 , b c + ab

82

Vasile Cîrtoaje

which is equivalent to (a − b)2 (a2 b + b2 c + c 2 a − a b2 − bc 2 ) ≥ 0, (a − b)2 [a b(a − b) + b2 c + c 2 (a − b)] ≥ 0. The equality holds for a = b = c.

P 1.56. If a, b, c are positive real numbers such that a4 + b4 + c 4 = 3, then (a)

a2 b2 c 2 + + ≥ 3; b c a

(b)

a2 b2 c2 3 + + ≥ . b+c c+a a+b 2 (Alexey Gladkich, 2005)

Solution. (a) By Hölder’s inequality, we have X 2  X 2  a a €X b

b

Š €X Š3 a2 b2 ≥ a2 .

Therefore, it suffices to show that €X

a2

Š3

≥9

X

a2 b2 ,

which has the homogeneous form €X

a2

Š3

≥3

€X

a2 b2

Šr X 3 a4 .

Using the notation x=

X

a2 ,

y=

X

a2 b2 ,

the inequality can be restated as x3 ≥ 3 y

Æ

3(x 2 − 2 y).

By squaring, this inequality has the form x 6 − 27x 2 y 2 + 54 y 3 ≥ 0, which is true because x 6 − 27x 2 y 2 + 54 y 3 = (x 2 − 3 y)2 (x 2 + 6 y) ≥ 0.

Cyclic Inequalities

83

The equality holds for a = b = c = 1. (b) By Hölder’s inequality, we have X 2  X 2  ”X — €X Š3 a a a2 (b + c)2 ≥ a2 . b+c b+c Thus, it suffices to prove that €X

a2

Š3

9X 2 a (b + c)2 . 4

≥

Using the inequality in (a), namely, €X

a2

Š3

≥9

X

a2 b2 ,

we still have to show that X

a2 b2 ≥

1X 2 a (b + c)2 . 4

This inequality is equivalent to X

a2 (b − c)2 ≥ 0.

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

P 1.57. If a, b, c are positive real numbers, then 3(a3 + b3 + c 3 ) a2 b2 c 2 + + ≥ . b c a a2 + b2 + c 2 (Vo Quoc Ba Can, 2010) Solution (by Ta Minh Hoang). Write the inequality as follows 3(a3 + b3 + c 3 ) a2 b2 c 2 + + −a−b−c ≥ − a − b − c, b c a a2 + b2 + c 2 X (a − b)2 b

≥

X 1 (a + b)(a − b)2 , a2 + b2 + c 2

(b − c)2 A + (c − a)2 B + (a − b)2 C ≥ 0, where A=

a2 + b2 − bc , c

B=

b2 + c 2 − ca , a

C=

c 2 + a2 − a b . b

84

Vasile Cîrtoaje

Assume that a = max{a, b, c}. Since A > 0 and C > 0, consider the nontrivial case B < 0; that is, ac − b2 − c 2 > 0. From ac − b2 − c 2 = c(a − 2b) − (b − c)2 , it follows that c(a − 2b) > (b − c)2 > 0, hence a > 2b. By the Cauchy-Schwarz inequality, we have AC (a − c)2 = (a − c)2 . 1 1 A+ C + A C AC 1 1 1 Therefore, it suffices to show that + B ≥ 0; that is, + + ≤ 0, or A+ C A B C (b − c)2 A + (a − b)2 C ≥

a2

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

Case 1: a ≥ b ≥ c. Since a2 + b2 − bc − (ca − b2 − c 2 ) > a2 + b2 − bc − ca = a(a − c) + b(b − c) ≥ 0, and c 2 + a2 − a b − (ca − b2 − c 2 ) > a2 + b2 − a(b + c) ≥ a2 + bc − a(b + c) = (a − b)(a − c) ≥ 0, it suffices to show that b + c ≤ a. Indeed, we have a > 2b ≥ b + c. Case 2: a ≥ c ≥ b. Replacing b and c by c and b, respectively, we need to show that a ≥ b ≥ c involves a2 c 2 b2 3(a3 + b3 + c 3 ) + + ≥ . c b a a2 + b2 + c 2 According to the preceding case, we have a2 b2 c 2 3(a3 + b3 + c 3 ) + + ≥ . b c a a2 + b2 + c 2 Therefore, it suffices to show that a2 c 2 b2 a2 b2 c 2 + + ≥ + + . c b a b c a This inequality is equivalent to (a + b + c)(a − b)(b − c)(a − c) ≥ 0, which is clearly true for a ≥ b ≥ c. The proof is completed. The equality holds for a = b = c = 1.

Cyclic Inequalities

85

P 1.58. If a, b, c are positive real numbers, then v  ‹ t a2 b2 c 2 a b c + + + a + b + c ≥ 2 (a2 + b2 + c 2 ) + + . b c a b c a (Pham Huu Duc, 2006) Solution. Without loss of generality, we may assume that b is between a and c; that is, (b − a)(b − c) ≤ 0. Since 2

v t

v ‹ ‹ t a2 + b2 + c 2  a b c b2 bc 2 2 2 + + + =2 a+ (a + b + c ) b c a b c a 2 2 2 2 b bc a +b +c +a+ + ≤ b c a 2 2 b bc c 2 a + +a+b+ + , = b c a b 

it suffices to prove that c2 bc c 2 +c ≥ + . a a b This is true because c2 bc c 2 c(b − a)(b − c) +c− − =− ≥ 0. a a b ab The proof is completed. The equality holds for a = b = c.

P 1.59. If a, b, c are positive real numbers, then  ‹ a b c a b c + + + 32 + + ≥ 51. b c a a+b b+c c+a (Vasile Cîrtoaje, 2009) Solution. Write the inequality as  ‹ a b c b c a + + + 45 ≥ 32 + + . b c a a+b b+c c+a a b c , y = , z = , which involves x yz = 1, the inequality b c a  ‹ 1 1 1 x + y + z + 45 − 32 + + ≥ 0. x +1 y +1 z+1

Using the substitution x = becomes

86

Vasile Cîrtoaje

We get this inequality by summing the inequalities x−

32 + 15 ≥ 9 ln x, x +1

y−

32 + 15 ≥ 9 ln y, y +1

z−

32 + 15 ≥ 9 ln z. z+1

f (x) = x −

32 + 15 − 9 ln x, x +1

Let

From f 0 (x) = 1 +

x > 0.

32 9 (x − 1)(x − 3)2 − = , (x + 1)2 x x(x + 1)2

it follows that f (x) is decreasing for 0 < x ≤ 1 and increasing for x ≥ 1. Therefore, we have f (x) ≥ f (1) = 0. The equality holds for a = b = c.

P 1.60. Find the greatest positive real number K such that the inequalities below hold for any positive real numbers a, b, c: (a) (b)

 ‹ a b c 3 a b c + + −3≥ K + + − ; b c a b+c c+a a+b 2  ‹ a b c a b c + + −3+K + + − 1 ≥ 0. b c a 2a + b 2b + c 2c + a (Vasile Cîrtoaje, 2008)

Solution. (a) For a = x 3 , b = x and c = 1, the inequality becomes 1 x + x + 3 −3≥ K x 2



 x3 x 1 3 + + − , x + 1 1 + x3 x3 + x 2

 ‹ (1 − K)x 3 x2 1 x 1 3 + + x + 3 −3−K + − ≥ 0. x +1 x +1 x 1 + x3 x3 + x 2 For x → ∞, we get the necessary condition 1 − K ≥ 0. We will show that the original inequality is true for K = 1; that is, a b c 3 a b c + + ≥ + + + . b c a 2 b+c c+a a+b

Cyclic Inequalities

87

Write the inequality as c

c  a a  − + − + a a+b b b+c



‹ b b 3 − ≥ , c c+a 2

ca ab 3 bc + + ≥ . a(a + b) b(b + c) c(c + a) 2 By the Cauchy-Schwarz inequality, we have ca ab (bc + ca + a b)2 bc + + ≥ a(a + b) b(b + c) c(c + a) a bc(a + b) + a bc(b + c) + a bc(c + a) (bc + ca + a b)2 3 = ≥ . 2a bc(a + b + c) 2 The equality holds for a = b = c. (b) For b = 1 and c = a2 , the inequality becomes  ‹ 1 2a 1 2a + 2 − 3 + K + 2 − 1 ≥ 0, a 2a + 1 a + 2 (a − 1)2 (2a + 1) K(a − 1)2 − ≥ 0. a2 (2a + 1)(a2 + 2) This inequality holds for any positive a if and only if 2a + 1 K − ≥ 0. 2 a (2a + 1)(a2 + 2) For a = 1, this inequality involves K ≤ 27. We will show that the original inequality is true for K = 27. a b c First Solution. Using the substitution x = , y = , z = , which involves x yz = 1, b c a the inequality can be restated as  ‹ 1 1 1 27 x + y +z−3− + + − 1 ≥ 0. 2 2x + 1 2 y + 1 2z + 1 We get this inequality by summing the inequalities x−

7 27 + ≥ 4 ln x, 2(2x + 1) 2

y−

27 7 + ≥ 4 ln y, 2(2 y + 1) 2

z−

27 7 + ≥ 4 ln z. 2(2z + 1) 2

88

Vasile Cîrtoaje

Let

7 27 + − 4 ln x, 2(2x + 1) 2

f (x) = x − From

f 0 (x) = 1 +

x > 0.

27 4(x − 1)3 4 = − , (2x + 1)2 x x(2x + 1)2

it follows that f (x) is decreasing for 0 < x ≤ 1 and increasing for x ≥ 1. Therefore, we have f (x) ≥ f (1) = 0. The equality holds for a = b = c. a b c Second Solution. Using the substitution x = ln , y = ln , z = ln , which involves b c a x + y + z = 0, we can write the inequality as x + y +z , f (x) + f ( y) + f (z) ≥ 3 f 3 where

27eu . 2eu + 1 If f is convex on R, then the inequality is true (by Jensen’s inequality). Indeed, we have f (u) = eu +

e−u f 00 (u) = 1 +

27(1 − 2eu ) 4(eu − 1)2 (eu + 7) = ≥ 0. (2eu + 1)3 (2eu + 1)3

˜ 1 P 1.61. If a, b, c ∈ , 2 , then 2  ‹  ‹ a b c b c a 8 + + ≥5 + + + 9. b c a a b c •

Solution. Let E(a, b, c) = 8



‹  ‹ a b c b c a + + −5 + + − 9. b c a a b c

Without loss of generality, assume that a = max{a, b, c}. We will show that p E(a, b, c) ≥ E(a, ac, c) ≥ 0. We have E(a, b, c) − E(a,

p

ac, c) = 8 =

(b −



p

a b + −2 b c

s ‹ s ‹  a b c c −5 + −2 c a b a

ac)2 (8a − 5c) ≥ 0. a bc

Cyclic Inequalities

89

Let t=

s

a , 1 ≤ t ≤ 2. c

We get E(a,

p

 s ‹  s ‹ a c c a ac, c) = 8 2 + −3 −5 2 + −3 c a a c ‹  ‹  2 1 + t2 − 3 = 8 2t + 2 − 3 − 5 t t 8 5 = 2 (t − 12 (2t + 1) − (t − 1)2 (t + 2) t t (t − 1)2 (4 + 5t)(2 − t) = ≥ 0. t2

The equality holds for a = b = c, and also for a = 2, b = 1 and c = permutation).

1 (or any cyclic 2

P 1.62. If a, b, c are positive real numbers such that a ≤ b ≤ c, then a b c 2a 2b 2c + + ≥ + + . b c a b+c c+a a+b First Solution. Since  ‹  ‹  b  a b c a c a b c + + − + + = −1 −1 − 1 ≥ 0, b c a a b c b c a it suffices to show that  ‹  ‹ a b c b c a 4a 4b 4c + + + + + ≥ + + . b c a a b c b+c c+a a+b This inequality is equivalent to  a

‹  ‹  ‹ 1 1 4 1 1 4 1 1 4 + − +b + − +c + − ≥ 0, b c b+c c a c+a a b a+b a2 (b − c)2 b2 (c − a)2 c 2 (a − b)2 + + ≥ 0. b+c c+a a+b

The equality holds for a = b = c.

90

Vasile Cîrtoaje

Second Solution. The inequality is equivalent to c(b − a) a(c − b) b(c − a) + − ≥ 0. a(a + b) b(b + c) c(c + a) Taking account of b(c − a) = c(b − a) + a(c − b), we may rewrite the inequality as • ˜ • ˜ 1 1 1 1 c(b − a) − + a(c − b) − ≥ 0. a(a + b) c(c + a) b(b + c) c(c + a) Since

and

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

it suffices to show that c(b − a)(c − b) a(c − b)(a − b) + ≥ 0. c(a + b)(c + a) b(b + c)(c + a) This inequality is true if 1 a − ≥ 0. a + b b(b + c) Indeed,

1 a 1 1 c−a − ≥ − = ≥ 0. a + b b(b + c) a+b b+c (a + b)(b + c)

P 1.63. Let a, b, c be positive real numbers such that a bc = 1. (a) If a ≤ b ≤ c, then a b c + + ≥ a3/2 + b3/2 + c 3/2 ; b c a (b) If a ≤ 1 ≤ b ≤ c, then p p p a b c + + ≥ a 3 + b 3 + c 3. b c a

(Vasile Cîrtoaje, 2008)

Cyclic Inequalities

91

Solution. (a) Since  ‹  ‹  b  a b c b c a c a + + − + + = −1 −1 − 1 ≥ 0, b c a a b c b c a it suffices to show that ‹  ‹  b c a a b c + + + + + ≥ 2(a3/2 + b3/2 + c 3/2 ). b c a a b c Indeed, by the AM-GM inequality, we have X a X b X  1 1 ‹ X 2a X + = a + ≥ a3/2 . p =2 b a b c bc The equality holds for a = b = c = 1. p (b) Let k = 3 and E(a, b, c) =

a b c + + − ak − bk − c k . b c a

We will show that E(a, b, c) ≥ E(a, that is, E( Substituting t =

p

Since

bc,

p

bc) ≥ 0;

1 1 p p , b, c) ≥ E( , bc, bc) ≥ 0. bc bc

bc, t ≥ 1, we rewrite the right inequality as f (t) ≥ 0, where f (t) =

We have

p

f 0 (t) = g(t), t2

1 1 + 1 + t 3 − 2k − 2t k . 3 t t g(t) =

−3 2k 2k + 3 + 2k+3 − 3−k . 6 t t t

1 2k+4 0 t g (t) = 9t 2k−3 − k(2k + 3) + k(3 − k)t 3k 2 ≥ 9 − k(2k + 3) + k(3 − k) = 9 − 3k2 = 0,

g(t) is increasing for t ≥ 1. Therefore, g(t) ≥ g(1) = 0, f 0 (t) ≥ 0, f (t) is increasing for t ≥ 1, and hence f (t) ≥ f (1) = 0. Substituting b = x 2 and c = y 2 , where 1 ≤ x ≤ y, the left inequality becomes  ‹  ‹ 1 1 2 2 E ,x ,y ≥ E , x y, x y , x2 y2 x2 y2

92

Vasile Cîrtoaje

or, equivalently, x2 1 1 + + x 2 y 4 − 3 3 − 1 − x 3 y 3 ≥ ( y k − x k )2 . x4 y2 y2 x y We write this inequality as  ( y − x) x 2 y 3 + and then show that  ( y − x) x 2 y 3 +

x+y 1 − x4 y3 y2

x+y 1 − 4 3 x y y2

‹

‹

≥ ( y k − x k )2 ,

≥ ( y − x)( y 3 − x 3 ) ≥ ( y k − x k )2 .

The left inequality (*) is true if f (x, y) ≥ 0, where f (x, y) = x 2 y 3 +

x+y 1 − − y 3 + x 3. x4 y3 y2

We will show that f (x, y) ≥ f (1, y) ≥ 0. Indeed, since 1 ≤ x ≤ y, we have  ‹ 1 1 1 f (x, y) − f (1, y) = x − 1 + y (x − 1) − 2 (x − 1) − 3 1 − 4 y y x  ‹ 1 ≥ x 3 − 1 + (x 2 − 1) − (x − 1) − 1 − 4 x ‹  ‹˜ • 1 1 ≥0 = (x 2 − 1) x − 2 + 1 − 4 x x 3

and f (1, y) =

3

2

1+ y (1 + y)(1 − y)2 1 − + 1 = ≥ 0. y3 y2 y3

In order to prove the right inequality (*), we will prove that ( y − x)( y 3 − x 3 ) ≥

3 2 ( y − x 2 )2 ≥ ( y k − x k )2 . 4

We have 4( y − x)( y 3 − x 3 ) − 3( y 2 − x 2 )2 = ( y − x)4 ≥ 0. To complete the proof, we only need to show that k 2 ( y − x 2) ≥ y k − x k , 2

k=

p 3.

(*)

Cyclic Inequalities

93

For fixed y, let k g(x) = x k − y k + ( y 2 − x 2 ), 1 ≤ x ≤ y. 2 Since g 0 (x) = k x(x k−2 − 1) ≤ 0, g(x) is decreasing, hence g(x) ≥ g( y) = 0. Thus, the proof is completed. The equality holds if and only if a = b = c = 1.

P 1.64. If k and a, b, c are positive real numbers, then 1 1 1 1 1 1 + + ≥ + + . (k + 1)a + b (k + 1)b + c (k + 1)c + a ka + b + c k b + c + a kc + a + b (Vasile Cîrtoaje, 2011) First Solution. For k = 1, we need to show that 1 1 1 3 + + ≥ . 2a + b 2b + c 2c + a a+b+c This follows immediately from the Cauchy-Schwarz inequality, as follows 1 1 1 9 + + ≥ 2a + b 2b + c 2c + a (2a + b) + (2b + c) + (2c + a) 3 = . a+b+c Further, consider two cases: k > 1 and 0 < k < 1. Case 1: k > 1. By the Cauchy-Schwarz inequality, we have k−1 1 [(k − 1) + 1]2 + ≥ (k + 1)a + b kc + a + b (k − 1)[(k + 1)a + b] + (kc + a + b) k = . ka + b + c Adding this inequality and the similar ones yields the desired inequality. Case 2: 0 < k < 1. By the Cauchy-Schwarz inequality, we have 1−k k [(1 − k) + k]2 + ≥ (k + 1)a + b ka + b + c (1 − k)[(k + 1)a + b] + k(ka + b + c) 1 = . kc + a + b

94

Vasile Cîrtoaje

Adding this inequality and the similar ones yields the desired inequality. The equality holds for a = b = c. Second Solution (by Vo Quoc Ba Can). By the Cauchy-Schwarz inequality, we have k k2 1 + + ≥ (k + 1)a + b (k + 1)b + c (k + 1)c + a ≥

(1 + k + k2 )2 [(k + 1)a + b] + k[(k + 1)b + c] + k2 [(k + 1)c + a] =

1 + k + k2 . kc + a + b

Therefore, we get in succession X

X X X 1 + k + k2 1 k k2 + + ≥ , (k + 1)a + b (k + 1)b + c (k + 1)c + a kc + a + b (1 + k + k2 )

X 1 1 ≥ (1 + k + k2 ) , (k + 1)a + b ka + b + c X X 1 1 ≥ . (k + 1)a + b ka + b + c X

Third Solution. We have 1 1 c−a − = (k + 1)a + b ka + b + c (ka + a + b)(ka + b + c)  ‹ c−a 1 1 1 ≥ = − , (kc + a + b)(ka + b + c) k − 1 ka + b + c kc + a + b and hence X

X ‹ X X 1 1 1 1 1 − ≥ − = 0. (k + 1)a + b ka + b + c k−1 ka + b + c kc + a + b

P 1.65. If a, b, c are positive real numbers, then (a) (b)

p a b c +p +p ≤ a + b + c; p 2c + a 2a + b 2b + c p a b c ≥ a + b + c. +p +p p c + 2a a + 2b b + 2c

Cyclic Inequalities

95

Solution. (a) By the Cauchy-Schwarz inequality, we have X X p s a ‹ s€X Š X a  a = a· ≤ a . p 2a + b 2a + b 2a + b Therefore, it suffices to show that X

a ≤ 1. 2a + b

This inequality is equivalent to b ≥ 1. 2a + b Applying the Cauchy-Schwarz inequality, we get P X b ( b)2 ≥P = 1. 2a + b b(2a + b) X

The equality holds for a = b = c. (b) By Hölder’s inequality, we have X

p

a a + 2b

‹2

≥P

P ( a)3 a(a + 2b)

=

X

a.

From this, the desired inequality follows. The equality holds for a = b = c.

P 1.66. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that v v v t b + 2c t a + 2b t c + 2a a +b +c ≤ 3. 3 3 3 First Solution. By the Cauchy-Schwarz inequality, we have v P v v X t a + 2b t€X Š •X a(a + 2b) ˜ t ( a)3 a ≤ a = = 3. 3 3 3 The equality holds for a = b = c = 1, and also for a = 3 and b = c = 0 (or any cyclic permutation). p Second Solution. Applying Jensen’s inequality to the concave function f (x) = x, x ≥ 0, we have p p p a a + 2b + b b + 2c + c c + 2a ≤

96

Vasile Cîrtoaje v t a(a + 2b) + b(b + 2c) + c(c + 2a) ≤ (a + b + c) a+b+c p p = (a + b + c) a + b + c = 3 3.

P 1.67. If a, b, c are nonnegative real numbers such that a + b + c = 3, then a

p

1 + b3 + b

p

1 + c3 + c

p

1 + a3 ≤ 5. (Pham Kim Hung, 2007)

Solution. Using the AM-GM inequality yields p

1 + b3 =

Æ

(1 + b)(1 − b + b2 ) ≤

b2 (1 + b) + (1 − b + b2 ) =1+ . 2 2

Therefore,  2 X p X  b a b2 + bc 2 + ca2 a 1 + b3 ≤ a 1+ . =3+ 2 2 To complete the proof, it remains to show that a b2 + bc 2 + ca2 ≤ 4. But this is just the inequality in P 1.2. The equality occurs for a = 0, b = 1 and c = 2 (or any cyclic permutation).

P 1.68. If a, b, c are positive real numbers such that a bc = 1, then v s s t b a c 3 (a) + + ≥ ; b+3 c+3 a+3 2

(b)

s 3

v s t a b c 3 3 + + 3 ≥ . b+7 c+7 a+7 2

Cyclic Inequalities

97

Solution. (a) Putting a=

x , y

b=

c=

y , z

+p

z

z , x

the inequality can be restated as x p

y(3x + z)

+p

y z(3 y + x)

x(3z + y)

≥

3 . 2

By Hölder’s inequality, we have X 2 ”X — €X Š3 x x y(3x + z) ≥ x . p y(3x + z) Therefore, it suffices to show that 4(x + y + z)3 ≥ 27(x 2 y + y 2 z + z 2 x + x yz). This is just the inequality (a) in P 1.28. The equality holds for a = b = c = 1. (b) Putting a=

x4 , y4

b=

z4 , x4

c=

y4 , z4

the inequality becomes v Xt 3

x8 3 ≥ . y 4 (7x 4 + z 4 ) 2

By Hölder’s inequality, we have – ™3 v X 2 4 Xt ”X — x x8 3 4 4 (7x + z ) ≥ . 4 4 4 y (7x + z ) y P P Since (7x 4 + z 4 ) = 8 x 4 , it is enough to show that 

y 2 z2 x2 + + y z x

4

≥ 27(x 4 + y 4 + z 4 ),

which is just the inequality in P 1.56-(a). The equality holds for a = b = c = 1.

P 1.69. If a, b, c are positive real numbers, then ‹  ‹  ‹  4b 2 4c 2 4a 2 1+ + 1+ + 1+ ≥ 27. a+b b+c c+a (Vasile Cîrtoaje, 2012)

98

Vasile Cîrtoaje

Solution. Let x=

a−b , a+b

y=

b−c c−a , z= . b+c c+a

We have −1 < x, y, z < 1 and x + y + z + x yz = 0. Since

2a 2b = x + 1, = y + 1, a+b b+c we can write the inequality as follows

2c = z + 1, c+a

(2x + 3)2 + (2 y + 3)2 + (2z + 3)2 ≥ 27, x 2 + y 2 + z 2 + 3(x + y + z) ≥ 0, x 2 + y 2 + z 2 ≥ 3x yz. By the AM-GM inequality, we have x 2 + y 2 + z2 ≥ 3

Æ 3

x 2 y 2z2.

Thus, it suffices to show that |x yz| ≤ 1, which is clearly true. The equality holds for a = b = c.

P 1.70. If a, b, c are positive real numbers, then v v v t 2b t 2c t 2a + + ≤ 3. a+b b+c c+a (Vasile Cîrtoaje, 1992) First Solution. By the Cauchy-Schwarz inequality, we have v v• ˜ ”X X t 2a — t X 2a ≤ (a + c) . a+b (a + b)(a + c) Thus, it suffices to show that X

a 9 ≤ . (a + b)(a + c) 4(a + b + c)

By expanding, this inequality becomes a(b − c)2 + b(c − a)2 + c(a − b)2 ≥ 0.

Cyclic Inequalities

99

The equality occurs for a = b = c. Second Solution. By the Cauchy-Schwarz inequality, we have v v• ˜ ”X X t 2a — t X 1 ≤ 2a(b + c) . a+b (a + b)(b + c) Thus, it suffices to show that X

1 9 ≤ , (a + b)(b + c) 4(a b + bc + ca)

which is equivalent to a(b − c)2 + b(c − a)2 + c(a − b)2 ≥ 0.

P 1.71. If a, b, c are nonnegative real numbers, then v s s t a b c + + ≤ 1. 4a + 5b 4b + 5c 4c + 5a (Vasile Cîrtoaje, 2004) Solution. If one of a, b, c is zero, then the inequality is clearly true. Otherwise, using the substitution b c a u= , v= , w= , a b c we need to show that uvw = 1 involves 1 1 1 +p +p ≤ 1. p 4 + 5u 4 + 5v 4 + 5w Using the contradiction method, it suffices to show that 1 1 1 +p +p >1 p 4 + 5u 4 + 5v 4 + 5w involves uvw < 1. Let 1 x=p , 4 + 5u  ‹ 1 where x, y, z ∈ 0, . Since 2 u=

1 − 4x 2 , 5x 2

1 1 y=p , z=p , 4 + 5v 4 + 5w

v=

1 − 4 y2 , 5 y2

w=

1 − 4z 2 , 5z 2

100

Vasile Cîrtoaje

we have to prove that x + y + z > 1 involves (1 − 4x 2 )(1 − 4 y 2 )(1 − 4z 2 ) < 125x 2 y 2 z 2 . Since 1 − 4x 2 < (x + y + z)2 − 4x 2 = (−x + y + z)(3x + y + z), it suffices to prove the homogeneous inequality (3x + y + z)(3 y + z + x)(3z + x + y)(−x + y + z)(− y + z + x)(−z + x + y) ≤ 125x 2 y 2 z 2 . By the AM-GM inequality, we have (3x + y + z)(3 y + z + x)(3z + x + y) ≤ 125

 x + y + z 3 . 3

Therefore, it is enough to show that  x + y + z 3 (−x + y + z)(− y + z + x)(−z + x + y) ≤ x 2 y 2 z 2 . 3 Using the substitution a = −x + y + z,

b = − y + z + x,

c = −z + x + y,

where a, b, c > 0, the inequality can be restated as 64a bc(a + b + c)3 ≤ 27(b + c)2 (c + a)2 (a + b)2 . Since 9(b + c)(c + a)(a + b) ≥ 8(a + b + c)(a b + bc + ca), it suffices to show that 3a bc(a + b + c) ≤ (a b + bc + ca)2 , which is a very known inequality. Thus, the proof is completed. The equality occurs for a = b = c.

P 1.72. If a, b, c are positive real numbers, then a b c +p +p ≤ 1. p 2 2 2 2 2 4a + a b + 4b 4b + bc + 4c 4c + ca + 4a2 (Bin Zhao, 2006)

Cyclic Inequalities

101

Solution. By the AM-GM inequality, we have a b + 4b2 ≥ 5

p 5

a b · b8 = 5

p 5

a b9 , v t a a a9/5 ≤p . = p p 5 4a9/5 + 5b9/5 4a2 + a b + 4b2 4a2 + 5 a b9 Therefore, it suffices to show that v v v t t t a9/5 b9/5 c 9/5 + + ≤ 1. 9/5 9/5 9/5 9/5 9/5 4a + 5b 4b + 5c 4c + 5a9/5 Replacing a9/5 , b9/5 , c 9/5 by a, b, c, respectively, we get the inequality in P 1.71. The equality holds for a = b = c.

P 1.73. If a, b, c are positive real numbers, then v s s t b 3 a c + + ≥ ; (a) 3b + c 3c + a 3a + b 2 s (b)

v s t b p a c 4 + + ≥ 8. 2b + c 2c + a 2a + b (Vasile Cîrtoaje and Pham Kim Hung, 2006)

Solution. Consider the inequality v v v t (k + 1)a t (k + 1)b t (k + 1)c + + ≥ Ak , kb + c kc + a ka + b and use the substitution v t (k + 1)a x= , kb + c

y=

v t (k + 1)b kc + a

, z=

k > 0,

v t (k + 1)c ka + b

.

From the identity (k b + c)(kc + a)(ka + b) = (k3 + 1)a bc + k bc(k b + c) + kca(kc + a) + ka b(ka + b), written as • ˜ kb + c kc + a ka + b k2 − k + 1 k kb + c kc + a ka + b · · = + + + , (k + 1)a (k + 1)b (k + 1)c (k + 1)2 (k + 1)2 (k + 1)a (k + 1)b (k + 1)c

102

Vasile Cîrtoaje

we get 1 k2 − k + 1 k = + 2 2 2 2 x y z (k + 1) (k + 1)2



‹ 1 1 1 + 2+ 2 , x2 y z

which is equivalent to F (x, y, z) = 0, where F (x, y, z) = k(x 2 y 2 + y 2 z 2 + z 2 x 2 ) + (k2 − k + 1)x 2 y 2 z 2 − (k + 1)2 . So, we need to show that F (x, y, z) = 0 yields x + y + z ≥ Ak . To do this, we use the contradiction method. Assume that x + y + z < Ak and show that F (x, y, z) < 0. Since F (x, y, z) is strictly increasing in each of its variables, it suffices to prove that x + y + z = Ak involves F (x, y, z) ≤ 0. Let p 49 + 9 17 k1 = ≈ 2.691. 32 (a) We will show that F (x, y, z) ≤ 0 for Ak = 3 and k ∈ (0, 1/k1 ] ∪ [k1 , ∞). The p AM-GM inequality x + y + z ≥ 3 3 x yz involves x yz ≤ 1. On the other hand, by Schur’s inequality (x + y + z)3 + 9x yz ≥ 4(x + y + z)(x y + yz + z x) we get 4(x y + yz + z x) ≤ 9 + 3x yz, hence (x y + yz + z x)2 − 9 ≤

(9 + 3x yz)2 9 −9= (x yz − 1)(x yz + 7). 16 16

Therefore, F (x, y, z) = k[(x y + yz + z x)2 − 6x yz] + (k2 − k + 1)x 2 y 2 z 2 − (k + 1)2 = k[(x y + yz + z x)2 − 9] + (k2 − k + 1)(x 2 y 2 z 2 − 1) − 6k(x yz − 1) 9k (x yz − 1)(x yz + 7) + (k2 − k + 1)(x 2 y 2 z 2 − 1) − 6k(x yz − 1) 16   1 = (x yz − 1) (16k2 − 7k + 16)x yz + 16k2 − 49k + 16 ≤ 0. 16

≤

The equality occurs for a = b =p c. In addition, when k = k1 or k = 1/k1 , the equality occurs also for a = 0 and b/c = k (or any cyclic permutation). v t 2 4 (k + 1) (b) We will show that F (x, y, z) ≤ 0 for Ak = 2 and k ∈ [1/k1 , k1 ]. From k F (x, y, z) = k(x y + yz + z x)2 − 2kAk x yz + (k2 − k + 1)x 2 y 2 z 2 − (k + 1)2 , it follows that for fixed x yz, F (x, y, z) is maximal when x y + yz +z x is maximal; that is, according to P 3.83 in Volume 1, when two of x, y, z are equal. Due to symmetry, we only

Cyclic Inequalities

103

need to show that F (x, y, z) ≤ 0 for y = z. Write this inequality in the homogeneous form k

p

  k (x + y + z)2 (x + y + z)4 − 16(x 2 y 2 + y 2 z 2 + z 2 x 2 ≥ 64(k3 + 1)x 2 y 2 z 2 .

Due to homogeneity, we can only consider the cases y = z = 0 and y = z = 1. In the non-trivial case y = z = 1, the inequality becomes k

p

k x(x + 2)2 (x 3 + 8x 2 − 8x + 32) ≥ 64(k3 + 1)x 2 .

This is true because 297k

p

k ≥ 64(k3 + 1)

for k ∈ [1/k1 , k1 ], and x(x + 2)2 (x 3 + 8x 2 − 8x + 32) ≥ 297x 2 . Notice that x(x + 2)2 (x 3 + 8x 2 − 8x + 32) − 297x 2 = x(x − 1)2 (x 3 + 14x 2 + 55x + 128) ≥ 0. For k ∈ (1/k1 , k1 ), the equality occurs only when a = 0 and b/c = permutation).

p

k (or any cyclic

Remark. From the proof above, it follows that the following more general statement holds: • Let a, b, c be nonnegative real numbers, no two of which are zero. If k > 0, then s

v s § ª t b a c 3 2 + + ≥ min p ,p . 4 kb + c kc + a ka + b k+1 k

P 1.74. If a, b, c are positive real numbers, then s

v s t a b c + + ≥ 1. a + b + 7c b + c + 7a c + a + 7b (Vasile Cîrtoaje, 2006)

104

Vasile Cîrtoaje

Solution. Making the substitution x=

s

a , a + b + 7c

y=

v t

b , z= b + c + 7a

s

c , c + a + 7b

we have  (x 2 − 1)a + x 2 b + 7x 2 c = 0      ( y 2 − 1)b + y 2 c + 7 y 2 a = 0 ,   2 2 2    (z − 1)c + z a + 7z b = 0 which involves

x2 − 1 7 y2 z2

x2 7x 2 y2 − 1 y2 2 2 7z z −1

=0 ;

that is, F (x, y, z) = 0, where F (x, y, z) = 324x 2 y 2 z 2 + 6

X

x2 y2 +

X

x 2 − 1.

We need to show that F (x, y, z) = 0 involves x + y + z ≥ 1, where x, y, z > 0. To do this, we use the contradiction method. Assume that x + y + z < 1 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 = 1 involves F (x, y, z) ≤ 0. We have F (x, y, z) = 324x 2 y 2 z 2 + 6

€X

xy

= 324x 2 y 2 z 2 + 6

€X

xy

Š2

− 12x yz

X

Š2

x+ X

X €X Š2 x −2 xy −1

− 12x yz − 2 xy €X Š € X Š = 12x yz(27x yz − 1) + 2 xy 3 xy −1 . Because 27x yz ≤

€X Š3 x =1

X

€X Š2 x = 1,

and 3

xy ≤

the conclusion follows. The equality occurs for a = b = c.

Cyclic Inequalities

105

P 1.75. If a, b, c are positive real numbers such that a b + bc + ca = 3, then (a)

1 1 3 1 + + ≥ ; (a + b)(3a + b) (b + c)(3b + c) (c + a)(3c + a) 8

(b)

1 1 1 1 + + ≥ . 2 2 2 (2a + b) (2b + c) (2c + a) 3 (Vasile Cîrtoaje and Pham Kim Hung, 2007)

Solution. (a) Using the Cauchy-Schwarz inequality and the inequality in P 1.73-(a) gives X

X 1 1 = (a + b)(3a + b) (b + c)(3b + c) 9 ≥ = 8(a b + bc + ca)

PÆ ≥ P


2 a 3b+c

a(b + c)

3 . 8

The equality holds for a = b = c. (b) We consider two cases (Vo Quoc Ba Can). Case 1: 4(a b + bc + ca ≥ a2 + b2 + c 2 . By the Cauchy-Schwarz inequality, we get P X 9( a)2 1 ≥P . (2a + b)2 (2a + b)2 (b + 2c)2 Thus, it suffices to show that 9p2 q ≥

X (2a + b)2 (b + 2c)2 ,

where p = a + b + c, q = a b + bc + ca. Since (2a + b)(b + 2c) = p b + q + 3ac, we have X X X (2a + b)2 (b + 2c)2 = p2 a2 + 3q2 + 9 a2 b2 + 2p2 q + 18a bc p + 6q2 = p2 (p2 − 2q) + 9q2 + 9(q2 − 2a bc p) + 2p2 q + 18a bc p = p4 + 18q2 , and the inequality becomes 9p2 q ≥ p4 + 18q2 , (p2 − 3q)(6q − p2 ) ≥ 0. The last inequality is true since p2 − 3q ≥ 0 and 6q − p2 = 4(a b + bc + ca) − a2 − b2 − c 2 ≥ 0.

106

Vasile Cîrtoaje

Case 2: 4(a b + bc + ca < a2 + b2 + c 2 . Assume that a = max{a, b, c}. From a2 − 4(b + c)a + (b + c)2 > 6bc > 0, we get a > (2 +

p 3)(b + c) > 2(b + c).

Since 1 1 1 1 1 2 + + > + ≥ , 2 2 2 2 2 (2a + b) (2b + c) (2c + a) (2b + c) (2c + a) (2b + c)(2c + a) it suffices to show that 2 1 ≥ . (2b + c)(2c + a) a b + bc + ca This is equivalent to the obvious inequality c(a − 2b − 2c) ≥ 0. The proof is completed. The equality holds for a = b = c. Conjecture. Let a, b, c be nonnegative real numbers, no two of which are zero. If k > 0, then 1 1 9 1 + + ≥ ; (a) (a + b)(ka + b) (b + c)(k b + c) (c + a)(kc + a) 2(k + 1)(a b + bc + ca) (b)

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

For k = 1, from (a) and (b), we get the well-known inequality (Iran 96) 1 1 1 9 + + ≥ . 2 2 2 (a + b) (b + c) (c + a) 4(a b + bc + ca)

P 1.76. If a, b, c are nonnegative real numbers, then a4 + b4 + c 4 + 15(a3 b + b3 c + c 3 a) ≥

47 2 2 (a b + b2 c 2 + c 2 a2 ). 4 (Vasile Cîrtoaje, 2011)

Cyclic Inequalities

107

Solution. Without loss of generality, assume that a = min{a, b, c}. There are two cases to consider: a ≤ b ≤ c and a ≤ c ≤ b. Case 1: a ≤ b ≤ c. For a = 0, the inequality is true because is equivalent to b4 + c 4 + 15b3 c − 

b−

47 2 2 b c ≥ 0, 4

c 2 2 (b + 16bc + 4c 2 ) ≥ 0. 2

Based on this result, it suffices to prove that a4 + 15(a3 b + c 3 a) ≥

47 2 2 a (b + c 2 ). 4

This inequality is true if a3 b + c 3 a ≥ a2 (b2 + c 2 ). Indeed, a2 b + c 3 − a(b2 + c 2 ) = c 2 (c − a) − a b(b − a) ≥ c 2 (b − a) − a b(b − a) = (c 2 − a b)(b − a) ≥ 0. Case 2: a ≤ c ≤ b. It suffices to show that a3 b + b3 c + c 3 a ≥ a2 b2 + b2 c 2 + c 2 a2 . Since a b3 + bc 3 + ca3 − (a3 b + b3 c + c 3 a) = (a + b + c)(a − b)(b − c)(c − a) ≤ 0, we have X

a3 b ≥

X X 1 X 3 1X ( a b+ a b3 ) = a b(a2 + b2 ) ≥ a2 b2 . 2 2

The equality holds for a = 0 and 2b = c (or any cyclic permutation).

P 1.77. If a, b, c are nonnegative real numbers such that a + b + c = 4, then a3 b + b3 c + c 3 a ≤ 27.

108

Vasile Cîrtoaje

Solution. Assume that a = max{a, b, c}. There are two possible cases: a ≥ b ≥ c and a ≥ c ≥ b. Case 1: a ≥ b ≥ c. Using the AM-GM inequality gives 3(a3 b + b3 c + c 3 a) ≤ 3a b(a2 + ac + c 2 ) ≤ 3a b(a + c)2 • ˜ a + 3b + (a + c) + (a + c) 4 = a · 3b · (a + c) · (a + c) ≤ 4  ‹4  ‹4 3a + 3b + 2c 3a + 3b + 3c = ≤ = 81. 4 4 Case 2: a ≥ c ≥ b. Since a b3 + bc 3 + ca3 − (a3 b + b3 c + c 3 a) = (a + b + c)(a − b)(b − c)(c − a) ≥ 0, it suffices to prove that a3 b + b3 c + c 3 a + (a b3 + bc 3 + ca3 ) ≤ 54. Indeed, X

a3 b +

X

a b3 ≤ (a b + bc + ca)(a2 + b2 + c 2 ) ≤

1 (a + b + c)4 = 32 < 54. 8

The equality holds for a = 3, b = 1 and c = 0 (or any cyclic permutation). Remark. The following sharper inequality holds (Michael Rozenberg). • If a, b, c are nonnegative real numbers such that a + b + c = 4, then a3 b + b3 c + c 3 a +

473 a bc ≤ 27, 64

with equality for a = b = c = 4/3, and also for a = 3, b = 1 and c = 0 (or any cyclic permutation). Assuming that a = min{a, b, c} and using the substitution b = a + p,

c = a + q,

p, q ≥ 0,

this inequality can be restated as Aa2 + Ba + C ≥ 0, where A = 217(p2 − pq + q2 ) ≥ 0. B = 68p3 − 269p2 q + 499pq2 + 68q3 ≥ 60p(p2 − 5pq + 8q2 ) ≥ 0, C = (p − 3q)2 (27p2 + 14pq + 3q2 ) ≥ 0.

Cyclic Inequalities

109

P 1.78. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 =

10 (a b + bc + ca). 3

Prove that a4 + b4 + c 4 ≥

82 3 (a b + b3 c + c 3 a). 27 (Vasile Cîrtoaje, 2011)

Solution (by Vo Quoc Ba Can). We see that the equality holds for a = 3, b = 1, c = 0, when a bc = 0. Since a4 + b4 + c 4 ≥ (a2 + b2 + c 2 )2 − 2(a b + bc + ca)2 82 = (a b + bc + ca)2 , 9 it suffices to show that 3(a b + bc + ca)2 ≥ a3 b + b3 c + c 3 a. In addition, since a b + bc + ca =

 ‹ 3(a2 + b2 + c 2 ) + 6(a b + bc + ca) a+b+c 2 =3 , 16 4

it suffices to show that 27



a+b+c 4

‹4

≥ a3 b + b3 c + c 3 a,

which is the inequality from the preceding P 1.77. The equality holds for a = 3b and c = 0 (or any cyclic permutation).

P 1.79. If a, b, c are positive real numbers, then a3 b3 c3 a+b+c + + ≥ . 2 2 2 2 2 2 2a + b 2b + c 2c + a 3 (Vasile Cîrtoaje, 2005)

110

Vasile Cîrtoaje

Solution. We write the inequality as       a b c a3 b3 c3 − + − + − ≥ 0, 2a2 + b2 3 2b2 + c 2 3 2c 2 + a2 3 a(a2 − b2 ) b(b2 − c 2 ) c(c 2 − a2 ) + + ≥ 0. 2a2 + b2 2b2 + c 2 2c 2 + a2 Taking into account that a(a2 − b2 ) b(a2 − b2 ) (a + b)(a − b)2 (a2 − a b + b2 ) − = ≥ 0, 2a2 + b2 2b2 + a2 (2a2 + b2 )(2b2 + a2 ) it suffices to show that b(a2 − b2 ) b(b2 − c 2 ) c(c 2 − a2 ) + + ≥ 0. 2b2 + a2 2b2 + c 2 2c 2 + a2 Since

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

the last inequality is equivalent to (c 2 − a2 )(c − b)[a2 (3b2 + bc + c 2 ) + 2b2 c(c − 2b)] ≥ 0.

(*)

Similarly, the desired inequality is true if (a2 − b2 )(a − c)[b2 (3c 2 + ca + a2 ) + 2c 2 a(a − 2c)] ≥ 0.

(**)

Without loss of generality, assume that c = max{a, b, c}. According to (*), the desired inequality is true if a2 (3b2 + bc + c 2 ) + 2b2 c(c − 2b) ≥ 0. p Notice that this inequality is true for for a ≥ b, and also for 2ac ≥ 3 b2 . Indeed, if a ≥ b, then a2 (3b2 + bc + c 2 ) + 2b2 c(c − 2b) ≥ b2 (3b2 + bc + c 2 ) + 2b2 c(c − 2b)

and if 2ac ≥

p 2 3 b , then

= 3b2 [b2 + c(c − b)] > 0,

a2 (3b2 + bc + c 2 ) + 2b2 c(c − 2b) ≥ =

3b4 (3b2 + bc + c 2 ) + 2b2 c(c − 2b) 4c 2

b2 (8c 4 − 16bc 3 + 3b2 c 2 + 3b3 c + 9b4 ) 4c 2

Cyclic Inequalities

111

b2 [2c(c + b)(2c − 3b)2 + 9b2 (c − b)2 + 3b3 c] > 0. 4c 2 p Consequently, we only need to consider that a < b ≤ c and 3 b2 > 2ac. According to (**), the desired inequality is true if =

b2 (3c 2 + ca + a2 ) + 2c 2 a(a − 2c) ≥ 0. We have b2 (3c 2 +ca+a2 )+2c 2 a(a−2c) >

4ac 2a2 c(2a + 5c) (3c 2 +ca+a2 )+2c 2 a(a−2c) = > 0. 3 3

This completes the proof. The equality occurs for a = b = c.

P 1.80. If a, b, c are positive real numbers, then a4 b4 c4 a+b+c + + ≥ . 3 3 3 3 3 3 a +b b +c c +a 3 (Vasile Cîrtoaje, 2005) Solution (by Vo Quoc Ba Can). Multiplying by a3 + b3 + c 3 , the inequality becomes X

a4 +

X a4 c 3 1 €X Š €X 3 Š ≥ a a . a3 + b3 2

By the Cauchy-Schwarz inequality, we have P P X a4 c 3 ( a2 b2 )2 ( a2 c 2 )2 ≥P =P . a3 + b3 c(a3 + b3 ) a(b3 + c 3 ) According to the inequality y x2 ≥x− , y 4 we have

x, y > 0,

P X ( a2 b2 )2 1X P ≥ a2 b2 − a(b3 + c 3 ). 4 a(b3 + c 3 )

Therefore, it suffices to show that X

a4 +

X

a2 b2 −

1X 1 €X Š €X 3 Š a(b3 + c 3 ) ≥ a a , 4 2

112

Vasile Cîrtoaje

which is equivalent to 2

X

a4 + 4

X

a2 b2 ≥ 3

X

a b(a2 + b2 ),

X [a4 + b4 + 4a2 b2 − 3a b(a2 + b2 )] ≥ 0, X (a − b)2 (a2 − a b + b2 ) ≥ 0. This completes the proof. The equality occurs for a = b = c.

P 1.81. If a, b, c are positive real numbers such that a bc = 1, then   ‹ a2 b2 c 2 b c a 3 + + + 4 2 + 2 + 2 ≥ 7(a2 + b2 + c 2 ); b c a a b c  3   ‹ a b3 c 3 b c a 8 + + + 5 3 + 3 + 3 ≥ 13(a3 + b3 + c 3 ). b c a a b c 

(a) (b)

(Vasile Cîrtoaje, 1992) Solution. (a) We use the AM-GM inequality, as follows

3

v  u ‹3 X b X  a2 X t 7 c a a2 c  a 3 +4 = 3 + + 3 · 2 ≥ 7 · b a2 b b2 c2 b b2 c

X a2

v X Xt a9 7 = 7 a2 . =7 b5 c 5 The equality holds for a = b = c = 1. (b) By the AM-GM inequality, we have

8

v  u ‹8 X b X  a3 X t a3 13 c a c  a 4 · +5 = 8 + + 4 ≥ 13 · 3 b a3 b b3 c3 b b3 c

X a3

v t a28 X 13 X = 13 = 13 a3 . b11 c 11 The equality holds for a = b = c = 1.

Cyclic Inequalities

113

P 1.82. If a, b, c are positive real numbers, then bc ca a2 + b2 + c 2 ab + + ≤ . b2 + bc + c 2 c 2 + ca + a2 a2 + a b + b2 a b + bc + ca (Tran Quoc Anh, 2007) Solution. Write the inequality as follows  X a2 ab − ≥ 0, a b + bc + ca b2 + bc + c 2 X ca(ca − b2 ) ≥ 0, b2 + bc + c 2  X X  ca(ca − b2 ) + ca ≥ ca, b2 + bc + c 2 X c 2 a(a + b + c) b2 + bc + c 2

≥

X

ca,

a b + bc + ca c2 a ≥ . 2 2 b + bc + c a+b+c By the Cauchy-Schwarz inequality, we have P P X ( ca)2 ca c2 a ≥P = P . 2 2 2 2 b + bc + c a(b + bc + c ) a X

The equality holds for a = b = c.

P 1.83. If a, b, c are positive real numbers, then a−b b−c c−a + + ≥ 0. b(2b + c) c(2c + a) a(2a + b) First Solution. Write the inequality as follows X ac(a − b)

≥ 0, 2b + c ˜ X • ac(a − b) + ac ≥ a b + bc + ca, 2b + c X ac a b + bc + ca ≥ . 2b + c a+b+c

114

Vasile Cîrtoaje

By the Cauchy-Schwarz inequality, we have X

P P ( ac)2 ( a b)2 ac P ≥P = . 2b + c ac(2b + c) 6a bc + a2 b

Thus, it suffices to prove that P

ab 1 P ≥P , a 6a bc + a2 b which is equivalent to X

a b2 ≥ 3a bc.

Clearly, the last inequality follows immediately from the AM-GM inequality. The equality holds for a = b = c. Second Solution. Without loss of generality, assume that a = max{a, b, c}. Case 1: a ≥ b ≥ c. Let x = a − b ≥ 0 and y = b − c ≥ 0. Write the inequality as follows y x+y x + − ≥ 0. b(2b + c) c(2c + a) a(2a + b) ˜ • ˜ 1 1 1 1 x − +y − ≥ 0, b(2b + c) a(2a + b) c(2c + a) a(2a + b) •

x

2(a2 − c 2 ) + a(b − c) 2(a2 − b2 ) + b(a − c) +y ≥ 0. a b(2a + b)(2b + c) ca(2a + b)(2c + a)

The last inequality is clearly true. Case 2: a ≥ c ≥ b. Let x = a − c ≥ 0 and y = c − b ≥ 0. We write the inequality as follows x+y y x − − ≥ 0. b(2b + c) c(2c + a) a(2a + b) • ˜ • ˜ 1 1 1 1 x − +y − ≥ 0, b(2b + c) a(2a + b) b(2b + c) c(2c + a) x

2(a2 − b2 ) + b(a − c) 2(c 2 − b2 ) + c(a − b) +y ≥ 0. a b(2a + b)(2b + c) bc(2b + c)(2c + a)

The last inequality is true, and the proof is completed.

Cyclic Inequalities

115

P 1.84. If a, b, c are positive real numbers, then (a)

a2 + 6bc b2 + 6ca c 2 + 6a b + + ≥ 7; a b + 2bc bc + 2ca ca + 2a b

(b)

a2 + 7bc b2 + 7ca c 2 + 7a b + + ≥ 12. a b + bc bc + ca ca + a b (Vasile Cîrtoaje, 2012)

Solution. (a) Write the inequality as follows X ac(a2 + 6bc)(b + 2a)(c + 2b) ≥ 7a bc(a + 2c)(b + 2a)(c + 2b), X X X € Š a2 b4 + a bc 72a bc + 4 a3 + 26 a2 b + 7 a b2 ≥ X X € Š ≥ 7a bc 9a bc + 4 a2 b + 2 a b2 , X X Š €X Š € X 2 a2 b4 − a bc a2 b + a bc 4 a3 + 9a bc − 7 a b2 ≥ 0. 2

X

Since 2

€X

a2 b4 − a bc

it suffices to show that 4

X

X

Š X a2 b = (a b2 − bc 2 )2 ≥ 0,

a3 + 9a bc − 7

X

a b2 ≥ 0.

Assume that a = min{a, b, c}. Using the substitution b = a + x,

c = a + y,

x, y ≥ 0,

we have 4

X

a3 + 9a bc − 7

X

a b2 = 5(x 2 − x y + y 2 )a + 4x 3 + 4 y 3 − 7x y 2 ≥ 0,

since 4x 3 + 4 y 3 = 4x 3 + 2 y 3 + 2 y 3 ≥ 3

Æ 3

p 3 4x 3 · 2 y 3 · 2 y 3 = 6 2 x y 2 ≥ 7x y 2 .

The equality holds for a = b = c. (b) Write the inequality as follows X ac(a2 + 7bc)(b + a)(c + b) ≥ 12a bc(a + c)(b + a)(c + b), X

X X X € Š a2 b4 + a bc 21a bc + a3 + 15 a2 b + 8 a b2 ≥ X X € Š ≥ 12a bc 2a bc + a2 b + a b2 ,

116

Vasile Cîrtoaje €X

a2 b4 − a bc

X

X X Š Š €X a3 − 3a bc + 4 a2 b − 4 a b2 ≥ 0. a2 b + a bc

Since X

a2 b4 − a bc

X

a2 b =

1X (a b2 − bc 2 )2 ≥ 0, 2

it suffices to show that X

a3 − 3a bc + 4

X

a2 b − 4

X

a b2 ≥ 0,

which is equivalent to X 1 (a + b + c) (a − b)2 − 4(a − b)(b − c)(c − a) ≥ 0. 2 Assume that a = min{a, b, c}. Making the substitution b = a + x, we have

c = a + y,

x, y ≥ 0,

X 1 (a + b + c) (a − b)2 − 4(a − b)(b − c)(c − a) = 2 = (x 2 − x y + y 2 )(3a + x + y) + 4x y(x − y) = 3(x 2 − x y + y 2 )a + x 3 + y 3 + 4x y(x − y) = 3(x 2 − x y + y 2 )a + x 3 + y(2x − y)2 ≥ 0.

The equality holds for a = b = c.

P 1.85. If a, b, c are positive real numbers, then (a)

bc ca a2 + b2 + c 2 ab + + ≤ ; 2b + c 2c + a 2a + b a+b+c

(b)

ab bc ca 3(a2 + b2 + c 2 ) + + ≤ ; b+c c+a a+b 2(a + b + c)

(c)

ab bc ca a2 + b2 + c 2 + + ≤ . 4b + 5c 4c + 5a 4a + 5b 3(a + b + c) (Vasile Cîrtoaje, 2012)

Cyclic Inequalities

117

Solution. (a) First Solution. Since ac 2a b =a− , 2b + c 2b + c we can write the inequality as X

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

By the Cauchy-Schwarz inequality, Pp 2 p p p X ac ac) ( ( a b + bc + ca)2 ≥P = . 2b + c 3(a + b + c) (2b + c) Then, it suffices to show that p p p ( a b + bc + ca)2 + 6(a2 + b2 + c 2 ) P ≥ a + b + c, 3 a which is equivalent to 3(a2 + b2 + c 2 ) + 2 Using the substitution x =

p

p

a bc

a, y =

€p

p

a+

b, z =

p p

b+

p Š c ≥ 5(a b + bc + ca).

c, the inequality can be restated as

3(x 4 + y 4 + z 4 ) + 2x yz(x + y + z) ≥ 5(x 2 y 2 + y 2 z 2 + z 2 x 2 ). We can get it by summing Schur’s inequality of degree four X 2(x 4 + y 4 + z 4 ) + 2x yz(x + y + z) ≥ 2 x y(x 2 + y 2 ) and x 4 + y 4 + z4 + 2

X

x y(x 2 + y 2 ) ≥ 5(x 2 y 2 + y 2 z 2 + z 2 x 2 ),

which is equivalent to the obvious inequality (x 4 + y 4 + z 4 − x 2 y 2 − y 2 z 2 − z 2 x 2 ) + 2

X

x y(x − y)2 ≥ 0.

The equality holds for a = b = c. Second Solution. By the Cauchy-Schwarz inequality, we have 1 1 a2 /b + b + c a2 + b2 + bc = ≤ = , 2b + c b+b+c (a + b + c)2 b(a + b + c)2 ab a(a2 + b2 + bc) ≤ , 2b + c (a + b + c)2

118

Vasile Cîrtoaje X

Since 3a bc ≤

P

ab ≤ 2b + c

P

a3 +

P

a b2 + 3a bc

(a + b + c)2

.

a2 b (by the AM-GM inequality), we get X

ab ≤ 2b + c

P

a3 +

P

a b2 +

P

a2 b

(a + b + c)2

=

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

Third Solution. Write the inequality as X a b(a + b + c) 2b + c

≤ a2 + b2 + c 2 .

Since 2a b(a + b + c) = (a2 + 2a b)(2b + c) − 2a b2 − a2 c, we can write the inequality as X a2 c X 2a b2 + + p ≥ 2q, 2b + c 2b + c where p = a2 + b2 + c 2 , q = a b + bc + ca (p ≥ q). By the Cauchy-Schwarz inequality, we have P X a b2 ( a b)2 q ≥P = 2b + c a(2b + c) 3 and

P X a2 c ( ac)2 q2 ≥P = . 2b + c p + 2q c(2b + c)

Thus, it suffices to show that q2 2q + + p ≥ 2q, 3 p + 2q which is equivalent to the obvious inequality (p − q)(3p + 5q) ≥ 0. (b) Write the inequality as X a b(a + b + c) 3 2 (a + b2 + c 2 ) ≥ . 2 b+c Since

a b(a + b + c) a2 b a2 c = + a b = a2 + a b − , b+c b+c b+c

Cyclic Inequalities

119

the inequality can be restated as X a2 c 1 + (a2 + b2 + c 2 ) ≥ a b + bc + ca. b+c 2 By the Cauchy-Schwarz inequality, P X a2 c ( ac)2 q2 ≥P , = b+c c(b + c) q + p where p = a2 + b2 + c 2 , q = a b + bc + ca (p ≥ q). Therefore, it suffices to prove that q2 p + ≥ q. p+q 2 Indeed,

q2 p p(p − q) + −q = ≥ 0. p+q 2 2(p + q)

The equality holds for a = b = c. (c) Since

4a b 5ac =a− , 4b + 5c 4b + 5c

we can write the inequality as 5

X

ac 4(a2 + b2 + c 2 ) + ≥ a + b + c. 4b + 5c 3(a + b + c)

By the Cauchy-Schwarz inequality, P X ac ( ac)2 (a b + bc + ca)2 ≥P . = 4b + 5c ac(4b + 5c) 12a bc + 5(a2 b + b2 c + c 2 a) Therefore, it suffices to show that 5(a b + bc + ca)2 4(a2 + b2 + c 2 ) + ≥ a + b + c. 12a bc + 5(a2 b + b2 c + c 2 a) 3(a + b + c) Due to homogeneity, we may assume that a + b + c = 3. Using the notation q = a b + bc + ca (q ≤ 3), this inequality becomes 5q2 4(9 − 2q) + ≥ 3. 2 2 2 5(a b + b c + c a + a bc) + 7a bc 9 According to the inequality (a) in P 1.28, we have a2 b + b2 c + c 2 a + a bc ≤ 4.

120

Vasile Cîrtoaje

On the other hand, from (a b + bc + ca)2 ≥ 3a bc(a + b + c), we get a bc ≤

q2 . 9

Thus, it suffices to prove that 4(9 − 2q) 5q2 + ≥ 3, 20 + 7q2 /9 9 which is equivalent to (q − 3)(14q2 − 75q + 135) ≤ 0. This is true since q ≤ 3 and 14q2 − 75q + 135 > 6(2q2 − 13q + 22) > 0. The equality holds for a = b = c.

P 1.86. If a, b, c are positive real numbers, then p p p (a) a b2 + 8c 2 + b c 2 + 8a2 + c a2 + 8b2 ≤ (a + b + c)2 ; p p p (b) a b2 + 3c 2 + b c 2 + 3a2 + c a2 + 3b2 ≤ a2 + b2 + c 2 + a b + bc + ca. (Vo Quoc Ba Can, 2007) Solution. (a) By the AM-GM inequality, we have p p (b2 + 8c 2 )(b + 2c)2 (b2 + 8c 2 ) + (b + 2c)2 b2 + 8c 2 = ≤ b + 2c 2(b + 2c) b2 + 2bc + 6c 2 3bc = b + 3c − , b + 2c b + 2c p 3a bc a b2 + 8c 2 ≤ a b + 3ac − , b + 2c X p X X 1 a b2 + 8c 2 ≤ 4 a b − 3a bc . b + 2c Therefore, it suffices to show that X 1 X €X Š2 a + 3a bc ≥4 a b. b + 2c =

Cyclic Inequalities

121

Since X

1 9 3 ≥P =P , b + 2c (b + 2c) a

it is enough to prove that €X Š €X Š €X Š3 ab . a a + 9a bc ≥ 4 This is just Shur’s inequality of degree three. The equality holds for a = b = c. (b) Similarly, we have p

p b2

+ 3c 2

=

(b2 + 3c 2 )(b + c)2 (b2 + 3c 2 ) + (b + c)2 ≤ b+c 2(b + c)

b2 + bc + 2c 2 2bc = b + 2c − , b+c b+c p 2a bc , a b2 + 3c 2 ≤ a b + 2ac − b+c X p X X 1 a b2 + 3c 2 ≤ 3 . a b − 2a bc b+c Therefore, it suffices to show that X X 1 €X Š2 ≥4 a b. a + 2a bc b+c =

Since X

1 9 9 = P , ≥P b+c (b + c) 2 a

it is enough to prove that €X Š3 €X Š €X Š a + 9a bc ≥ 4 a ab , which is just Shur’s inequality of degree three. The equality holds for a = b = c.

P 1.87. If a, b, c are positive real numbers, then (a)

(b)

1

1

1

s

3 ; a bc

1

1

1

s

1 . a bc

+ p + p ≥ a a + 2b b b + 2c c c + 2a p

+ p + p ≥ a a + 8b b b + 8c c c + 8a p

(Vasile Cîrtoaje, 2007)

122

Vasile Cîrtoaje

Solution. (a) Write the inequality as v Xt

Replacing a, b, c by

bc ≥ 1. 3a(a + 2b)

1 1 1 , , , respectively, the inequality can be restated as x y z X x ≥ 1. p 3z(2x + y)

Since Æ

3z(2x + y) ≤

it suffices to show that X

3z + (2x + y) , 2

x 1 ≥ . 2x + y + 3z 2

Indeed, using the Cauchy-Schwarz inequality gives P X X ( x)2 x 1 P ≥ = . 2x + y + 3z x(2x + y + 3z) 2 The equality holds for a = b = c. (b) Write the inequality as v Xt

Replacing a, b, c by

bc ≥ 1. a(a + 8b)

1 1 1 , , , respectively, the inequality becomes x 2 y 2 z2 x2

X z

p

8x 2 + y 2

≥ 1.

Applying the Cauchy-Schwarz inequality yields X

P ( x)2 ≥P p . p z 8x 2 + y 2 z 8x 2 + y 2 x2

Therefore, it suffices to show that X Æ z 8x 2 + y 2 ≤ (x + y + z)2 , which is just the inequality in P 1.86-(a). The equality holds for a = b = c.

Cyclic Inequalities

123

P 1.88. If a, b, c are positive real numbers, then b

a

c

+p +p ≤ p 5c + 4a 5a + 4b 5b + 4c

v ta + b + c 3

.

(Vasile Cîrtoaje, 2012) Solution. By the Cauchy-Schwarz inequality, we have X

a

‹2

p 5a + 4b

≤

X

It suffices to show that X and

 X a(4a + 4b + c) ‹ a . 4a + 4b + c 5a + 4b

a 1 ≤ 4a + 4b + c 3

X a(4a + 4b + c) 5a + 4b

≤ a + b + c.

The first is just the inequality in P 1.11, while the second is equivalent to ‹ X  4a + 4b + c ≥ 0, a 1− 5a + 4b X a(a − c) 5a + 4b

≥ 0,

X

a(a − c)(5b + 4c)(5c + 4a) ≥ 0, X X X a2 b2 + 4 a b3 ≥ 5a bc a.

The last inequality follows from the well-known inequality X X a2 b2 ≥ a bc a and the known inequality X

a b3 ≥ a bc

X

a,

which can be proved by the Cauchy-Schwarz inequality, as follows €X Š €X Š €X p Š2 €X Š2 c a b3 ≥ a b3 c = a bc b . The equality holds for a = b = c.

124

Vasile Cîrtoaje

P 1.89. If a, b, c are positive real numbers, then p

(a)

p

(b)

a a+b

a a+b

+p

+p

b b+c

b b+c

+p

+p

p

c

≥

c+a

c c+a

≥

p

a+

b+

p 2

p

c

;

v t 4 27(a b + bc + ca) 4

.

(Lev Buchovsky - 1995, Pham Huu Duc - 2007) Solution. (a) By squaring, the inequality becomes X a2 X Xp ab 1X a b. +2 ≥ a+ p a+b (a + b)(b + c) 2 Since the sequences 1

§ p and

§

a+b

,

p

1 b+c

,

p

bc

ab

1

ª

c+a ª

ca

, , p p c+a a+b b+c are always reversely ordered, we can apply the rearrangement inequality. For example, assuming that a ≥ b ≥ c, we have p

p and p hence

1

ab

a+b ab

a+b +p

≤p

≥p 1

1 c+a ca

c+a ·p

≤p

≥p

ca

1 b+c bc

b+c

+p

,

1

·p

bc

≤ c+a c+a a+b a+b b+c b+c ca 1 bc 1 ab 1 ·p +p ·p +p ·p , ≤p c+a c+a a+b b+c b+c a+b X ab X ab ≤ . p a+b (a + b)(b + c) p

·p

1

Therefore, it suffices to show that X a2 X ab Xp 1X +2 ≥ a+ a b. a+b a+b 2 Since

X a2 X ab X + = a, a+b a+b

Cyclic Inequalities

125

the inequality becomes in succession X ab Xp 1X a b, ≥ a+ a+b 2 X a + b X 2a b Xp + ≥2 a b, 2 a+b v ‚v Œ X t a + b t 2a b 2 − ≥ 0. 2 a+b

X

a+

The equality holds for a = b = c. (b) By Hölder’s inequality, we have X ‹2 X €X Š3 a a(a + b) ≥ a . p a+b Thus, it suffices to show that X ŠÆ €X Š3 3 €X a ≥ a2 + ab 3(a b + bc + ca), 2 which is equivalent to 2p3 + q3 ≥ 3p2 q, where p = a + b + c and q = have

p

3(a b + bc + ca). Clearly, by the AM-GM inequality, we

2p3 + q3 ≥ 3

Æ 3

p6 q3 = 3p2 q.

The equality holds for a = b = c.

P 1.90. If a, b, c are nonnegative real numbers such that a + b + c = 3, then p p p 3a + b2 + 3b + c 2 + 3c + a2 ≥ 6. First Solution. Assume that a = max{a, b, c}. We can get the desired inequality by summing the inequalities p p p 3b + c 2 + 3c + a2 ≥ 3a + c 2 + b + c, p p 3a + b2 + 3a + c 2 ≥ 2a + b + c. By squaring two times, the first inequality becomes in succession p Æ (3b + c 2 )(3c + a2 ) ≥ (b + c) 3a + c 2 ,

126

Vasile Cîrtoaje [b(a + b + c) + c 2 ][c(a + b + c) + a2 ] ≥ (b + c)2 [a(a + b + c) + c 2 ], b(a − b)(a − c)(a + b + c) ≥ 0.

Similarly, the second inequality becomes Æ (3a + b2 )(3a + c 2 ) ≥ (a + b)(a + c), [a(a + b + c) + b2 ][a(a + b + c) + c 2 ] ≥ (a + b)2 (a + c)2 , a(a + b + c)(b − c)2 ≥ 0. The original inequality becomes an equality when a = b = c, and also when two of a, b, c are zero. Second Solution. Write the inequality as p p p p p p X + Y + Z ≤ A + B + C, where X = (b + c)2 ,

Y = (c + a)2 ,

A = 3a + b2 ,

B = 3b + c 2 ,

Z = (a + b)2 , C = 3c + a2 c.

According to Lemma from the proof of P 2.9 in Volume 2, since X + Y + Z = A + B + C, it suffices to show that max{X , Y, Z} ≥ max{A, B, C},

min{X , Y, Z} ≤ min{A, B, C}.

To show that max{X , Y, Z} ≥ max{A, B, C}, we assume that a = min{a, b, c}, when max{X , Y, Z} = X and A − X = (a2 − c 2 ) + b(a − c) + c(a − b) ≤ 0, B − X = b(a − c) ≤ 0, C − X = (a2 − b2 ) + c(a − b) ≤ 0. Similarly, to show that min{X , Y, Z} ≤ min{A, B, C}, we assume that a = max{a, b, c}, when min{X , Y, Z} = X and A − X = (a2 − c 2 ) + b(a − c) + c(a − b) ≥ 0, B − X = b(a − c) ≥ 0, C − X = (a2 − b2 ) + c(a − b) ≥ 0.

Cyclic Inequalities

127

P 1.91. If a, b, c are nonnegative real numbers, then p p p a2 + b2 + 2bc + b2 + c 2 + 2ca + c 2 + a2 + 2a b ≥ 2(a + b + c). (Vasile Cîrtoaje, 2012) First Solution (by Nguyen Van Quy). Assume that a = max{a, b, c}. We can get the desired inequality by summing the inequalities p p p a2 + b2 + 2bc + b2 + c 2 + 2ca ≥ a2 + b2 + 2ca + b + c, p p c 2 + a2 + 2a b + a2 + b2 + 2ca ≥ 2a + b + c. By squaring two times, the first inequality becomes p Æ (a2 + b2 + 2bc)(b2 + c 2 + 2ca) ≥ (b + c) a2 + b2 + 2ca, c(a − b)(a2 − c 2 ) ≥ 0. Similarly, the second inequality becomes Æ (c 2 + a2 + 2a b)(a2 + b2 + 2ca) ≥ (a + b)(a + c), a(b + c)(b − c)2 ≥ 0. The original inequality becomes an equality when a = b = c, and also when two of a, b, c are zero. Second Solution. Let {x, y, z} be a permutation of {a b, bc, ca}. We will prove that p p Æ 2(a + b + c) ≤ b2 + c 2 + 2x + c 2 + a2 + 2 y + a2 + b2 + 2z. Due to symmetry, assume that a ≥ b ≥ c. Using the substitution X = a2 + b2 + 2a b, A = b2 + c 2 + 2x,

Y = c 2 + a2 + 2ca, B = c 2 + a2 + 2 y,

Z = b2 + c 2 + 2bc, C = a2 + b2 + 2z,

we can write the inequality as p p p p p p X + Y + Z ≤ A + B + C. Since X + Y + Z = A + B + C, X ≥ Y ≥ Z and X ≥ max{A, B, C},

Z ≤ min{A, B, C},

the conclusion follow by Lemma from the proof of P 2.9 in Volume 2.

128

Vasile Cîrtoaje

P 1.92. If a, b, c are nonnegative real numbers, then p

a2 + b2 + 7bc +

p

b2 + c 2 + 7ca +

p

c 2 + a2 + 7a b ≥ 3

Æ

3(a b + bc + ca). (Vasile Cîrtoaje, 2012)

Solution. We will show that p p Æ Æ b2 + c 2 + 7a x + c 2 + a2 + 7b y + a2 + b2 + 7cz ≥ 3 3(a b + bc + ca), where (x, y, z) is a permutation of (a, b, c) such that x 6= a, y 6= b, z 6= c. For x = c, y = a and z = b, we get the original inequality. Due to symmetry, without loss of generality, assume that a ≥ b ≥ c. In addition, let us denote X = a2 + c 2 + 7a b,

Y = a2 + b2 + 7ac,

Z = b2 + c 2 + 7bc,

A = b2 + c 2 + 7a x,

B = c 2 + a2 + 7b y,

C = a2 + b2 + 7cz.

Since X + Y + Z = A + B + C, X ≥ Y ≥ Z and X ≥ max{A, B, C},

Z ≤ min{A, B, C},

according to Lemma from the proof of P 2.9 in Volume 2, we have p

X+

p

Y+

p

Z≤

p p p A + B + C.

Therefore, it suffices to prove that p

X+

p

Y+

p

Æ Z ≥ 3 3(a b + bc + ca)

for a ≥ b ≥ c. Rewrite this inequality as p

a2 + c 2 + 7a b +

p

a2 + b2 + 7ac ≥ 3

Æ

3(a b + bc + ca) −

p

b2 + c 2 + 7bc,

and consider the non-trivial case b + c > 0. By squaring, the inequality can be restated as p p a2 + E + 3 3F ≥ 10a(b + c) + 17bc, where E = (a2 + c 2 + 7a b)(a2 + b2 + 7ac) = a4 + 7(b + c)a3 + (b2 + c 2 + 49bc)a2 + 7(b3 + c 3 )a + b2 c 2

Cyclic Inequalities

129

and F = (a b + bc + ca)(b2 + c 2 + 7bc). Due to homogeneity, we may assume that b + c = 1. Let us denote x = bc. We need to 1 1 show that f (x) ≥ 0 for 0 ≤ x ≤ and a ≥ , where 4 2 Æ Æ f (x) = a2 − 10a − 17x + g(x) + 3 3h(x), with g(x) = a4 + 7a3 + (1 + 47x)a2 + 7(1 − 3x)a + x 2 = x 2 + a(47a − 21)x + a4 + 7a3 + a2 + 7a, h(x) = (a + x)(1 + 5x) = 5x 2 + (5a + 1)x + a. We have p g0 3 3h0 f (x) = −17 + p + p 2 g 2 h p 2x + a(47a − 21) 3 3(10x + 5a + 1) = −17 + + , p p 2 g(x) 2 h(x) 0

p 2g 00 g − (g 0 )2 3 3[2h00 h − (h0 )2 ] f (x) = + p p 4g g 4h h p a(28 − 45a)(7a − 1)2 3 3(5a − 1)2 = − . p p 4g g 4h h 00

We will show that g ≥ 3h. Since 0 ≤ x ≤

1 1 and a ≥ , we have 4 2

g − 3h = −14x 2 + (47a2 − 36a − 3)x + a4 + 7a3 + a2 + 4a 7 ≥ − + (47a2 − 36a − 3)x + a4 + 7a3 + a2 + 4a. 8 For the non-trivial case 47a2 − 36a − 3 < 0, we get 7 47a2 − 36a − 3 g − 3h ≥ − + + a4 + 7a3 + a2 + 4a 8 4 (2a − 1)(4a3 + 30a2 + 66a + 13) = ≥ 0. 8

130

Vasile Cîrtoaje

We will prove now that f 00 (x) < 0. This is clearly true for a ≥ 1 28 ≤a≤ , we have 2 45 f 00 (x) ≤

28 . Otherwise, for 45

a(28 − 45a)(7a − 1)2 − 27(5a − 1)2 < 0, p 4g g

since  ‹ 45 a(28 − 45a)(7a − 1)2 − 27(5a − 1)2 < 28 − (7a − 1)2 − 27(5a − 1)2 2 27(1 − 3a)(17a − 3) 27 (7a − 1)2 − 27(5a − 1)2 = < 0. 4 4  ‹ 1 Since f is concave, it suffices to show that f (0) ≥ 0 and f ≥ 0. 4 From Š p p p € p p f (0) = a a a − 10 a + 3 3 + a3 + 7a2 + a + 7 ,
a(4a2 + 3a b − 5bc) > a(4c 2 + 3b2 − 5bc) > 0, C > a(3c 2 + 2ca + a2 − 5bc) > a(3c 2 − 3ca + a2 ) > 0, A + B > 4a3 + 5b3 + c 3 + 3a2 b + 2bc 2 − 10a bc p p 3 ≥ 3 4a3 · 5b3 · c 3 + 2 3a2 b · 2bc 2 − 10a bc p p 3 = (3 20 + 2 6 − 10)a bc > 0, B + C > a3 + 4b3 + 5c 3 + 3b2 c + 2ca2 − 10a bc p p 3 ≥ 3 a3 · 4b3 · 5c 3 + 2 3b2 c · 2ca2 − 10a bc p p 3 = (3 20 + 2 6 − 10)a bc > 0. If a ≥ b ≥ c, then X A(a − b)2 ≥ B(b − c)2 + C(a − c)2 ≥ (B + C)(b − c)2 ≥ 0. If a ≥ c ≥ b, then X A(a − b)2 ≥ A(a − b)2 + B(c − b)2 ≥ (A + B)(c − b)2 ≥ 0. The equality holds for a = b = c.

Cyclic Inequalities

133

P 1.95. If a, b, c are real numbers, then (a2 + b2 + c 2 )2 ≥ 3(a3 b + b3 c + c 3 a). (Vasile Cîrtoaje, 1992) First Solution. Write the inequality as E1 − 2E2 ≥ 0, where E1 = a3 (a − b) + b3 (b − c) + c 3 (c − a), E2 = a2 b(a − b) + b2 c(b − c) + c 2 a(c − a). Using the substitution b = a + p,

c = a + q,

we have E1 = a3 (a − b) + b3 [(b − a) + (a − c)] + c 3 (c − a) = (a − b)2 (a2 + a b + b2 ) + (a − c)(b − c)(b2 + bc + c 2 ) = p2 (a2 + a b + b2 ) − q(p − q)(b2 + bc + c 2 ) = 3(p2 − pq + q2 )a2 + 3(p3 − p2 q + q3 )a + p4 − p3 q + q4 and E2 = a2 b(a − b) + b2 c[(b − a) + (a − c)] + c 2 a(c − a) = (a − b)b(a2 − bc) + (a − c)c(b2 − ca) = p b(bc − a2 ) + qc(ca − b2 ) = (p2 − pq + q2 )a2 + (p3 + p2 q − 2pq2 + q3 )a + p3 q − p2 q2 . Thus, the inequality can be rewritten as Aa2 + Ba + C ≥ 0, where A = p2 − pq + q2 , B = p3 − 5p2 q + 4pq2 + q3 , C = p4 − 3p3 q + 2p2 q2 + q4 . For the non-trivial case A > 0, it is enough to show that δ ≤ 0, where δ is the discriminant of the quadratic function Aa2 + Ba + C. Indeed, we have δ = B 2 − 4AC = −3(p6 − 2p5 q − 3p4 q2 + 6p3 q3 + 2p2 q4 − 4pq5 + q6 = −3(p3 − p2 q − 2pq2 + q3 )2 ≤ 0.

134

Vasile Cîrtoaje

The equality holds for a = b = c, and also for a sin2 4π 7

=

b sin2 2π 7

c

=

sin2

π 7

(or any cyclic permutation). Second Solution. Let us denote x = a2 − a b + b2 , y = b2 − bc + c 2 , z = c 2 − ca + a2 . We have x 2 + y 2 + z2 =

X

a4 + 2

X

and x y + yz + z x =

a2 b2 − 2

X

X

a3 b

a3 b.

From the known inequality x 2 + y 2 + z 2 ≥ x y + yz + z x, the desired inequality follows. Third Solution. Let us denote x = a(a − 2b − c), y = b(b − 2c − a), z = c(c − 2a − b). We have x 2 + y 2 + z2 =

X

a4 + 5

and x y + yz + z x = 3

X

X

a2 b2 + 4a bc

a2 b2 + 4a bc

X

X

a−4

a−

X

X

a3 b − 2

a3 b − 2

From the known inequality x 2 + y 2 + z 2 ≥ x y + yz + z x, the desired inequality follows. Remark 1. Using the formula 4A(Aa2 + Ba + C) = (2Aa + B)2 − δ,

X

X

a b3 .

a b3

Cyclic Inequalities

135

from the first solution, we can deduce the following identity (a2 + b2 + c 2 )2 − 3(a3 b + b3 c + c 3 a) =

(A1 − 5B1 + 4C1 )2 + 3(A1 − B1 − 2C1 + 2D1 )2 , 4E1

where A1 = a 3 + b 3 + c 3 ,

B1 = a2 b + b2 c + c 2 a,

C1 = a b2 + bc 2 + ca2 ,

D1 = 3a bc,

E1 = a2 + b2 + c 2 − a b − bc − ca. Remark 2. Since x 2 + y 2 + z 2 − x y − yz − z x =

1X (x − y)2 , 2

from the second or third solution, we can deduce the following identity (a2 + b2 + c 2 )2 − 3(a3 b + b3 c + c 3 a) =

1X 2 (a − b2 − a b + 2bc − ca)2 . 2

In addition, we have also (a2 + b2 + c 2 )2 − 3(a3 b + b3 c + c 3 a) =

1X (2a2 − b2 − c 2 − 3a b + 3bc)2 6

and (a2 + b2 + c 2 )2 − 3(a3 b + b3 c + c 3 a) = =

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

P 1.96. If a, b, c are real numbers, then a4 + b4 + c 4 + a b3 + bc 3 + ca3 ≥ 2(a3 b + b3 c + c 3 a). (Vasile Cîrtoaje, 1992) First Solution. Making the substitution b = a + p,

c = a + q,

the inequality turns into Aa2 + Ba + C ≥ 0, where A = 3(p2 − pq + q2 ),

B = 3(p3 − 2p2 q + pq2 + q3 ),

C = p4 − 2p3 q + pq3 + q4 .

136

Vasile Cîrtoaje

Since the discriminant of the quadratic Aa2 + Ba + C is δ = B 2 − 4AC = −3(p6 − 6p4 q + 2p3 q3 + 9p2 q4 − 6pq5 + q6 ) = −3(p3 − 3pq2 + q3 )2 ≤ 0, the conclusion follows. The equality holds for a = b = c, and also for b c a = π = 7π sin 9 sin 9 sin 13π 9 (or any cyclic permutation). Second Solution. Let us denote x = a(a − b), y = b(b − c), z = c(c − a). We have x 2 + y 2 + z2 =

X

a4 +

and x y + yz + z x =

X

X

a2 b2 − 2

a2 b2 −

X

X

a3 b

a b3 .

From the known inequality x 2 + y 2 + z 2 ≥ x y + yz + z x, the desired inequality follows. Third Solution. Let us denote x = a2 + bc + ca, y = b2 + ca + a b, z = c 2 + a b + bc. We have x 2 + y 2 + z2 =

X

and x y + yz + z x = 2

a4 + 2

X

X

a2 b2 + 4a bc

a2 b2 + 4a bc

X

a+2

X

a+2

X

a3 b +

From the known inequality x 2 + y 2 + z 2 ≥ x y + yz + z x, the desired inequality follows.

X

a b3

X

a b3 .

Cyclic Inequalities

137

Remark 1. The inequality is more interesting in the case a bc < 0. If a, b, c are positive, then the inequality is less sharp than the inequality in P 1.95, because it can be obtained by adding the last inequality to a b(a − b)2 + bc(b − c)2 + ca(c − a)2 ≥ 0. In addition, if a, b, c are positive, then the inequality 3(a4 + b4 + c 4 ) + 4(a b3 + bc 3 + ca3 ) ≥ 7(a3 b + b3 c + c 3 a) is a refinement of the inequality in P 1.96. To prove this inequality, we consider a = min{a, b, c} and use the substitution b = a + p and c = a + q (a > 0, p ≥ 0, q ≥ 0). Since X X X a4 − a3 b = a3 (a − b) = 3(p2 − pq + q2 )a2 + 3(p3 − p2 q + q3 )a + p4 − p3 q + q4 and X

a b3 −

X

a3 b = (a + b + c)(a − b)(b − c)(c − a) = pq(q − p)(3a + p + q),

the inequality becomes Aa2 + Ba + C ≥ 0, where A = 9(p2 − pq + q2 ),

B = 3(3p3 − 7p2 q + 4pq2 + 3q3 ),

C = 3p4 − 7p3 q + 4pq3 + 3q4 . This inequality is true because A ≥ 0, B = p(3p − 4q)2 + q(p − 3q)2 + 2pq(p + q) ≥ 0 and

 ‹ 13q 2 11 4 3C = p(p + q)(3p − 5q)2 + 5q2 p − + q ≥ 0. 10 20

Remark 2. Using the formula 4A(Aa2 + Ba + C) = (2Aa + B)2 − δ, from the first solution, we can deduce the following identity X

a4 +

X

a b3 − 2

X

a3 b =

(A1 − 3C1 + 2D1 )2 + 3(A1 − 2B1 + C1 )2 , 4E1

138

Vasile Cîrtoaje

where A1 = a 3 + b 3 + c 3 ,

B1 = a2 b + b2 c + c 2 a,

C1 = a b2 + bc 2 + ca2 ,

D1 = 3a bc,

E1 = a2 + b2 + c 2 − a b − bc − ca. Remark 3. Since x 2 + y 2 + z 2 − x y − yz − z x =

1X (x − y)2 , 2

from the second or third solution, we can deduce the following identity X

a4 +

X

a b3 − 2

X

a3 b =

1X 2 (a − b2 − a b + bc)2 . 2

In addition, we have also X

a4 +

X

a b3 − 2

and

X

X =

a3 b =

a4 +

1X (2a2 − b2 − c 2 − 2a b + bc + ca)2 6

X

a b3 − 2

X

a3 b =

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

Remark 4. The inequalities in P 1.95 and P 1.96 are particular cases of the following more general statement (Vasile Cîrtoaje, 2007). • Let f4 (a, b, c) =

X

a4 + A

X

a2 b2 + Ba bc

X

a+C

X

a3 b + D

X

a b3 ,

where A, B, C, D are real constants such that 1 + A + B + C + D = 0,

3(1 + A) ≥ C 2 + C D + D2 .

If a, b, c are real numbers, then f4 (a, b, c) ≥ 0. Notice that  ‹ C−D 2 4 4 2 f4 (a, b, c) = (U + V + C + D) + 3 U − V + + (3 + 3A − C 2 − C D − D2 ), S 3 3 where X X S= a2 b2 − a2 bc,

U=

X Š 1 €X 3 a b− a2 bc , S

V=

1 €X 3 X 2 Š ab − a bc . S

Cyclic Inequalities

139

Also, for the main case 3(1+A) = C 2 +C D+D2 , the inequality f4 (a, b, c) ≥ 0 is equivalent to each of the following two inequalities X [2a2 − b2 − c 2 + C a b − (C + D)bc + Dca]2 ≥ 0, X [3b2 − 3c 2 + (C + 2D)a b + (C − D)bc − (2C + D)ca]2 ≥ 0.

P 1.97. If a, b, c are positive real numbers, then a2 b2 c2 + + ≥ 1. a b + 2c 2 bc + 2a2 ca + 2b2 Solution. By the Cauchy-Schwarz inequality, we have X

P P ( a2 )2 ( a2 )2 a2 P . ≥P =P a b + 2c 2 a2 (a b + 2c 2 ) a3 b + 2 a2 b2

Therefore, it suffices to show that X X X ( a2 )2 ≥ 2 a2 b2 + a3 b. We get this inequality by summing the known inequality X 2 X 2 2 ( a ) ≥2 a2 b2 3 and

1 X 2 2 X 3 ( a ) ≥ a b, 3

which is just the inequality in P 1.95. The equality holds for a = b = c = 1.

P 1.98. If a, b, c are positive real numbers such that a + b + c = 3, then a b c 3 + + ≥ . a b + 1 bc + 1 ca + 1 2

140

Vasile Cîrtoaje

Solution. We use the following hint a2 b a =a− , ab + 1 ab + 1

b b2 c = b− , bc + 1 bc + 1

c c2 a =c− , ca + 1 ca + 1

which transforms the desired inequality into b2 c c2 a 3 a2 b + + ≤ . a b + 1 bc + 1 ca + 1 2 By the AM-GM inequality, we have p p a b + 1 ≥ 2 a b, bc + 1 ≥ 2 bc,

p ca + 1 ≥ 2 ca.

Consequently, it suffices to show that b2 c a2 b c2 a 3 p + p + p ≤ , 2 2 a b 2 bc 2 ca which is equivalent to

p p a b + b bc + c ca ≤ 3, p p p 3(a a b + b bc + c ca) ≤ (a + b + c)2 . a

p

p p p Replacing a, b, c by a, b, c, respectively, we get the inequality in P 1.95. The equality holds for a = b = c = 1.

P 1.99. If a, b, c are positive real numbers such that a + b + c = 3, then a b c 3 + + ≤ . 3a + b2 3b + c 2 3c + a2 2 (Vasile Cîrtoaje, 2007) Solution. Since a 1 b2 = − , 3a + b2 3 3(3a + b2 )

b 1 c2 = − , 3b + c 2 3 3(3b + c 2 )

c 1 a2 = − , 3c + a2 3 3(3c + a)

the desired inequality can be rewritten as b2 c2 a2 3 + + ≥ . 2 2 2 3a + b 3b + c 3c + a 2 By the Cauchy-Schwarz inequality, we have P P X b2 ( a2 )2 ( b2 )2 P P ≥P =P 3a + b2 b2 (3a + b2 ) a4 + ( a)( a b2 )

Cyclic Inequalities

141

P ( a2 )2 P P P P . ≥ P =P ( a2 )2 + a b3 a4 + a2 b2 + a bc a + a b3 P ( a2 )2

Thus, it is enough to show that X X ( a2 )2 ≥ 3 a b3 . Taking into account the inequality in P 1.95, the conclusion follows. The equality holds for a = b = c = 1.

P 1.100. If a, b, c are positive real numbers such that a + b + c = 3, then b2

a b c 3 + 2 + 2 ≥ . +c c +a a +b 2 (Pham Kim Hung, 2007)

Solution. By the Cauchy-Schwarz inequality, we have P P 3 P X a ( a3/2 )2 a + 2 a3/2 b3/2 P ≥P = P . b2 + c a2 (b2 + c) a2 b2 + a b2 Thus, it is enough to show that X X X X 2 a3 + 4 a3/2 b3/2 ≥ 3 a2 b2 + 3 a b2 , which is equivalent to the homogeneous inequality X X X X X X X 2( a)( a3 ) + 4( a)( a3/2 b3/2 ) ≥ 9 a2 b2 + 3( a)( a b2 ). In order to get a symmetric inequality, we use the inequality in P 1.95. We have X X X X X 3( a)( a b2 ) = 3 a2 b2 + 3a bc a+3 a b3 ≤3

X

a2 b2 + 3a bc

X

X X X X a+( a2 )2 = a4 + 5 a2 b2 + 3a bc a.

Therefore, it suffices to prove the symmetric inequality X X X X X X X X 2( a)( a3 ) + 4( a)( a3/2 b3/2 ) ≥ 9 a2 b2 + a4 + 5 a2 b2 + 3a bc a, which is equivalent to X X Xp X X a4 + 2 a b(a2 + b2 ) + 4a bc a b + 4A ≥ 14 a2 b2 + 3a bc a,

142

Vasile Cîrtoaje

where A=

X (a b)3/2 (a + b).

Since A≥2

X

a2 b2 ,

it suffices to prove that X X X X Xp a4 + 2 a b(a2 + b2 ) + 4a bc ab ≥ 6 a2 b2 + 3a bc a. According to Schur’s inequality of degree four, X X X a4 ≥ a b(a2 + b2 ) − a bc a, it is enough to show that X X X Xp ab ≥ 6 a2 b2 + 4a bc a. 3 a b(a2 + b2 ) + 4a bc Write this inequality as 3 X

X

a b(a − b)2 ≥ 2a bc

Xp p ( a − b)2 ,

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

We will prove the stronger inequality X p p p p a b( a − b)2 [( a + b)2 − c] ≥ 0, which is equivalent to X  pa − p b 2 p p p ( a + b − c) ≥ 0. p c Substituting x =

p

a, y =

p

b, z =

p

c, the inequality becomes

X  x − y 2 (x + y − z) ≥ 0. z Without loss of generality, assume that x ≥ y ≥ z. It suffices to show that  y − z 2 x Since

x −z ( y + z − x) + y 

‹2

(z + x − y) ≥ 0.

y −z x −z ≥ ≥ 0, y x

Cyclic Inequalities we have

143

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

( y + z − x) +



P 1.101. If a, b, c are nonnegative real numbers such that a + b + c = 3, then p p p p a a + b + b b + c + c c + a ≥ 3 2. (Hong Ge Chen, 2011) Solution. Using the substitution v v s ta + b x + y tb+c y +z c+a z+x = , = , = 2 2 2 2 2 2 gives v ta + b

v a+c tb+c x= + − ≥ 0, 2 2 2  x + y 2  x + z 2  y + z 2 x(x + y + z) − yz − . a= + = 2 2 2 2 In addition, from a + b + c = 3, we get s

x 2 + y 2 + z 2 + x y + yz + z x = 6. Let p = x + y + z, q = x y + yz + z x,

p2 − q = 6.

From 18 − 2p2 = 3(x 2 + y 2 + z 2 + x y + yz + z x) − 2p2 = x 2 + y 2 + z 2 − x y − yz − z x ≥ 0, it follows that p ≤ 3. The desired inequality is equivalent to X (x p − yz)(x + y) ≥ 12, p

X X y 2 z ≥ 12, (x 2 + x y) − 3x yz −

144

Vasile Cîrtoaje 6p +

X

X X yz 2 ≥ 3x yz + y 2z + yz 2 + 12, X 6p + yz 2 ≥ pq + 12.

Since €X

yz 2

Š €X Š €X Š2 y ≥ yz

(by the Cauchy-Schwarz inequality), it suffices to show that 6p +

q2 ≥ pq + 12. p

Indeed, 6p +

p2 (6 − q) + q2 − 12p (6 + q)(6 − q) + q2 − 12p 12(3 − p) q2 − pq −12 = = = ≥ 0. p p p p

The proof is completed. The equality holds for a = b = c = 1.

P 1.102. If a, b, c are positive real numbers such that a + b + c = 3, then a b c + + ≥ 1. 2b2 + c 2c 2 + a 2a2 + b (Vasile Cîrtoaje and Nguyen Van Quy, 2007) Solution. By the Cauchy-Schwarz inequality, we have X

Since


2 P p a a+c a ≥P . 2b2 + c a(a + c)(2b2 + c)

P p p a a + c ≥ 3 2 (see the preceding P 1.101), it suffices to prove that X a(a + c)(2b2 + c) ≤ 18,

which is equivalent to 2 2

X

X

a2 b2 + 6a bc +

a2 b2 + 3a bc +

X

ac(a + c) ≤ 18,

€X Š €X Š a a b ≤ 18.

Denoting q = a b + bc + ca and r = a bc (q ≤ 3, r ≤ 3), the inequality becomes 9(r + 2) ≥ 2q2 + 3q.

Cyclic Inequalities

145

This inequality is true for q ≤ 2, because 18 > 2q2 + 3q. Consider further the case 2 ≤ q ≤ 3. By Schur’s inequality of degree three, we have 9r ≥ 4pq − p3 = 12q − 27. Therefore, 9(r + 2) − (2q2 + 3q) ≥ 12q − 27 + 18 − (2q2 + 3q) = −2q2 + 9q − 9 = (3 − q)(2q − 3) ≥ 0. This completes the proof. The equality holds for a = b = c = 1.

P 1.103. If a, b, c are positive real numbers such that a2 + b2 + c 2 = 3, then a3 b3 c3 3 + + ≥ . 5 5 5 a+b b+c c+a 2 (Marin Bancos, 2010) Solution. Write the inequality as  X  a3 3 2 − a + ≥ 0, 5 a+b 2 X a2 b5 3 ≤ . 5 a+b 2 Since a + b5 ≥ 2 it suffices to show that

X

a b2

p

p

a b5 ,

a b ≤ 3.

p In addition, since 2 a b ≤ a + b, it suffices to prove that X X a2 b2 + a b3 ≤ 6. This is true since X

a2 b2 ≤

1 2 (a + b2 + c 2 )2 = 3, 3

and, according to P 1.95, X

a b3 ≤

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

1 2 (a + b2 + c 2 )2 = 3. 3

146

Vasile Cîrtoaje

P 1.104. If a, b, c are positive real numbers such that a2 + b2 + c 2 = 3, then b c 3 a + + ≥ . 1+ b 1+c 1+a 2 (Vasile Cîrtoaje, 2005) Solution. Let p = a + b + c,

q = a b + bc + ca,

p2 = 3 + 2q.

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

P ( a)2 3 + 2q a = ≥P . 1+ b p+q a(1 + b)

Thus, it suffices to prove that 6 + q ≥ 3p. Indeed, 2(6 + q − 3p) = 12 + (p2 − 3) − 6p = (p − 3)2 ≥ 0. The equality holds for a = b = c = 1. Second Solution. By the AM-GM inequality, we have X

X X a a(a + c) 4a(a + c) = ≥ 1+ b (1 + b)(a + c) [(1 + b) + (a + c)]2 P 2 P
4 a + ac 4(3 + q) 6 + 2p2 = = = . (1 + p)2 (1 + p)2 (1 + p)2

Therefore, it suffices to show that 6 + 2p2 3 ≥ , 2 (1 + p) 2 which is equivalent to (p − 3)2 ≥ 0.

P 1.105. If a, b, c are real numbers such that a2 + b2 + c 2 = 3, then a2 b + b2 c + c 2 a + 9 ≥ 4(a + b + c). (Vasile Cîrtoaje, 2007)

Cyclic Inequalities

147

First Solution (by Nguyen Van Quy). Since 2a2 b = a2 (b2 + 1) − a2 (b − 1)2 , we have 4

X

X X X a2 b = 2 a2 b2 + 2 a2 − 2 a2 (b − 1)2 X X €X Š2 X = a2 − a4 + 2 a2 − 2 c 2 (a − 1)2 X X = 15 − a4 − 2 c 2 (a − 1)2 .

Therefore, we can write the desired inequality as follows: X X X ” — 15 − a4 − 2 c 2 (a − 1)2 + 36 ≥ 16 a, X X (17 − 16a − a4 ) ≥ 2 c 2 (a − 1)2 , X X X (17 − 16a − a4 ) + 10 (a2 − 1) ≥ 2 c 2 (a − 1)2 , X X (7 − 16a + 10a2 − a4 ) ≥ 2 c 2 (a − 1)2 , X X (a − 1)2 (7 − 2a − a2 ) ≥ 2 c 2 (a − 1)2 , X (a − 1)2 (7 − 2a − a2 − 2c 2 ) ≥ 0. Since 7 − 2a − a2 − 2c 2 = (a − 1)2 + 2(3 − a2 − c 2 ) = (a − 1)2 + 2b2 ≥ 0, the conclusion follows. The equality holds for a = b = c = 1. Second Solution. Consider only the case where a, b, c are nonnegative numbers and a + b + c > 0. Multiplying both sides by a + b + c, the inequality can be restated as (a + b + c)(a2 b + b2 c + c 2 a) + 9(a + b + c) ≥ 4(a + b + c)2 . Using the known inequality 3

P

a2 b2 ≥

P

ab


2

and the inequality from P 1.95, namely

(a2 + b2 + c 2 )2 ≥ 3(a b3 + bc 3 + ca3 ), we have X €X Š €X Š €X Š €X Š X a a2 b = a2 ab + a2 b2 − a b3 X 1 €X Š2 a b − 3. ≥3 ab + 3

148

Vasile Cîrtoaje

Therefore, it suffices to show that 3 Setting

P

X

ab +

X €X Š2 1 €X Š2 ab − 3 + 9 a≥4 a . 3

a = p, which involves X

ab =

p2 − 3 , 2

the inequality becomes 3(p2 − 3) (p2 − 3)2 + − 3 + 9p ≥ 4p2 , 2 12 (p − 3)2 (p2 + 6p − 9) ≥ 0. The last inequality is true since p2 + 6p − 9 > 6p − 9 ≥ 6

p

p a2 + b2 + c 2 − 9 = 6 3 − 9 > 0.

P 1.106. If a, b, c are real numbers such that a2 + b2 + c 2 = 3, then (1 − a)(1 − a b) + (1 − b)(1 − bc) + (1 − c)(1 − ca) ≥ 0. (Vasile Cîrtoaje, 2007) Solution. Write the inequality as follows: X (1 − a)(a2 + b2 + c 2 − 3a b) ≥ 0, X X (a2 + b2 + c 2 − 3a b) − a(a2 + b2 + c 2 − 3a b) ≥ 0, X Š X X €X 3 a2 − ab − a(a − b)2 − a(c 2 − a b) ≥ 0, X 3X (a − b)2 − a(a − b)2 ≥ 0, 2 X (a − b)2 (3 − 2a) ≥ 0. Assume that a = max{a, b, c}. For 3 − 2a ≥ 0, the inequality is clearly true. Consider now that 3 − 2a < 0. Since (a − b)2 = [(a − c) + (c − b)]2 ≤ 2[(a − c)2 + (c − b)2 ],

Cyclic Inequalities

149

it suffices to show that 2[(a − c)2 + (c − b)2 ](3 − 2a) + (b − c)2 (3 − 2b) + (c − a)2 (3 − 2c) ≥ 0, which can be rewritten as (a − c)2 (9 − 4a − 2c) + (b − c)2 (9 − 4a − 2b) ≥ 0. This inequality is true since 9 > 4a+2c and 9 > 4a+2b. For instance, the last inequality holds if 81 > 4(2a + b)2 ; indeed, we have 81 − (2a + b)2 > 15 − (2a + b)2 = 5(a2 + b2 + c 2 ) − (2a + b)2 = (a − 2b)2 + c 2 ≥ 0. 4 The equality holds for a = b = c = 1. Remark. This inequality, written as X

a2 b + 3 ≥

X

a+

X

a b,

is sharper than the inequality in P 1.105, namely X

a2 b + 9 ≥ 4

X

a.

This claim is true if X

a+

X

ab − 3 ≥ 4

X

a − 9;

that is, q + 6 ≥ 3p, where p = a + b + c and q = a b + bc + ca. Indeed, we have q + 6 − 3p =

p2 − 3 (p − 3)2 + 6 − 3p = ≥ 0. 2 2

P 1.107. If a, b, c are positive real numbers such that a + b + c = a b + bc + ca, then

a2

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

150

Vasile Cîrtoaje

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

X 1 + b + c2 1 3 + a + b + c + a2 + b2 + c 2 ≤ = . a2 + b + 1 (a + b + c)2 (a + b + c)2

Therefore, it suffices to show that 3 + a + b + c ≤ 2(a b + bc + ca), which is equivalent to a + b + c ≥ 3. We can get this inequality from the known inequality (a + b + c)2 ≥ 3(a b + bc + ca). The equality holds for a = b = c = 1.

P 1.108. Let m > n > 0, and let a, b, c be positive real numbers such that a m+n + b m+n + c m+n = 3. Then,

am bm c m + n + n ≥ 3. bn c a (Vasile Cîrtoaje, 2005)

Solution. Making the substitution x=a and k =

m+n 2

,

y=b

m+n 2

, z=c

m+n 2

2n , 0 < k < 1, we have to show that x 2 + y 2 + z 2 = 3 yields m+n y 2−k z 2−k x 2−k + + k ≥ 3, yk zk x

or, equivalently, y2 x2 z2 + + ≥ 3. (x y)k ( yz)k (z x)k Applying Jensen’s inequality to the convex function f (u) =

1 , we get uk

y2 x 2 + y 2 + z2 x2 z2 + + ≥  2 k (x y)k ( yz)k (z x)k x · x y + y 2 · yz + z 2 · z x x 2 + y 2 + z2

Cyclic Inequalities

151

=

3k+1 . (x 3 y + y 3 z + z 3 x)k

Thus, it suffices to show that x 3 y + y 3 z + z 3 x ≤ 3. This is just the inequality in P 1.95. The equality holds for a = b = c = 1.

P 1.109. If a, b, c are positive real numbers, then  ‹ 1 1 1 1 1 1 1 1 1 (a) + + + + + ≥3 + + ; 4a 4b 4c a + b b + c c + a 3a + b 3b + c 3c + a  ‹ 1 1 1 1 1 1 1 1 1 (b) + + + + + ≥2 + + . 4a 4b 4c a + 3b b + 3c c + 3a 3a + b 3b + c 3c + a (Gabriel Dospinescu and Vasile Cîrtoaje, 2004) Solution. We will prove that the following more general inequalities hold for t ≥ 0:  3a+b  t 4a t 4b t 4c t 2a+2b t 2b+2c t 2c+2a t t 3b+c t 3c+a + + + + + ≥3 + + , 4a 4b 4c a+b b+c c+a 3a + b 3b + c 3c + a  3a+b  t 4a t 4b t 4c t a+3b t b+3c t c+3a t t 3b+c t 3c+a + + + + + ≥2 + + . 4a 4b 4c a + 3b b + 3c c + 3a 3a + b 3b + c 3c + a For t = 1, we get the desired inequalities. (a) Denoting the left hand side of the former inequality by f (t), the inequality becomes f (t) ≥ f (0). This is true if f 0 (t) ≥ 0 for t > 0. We have t f 0 (t) = t 4a + t 4b + t 4c + 2(t 2a+2b + t 2b+2c + t 2c+2a ) − 3(t 3a+b + t 3b+c + t 3c+a ). Using the substitution x = t a , y = t b , z = t c , the inequality f 0 (t) ≥ 0 turns into x 4 + y 4 + z 4 + 2(x 2 y 2 + y 2 z 2 + z 2 x 2 ) ≥ 3(x 3 y + y 3 z + z 3 x), which is just the the inequality in P 1.95. The equality holds for a = b = c. (b) Similarly, we have t f 0 (t) = t 4a + t 4b + t 4c + t a+3b + t b+3c + t c+3a − 2(t 3a+b + t 3b+c + t 3c+a ). Denoting x = t a , y = t b , z = t c , the inequality f 0 (t) ≥ 0 turns into x 4 + y 4 + z 4 + x y 3 + yz 3 + z x 3 ≥ 2(x 3 y + y 3 z + z 3 x), which is just the the inequality in P 1.96. The equality holds for a = b = c.

152

Vasile Cîrtoaje

P 1.110. If a, b, c are positive real numbers such that a6 + b6 + c 6 = 3, then a5 b5 c 5 + + ≥ 3. b c a (Tran Quoc Anh, 2007) Solution. By Hölder’s inequality, we have 

a5 b5 c 5 + + b c a

3 ≥

(a6 + b6 + c 6 )4 81 = 9 3 . 9 3 9 3 9 3 a b +b c +c a a b + b9 c 3 + c 9 a3

Therefore, it suffices to show that a9 b3 + b9 c 3 + c 9 a3 ≤ 3. This is equivalent to 3(a9 b3 + b9 c 3 + c 9 a3 ) ≤ (a6 + b6 + c 6 )2 , which is just the inequality in P 1.95. The equality holds for a = b = c.

P 1.111. If a, b, c are positive real numbers, then 1 1 1 1 + + ≤ . 2 2 2 (a + 2b + 3c) (b + 2c + 3a) (c + 2a + 3b) 4(a b + bc + ca) Solution. By the AM-GM inequality, we have (a + 2b + 3c)2 = [(a + c) + 2(b + c)]2 = (a + c)2 + 4(b + c)2 + 4(a + c)(b + c) ≥ 3(b + c)2 + 6(a + c)(b + c) = 3(b + c)(2a + b + 3c). Thus, it suffices to show that X

1 3 ≤ . (b + c)(2a + b + 3c) 4(a b + bc + ca)

Write this inequality as follows 3 X a b + bc + ca − ≥ 0, 4 (b + c)(2a + b + 3c) ˜ X• 2(a b + bc + ca) 3 1− ≥ , (b + c)(2a + b + 3c) 2

Cyclic Inequalities

153 3 (b + c)2 + 2c 2 ≥ , (b + c)(2a + b + 3c) 2 X X b+c 2c 2 3 + ≥ . 2a + b + 3c (b + c)(2a + b + 3c) 2 X

Applying the Cauchy-Schwarz inequality, we get P X [ (b + c)]2 b+c ≥P =1 2a + b + 3c (b + c)(2a + b + 3c) and X

P ( c)2 1 c2 ≥P = , (b + c)(2a + b + 3c) (b + c)(2a + b + 3c) 4

from where the conclusion follows. The equality holds for a = b = c.

P 1.112. If a, b, c ∈ [0, 1], then a(1 − b2 ) + b(1 − c 2 ) + c(1 − a2 ) ≤

5 . 4 (Ji Chen, 2007)

Proof. We use the substitution a = 1 − x,

b = 1 − y,

c = 1 − z,

where x, y, z ∈ [0, 1]. Since X X X a(1 − b2 ) = y(1 − x)(2 − y) = y(2 − 2x − y + x y) X €X Š2 X =2 x− x + x y 2, the inequality can be written as 2

X

x−

€X Š2 X 5 x + x y2 ≤ . 4

According to the known inequality in P 1.2, we have X

x y2 ≤

4 €X Š3 x . 27

Thus, it suffices to prove the following inequality 2t − t 2 +

4 3 5 t ≤ , 27 4

154

Vasile Cîrtoaje

where t = x + y + z ≤ 3. This inequality is equivalent to 1 (15 − 4t)(3 − 2t)2 ≥ 0, 108 which is obviously true for t ≤ 3. The proof is completed. The equality occurs for a = 0, 1 b = 1 and c = (or any cyclic permutation thereof). 2

P 1.113. Let a, b, c be nonnegative real numbers such that a ≤ 1 ≤ b ≤ c. (a) If a + b + c = 3, then a2 b + b2 c + c 2 a ≥ a bc + 2; (b) If a b + bc + ca = 3, then a2 b + b2 c + c 2 a ≥ 3. (Vasile Cîrtoaje, 2008) Proof. Since a(b − a)(b − c) ≤ 0, we have a2 b + b2 c + c 2 a ≥ a2 b + b2 c + c 2 a + a(b − a)(b − c) = b2 (a + c) + ac(a + c − b). We will use this inequality to prove (a) and (b). (a) It suffices to prove that b2 (a + c) + ac(a + c − b) ≥ a bc + 2. This inequality is equivalent to b2 (a + c) − 2 ≥ ac(2b − a − c), b2 (3 − b) − 2 ≥ ac(3b − 3). From the inequality (b − a)(b − c) ≤ 0, we deduce that ac ≤ b(a + c − b) = b(3 − 2b). Therefore, since b ≥ 1, we have ac(3b − 3) ≤ b(3 − 2b)(3b − 3).

Cyclic Inequalities

155

Thus, it suffices to show that b2 (3 − b) − 2 ≥ b(3 − 2b)(3b − 3), which is equivalent to the obvious inequality (5b − 2)(b − 1)2 ≥ 0. The equality holds for a = b = c = 1, and also for a = 0, b = 1 and c = 2. (b) It suffices to prove that b2 (a + c) + ac(a + c − b) ≥ 3. Let x = a + c. From the given conditions, we get 1 ≤ x ≤ 3 and ac by 3 − b x. Thus, the inequality becomes b2 x + (3 − b x)(x − b) ≥ 3, 2b2 x − (x 2 + 3)b + 3x − 3 ≥ 0. Since 2b2 x − (x 2 + 3)b + 3x − 3 = 2(b2 − 2b + 1)x + 2(2b − 1)x − (x 2 + 3)b + 3x − 3 = 2(b − 1)2 x + (3 − x)(b x − b − 1) ≥ (3 − x)(b x − b − 1), it is enough to prove that b x − b − 1 ≥ 0. From the inequality (b − a)(b − c) ≤ 0, we get b x ≥ b2 + ac = b2 + 3 − b x,

bx ≥

b2 + 3 . 2

Therefore,

b2 + 3 (b − 1)2 − b−1= ≥ 0. 2 2 The proof is completed. The equality holds for a = b = c = 1, and for a = 0, b = 1 and c = 3. bx − b − 1 ≥

P 1.114. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 = Prove that a4 + b4 + c 4 ≥

5 (a b + bc + ca). 2

17 3 (a b + b3 c + c 3 a). 8 (Vasile Cîrtoaje, 2011)

156

Vasile Cîrtoaje

Solution. We can see that the equality holds for a = 2, b = 1, c = 0, when a bc = 0. Let us denote x = a2 + b2 + c 2 , y = a b + bc + ca. Since 2x = 5 y and a4 + b4 + c 4 − (a2 + b2 + c 2 )2 + 2(a b + bc + ca)2 = 4a bc(a + b + c) ≥ 0, we have a4 + b4 + c 4 ≥ x 2 − 2 y 2 =

17 (2x + y)2 . 144

Therefore, it suffices to prove that (2x + y)2 ≥ 18(a3 b + b3 c + c 3 a). We will show that this inequality holds for all nonnegative real numbers a, b, c. Assume that a = max{a, b, c}. There are two possible cases: a ≥ b ≥ c and a ≥ c ≥ b. Case 1: a ≥ b ≥ c. Using the AM-GM inequality gives  2 2a b + (a2 + bc + c 2 ) 3 3 3 2 2 2(a b + b c + c a) ≤ 2a b(a + bc + c ) ≤ . 2 Therefore, it suffices to show that 2x + y ≥

3 (2a b + a2 + bc + c 2 ), 2

which is equivalent to the obvious inequality (a − 2b)2 + c(2a − b + c) ≥ 0. Case 2: a ≥ c ≥ b. Since a b3 + bc 3 + ca3 − (a3 b + b3 c + c 3 a) = (a + b + c)(a − b)(b − c)(c − a) ≥ 0, we have 2(a3 b + b3 c + c 3 a) ≤ (a3 b + b3 c + c 3 a) + (a b3 + bc 3 + ca3 ) ≤ x y. Thus, it suffices to prove that (2x + y)2 ≥ 9x y. Indeed, since x ≥ y, we get (2x + y)2 − 9x y = (x − y)(4x − y) ≥ 0. Thus, the proof is completed. The equality holds for a = 2b and c = 0 (or any cyclic permutation).

Cyclic Inequalities

157

P 1.115. Let a, b, c be nonnegative real numbers such that a2 + b2 + c 2 =

5 (a b + bc + ca). 2

Prove that (a)

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

(b)

11(a4 + b4 + c 4 ) ≥ 17(a3 b + b3 c + c 3 a) + 129a bc(a + b + c);

(c)

a b+b c+c a≤ 3

3

3

14 +

p 102 2 2 (a b + b2 c 2 + c 2 a2 ). 8

Solution. For a = b = c = 0, the inequalities are trivial. Otherwise, let us denote p = a + b + c, q = a b + bc + ca,

r = a bc.

Due to homogeneity, we may and assume that p = 3, which involves q = 2. (a) Write the inequality as X X X X X a b(a2 + b2 ) + a3 b − a b3 ≥ a2 b2 + a bc a. Since

X

a b(a2 + b2 ) = q(p2 − 2q) − pr = 10 − 3r, X a2 b2 = q2 − 2pr = 4 − 6r, X a bc a = 3r, X X Æ a3 b − a b3 = −p(a − b)(b − c)(c − a) ≥ −p (a − b)2 (b − c)2 (c − a)2 v t 4(p2 − 3q)3 − (2p3 − 9pq + 27r)2 p = −p = −3 4 − 27r 2 , 27 it suffices to prove that p 10 − 3r − 3 4 − 27r 2 ≥ 4 − 6r + 3r, which is equivalent to the obvious inequality 2≥

p

4 − 27r 2 .

The equality holds for a = 0 and 2b = c (or any cyclic permutation).

158

Vasile Cîrtoaje (b) Write the inequality as X X X X X 22 a4 ≥ 17 a b(a2 + b2 ) + 17( a3 b − a b3 ) + 258a bc a.

Since

X

a4 = p4 − 4p2 q + 2q2 + 4pr = 17 + 12r,

X

a b(a2 + b2 ) = q(p2 − 2q) − pr = 10 − 3r, X a bc a = 3r, X X Æ a3 b − a b3 = −p(a − b)(b − c)(c − a) ≤ p (a − b)2 (b − c)2 (c − a)2 v t 4(p2 − 3q)3 − (2p3 − 9pq + 27r)2 p =p = 3 4 − 27r 2 , 27 it suffices to prove that 22(17 + 12r) ≥ 17(10 − 3r) + 51

p

4 − 27r 2 + 774r

2 for 0 ≤ r ≤ p . Write this inequality as 3 3 4 − 9r ≥

p

4 − 27r 2 .

p We have 4 − 9r ≥ 4 − 2 3 > 0. By squaring, we get the obvious inequality (3r − 1)2 ≥ 0. 1 and (a − b)(b − c)(c − a) ≤ 0. In 3 general, the equality holds when a, b, c are proportional to the roots of the equation

For p = 3, the equality holds when q = 2, r =

3x 3 − 9x 2 + 6x − 1 = 0 and satisfy (a − b)(b − c)(c − a) ≤ 0. This occurs when (Wolfgang Berndt) a sin2

2π 4π π = b sin2 = c sin2 . 9 9 9

(c) Write the inequality as X X X a b(a2 + b2 ) + ( a3 b − a b3 ) ≤ k(a2 b2 + b2 c 2 + c 2 a2 ),

Cyclic Inequalities

159

where

14 +

k= Since

p 102 . 4

X

a b(a2 + b2 ) = q(p2 − 2q) − pr = 10 − 3r, X a2 b2 = q2 − 2pr = 4 − 6r, X X Æ a3 b − a b3 = −p(a − b)(b − c)(c − a) ≤ p (a − b)2 (b − c)2 (c − a)2 v t 4(p2 − 3q)3 − (2p3 − 9pq + 27r)2 p =p = 3 4 − 27r 2 , 27 it suffices to prove that 10 − 3r + 3

p

4 − 27r 2 ≤ k(4 − 6r),

2 where r ≤ p . Write this inequality as 3 3 p 3 4 − 27r 2 ≤ 4k − 10 − 3(2k − 1)r. We have  ‹ 1 2 2(2k − 1) = 4 1 − p k − 10 + p > 0. 4k − 10 − 3(2k − 1)r ≥ 4k − 10 − p 3 3 3 Thus, by squaring, we get the obvious inequality (r − k1 )2 ≥ 0, where 2 k1 = 129

v t

p 787 + 72 102 . 3

For p = 3, the equality holds when q = 2, r = k1 and (a − b)(b − c)(c − a) ≤ 0. In general, the equality holds when a, b, c are proportional to the roots of the equation x 3 − 3x 2 + 2x − k1 = 0 and satisfy (a − b)(b − c)(c − a) ≤ 0.

160

Vasile Cîrtoaje

P 1.116. If a, b, c are real numbers such that a3 b + b3 c + c 3 a ≤ 0, then a2 + b2 + c 2 ≥ k(a b + bc + ca), where k=

1+

p

p 21 + 8 7 . 2 (Vasile Cîrtoaje, 2012)

Solution. Let us denote p = a + b + c, q = a b + bc + ca,

r = a bc.

For p = 0, we have a2 + b2 + c 2 − k(a b + bc + ca) = −(2 + k)(a b + bc + ca) ≥ 0, since a b + bc + ca = a(b + c) + bc = −(b + c)2 + bc ≤ 0. Consider now that p 6= 0 and use the contradiction method. It suffices to prove that a2 + b2 + c 2 < k(a b + bc + ca) involves a3 b + b3 c + c 3 a > 0. Since the statement remains unchanged by replacing a, b, c with −a, −b, −c, respectively, we can consider that p > 0. In addition, due to homogeneity, we may assume that p = 1. From the hypothesis a2 + b2 + c 2 < k(a b + bc + ca), we get q>

1 . k+2

Write the desired inequality as X X X a b(a2 + b2 ) + a3 b − a b3 > 0. Since

X

a b(a2 + b2 ) = q(p2 − 2q) − pr = q − 2q2 − r

and X

a3 b −

X

a b3 = −p(a − b)(b − c)(c − a) ≥ −p

Æ

(a − b)2 (b − c)2 (c − a)2

Cyclic Inequalities

= −p

161

v t 4(p2 − 3q)3 − (2p3 − 9pq + 27r)2 27

=−

v t 4(1 − 3q)3 − (2 − 9q + 27r)2 27

,

it suffices to prove that 2

q − 2q − r >

v t 4(1 − 3q)3 − (2 − 9q + 27r)2 27

.

From p2 ≥ 3q, we get

1 1 0. 3 3

By squaring, the desired inequality can be restated as 7r 2 + (1 − 5q + q2 )r + q4 > 0. This is true if the discriminant p p D = (1 − 5q + q2 )2 − 28q4 = [1 − 5q + (1 + 2 7)q2 ][1 − 5q + (1 − 2 7)q2 ] is negative. Since p ‹  p 2 5q 2 8 7 − 21 2 1 − 5q + (1 + 2 7)q = 1 − + q > 0, 2 4 we only need to show that f (q) > 0, where p f (q) = (2 7 − 1)q2 + 5q − 1. Indeed, since q >

1 , we have k+2

p 5 2 7−1 − 1 = 0. f (q) > + 2 (k + 2) k+2

For p = 1, the equality holds when (a − b)(b − c)(c − a) > 0 and q=

−q2 1 r = p = −p . 7 7(k + 2)2

1 , k+2

In general, the equality holds when a, b, c are proportional to the roots of the equation w3 − w2 +

1 1 w+ p =0 k+2 7(k + 2)2

and satisfy (a − b)(b − c)(c − a) > 0.

162

Vasile Cîrtoaje

P 1.117. If a, b, c are real numbers such that a3 b + b3 c + c 3 a ≥ 0, then a2 + b2 + c 2 + k(a b + bc + ca) ≥ 0, where k=

−1 +

p

p 21 + 8 7 . 2 (Vasile Cîrtoaje, 2012)

Solution. Let us denote p = a + b + c, q = a b + bc + ca,

r = a bc.

At least two of a, b, c have the same sign; let b and c be these numbers. If p = 0, then a = −b − c and a3 b + b3 c + c 3 a = −(b + c)3 b + b3 c − c 3 (b + c). Clearly, the hypothesis a3 b + b3 c + c 3 a ≥ 0 is satisfied only if a = b = c = 0, when the statement is trivial. Consider now that p 6= 0 and use the contradiction method. It suffices to prove that a2 + b2 + c 2 + k(a b + bc + ca) < 0 involves a3 b + b3 c + c 3 a < 0. Since the statement remains unchanged by replacing a, b, c with −a, −b, −c, respectively, we can consider that p > 0. In addition, due to homogeneity, we may assume that p = 1. From the hypothesis, we get q
1, 3

and hence 2

2 2 2

r =a b c ≤



a2 + b2 + c 2 3

3

1 − 2q = 3 

‹3

1 − 2q < 3 

‹4

,

which implies r >−



1 − 2q 3

‹2

.

Therefore, 1 − 2q r + 2q − q > − 3 2



‹2

+ 2q2 − q =

(2q − 1)(7q + 1) > 0. 9

By squaring, the desired inequality becomes 7r 2 + (1 − 5q + q2 )r + q4 > 0. This is true if the discriminant p p D = (1 − 5q + q2 )2 − 28q4 = [1 − 5q + (1 + 2 7)q2 ][1 − 5q + (1 − 2 7)q2 ] is negative. Since

p 1 − 5q + (1 + 2 7)q2 > 0,

we only need to show that f (q) > 0, where p f (q) = (2 7 − 1)q2 + 5q − 1. Since

p p p −2(2 7 − 1) 5k − 8 − 4 7 f (q) = 2(2 7 − 1)q + 5 < +5= < 0, k−2 k−2 f (q) is strictly decreasing, and hence 0

f (q) > f (

−1 ) = 0. k−2

164

Vasile Cîrtoaje

For p = 1, the equality holds when (a − b)(b − c)(c − a) < 0 and q=

−q2 −1 r= p =p . 7 7(k − 2)2

−1 , k−2

In general, the equality holds when a, b, c are proportional to the roots of the equation w3 − w2 −

1 1 w+ p =0 k−2 7(k − 2)2

and satisfy (a − b)(b − c)(c − a) > 0.

P 1.118. If a, b, c are real numbers such that k(a2 + b2 + c 2 ) = a b + bc + ca, where k ∈



‹ −1 , 1 , then 2 αk ≤

where

a3 b + b3 c + c 3 ≤ βk , (a2 + b2 + c 2 )2

v t 7(1 − k) 27αk = 1 + 13k − 5k − 2(1 − k)(1 + 2k) , 1 + 2k v t 7(1 − k) 2 27βk = 1 + 13k − 5k + 2(1 − k)(1 + 2k) . 1 + 2k 2

(Vasile Cîrtoaje, 2012) Solution. Let us denote p = a + b + c, q = a b + bc + ca,

r = a bc.

For p = 0, from the hypothesis k(a2 + b2 + c 2 ) = a b + bc + ca, we get a b + bc + ca = 0. Since a b + bc + ca = a(b + c) + bc = −(b + c)2 + bc = −b2 − bc − c 2 , it follows that b2 + bc + c 2 = 0, which involves a = b = c = 0. Consider now that p 6= 0. Since the statement remains unchanged by replacing a, b, c with −a, −b, −c, respectively, it suffices to consider the case p > 0. In addition, due to homogeneity, we may assume that p = 1, which implies q=

k . 1 + 2k

Cyclic Inequalities

165

(a) Write the desired inequality as X X X 2αk (a2 + b2 + c 2 )2 ≤ a b(a2 + b2 ) + ( a3 b − a b3 ). Since

X

a2 = p2 − 2q = 1 − 2q,

X

a b(a2 + b2 ) = q(p2 − 2q) − pr = q − 2q2 − r, X X Æ a3 b − a b3 = −p(a − b)(b − c)(c − a) ≥ −p (a − b)2 (b − c)2 (c − a)2 v v t 4(p2 − 3q)3 − (2p3 − 9pq + 27r)2 t 4(1 − 3q)3 − (2 − 9q + 27r)2 = −p =− , 27 27 it suffices to prove that 2K(1 − 2q)2 ≤ q − 2q2 − r −

v t 4(1 − 3q)3 − (2 − 9q + 27r)2 27

.

Applying Lemma below for −1 1 , α= p , β = 27 27

p a = 2(1 − 3q) 1 − 3q,

b = 2 − 9q + 27r,

we get v t 4(1 − 3q)3 − (2 − 9q + 27r)2 27

p 4(1 − 3q) 7(1 − 3q) 2 − 9q +r+ ≤ , 27 27

with equality for v t 1 − 3q (1 − 3q) − 2 + 9q − 27r = 0. 7 Thus, it suffices to show that p 2 − 9q 4(1 − 3q) 7(1 − 3q) 2αk (1 − 2q) ≤ q − 2q + − , 27 27 2

2

which is equivalent to v t 7(1 − k) 27αk ≤ 1 + 13k − 5k2 − 2(1 − k)(1 + 2k) . 1 + 2k Since this inequality is an identity, the proof is completed. For p = 1, the equality holds when (a − b)(b − c)(c − a) ≥ 0, q = k/(1 + 2k) and v t 1 − 3q r1 27r = (1 − 3q) − 2 + 9q = , 7 1 + 2k

166

Vasile Cîrtoaje

where

v t r1 = 5k − 2 + (1 − k)

1−k . 7(1 + 2k)

Therefore, the equality holds when a, b, c are proportional to the roots of the equation w3 − w2 +

r1 k w− =0 1 + 2k 27(1 + 2k)

and satisfy (a − b)(b − c)(c − a) ≥ 0. (b) Write the desired inequality βk (a2 + b2 + c 2 )2 ≥ (a3 b + b3 c + c 3 a) as X X X a b(a2 + b2 ) + ( a3 b − a b3 ). 2βk (a2 + b2 + c 2 )2 ≥ Since

X

a2 = p2 − 2q = 1 − 2q,

X

a b(a2 + b2 ) = q(p2 − 2q) − pr = q − 2q2 − r, X X Æ a3 b − a b3 = −p(a − b)(b − c)(c − a) ≤ p (a − b)2 (b − c)2 (c − a)2 v v t 4(p2 − 3q)3 − (2p3 − 9pq + 27r)2 t 4(1 − 3q)3 − (2 − 9q + 27r)2 =p = , 27 27 it suffices to prove that 2βk (1 − 2q)2 ≥ q − 2q2 − r +

v t 4(1 − 3q)3 − (2 − 9q + 27r)2 27

.

Applying Lemma below for 1 1 α= p , β = , 27 27

p a = 2(1 − 3q) 1 − 3q,

b = 2 − 9q + 27r,

we get v t 4(1 − 3q)3 − (2 − 9q + 27r)2 27

p 4(1 − 3q) 7(1 − 3q) 2 − 9q −r− ≤ , 27 27

with equality for v t 1 − 3q (1 − 3q) + 2 − 9q + 27r = 0. 7 Thus, it suffices to show that p 2 − 9q 4(1 − 3q) 7(1 − 3q) 2βk (1 − 2q) ≥ q − 2q + + , 27 27 2

2

Cyclic Inequalities

167

which is equivalent to v t 7(1 − k) 27βk ≥ 1 + 13k − 5k2 + 2(1 − k)(1 + 2k) . 1 + 2k Clearly, this inequality is an identity. For p = 1, the equality holds when (a − b)(b − c)(c − a) ≤ 0, q = k/(1 + 2k) and v t 1 − 3q r0 = , 27r = 9q − 2 − (1 − 3q) 7 1 + 2k where

v t r0 = 5k − 2 − (1 − k)

1−k . 7(1 + 2k)

Therefore, the equality holds when a, b, c are proportional to the roots of the equation w3 − w2 +

r0 k w− =0 1 + 2k 27(1 + 2k)

and satisfy (a − b)(b − c)(c − a) ≤ 0. Lemma. If α, β, a, b are real numbers, α ≥ 0, a ≥ 0 and a2 ≥ b2 , then p Æ α a2 − b2 ≤ a α2 + β 2 + β b, with equality if and only if βa + b

Æ

α2 + β 2 = 0.

Proof. Since a

Æ

α2 + β 2 + β b ≥ |β|a + β b ≥ |β||b| + β b ≥ 0,

we can write the inequality as α2 (a2 − b2 ) ≤ (a

Æ

α2 + β 2 + β b)2 ,

which is equivalent to the obvious inequality Æ (β a + b α2 + β 2 )2 ≥ 0.

P 1.119. If a, b, c are positive real numbers, then a2 + 3a b b2 + 3bc c 2 + 3ca + + ≥ 3. (b + c)2 (c + a)2 (a + b)2

168

Vasile Cîrtoaje

Solution. Write the inequality as X a(a + b) (b + c)2

+2

X

ab ≥ 3. (b + c)2

Since the sequences {bc, ca, a b} and

§

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

ª

are opposite arranged, by the rearrangement inequality, we have X

X ab bc ≤ . (b + c)2 (b + c)2

Therefore, it suffices to show that X a(a + b) (b + c)2

+

X b(a + c) ≥ 3, (b + c)2

which is equivalent to X • a+b X b+c ˜ a + ≥ 3. (b + c)2 (a + b)2 By the AM-GM inequality, we have b+c 2 4 a+b + ≥p ≥ . (b + c)2 (a + b)2 (a + b) + (b + c) (a + b)(b + c) Thus, it is enough to prove that X

a 3 ≥ . a + 2b + c 4

Indeed, by the Cauchy-Schwarz inequality, we get P P 2 P X ( a)2 a + 2 ab a P =P ≥P a + 2b + c a(a + 2b + c) a2 + 3 a b P 2 P a − ab 3 3 P = + P ≥ . 2 4 4( a + 3 a b) 4 The equality holds for a = b = c.

Cyclic Inequalities

169

P 1.120. If a, b, c are positive real numbers, then a2 b + 1 b2 c + 1 c2 a + 1 + + ≥ 3. a(b + 1) b(c + 1) c(a + 1) Solution. By the Cauchy-Schwarz inequality, we have ‹  1 + 1 ≥ (a + 1)2 , (a2 b + 1) b hence

a2 b + 1 b(a + 1)2 ≥ . a(b + 1) a(b + 1)2

Therefore, it suffices to prove that X b(a + 1)2 ≥ 3. a(b + 1)2 This inequality follows immediately from the AM-GM inequality: v Y b(a + 1)2 X b(a + 1)2 t 3 ≥ 3 = 3. a(b + 1)2 a(b + 1)2 The equality holds for a = b = c = 1.

P 1.121. If a, b, c are positive real numbers such that a + b + c = 3, then p

a3 + 3b +

p

b3 + 3c +

p

c 3 + 3a ≥ 6.

Solution. By the Cauchy-Schwarz inequality, we have (a3 + 3b)(a + 3b) ≥ (a2 + 3b)2 . Thus, it suffices to show that X a2 + 3b ≥ 6. p a + 3b By Hölder’s inequality, we have X 2 2 ”X — ”X —3 €X Š3 a + 3b (a2 + 3b)(a + 3b) ≥ (a2 + 3b) = a2 + 9 . p a + 3b

170

Vasile Cîrtoaje

Therefore, it is enough to show that €X

a2 + 9

Š3

≥ 36

X (a2 + 3b)(a + 3b).

Let p = a + b + c = 3 and q = a b + bc + ca, q ≤ 3. We have X a2 + 9 = p2 − 2q + 9 = 2(9 − q), X X X X X (a2 + 3b)(a + 3b) = a3 + 3 a2 b + 9 a2 + 3 ab X = (p3 − 3pq + 3a bc) + 3 a2 b + 9(p2 − 2q) + 3q X = 108 − 24q + 3(a bc + a2 b). Since a bc +

P

a2 b ≤ 4 (see the inequality (a) in P 1.28), we get X (a2 + 3b)(a + 3b) ≤ 24(5 − q).

Thus, it suffices to show that (9 − q)3 ≥ 108(5 − q), which is equivalent to the obvious inequality (3 − q)2 (21 − q) ≥ 0. The equality holds for a = b = c = 1.

P 1.122. If a, b, c are positive real numbers such that a bc = 1, then v s s t a b c + + ≥ 1. a + 6b + 2bc b + 6c + 2ca c + 6a + 2a b (Nguyen Van Quy and Vasile Cîrtoaje, 2013) Solution. By Hölder’s inequality, we have X s

a a + 6b + 2bc

‹2 ”X

— €X Š3 a(a + 6b + 2bc) ≥ a2/3 .

Therefore, it suffices to show that €X

a2/3

Š3

≥

X

a2 + 6

X

a b + 6,

Cyclic Inequalities

171

which is equivalent to 3

X X (a b)2/3 (a2/3 + b2/3 ) ≥ 6 a b.

Indeed, 3

X X X (a b)2/3 (a2/3 + b2/3 ) − 6 ab = 3 (a b)2/3 (a1/3 − b1/3 )2 ≥ 0.

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

P 1.123. If a, b, c are positive real numbers such that a + b + c = 3, then a2 b2 c2 3 + + ≥ . 2 2 2 4a + b 4b + c 4c + a 5 (Michael Rozenberg, 2008) Solution. By the Cauchy-Schwarz inequality, we have P P P X a2 [ a(2a + c)]2 (2 a2 + a b)2 ≥P =P . 4a + b2 (4a + b2 )(2a + c)2 (4a + b2 )(2a + c)2 Therefore, it suffices to show that X X X 5(2 a2 + a b)2 ≥ 3 (4a + b2 )(2a + c)2 , which is equivalent to the homogeneous inequalities X X X 5(2 a2 + a b)2 ≥ [4a(a + b + c) + 3b2 ](2a + c)2 , X X X 5(2 a2 + a b)2 ≥ (4a2 + 3b2 + 4a b + 4ac)(4a2 + c 2 + 4ac), X X X X 2 a4 + 5 a2 b2 ≥ a bc a+6 a b3 . Using the known inequality 3

X

X a b3 ≤ ( a2 )2

from P 1.95, it is enough to show that X X X X 2 a4 + 5 a2 b2 ≥ a bc a + 2( a2 )2 , which is equivalent to the well know inequality X X a2 b2 ≥ a bc a. The equality holds for a = b = c = 1.

172

Vasile Cîrtoaje

P 1.124. If a, b, c are positive real numbers, then (a + b + c)3 a2 + bc b2 + ca c 2 + a b + + ≤ . a+b b+c c+a 3(a b + bc + ca) (Michael Rozenberg, 2013) Solution (by Manlio Marangelli). Write the inequality as 3

X a2 + bc

3

a+b P

−3

a3 b − 3a bc

X P

a≤

a

(a + b)(b + c)(c + a)

≤

(a + b + c)3 − 3(a + b + c), a b + bc + ca (a + b + c)3 − 3(a + b + c). a b + bc + ca

By the known inequality 3

X

X a3 b ≤ ( a2 )2 ,

it suffices to prove the symmetric inequality P P ( a2 )2 − 3a bc a (a + b + c)3 ≤ − 3(a + b + c). (a + b)(b + c)(c + a) a b + bc + ca Using the notation p = a + b + c, q = a b + bc + ca,

r = a bc,

this inequality can be written as (p2 − 2q)2 − 3pr p3 ≤ − 3p, pq − r q which is equivalent to q2 (p2 − 4q) − (p2 − 6q)pr ≥ 0. For p2 − 6q ≥ 0, since 3pr ≤ q2 , we have q2 (p2 − 4q) − (p2 − 6q)pr ≥ q2 (p2 − 4q) −

q2 (p2 − 6q) 2q2 (p2 − 3q) = ≥ 0. 3 3

For p2 − 6q ≤ 0, using Schur’s inequality of fourth degree 6pr ≥ (p2 − q)(4q − p2 ), we get (p2 − 6q)(p2 − q)(4q − p2 ) 6 (p2 − 3q)(p2 − 4q)2 = ≥ 0. 6

q2 (p2 − 4q) − (p2 − 6q)pr ≥ q2 (p2 − 4q) −

Cyclic Inequalities

173

The equality holds for a = b = c = 1. Remark. We can also prove the inequality q2 (p2 − 4q) − (p2 − 6q)pr ≥ 0 writing it in the form q(p2 q − 4q2 + 3pr) ≥ pr(p2 − 3q). Since p2 q − 4q2 + 3pr = q(p2 − 2q) − 2(q2 − 2pr) − pr X X €X Š €X Š = ab a2 − 2 a2 b2 − a bc a X X X = a b(a2 + b2 ) − 2 a2 b2 = a b(a − b)2 , the inequality becomes pr(p2 − 3q) X ≤ a b(a − b)2 . q Without loss of generality, assume that a ≥ b ≥ c. We have pr(p2 − 3q) a bc(a + b + c)[(a − c)2 + (b − a)(b − c)] ≤ q b(a + c) ≤

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

Thus, it suffices to show that a bc(a − c)2 ≤ a b(a − b)2 + bc(b − c)2 ; a+c that is,   ac(a − c)2 ≤ (a + c) a(a − b)2 + c(b − c)2 ,  ‹  1 1  2 a(a − b)2 + c(b − c)2 . [(a − b) + (b − c)] ≤ + a c Clearly, the last inequality follows immediately from the Cauchy-Schwarz inequality.

P 1.125. If a, b, c are positive real numbers such that a + b + c = 3, then p

a b2 + bc 2 +

p

bc 2 + ca2 +

p

p ca2 + a b2 ≤ 3 2. (Nguyen Van Quy, 2013)

174

Vasile Cîrtoaje

Solution (by Michael Rozenberg). By the Cauchy-Schwarz inequality, we have €X p

a b2 + bc 2

Š2

X a b + c2 X · b(a + c). a+c

≤

Therefore, it suffices to show that X a b + c2 a+c

≤

9 , a b + bc + ca

which is equivalent to the homogeneous inequality X a b + c2 a+c

≤

(a + b + c)3 , 3(a b + bc + ca)

which is just the inequality from the preceding P 1.124. The equality holds for a = b = c = 1.

P 1.126. If a, b, c are positive real numbers such that a bc = 1, then 

a+

1 b

‹2

 ‹  ‹ 1 2 1 2 + b+ + c+ ≥ 6(a + b + c − 1). c a (Marius Stanean, 2014)

Solution (by Michael Rozenberg). By the AM-GM inequality, we have X

1 a+ b

‹2

≥

+6=

X X (a + ac)2 + 6 = (a2 + a2 b2 + 2a2 c + 2)

X Æ X 6 6 a2 · a2 b2 · (a2 c)2 · 1 · 1 = 6 a.

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

P 1.127. If a, b, c are positive real numbers such that a + b + c = 3, then 12 1 ≤3+ . a2 b + b2 c + c 2 a a bc (Vasile Cîrtoaje and ShengLi Chen, 2009)

Cyclic Inequalities

175

Solution. Let us denote p = a + b + c (p = 3), q = a b + bc + ca,

r = a bc (r ≤ 1).

Write the inequality as (3a bc + 1)(a2 b + b2 c + c 2 a) ≥ 12a bc. From (a − b)2 (b − c)2 (c − a)2 = −27r 2 + 2(9pq − 2p3 )r + p2 q2 − 4q3 = −27r 2 + 54(q − 2)r + 9q2 − 4q3 , we get (a − b)(b − c)(c − a) ≤ hence 2(a2 b + b2 c + c 2 a) =

Æ

−27r 2 + 54(q − 2)r + 9q2 − 4q3 ,

X

a b(a + b) − (a − b)(b − c)(c − a) Æ = pq − 3r − (a − b)(b − c)(c − a) ≥ 3q − 3r − −27r 2 + 54(q − 2)r + 9q2 − 4q3 . Therefore, it suffices to show that ” — Æ (3r + 1) 3q − 3r − −27r 2 + 54(q − 2)r + 9q2 − 4q3 ≥ 24r, which is equivalent to Æ 3[(3r + 1)q − 3r 2 − 9r] ≥ (3r + 1) −27r 2 + 54(q − 2)r + 9q2 − 4q3 . Before squaring this inequality, we need to show that (3r + 1)q − 3r 2 − 9r ≥ 0. Using the known inequality q2 ≥ 3pr, we have p p p (3r + 1)q − 3r 2 − 9r ≥ 3(3r + 1) r − 3r 2 − 9r = 3 r (1 − r)3 ≥ 0. By squaring, the desired inequality can be restated as Aq3 + C ≥ 3Bq, where A = 4(3r + 1)2 ,

B = 72r(3r + 1)(r + 1),

C = 108r(r + 1)(3r 2 + 12r + 1).

Since, by the AM-GM inequality, Aq3 + C ≥ 3

Æ 3

Aq3 · (C/2)2 = 3q

Æ 3

A(C/2)2 ,

176

Vasile Cîrtoaje

it is enough to show that AC 2 ≥ 4B 3 , which is equivalent to (3r 2 + 12r + 1)2 ≥ 32r(3r + 1)(r + 1). Indeed, (3r 2 + 12r + 1)2 − 32r(3r + 1)(r + 1) = (r − 1)2 (3r − 1)2 ≥ 0. 1 C The equality holds for a = b = c = 1, and also for r = and Aq3 = ; that is, 3 2 s 3 C q = = 2. Therefore, the equality holds also when a, b, c are the roots of the 2A equation 1 x 3 − 3x 2 + 2x − = 0 3 such that a ≤ b ≤ c or b ≤ c ≤ a or c ≤ a ≤ b.

P 1.128. If a, b, c are positive real numbers such that a + b + c = 3, then 24 1 + ≥ 9. a2 b + b2 c + c 2 a a bc (Vasile Cîrtoaje, 2009) Solution (by Vo Quoc Ba Can). Let us denote p = a + b + c (p = 3), q = a b + bc + ca,

r = a bc.

Write the inequality as 24r ≥ (9r − 1)(a2 b + b2 c + c 2 a), and consider further the nontrivial case 9r ≥ 1. From (a − b)2 (b − c)2 (c − a)2 = −27r 2 + 2(9pq − 2p3 )r + p2 q2 − 4q3 = −27r 2 + 54(q − 2)r + 9q2 − 4q3 , we get 2(a2 b + b2 c + c 2 a) =

X

a b(a + b) − (a − b)(b − c)(c − a) Æ = pq − 3r − (a − b)(b − c)(c − a) ≤ 3q − 3r + −27r 2 + 54(q − 2)r + 9q2 − 4q3 .

Cyclic Inequalities

177

Therefore, it suffices to show that ” — Æ 48r ≥ (9r − 1) 3q − 3r + −27r 2 + 54(q − 2)r + 9q2 − 4q3 , which is true if Æ 3[9r 2 + 15r − (9r − 1)q] ≥ (9r − 1) −27r 2 + 54(q − 2)r + 9q2 − 4q3 . Before squaring this inequality, we need to show that 9r 2 + 15r − (9r − 1)q ≥ 0. From Schur’s inequality p3 + 9r ≥ 4pq, we get q≤

3(r + 3) , 4

hence 9r 2 + 15r − (9r − 1)q ≥ 9r 2 + 15r −

3(r + 3)(9r − 1) 9(r − 1)2 = ≥ 0. 4 4

By squaring, the desired inequality can be written as Aq3 + C ≥ 3Bq, where A = (9r − 1)2 ,

B = 18r(9r − 1)(3r + 1),

C = 27r(27r 3 + 99r 2 + r + 1).

Since, by the AM-GM inequality, Aq3 + C ≥ 3

Æ 3

Aq3 · (C/2)2 = 3q

Æ 3

A(C/2)2 ,

it is enough to show that AC 2 ≥ 4B 3 , which is equivalent to (27r 3 + 99r 2 + r + 1)2 ≥ 32r(9r − 1)(3r + 1)3 , 729r 6 − 2430r 5 + 2943r 4 − 1476r 3 + 199r 2 + 34r + 1 ≥ 0, (r − 1)2 (27r 2 − 18r − 1)2 ≥ 0.

p p 3+2 3 The equality holds for a = b = c = 1, and also for r = and q = 1 + 3; that is, 9 when a, b, c are the roots of the equation p p 3+2 3 3 2 x − 3x + (1 + 3)x − =0 9 such that a ≥ b ≥ c or b ≥ c ≥ a or c ≥ a ≥ b.

178

Vasile Cîrtoaje

P 1.129. If a, b, c are positive real numbers, then b c a+b+c a + + ≥ . p 3 a+b b+c c+a a + b + c − a bc (Michael Rozenberg, 2014) Solution. By the Cauchy-Schwarz inequality, we have P X a ( a)2 (a + b + c)2 ≥P . = a+b a(a + b) (a + b + c)2 − (a b + bc + ca) Therefore, it suffices to show that (a + b + c)2 a+b+c ≥ , p 3 2 (a + b + c) − (a b + bc + ca) a + b + c − a bc which is equivalent to (a + b + c)[a b + bc + ca −

p 3

a bc (a + b + c)] ≥ 0. p 3 Since this inequality is true for a b + bc + ca − a bc (a + b + c) ≥ 0, consider further that p 3 a bc (a + b + c) ≥ a b + bc + ca. By the Cauchy-Schwarz inequality, we have P X a ( ac)2 (a b + bc + ca)2 ≥P = . a+b ac 2 (a + b) (a b + bc + ca)2 − a bc(a + b + c) Thus, it suffices to show that (a b + bc + ca)2 a+b+c , ≥ p 3 (a b + bc + ca)2 − a bc(a + b + c) a + b + c − a bc which is equivalent to the obvious inequality p 3 [ a bc (a + b + c)]2 ≥ (a b + bc + ca)2 . The equality holds for a = b = c.

P 1.130. If a, b, c are positive real numbers such that a + b + c = 3, then p p p p a b2 + b + 1 + b c 2 + c + 1 + c a2 + a + 1 ≤ 3 3. (Nguyen Van Quy, 2014)

Cyclic Inequalities

179

Solution. From 4(b2 + b + 1) = 2(b + 1)2 + 2(b2 + 1) ≥ 3(b + 1)2 , we get p hence

b2

p 3 + b+1≥ (b + 1), 2

X p X a(b2 + b + 1) X 2a(b2 + b + 1) a b2 + b + 1 = ≤ . p p 3(b + 1) b2 + b + 1

Therefore, it suffices to prove that X a(b2 + b + 1) b+1

≤

9 , 2

which is equivalent to X a b2 3 ≤ . b+1 2

p In addition, since b + 1 ≥ 2 b, it is enough to show that X a b3/2 ≤ 3. Replacing a, b, c by a2 , b2 , c 2 , respectively, we need to show that a2 + b2 +c 2 = 3 involves a2 b3 + b2 c 3 + c 2 a3 ≤ 3, which is just the inequality in P 1.34. The equality holds for a = b = c.

P 1.131. If a, b, c are positive real numbers, then 1 1 1 1 + + ≤ . 2 2 2 b(a + 2b + 3c) c(b + 2c + 3a) a(c + 2a + 3b) 12a bc (Vo Quoc Ba Can, 2012) Solution (by Leviethai). Assume that c = min{a, b, c}, and write the inequality as X

ca 1 ≤ . a + 2b + 3c 12

Case 1: a ≥ b ≥ c. Since, by the AM-GM inequality, (a + 2b + 3c)2 ≥ 4(2b + c)(2c + a),

180

Vasile Cîrtoaje

it suffices to show that X

ca 1 ≤ , (2b + c)(2c + a) 3

which is equivalent to 3

X

ca(2a + b) ≤ (2a + b)(2b + c)(2c + a),

a b2 + bc 2 + ca2 ≤ a2 b + b2 c + c 2 a, (a − b)(b − c)(c − a) ≤ 0. Clearly, the last inequality is obvious. Case 2: a ≥ c ≥ b. Since, by the AM-GM inequality, (a + 2b + 3c)2 ≥ 12c(a + 2b), (b + 2c + 3a)2 ≥ 4(2a + b)(2c + a), (c + 2a + 3b)2 ≥ 4(a + 2b)(a + b + c), it suffices to show that a ab bc 1 + + ≤ , 3(a + 2b) (2a + b)(2c + a) (a + 2b)(a + b + c) 3 which is equivalent to bc 2b ab + ≤ , (2a + b)(2c + a) (a + 2b)(a + b + c) 3(a + 2b) a c 2 + ≤ , (2a + b)(2c + a) (a + 2b)(a + b + c) 3(a + 2b) a(a + 2b) c 2 + ≤ , (2a + b)(2c + a) a + b + c 3 a(a + 2b) c(2c + a) 2(2c + a) + ≤ . 2a + b a+b+c 3 For fixed a and b, let f (x) =

2(2x + a) x(2x + a) − . 3 a+b+x

We need to show that f (c) ≥

a(a + 2b) . 2a + b

Cyclic Inequalities

181

which is equivalent to f (c) ≥ f (a). We have   2 3a + 4ac + b(3a + 2c) 4 − f (c) − f (a) =(a − c) (a + b + c)(2a + b) 3 =

(a − c)[a2 − 3a b − 4b2 + 2c(2a + b)] ≥ 0, 3(a + b + c)(2a + b)

because a2 − 3a b − 4b2 + 2c(2a + b) ≥ a2 − 3a b − 4b2 + 2b(2a + b) = (a − b)(a + 2b) ≥ 0. The equality holds for a = b = c.

P 1.132. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that (a)

a2 + 9b b2 + 9c c 2 + 9a + + ≥ 15; b+c c+a a+b

(b)

a2 + 3b b2 + 3c c 2 + 3a + + ≥ 6. a+b b+c c+a

Solution. (a) Write the inequality as follows: X a2 + 3b(a + b + c) b+c X  a2 + 3b(a + b + c) b+c X a2 + 3a b

≥ 5(a + b + c), 

− 3b ≥ 2(a + b + c),

≥ 2(a + b + c), b+c  X  a2 + 3a b − 2a ≥ 0, b+c X a(a + b − 2c) ≥ 0, b+c X a(a − c) X a(b − c) + ≥ 0, b+c b+c X a(a − c) X b(c − a) + ≥ 0, b+c c+a  ‹ X a b (a − c) − ≥ 0, b+c c+a

182

Vasile Cîrtoaje (a + b + c)

X (a − b)(a − c) (b + c)(c + a)

≥ 0.

Therefore, it remains to show that X (a2 − b2 )(a − c) ≥ 0, which is equivalent to the obvious inequality X a(a − c)2 ≥ 0. The equality holds for a = b = c. (b) Write the inequality as follows: X a2 + b(a + b + c) a+b

≥ 2(a + b + c),

X a2 + bc

≥ a + b + c, a+b  X  a2 + bc − a ≥ 0, a+b X b(c − a) ≥ 0, a+b X bc X ab ≥ . a+b a+b

Since the sequences {a b, and §

1 , a+b

bc, 1 , b+c

ca} 1 c+a

ª

are reversely ordered, the inequality follows from the rearrangement inequality. The equality holds for a = b = c.

P 1.133. If a, b, c are positive real numbers such that a5 + b5 + c 5 = 3, then a2 b2 c 2 + + ≥ 3. b c a

Cyclic Inequalities

183

Solution. We will prove the inequality under the more large condition a m + b m +c m = 3, where 0 < m ≤ 21/4. First write the inequality in the homogeneous form  m ‹ a2 b2 c 2 a + b m + c m 1/m + + ≥ . b c a 3 By the Power Mean inequality, we have 

am + bm + c m 3

‹1/m

 ≤

a21/4 + b21/4 + c 21/4 3

4/21 .

Thus, it suffices to show that a2 b2 c 2 + + ≥ b c a



a21/4 + b21/4 + c 21/4 3

4/21 .

By the known inequality in P 1.95, namely x 4 + y 4 + z 4 ≥ 3(x 3 y + y 3 z + z 3 x), we have

x, y, z ∈ R,

 3  a c3 a2 b2 c 2 b3 + + ≥3 p +p +p . b c a ca bc ab

Therefore, it suffices to prove that c3 a3 b3 ≥ p +p +p ca bc ab



a21/4 + b21/4 + c 21/4 3

8/21 .

Setting a = x 7/2 ,

b = y 7/2 ,

c = z 7/2 ,

where x, y, z > 0, the inequality becomes  3/2 2  x + y + z 21/4 + y 3/2 + z 3/2 3/4 x ≥ (x yz) . 3 3 By the Cauchy-Schwarz inequality, we have (x + y + z)(x 2 + y 2 + z 2 ) ≥ (x 3/2 + y 3/2 + z 3/2 )2 . Thus, it is enough to prove that  x + y + z 17/4 1 ≥ (x yz)3/4 (x 2 + y 2 + z 2 ). 3 3

184

Vasile Cîrtoaje

Due to homogeneity, we may assume that x + y + z = 3, when the inequality becomes (x yz)3/4 (x 2 + y 2 + z 2 ) ≤ 3. p Since 3/4 > 1/ 2 and x yz ≤

 x + y + z 3 = 1, 3

this inequality follows from (x yz)1/

p 2

(x 2 + y 2 + z 2 ) ≤ 3,

which is just the inequality in P 2.87 in Volume 2. The proof is completed. The equality holds for a = b = c = 1.

P 1.134. Let P(a, b, c) be a cyclic homogeneous polynomial of degree three. The inequality P(a, b, c) ≥ 0 holds for all a, b, c ≥ 0 if and only if the following two conditions are fulfilled: (a) P(1, 1, 1) ≥ 0; (b) P(0, b, c) ≥ 0 for all b, c ≥ 0. (Pham Kim Hung, 2007) Solution. The conditions (a) and (b) are clearly necessary. Therefore, we will prove further that these conditions are also sufficient to have P(a, b, c) ≥ 0. The polynomial P(a, b, c) has the general form P(a, b, c) = A(a3 + b3 + c 3 ) + B(a2 b + b2 c + c 2 a) + C(a b2 + bc 2 + ca2 ) + 3Da bc. Since P(1, 1, 1) = 3(A + B + C + D),

P(0, 1, 1) = 2A + B + C,

P(0, 0, 1) = A,

the conditions (a) and (b) involves A + B + C + D ≥ 0, 2A + B + C ≥ 0, A ≥ 0. Assume that a = min{a, b, c}, and denote b = a + p,

c = a + q,

p, q ≥ 0.

Cyclic Inequalities

185

For fixed p and q, define the function f (a) = P(a, a + p, a + q),

a ≥ 0.

We have f 0 (a) = 3A(a2 + b2 + c 2 ) + (B + C)(a + b + c)2 + 3D(a b + bc + ca) = (3A + B + C)(a2 + b2 + c 2 ) + (2B + 2C + 3D)(a b + bc + ca) = (3A + B + C)(a2 + b2 + c 2 − a b − bc − ca) + 3(A + B + C + D)(a b + bc + ca). Since f 0 ≥ 0, it follows that f is increasing, therefore f (a) ≥ f (0), which is equivalent to P(a, b, c) ≥ P(0, p, q). According to the condition (b), we have P(0, p, q) ≥ 0, and hence P(a, b, c) ≥ P(0, p, q) ≥ 0.

P 1.135. If a, b, c are nonnegative real numbers, then a3 + b3 + c 3 − 3a bc ≥

Æ

p 9 + 6 3 (a − b)(b − c)(c − a).

First Solution. Write the inequality as P(a, b, c) ≥ 0. According to P 1.134, it suffices to show that P(1, 1, 1) ≥ 0 and P(0, b, c) ≥ 0 for all b, c ≥ 0. We have P(1, 1, 1) = 0 and Æ p P(0, b, c) = b3 + c 3 + 9 + 6 3 bc(b − c). The inequality P(0, b, c) ≥ 0 is true if p (b3 + c 3 )2 ≥ (9 + 6 3 )b2 c 2 (b − c)2 , which is equivalent to p (b + c)2 (b2 − bc + c 2 )2 ≥ (9 + 6 3 )b2 c 2 (b − c)2 . For the non-trivial case bc > 0, denoting x = b/c + c/b − 1, we can write this inequality as p (x + 3)x 2 ≥ (9 + 6 3 )(x − 1), p p (x − 3 )2 (x + 3 + 2 3 ) ≥ 0. p The equality holds for a = b = c, and also for a = 0 and b/c + c/b = 1 + 3, b < c (or any cyclic permutation).

186

Vasile Cîrtoaje

Second Solution. Assume that a = min{a, b, c}. Since the case a ≤ c ≤ b is trivial, consider further that a ≤ b ≤ c. Write the inequality as (a + b + c)[(a − b)2 + (b − c)2 + (c − a)2 ] ≥ 2

Æ

p 9 + 6 3 (a − b)(b − c)(c − a).

Using the substitution b = a + p, c = a + q, where q ≥ p ≥ 0, the inequality becomes (3a + p + q)(p2 − pq + q2 ) ≥

Æ

p 9 + 6 3 pq(q − p).

Since p2 − pq + q2 ≥ 0, it suffices to consider the case a=0 (as in the first solution).

P 1.136. If a, b, c are the lengths of the sides of a triangle, then b c a + + ≥ 1. 3a + b − c 3b + c − a 3c + a − b

Solution. Write the inequality as follows X

‹ a 1 1 − ≥ , 3a + b − c 4 4

X a−b+c ≥ 1. 3a + b − c Applying the Cauchy-Schwarz inequality, we get P P X a−b+c [ (a − b + c)]2 ( a)2 P ≥P =P = 1. 3a + b − c (a − b + c)(3a + b − c) a2 + 2 a b The equality holds for a = b = c.

P 1.137. If a, b, c are the lengths of the sides of a triangle, then a2 − b2 b2 − c 2 c 2 − a2 + + ≤ 0. a2 + bc b2 + ca c 2 + a b (Vasile Cîrtoaje, 2007)

Cyclic Inequalities

187

First Solution. Suppose that a = max{a, b, c}. Since c 2 − a2 = −(a2 − b2 ) − (b2 − c 2 ), the inequality can be written as follows  ‹  ‹ 1 1 1 1 (a2 − b2 ) 2 − 2 + (b2 − c 2 ) 2 − 2 ≤ 0, a + bc c + a b b + ca c + a b (a2 − b2 )(a − c)(a − b + c) (b2 − c 2 )(b − c)(b + c − a) − ≤ 0. a2 + bc a2 + bc The equality holds for an equilateral triangle, and also for a degenerate triangle having a side equal to zero. −

Second Solution. The sequences {a2 , and

c2}

ª 1 c2 + a b 1 1 1 are opposite arranged. Indeed, if a ≥ b ≥ c, then 2 ≤ 2 ≤ 2 , because a + bc b + ca c + ab §

1 , 2 a + bc

b2 ,

1 , 2 b + ca

b2

1 (a − b)(a + b − c) 1 − 2 = 2 ≥ 0, + ca a + bc (b + ca)(a2 + bc)

c2

1 1 (b − c)(b + c − a) − 2 = 2 ≥ 0. + a b b + ca (c + a b)(b2 + ca)

Then, by the rearrangement inequality, we have a2 ·

a2

1 1 1 + b2 · 2 + c2 · 2 ≤ + bc b + ca c + ab

1 1 1 + y2 · 2 + z2 · 2 , + bc b + ca c + ab for any permutation (x, y, z) of (a, b, c). Putting x = b, y = c, z = a, we get the desired inequality. ≤ x2 ·

a2

P 1.138. If a, b, c are the lengths of the sides of a triangle, then a2 (a + b)(b − c) + b2 (b + c)(c − a) + c 2 (c + a)(a − b) ≥ 0. (Vasile Cîrtoaje, 2006)

188

Vasile Cîrtoaje

First Solution. We write the inequality as a2 b2 + b2 c 2 + c 2 a2 − a bc(a + b + c) ≥ a b3 + bc 3 + ca3 − a3 b − b3 c − c 3 a, a2 (b − c)2 + b2 (c − a)2 + c 2 (a − b)2 ≥ 2(a + b + c)(a − b)(b − c)(c − a). Using the classical substitution a = y + z, b = z + x, c = x + y (x, y, z ≥ 0), we have a2 (b − c)2 + b2 (c − a)2 + c 2 (a − b)2 = ( y 2 − z 2 )2 + (z 2 − x 2 )2 + (x 2 − y 2 )2 = 2(x 4 + y 4 + z 4 − x 2 y 2 − y 2 z 2 − z 2 x 2 ) and 2(a + b + c)(a − b)(b − c)(c − a) = 4(x + y + z)( y − x)(z − y)(x − z) = 4(x 3 y + y 3 z + z 3 x − x y 3 − yz 3 − z x 3 ). Thus, the desired inequality can be restated as x 4 + y 4 + z 4 − x 2 y 2 − y 2 z 2 − z 2 x 2 ≥ 2(x 3 y + y 3 z + z 3 x − x y 3 − yz 3 − z x 3 ), which is just the inequality in P 3.93 from Volume 1. The equality holds for an equilateral a b c triangle, and also for a degenerate triangle with = p = p (or any cyclic 2 1+ 5 3+ 5 permutation). Second Solution. Write the inequality as follows b2 c 2 + c 2 a2 + a2 b2 ≥ a b(b2 + c 2 − a2 ) + bc(c 2 + a2 − b2 ) + ca(a2 + b2 − c 2 ), bc ca a b + + ≥ 2b cos A + 2c cos B + 2a cos C, a b c x 2 + y 2 + z 2 ≥ 2 yz cos A + 2z x cos B + 2x y cos C, (x − y cos C − z cos B)2 + ( y sin C − z sin B)2 ≥ 0, where x=

s

ca , b

y=

v t ab c

, z=

v t bc a

.

P 1.139. If a, b, c are the lengths of the sides of a triangle, then  ‹     2 c 2 b 2 a a −1 + b −1 +c − 1 ≥ 0. c a b (Vasile Cîrtoaje, Moldova TST, 2006)

Cyclic Inequalities

189

First Solution. Using the substitution a =

1 1 1 , b = and c = , the inequality becomes x y z

E(x, y, z) ≥ 0, where E(x, y, z) = yz 2 (z − y) + z x 2 (x − z) + x y 2 ( y − x). Without loss of generality, assume that x = max{x, y, z}; that is, a = min{a, b, c}. We will show that E(x, y, z) ≥ E( y, y, z) ≥ 0. We have E(x, y, z) − E( y, y, z) = z(x 3 − y 3 ) − z 2 (x 2 − y 2 ) + y 3 (x − y) − y 2 (x 2 − y 2 ) = (x − y)(x − z)(xz + yz − y 2 ) ≥ 0, because xz + yz − y 2 ≥ y(2z − y) =

(b − a) + (a + b − c) 2b − c = > 0. 2 b c b2 c

Also, E( y, y, z) = yz( y − z)2 ≥ 0. The equality holds for a = b = c. Second Solution. Write the inequality as F (a, b, c) ≥ 0, where F (a, b, c) = a3 b2 + b3 c 2 + c 3 a2 − a bc(a2 + b2 + c 2 ). Since 2E(a, b, c) =

€X

a3 b2 + =

and

X

X

X

a3 c 2 − 2a bc

X

X Š €X Š a2 − a2 b3 − a2 c 3

a3 (b − c)2 −

X

a2 (b3 − c 3 )

a2 (b3 − c 3 ) =

X

a2 (b − c)3 ,

we get E(a, b, c) =

X

a2 (b − c)2 (a − b + c) ≥ 0.

Third Solution. Without loss of generality, assume that b is between a and c; that is (a − b)(b − c) ≥ 0. By the Cauchy-Schwarz inequality, we have X a2 b c

P ≥ P

a2 b


2

a2 bc

.

190

Vasile Cîrtoaje

Therefore, it suffices to show that €X

a2 b

Š2

≥ a bc(a + b + c)(a2 + b2 + c 2 ).

On the other hand, by the AM-GM inequality, we have 4a bc(a + b + c)(a2 + b2 + c 2 ) ≤ [ac(a + b + c) + b(a2 + b2 + c 2 )]2 . Thus, we only need to show that X 2 a2 b ≥ ac(a + b + c) + b(a2 + b2 + c 2 ), which is equivalent to b[a2 − (b − c)2 ] − ac(a + b − c) ≥ 0, (a + b − c)(a − b)(b − c) ≥ 0.

P 1.140. If a, b, c are the lengths of the sides of a triangle, then (a)

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

(b)

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

(c)

 ‹ a3 b + b3 c + c 3 a+b+c 4 ≥ . 3 3

Solution. (a) First Solution. Write the inequality as a2 b(a − b) + b2 c(b − c) + c 2 a(c − a) ≥ 0. Using the classical substitution a = y +z, b = z+ x, c = x + y (x, y, z ≥ 0), the inequality turns into x y 3 + yz 3 + z x 3 ≥ x yz(x + y + z), which follows from the Cauchy-Schwarz inequality (z x 3 + x y 3 + yz 3 )( y + z + x) ≥ x yz(x + y + z)2 . The equality holds for an equilateral triangle, and also for a degenerate triangle with a = 0 and b = c (or any cyclic permutation).

Cyclic Inequalities

191

Second Solution. Multiplying by a + b + c, the inequality becomes as follows X X X X a4 b + a bc a2 ≥ a2 b3 + a bc a b, X

b4 c + a bc

X

a2 ≥

X

b2 c 3 + a bc

X

a b,

X b3 X X bc 2 X + a2 ≥ + a b, a a X Xb (c 2 + a2 − b2 ), a2 ≥ a a2 + b2 + c 2 ≥ 2bc cos B + 2ca cos C + 2a b cos A, (a − b cos A − c cos C)2 + (b sin A − c sin C)2 ≥ 0.

(b) Write the inequality as X X a2 b(a − b) + b2 (a − b)(a − c) ≥ 0. Since

P

a2 b(a − b) ≥ 0 (according to the inequality in (a)), it suffices to show that X b2 (a − b)(a − c) ≥ 0.

This is a particular case of the following inequality (x − y)(x − z)a2 + ( y − z)( y − x)b2 + (z − x)(z − y)c 2 ≥ 0, where x, y, z are real numbers. If two of x, y, z are equal, then the inequality is trivial. Otherwise, assume that x > y > z and write the inequality as a2 c2 b2 + ≥ . y −z x − y x −z Applying the Cauchy-Schwarz inequality, we get a2 c2 (a + c)2 (a + c)2 b2 + ≥ = ≥ . y −z x − y ( y − z) + (x − y) x −z x −z The equality holds for a = b = c. (c) According to (a), it suffices to show that 9(a b + bc + ca)(a2 + b2 + c 2 ) ≥ (a + b + c)4 . This is equivalent to (A − B)(4B − A) ≥ 0,

192

Vasile Cîrtoaje

where A = a2 + b2 + c 2 and B = a b + bc + ca. Since A ≥ B and 4B − A > 2(a b + bc + ca) − a2 − b2 − c 2 p p p p p p p p p p p p = ( a + b + c)(− a + b + c)( a − b + c)( a + b − c) ≥ 0, the conclusion follows. The equality holds for a = b = c.

P 1.141. If a, b, c are the lengths of the sides of a triangle, then  2  a b2 c 2 a2 b2 c 2 2 ≥ + + + + + 3. b2 c 2 a2 a2 b2 c 2 Solution. Write the inequality as follows X a2 b2

≥3+

X b2 X a2 − , a2 b2

X b2 X c 2 X a2 ≥ 3 + − , c2 b2 b2  X b2 X  c 2 a2 1+ 2 − 2 , ≥ c2 b b X b2 Xc ≥2 cos A. c2 b Putting x=

b , c

y=

c a , z= , a b

we have x yz = 1 and c 1 = = yz, b x

a 1 = = z x, c y

b 1 = = x y. a z

Therefore, we can write the inequality as x 2 + y 2 + z 2 ≥ 2 yz cos A + 2z x cos B + 2x y cos C, which is equivalent to the obvious inequality (x − y cos C − z cos B)2 + ( y sin C − z sin B)2 ≥ 0. The equality occurs for a = b = c.

Cyclic Inequalities

193

P 1.142. If a, b, c are the lengths of the sides of a triangle such that a < b < c, then a2 b2 c2 + + ≤ 0. a2 − b2 b2 − c 2 c 2 − a2 (Vasile Cîrtoaje, 2003) Solution. Write the inequality as b2 c2 a2 + ≥ . b2 − a2 c 2 − b2 c 2 − a2 Since c ≤ a + b, it suffices to show that a2 b2 (a + b)2 + ≥ . b2 − a2 c 2 − b2 c 2 − a2 We can rewrite this inequality as follows   ‹ ‹ 1 1 1 1 2a b 2 2 a − 2 +b − 2 ≥ 2 , 2 2 2 2 2 2 b −a c −a c −b c −a c − a2 a2 (c 2 − b2 ) b2 (b2 − a2 ) + ≥ 2a b, b2 − a2 c 2 − b2 v ‚ v Œ t c 2 − b2 t b2 − a2 2 a −b ≥ 0. b2 − a2 c 2 − b2 The equality occurs for a degenerate triangle with c = a + b and b = x a, where x ≈ 0.53209 is the positive root of the equation x 3 + 3x 2 − 1 = 0.

P 1.143. If a, b, c are the lengths of the sides of a triangle, then  ‹ a+b b+c c+a a b c + + +3≥2 + + . b c a b+c c+a a+b (Manlio Marangelli, 2008) Solution (by Vo Quoc Ba Can). Since X a+b b+c

=

X X a−c a−c 1+ =3+ , b+c b+c

we can write the desired inequality as Xa X a−c −3≥2 . b b+c

194

Vasile Cîrtoaje

Since (a b + bc + ca)

 X X X a2 c −3 = a2 − 2 ab + b b

X a

and (a b + bc + ca)

X a−c b+c

= [a(b + c) + bc]

X a−c b+c

=

X

a2 −

X

ab +

X bc(a − c) , b+c

the inequality becomes X a2 c b Since

+2

X bc(c − a) X ≥ a2 . b+c

X a2 c

X

a2 b (see the inequality from P 1.139), we only need to show that ≥

X bc(c − a) ≥ 0, b+c which is equivalent to X

bc(c 2 − a2 )(a + b) ≥ 0, ‹  X b ≥ 0, (c 2 − a2 ) 1 + a X b (c 2 − a2 ) ≥ 0, a X bc 2 X ≥ a b. a Indeed, according to P 1.139, we have X bc 2 X X ≥ a2 ≥ a b. a The equality holds for a = b = c.

P 1.144. Let a, b, c be the lengths of the sides of a triangle. If k ≥ 2, then a k b(a − b) + b k c(b − c) + c k a(c − a) ≥ 0. (Vasile Cîrtoaje, 1986)

Cyclic Inequalities

195

Solution (by Darij Grinberg). We will prove that if f is an increasing nonnegative function on [0, ∞), then E(a, b, c) ≥ 0, where E(a, b, c) = a2 b f (a)a − b) + b2 c f (b)(b − c) + c 2 a f (c)(c − a). For f (x) = x k−2 , k ≥ 2, we get the original inequality. In addition, for k = 2, we get the known inequality (a) in P 1.140. In order to prove the above generalization, assume that a = max{a, b, c}. There are two cases to consider. Case 1: a ≥ b ≥ c. Since f (a) ≥ f (b) ≥ f (c) ≥ 0, we have E(a, b, c) ≥ a2 b f (c)(a − b) + b2 c f (c)(b − c) + c 2 a f (c)(c − a) = f (c)[a2 b(a − b) + b2 c(b − c) + c 2 a(c − a)] ≥ 0.

Case 2: a ≥ c ≥ b. Since f (a) ≥ f (c) ≥ f (b) ≥ 0, we have E(a, b, c) ≥ a2 b f (a)(a − b) + b2 c f (a)(b − c) + c 2 a f (a)(c − a) = f (a)[a2 b(a − b) + b2 c(b − c) + c 2 a(c − a)] ≥ 0. The equality holds for a = b = c, and also for a degenerate triangle with a = 0 and b = c (or any cyclic permutation).

P 1.145. Let a, b, c be the lengths of the sides of a triangle. If k ≥ 2, then 3(a k b + b k c + c k a) ≥ (a + b + c)(a k−1 b + b k−1 c + c k−1 a). Solution. For k = 2, the inequality is equivalent to ‹  b c a a b c 2 + + ≥ + + + 3. a b c b c a Assuming that a = min{a, b, c} and making the substitution b = x+ becomes

a+c , this inequality 2

‹ 3a (2c − a)x + x + (a − c)2 ≥ 0. 4 2



It is true since 4x + 3a = a + 4b − 2c = 2(a + b − c) + (2b − a) > 0.

196

Vasile Cîrtoaje

In order to prove the desired inequality for k > 2, we rewrite it as a k−1 b(2a − b − c) + b k−1 c(2b − c − a) + c k−1 a(2c − a − b) ≥ 0. We will prove that if f is an increasing nonnegative function on [0, ∞), then E(a, b, c) ≥ 0, where E(a, b, c) = a b(2a − b − c) f (a) + bc(2b − c − a) f (b) + ca(2c − a − b) f (c). For f (x) = x k−2 , k ≥ 2, we get the original inequality. In order to prove this generalization, assume that a = max{a, b, c}. There are two cases to consider. Case 1: a ≥ b ≥ c. Since f (a) ≥ f (b) ≥ f (c) ≥ 0, we have E(a, b, c) ≥ a b(2a − b − c) f (b) + bc(2b − c − a) f (b) + ca(2c − a − b) f (c) = b[2(a − b)(a − c) + a b − c 2 ] f (b) + ca(2c − a − b) f (c) ≥ b[2(a − b)(a − c) + a b − c 2 ] f (c) + ca(2c − a − b) f (c) •  ‹  ‹ ˜ b c a a b c = a bc 2 + + − + + − 3 f (c) ≥ 0. a b c b c a Case 2: a ≥ c ≥ b. Since f (a) ≥ f (c) ≥ f (b) ≥ 0, we have E(a, b, c) ≥ a b(2a − b − c) f (c) + bc(2b − c − a) f (b) + ca(2c − a − b) f (c) = a[(c − b)(2c − a) + b(a − b)] f (c) + bc(2b − c − a) f (b). Since (c − b)(2c − a) + b(a − b) ≥ (c − b)(b + c − a) + b(a − b) ≥ 0, we get E(a, b, c) ≥ a[(c − b)(2c − a) + b(a − b)] f (b) + bc(2b − c − a) f (b)  ‹  ‹ b c a a b c = a bc[2 + + − + + − 3] f (b) ≥ 0. a b c b c a The equality holds for a = b = c. Remark. The following inequality  ‹  ‹ b c a a b c 3 + + ≥2 + + +3 a b c b c a is sharper than the original inequality for k = 2. Using the substitution b = x + we obtain the inequality c (3c − 2a)x + x + a − (a − c)2 ≥ 0, . 4 2



a+c , 2

Cyclic Inequalities

197

which is true since, on the assumption a = min{a, b, c}, we have 3c − 2a > 0 and 4x + 4a − c = 2a + 4b − 3c = 3(a + b − c) + (b − a) > 0.

P 1.146. Let a, b, c, d be positive real numbers such that a + b + c + d = 4. Prove that b c d a + + + ≥ 1. 3+ b 3+c 3+d 3+a Solution. By the Cauchy-Schwarz inequality, we have P X a ( a)2 16 P . ≥P = 3+ b a(3 + b) 12 + a b Therefore, it suffices to show that a b + bc + cd + d a ≤ 4. Indeed, (a + c) + (b + d) a b + bc + cd + d a = (a + c)(b + d) ≤ 2 •

˜2

= 2.

The equality occurs for a = b = c = d = 1.

P 1.147. Let a, b, c, d be positive real numbers such that a + b + c + d = 4. Prove that a b c d + + + ≥ 2. 2 2 2 1+ b 1+c 1+d 1 + a2 Solution. Since

a a b2 = a − , 1 + b2 1 + b2

the inequality is equivalent to a b2 bc 2 cd 2 d a2 + + + ≤ 2. 1 + b2 1 + c 2 1 + d 2 1 + a2 Since

a b2 a b2 ab ≤ = , 2 1+ b 2b 2

198

Vasile Cîrtoaje

it suffices to show that a b + bc + cd + d a ≤ 4. Indeed, we have (a + c) + (b + d) a b + bc + cd + d a = (a + c)(b + d) ≤ 2 •

˜2

= 2.

The equality occurs for a = b = c = d = 1.

P 1.148. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then a2 bc + b2 cd + c 2 d a + d 2 a b ≤ 4. (Song Yoon Kim, 2006) Solution. Let (x, y, z, t) be a permutation of (a, b, c, d) such that x ≥ y ≥ z ≥ t. Since x yz ≥ x y t ≥ xz t ≥ yz t, by the rearrangement inequality, we have a2 bc + b2 cd + c 2 d a + d 2 a b ≤ x · x yz + y · x y t + z · xz t + t · yz t = (x y + z t)(xz + y t). Consequently, it suffices to show that x + y + z + t = 4 involves (x y + z t)(xz + y t) ≤ 4. Indeed, by the AM-GM inequality, we have (x y + z t)(xz + y t) ≤

1 1 (x y + z t + xz + y t)2 = (x + t)2 ( y + z)2 ≤ 4, 4 4

because

1 (x + t + y + z)2 = 4. 4 The equality holds for a = b = c = d = 1, and also for a = 2, b = c = 1 and d = 0 (or any cyclic permutation). (x + t)( y + z) ≤

P 1.149. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then a(b + c)2 + b(c + d)2 + c(d + a)2 + d(a + b)2 ≤ 16.

Cyclic Inequalities

199

Solution (by Vo Quoc Ba Can). Write the inequality as (a + b + c + d)3 ≥ 4[a(b + c)2 + b(c + d)2 + c(d + a)2 + d(a + b)2 ]. Since (a + b + c + d)2 ≥ 4(a + b)(c + d), we have (a + b + c + d)3 ≥ 4(a + b)(c + d)(a + b + c + d) = 4(c + d)(a + b)2 + 4(a + b)(c + d)2 . Therefore, it suffices to show that 4(c + d)(a + b)2 + 4(a + b)(c + d)2 ≥ 4[a(b + c)2 + b(c + d)2 + c(d + a)2 + d(a + b)2 ], which is equivalent to c(a + b)2 + a(c + d)2 ≥ a(b + c)2 + c(d + a)2 , a[(c + d)2 − (b + c)2 ] + c[(a + b)2 − (d + a)2 ] ≥ 0, (b + d)(b − d)(c − a) ≥ 0. Similarly, since (a + b + c + d)2 ≥ 4(b + c)(d + a), we can show that it suffices to show that (c + a)(c − a)(d − b) ≥ 0. Therefore, the desired inequality holds if one of the inequalities (b − d)(c − a) ≥ 0 and (c − a)(d − b) ≥ 0, which is obvious. The equality holds for a = c and b = d.

P 1.150. If a, b, c, d are positive real numbers, then a−b b−c c−d d−a + + + ≥ 0. b+c c+d d+a a+b Solution. We have  ‹ a−b c−d a+c a+c 1 1 + = + − 2 = (a + c) + −2 b+c d+a b+c d+a b+c d+a ≥ (a + c)

4 4(a + c) −2= − 2. (b + c) + (d + a) a+b+c+d

200

Vasile Cîrtoaje

Similarly,

b−c d−a 4(b + d) + ≥ − 2. c+d a+b a+b+c+d Adding these inequalities yields the desired inequality. The equality holds for a = c and b = d. Conjecture. If a, b, c, d, e are positive real numbers, then a−b b−c c−d d−e e−a + + + + ≥ 0. b+c c+d d+e e+a a+b

P 1.151. If a, b, c, d are positive real numbers, then (a)

a−b b−c c−d d−a + + + ≥ 0; a + 2b + c b + 2c + d c + 2d + a d + 2a + b

(b)

a b c d + + + ≤ 1. 2a + b + c 2b + c + d 2c + d + a 2d + a + b

Solution. (a) Write the first inequality as ‹ X a − b 1 + ≥ 2, a + 2b + c 2 X 3a + c ≥ 4. a + 2b + c By the Cauchy-Schwarz inequality, we get P P X 3a + c [ (3a + c)]2 16( a)2 P ≥P = = 4. a + 2b + c (3a + c)(a + 2b + c) 4( a)2 The equality holds for a = b = c = d. (b) Write the inequality as b+c c+d d+a a+b + + + ≥ 2. 2a + b + c 2b + c + d 2c + d + a 2d + a + b By the Cauchy-Schwarz inequality, we get P P X b+c [ (b + c)]2 4( a)2 ≥P = P = 1. 2a + b + c (b + c)(2a + b + c) 4( a)2 The equality holds for a = b = c = d.

Cyclic Inequalities

201

Conjecture 1. If a, b, c, d, e are positive real numbers, then b−c c−d d−e e−a a−b + + + + ≥ 0. a + 2b + c b + 2c + d c + 2d + e d + 2e + a e + 2a + b

Conjecture 2 (by Ando). If a1 , a2 , . . . , an (n ≥ 4) are positive real numbers, then an a1 a2 + + ··· + ≤ 1. (n − 2)a1 + a2 + a3 (n − 2)a2 + a3 + a4 (n − 2)an + a1 + a2

P 1.152. If a, b, c, d are positive real numbers such that a bcd = 1, then 1 1 1 1 + + + ≥ 2. a(a + b) b(b + c) c(c + d) d(d + a) (Vasile Cîrtoaje, 2007) Solution. Making the substitution a=

s

y , x

b=

v tz y

c=

,

s

t , z

d=

s

x , t

where x, y, z, t are positive real numbers, the inequality can be rewritten as y+

x p

xz

Since

+

z+

y p

yt

+

z t ≥ 2. p + p t + zx x + t y

p p 2 xz ≤ x + z, 2 y t ≤ y + t,

it suffices to show that y x z t + + + ≥ 1. x + 2y + z y + 2z + t z + 2t + x t + 2x + y By the Cauchy-Schwarz inequality, we have X

P ( x)2 x ≥P = 1. z + 2y + z x(x + 2 y + z)

The equality holds for a = c =

1 1 = . b d

202

Vasile Cîrtoaje

Conjecture 1. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then 1 1 n 1 + + ··· + ≥ . a1 (a1 + a2 ) a2 (a2 + a3 ) an (an + a1 ) 2 Remark 1. Using the substitution a1 =

x3 x1 x2 , a2 = , · · · , an = , x1 x2 xn

the above inequality becomes x 12 x 22 + x 1 x 3

+

x 22 x 32 + x 2 x 4

+ ··· +

x n2 x 12 + x n x 2

≥

n , 2

where x 1 , x 2 , · · · , x n > 0. This cyclic inequality is similar to Shapiro’s inequality xn x1 x2 n + + ··· + ≥ , x2 + x3 x3 + x4 x1 + x2 2 which is true for even n ≤ 12 and for odd n ≤ 23. Conjecture 2. If a1 , a2 , . . . , an are positive real numbers, then 1 a12 + a1 a2

+

1 a22 + a2 a3

+ ··· +

n2 1 ≥ . an2 + an a1 2(a1 a2 + a2 a3 + · · · + an a1 )

Remark 2. By the AM-GM inequality, we have a1 a2 + a2 a3 + · · · + an a1 ≥ n

q n

a12 a22 · · · an2 .

Thus, the inequality in Conjecture 1 is sharper than the inequality in Conjecture 2. Therefore, if Conjecture 1 is true, then Conjecture 2 is also true.

P 1.153. If a, b, c, d are positive real numbers, then (a)

1 1 1 1 16 + + + ≥ ; p a(1 + b) b(1 + c) c(1 + d) d(1 + a) 1 + 8 a bcd

(b)

1 1 1 1 16 + + + ≥ . p a(1 + b) b(1 + a) c(1 + d) d(1 + c) 1 + 8 a bcd (Pham Kim Hung and Vasile Cîrtoaje, 2007)

Cyclic Inequalities Solution. (a) Let p =

203 p 4

a bcd. Putting

a=p

x2 , x1

b=p

x3 , x2

c=p

x4 , x3

d=p

x1 , x4

where x 1 , x 2 , x 3 , x 4 are positive real numbers, the inequality turns into X

x1 16p ≥ . x2 + p x3 1 + 8p2

By the Cauchy-Schwarz inequality, we have P P X ( x 1 )2 ( x 1 )2 x1 ≥P = . x2 + p x3 x 1 (x 2 + p x 3 ) (x 1 + x 3 )(x 2 + x 4 ) + 2p(x 1 x 3 + x 2 x 4 ) Since x1 x3 + x2 x4 ≤ it suffices to show that

 x + x 2  x + x 2 2 4 1 3 , + 2 2

8p (A + B)2 ≥ , 2AB + p(A2 + B 2 ) 1 + 8p2

where A = x 1 + x 3 , B = x 2 + x 4 . This inequality is equivalent to A2 + B 2 + 2(8p2 − 8p + 1)AB ≥ 0, which is true because A2 + B 2 + 2(8p2 − 8p + 1)AB ≥ 2AB + 2(8p2 − 8p + 1)AB = 4(2p − 1)2 AB ≥ 0. The equality holds for a = b = c = d =

1 . 2

(b) Write the inequality as a + b + 2a b c + d + 2cd 16 + ≥ . p a b(1 + a)(1 + b) cd(1 + c)(1 + d) 1 + 8 a bcd We claim that a b ≥ 1 involves a + b + 2a b 1 ≥ , a b(1 + a)(1 + b) ab and a b ≤ 1 involves

a + b + 2a b 2 ≥p . a b(1 + a)(1 + b) ab + ab The first inequality is obvious, while the second inequality is equivalent to p p p (1 − a b)( a − b)2 ≥ 0.

204

Vasile Cîrtoaje

Similarly, we have

1 c + d + 2cd ≥ cd(1 + d)(1 + d) cd

for cd ≥ 1, and

2 c + d + 2cd ≥p cd(1 + d)(1 + d) cd + cd

for cd ≤ 1. Let

x=

p

a b,

y=

p

cd.

There are four cases to consider. Case 1: x ≥ 1, y ≥ 1. It suffices to show that 1 16 1 + 2≥ . 2 x y 1 + 8x y Indeed, we have 1 1 2 16 + 2≥ > . x2 y xy 1 + 8x y Case 2: x ≤ 1, y ≤ 1. It suffices to show that 1 1 8 + ≥ . 2 2 x+x y+y 1 + 8x y Putting s = x + y and p =

p

x y, this inequality becomes s2 + s − 2p2 8 ≥ , 2 2 p (s + p + 1) 1 + 8p2

(1 + 8p2 )s2 + s − 24p4 − 10p2 ≥ 0. Since s ≥ 2p, we get (1 + 8p2 )s2 + s − 24p4 − 10p2 ≥ 4(1 + 8p2 )p2 + 2p − 24p4 − 10p2 = p(p + 1)(2p − 1)2 ≥ 0. Case 3: x ≥ 1, y ≤ 1. It suffices to show that 1 2 16 + ≥ . 2 2 x y+y 1 + 8x y This inequality is equivalent in succession to (1 + 8x y)(2x 2 + y 2 + y) ≥ 16x 2 y(1 + y), (1 + 8x y)(x − y)2 + 8x 3 y + 8x y 2 − 16x 2 y + 2x y + x 2 + y ≥ 0,

Cyclic Inequalities

205

(1 + 8x y)(x − y)2 + 8x y(x − 1)2 + 8x y 2 + x 2 + y ≥ 6x y. The last inequality is true since the AM-GM inequality yields 8x y 2 + x 2 + y ≥ 3

Æ 3

8x y 2 · x 2 · y = 3

Æ 3

8x 3 y 3 = 6x y.

Case 4: x ≤ 1, y ≥ 1. This case is similar to the case 3. The proof is completed. The equality holds for a = b = c = d =

1 . 2

P 1.154. If a, b, c, d are nonnegative real numbers such that a2 + b2 + c 2 + d 2 = 4, then (a) (b)

3(a + b + c + d) ≥ 2(a b + bc + cd + d a) + 4; a + b + c + d − 4 ≥ (2 −

p 2)(a b + bc + cd + d a − 4). (Vasile Cîrtoaje, 2006)

Solution. Let p = a + b + c + d. By the Cauchy-Schwarz inequality (1 + 1 + 1 + 1)(a2 + b2 + c 2 + d 2 ) ≥ (a + b + c + d)2 , we get p ≤ 4, and by the inequality (a + b + c + d)2 ≥ a2 + b2 + c 2 + d 2 , we get p ≥ 2. In addition, we have a b + bc + cd + d a = (a + c)(b + d) ≤

p2 (a + c + b + d)2 = . 4 4

(a) It suffices to show that 3p ≥

p2 + 4. 2

Indeed, 3p −

p2 (4 − p)(p − 2) −4= ≥ 0. 2 2

The equality holds for a = b = c = d = 1. (b) It suffices to show that  2  p p p − 4 ≥ (2 − 2) −4 . 4

206

Vasile Cîrtoaje

This inequality is equivalent to p (4 − p)(p − 2 2) ≥ 0, p p which is true for p ≥ 2 2. So, it remains to consider the case 2 ≤ p < 2 2. Since 2(a b + bc + cd + d a) ≤ (a + b + c + d)2 − (a2 + b2 + c 2 + d 2 ) = p2 − 4, it is enough to prove that  2  p p −4 −4 . p − 4 ≥ (2 − 2) 2 Write this inequality as 1 p (p2 − 12), 2+ 2 which is equivalent to the obvious inequality p p (2 2 − p)(p − 2 + 2) ≥ 0. p−4≥

The equality holds for a = b = c = d = 1, and also for a = b = 0 and c = d = any cyclic permutation).

p 2 (or

P 1.155. Let a, b, c, d be positive real numbers. (a) If a, b, c, d ≥ 1, then  ‹ ‹ ‹ ‹  ‹ 1 1 1 1 1 1 1 1 b+ c+ d+ ≥ (a + b + c + d) + + + ; a+ b c d a a b c d (b) If a bcd = 1, then  ‹ ‹ ‹ ‹  ‹ 1 1 1 1 1 1 1 1 a+ b+ c+ d+ ≤ (a + b + c + d) + + + . b c d a a b c d (Vasile Cîrtoaje and Ji Chen, 2011) Solution. Let A =(1 + a b)(1 + bc)(1 + cd)(1 + d a) X X X =1 + ab + a2 bd + 2a bcd + a bcd a b + a2 b2 c 2 d 2 X X =(1 − a bcd)2 + 4a bcd + (1 + a bcd) ab + a2 bd X =(1 − a bcd)2 + 4a bcd + (1 + a bcd)(a + c)(b + d) + a2 bd

Cyclic Inequalities

207

and B =(a + b + c + d)(a bc + bcd + cd a + d a b) X =4a bcd + a2 (bc + cd + d b) X X =4a bcd + a2 c(b + d) + a2 bd X =4a bcd + (ac + bd)(a + c)(b + d) + a2 bd. Thus, A − B = (1 − a bcd)2 + (1 + a bcd)(a + c)(b + d) − (ac + bd)(a + c)(b + d) = (1 − a bcd)2 + (1 − ac)(1 − bd)(a + c)(b + d).

(a) The inequality A ≥ B is clearly true for a, b, c, d ≥ 1. The equality holds for a = b = c = d = 1. (b) For a bcd = 1, we have B−A=

1 (1 − ac)2 (a + c)(b + d) ≥ 0. ac

The equality holds for ac = bd = 1.

P 1.156. If a, b, c, d are positive real numbers, then  1+

 ‹2   ‹2 b c 2 d a 2 + 1+ + 1+ + 1+ > 7. a+b b+c c+d d+a (Vasile Cîrtoaje, 2012)

First Solution. Assume that d = max{a, b, c, d}. We get the desired inequality by summing the inequalities  1+

 ‹2  a 2 b c 2 + 1+ + 1+ >6 a+b b+c c+a

 1+

 ‹2  c 2 d c 2 >1+ 1+ + 1+ . c+d d+a c+a

and

Let x=

a−b , a+b

y=

b−c c−a , z= . b+c c+a

208

Vasile Cîrtoaje

We have −1 < x, y, z < 1 and x + y + z + x yz = 0. Since

y +1 a x +1 b = , = , a+b 2 b+c 2 we can write the first inequality as follows

c z+1 = , c+a 2

(x + 3)2 + ( y + 3)2 + (z + 3)2 > 24, x 2 + y 2 + z 2 + 6(x + y + z) + 3 > 0, x 2 + y 2 + z 2 + 3 > 6x yz. By the AM-GM inequality, we have x 2 + y 2 + z2 + 3 ≥ 6

Æ 6

x 2 y 2 z 2 > 6x yz.

Write now the second inequality as  ‹ ‹  c 2 c d c d 1+ −1> − 2+ + . c+d c+a d+a c+a d+a Since

c d a(c − d) − = ≤ 0, c+a d+a (c + a)(d + a)

we have  1+

 ‹ ‹ c d c d c 2 −1>0≥ − 2+ + . c+d c+a d+a c+a d+a

Second Solution. Making the inequality (1 + x)2 > 1 + 3x 2 , 0 < x < 1, we have  ‹2   ‹2 a 2 b c 2 d + 1+ + 1+ + 1+ > a+b b+c c+d d+a   ‹2   ‹2  b d a 2 c 2 >4+3 + + + . a+b b+c c+d d+a

 1+

Therefore, it suffices to prove that  a 2  b ‹2  c 2  d ‹2 + + + ≥ 1, a+b b+c c+d d+a which is just the known inequality in P 1.176 from Volume 2.

Cyclic Inequalities

209

P 1.157. If a, b, c, d are positive real numbers, then b2 − ca c2 − d b d 2 − ac a2 − bd + + + ≥ 0. b + 2c + d c + 2d + a d + 2a + b a + 2b + c (Vo Quoc Ba Can, 2009) Solution. Write the inequality as follows  X  4a2 − 4bd + b + d − 2a ≥ 0, b + 2c + d X (b − d)2 + 2(a − c)(2a − b − d) b + 2c + d

≥ 0.

Clearly, it suffices to show that X (a − c)(2a − b − d) b + 2c + d

≥ 0.

This inequality is equivalent to  ‹  ‹ 2a − b − d 2c − d − b 2b − c − a 2d − a − c (a − c) − + (b − d) − ≥ 0, b + 2c + d d + 2a + b c + 2d + a a + 2b + c which can be written in the obvious form (b − d)(b2 − d 2 ) (a − c)(a2 − c 2 ) + ≥ 0. (b + 2c + d)(d + 2a + b) (c + 2d + a)(a + 2b + c) The equality occurs for a = c and b = d.

P 1.158. If a, b, c, d are positive real numbers such that a ≤ b ≤ c ≤ d, then v v v v t 2b t 2d t 2a t 2c + + + ≤ 4. a+b b+c c+d d+a (Vasile Cîrtoaje, 2009) Solution. According to the inequality in P 1.70, we have v v v t 2b t 2a t 2c + + ≤ 3. a+b b+c c+a Therefore, it suffices to show that v v v t 2d t 2c t 2c + ≤1+ . c+d d+a c+a

210

Vasile Cîrtoaje

By squaring, this inequality becomes v v t 2c t 2d 2c 2c 4cd + +2 ≤1+ +2 . c+d d+a (c + d)(d + a) c+a c+a We can get it by summing the inequalities 2c 2d 2c + ≤1+ , c+d d+a c+a v v t 2c t 4cd 2 ≤2 . (c + d)(d + a) c+a The former inequality is true since 2d 2c (a − d)(d − c)(c − a) 2c + −1− = ≤ 0, c+d d+a c+a (c + d)(d + a)(a + c) while the second inequality reduces to c(a − d)(d − c) ≤ 0. The equality holds for a = b = c = d.

P 1.159. Let a, b, c, d be nonnegative real numbers, and let x=

a , b+c

y=

b c , z= , c+d d+a

t=

d . a+b

Prove that p

(a) (b)

xz +

p

y t ≤ 1;

x + y + z + t + 4(xz + y t) ≥ 4. (Vasile Cîrtoaje, 2004)

Solution. (a) Using the Cauchy-Schwarz inequality, we have p p p p ac bd xz + y t = + (b + c)(d + a) (c + d)(a + b) p p ac bd ≤p +p = 1. p p ac + bd ac + bd The equality holds for a = b = c = d, for a = c = 0, and for b = d = 0 (b) Write the inequality as A + B ≥ 6,

Cyclic Inequalities

211

where (a + b)(c + d) + (a + c)2 + a b + 2ac + cd (b + c)(d + a) (a + c)2 a c (a + b)(c + d) + + + , = (b + c)(d + a) (b + c)(d + a) d + a b + c

A = x + z + 4xz + 1 =

B = y + t + 4yt + 1 = Since

(b + c)(d + a) (b + d)2 b d + + + . (c + d)(a + b) (c + d)(a + b) a + b c + d

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

it suffices to show that X a (b + d)2 (a + c)2 + + ≥ 4. (b + c)(d + a) (c + d)(a + b) a+d By the Cauchy-Schwarz inequality, we have (a + c)2 (b + d)2 (a + b + c + d)2 + ≥ , (b + c)(d + a) (c + d)(a + b) C X

(a + b + c + d)2 a ≥ , a+d D

where C = (b + c)(d + a) + (c + d)(a + b), D=

X

a(d + a) = a2 + b2 + c 2 + d 2 + a b + bc + cd + d a.

Since C + D = (a + b + c + d)2 , it is enough to show that (C + D)



‹ 1 1 + ≥ 4, C D

which is clearly true. The equality holds for a = b = c = d.

P 1.160. If a, b, c, d are nonnegative real numbers, then  ‹ ‹ ‹ ‹ 2b 2c 2d 2a 1+ 1+ 1+ 1+ ≥ 9. b+c c+d d+a a+b (Vasile Cîrtoaje, 2004)

212

Vasile Cîrtoaje

Solution. We can rewrite the inequality as  ‹ ‹  a + c  a+c b+d b+d 1+ 1+ 1+ 1+ ≥ 9. a+b c+d b+c d+a Using the Cauchy-Schwarz inequality and the AM-GM inequality yields  2  ‹2  a + c  a+c a+c 2a + 2c 1+ 1+ ≥ 1+ p , ≥ 1+ a+b c+d a+b+c+d (a + b)(c + d) 2   ‹ ‹  ‹2 b+d b+d b+d 2b + 2d 1+ 1+ ≥ 1+ p ≥ 1+ . b+c d+a a+b+c+d (b + c)(d + a) Thus, it suffices to show that ‹ ‹  2b + 2d 2a + 2c 1+ ≥ 3. 1+ a+b+c+d a+b+c+d This is equivalent to the obvious inequality 4(a + c)(b + d) ≥ 0. (a + b + c + d)2 The equality holds for a = c = 0 and b = d, as well as for b = d = 0 and a = c.

P 1.161. Let a, b, c, d be nonnegative real numbers. If k > 0, then  ‹ ‹ ‹ ‹ ka kb kc kd 1+ 1+ 1+ 1+ ≥ (1 + k)2 . b+c c+d d+a a+b (Vasile Cîrtoaje, 2004) Solution. Let us denote x=

a , b+c

y=

b c , z= , c+d d+a

t=

d . a+b

Since Y (1 + k x) ≥ 1 + k(x + y + z + t) + k2 (x y + yz + z t + t x + xz + y t), it suffices to show that x + y +z+ t ≥2 and x y + yz + z t + t x + xz + y t ≥ 1.

Cyclic Inequalities

213

The inequality x + y + z + t ≥ 2 is the well-known Shapiro’s inequality for 4 positive real numbers. This can be proved by the Cauchy-Schwarz inequality, as follows a b c d (a + b + c + d)2 + + + ≥ ≥ 2. b+c c+d d+a a+b a(b + c) + b(c + d) + c(d + a) + d(a + b) The right inequality reduces to the obvious inequality (a − c)2 + (b − d)2 ≥ 0. In order to get the inequality x y + yz +z t + t x + xz + y t ≥ 1, we will use the inequalities x +z ≥ xz, 2 y+t ≥ y t, 2 and the identity xz(1 + y + t) + y t(1 + x + z) = 1. If these are true, then x y + yz + z t + t x + xz + y t =

y+t x +z ( y + t) + (x + z) + xz + y t 2 2

≥ xz( y + t) + y t(x + z) + xz + y t = xz(1 + y + t) + y t(1 + x + z) = 1. We have

x +z bc + d a + (a − c)2 − xz = ≥0 2 2(b + c)(d + a)

and

y+t a b + cd + (b − d)2 − yt = ≥ 0. 2 2(a + b)(c + d)

To prove the identity above, we rewrite it as X x yz + xz + y t = 1, and see that X

P x yz =

a bc(a + b) A

P =

a2 bc +

P

a2 bd

A

and ac(a + b)(c + d) + bd(b + c)(d + a) xz + y t = = A where A=

P

a2 cd + (ac + bd)2 A

Y X X X (a + b) = a2 bc + a2 bd + a2 cd + (ac + bd)2 .

,

214

Vasile Cîrtoaje

Thus, the proof is completed. The equality holds for a = c = 0 and b = d, as well as for b = d = 0 and a = c. Remark. For k = 2, we get the inequality in P 1.160, while for k = 1, we get the following known inequality (a + b + c)(b + c + d)(c + d + a)(d + a + b) ≥ 4(a + b)(b + c)(c + d)(d + a). There is a simple proof of this inequality. Since (a + b + c)2 ≥ (2a + b)(2c + b) and (2a + b)(2b + a) ≥ 2(a + b)2 , we have Y Y Y Y Y (a + b + c)2 ≥ (2a + b) · (2c + b) = (2a + b) · (2b + a) = and hence

Y Y (2a + b)(2b + a) ≥ 24 (a + b)2 , Y (a + b + c) ≥ 4pr od(a + b).

P 1.162. If a, b, c, d are positive real numbers such that a + b + c + d = 4, then 1 1 1 1 + + + ≥ a2 + b2 + c 2 + d 2 . a b bc cd d a (Vasile Cîrtoaje, 2007) Solution. Write the inequality as (a + c)(b + d) ≥ a bcd(a2 + b2 + c 2 + d 2 ). Since (a + c)4 ≥ 8ac(a2 + c 2 ), (b + d)4 ≥ 8bd(b2 + d 2 ), we get bd(a + c)4 + ac(b + d)4 ≥ 8a bcd(a2 + b2 + c 2 + d 2 ). Therefore, it suffices to show that 8(a + c)(b + d) ≥ bd(a + c)4 + ac(b + d)4 .

Cyclic Inequalities

215

Since 4bd ≤ (b + d)2 and 4ac ≤ (a + c)2 , it is enough to show that 32(a + c)(b + d) ≥ (b + d)2 (a + c)4 + (a + c)2 (b + d)4 . This inequality is true if 32 ≥ x y(x 2 + y 2 ) for all positive x, y satisfying x + y = 4. Indeed, 8[32 − x y(x 2 + y 2 )] = (x + y)4 − 8x y(x 2 + y 2 ) = (x − y)4 ≥ 0. The equality occurs for a = b = c = d = 1.

P 1.163. If a, b, c, d are positive real numbers, then a2 b2 c2 d2 4 + + + ≥ . 2 2 2 2 (a + b + c) (b + c + d) (c + d + a) (d + a + b) 9 (Pham Kim Hung, 2006) First Solution. By Hölder’s inequality, we have X

Since

X

P 4/3
3 a a2 ≥ P . 2 (a + b + c) [ a(a + b + c)]2

a(a + b + c) = (a + c)2 + (b + d)2 + (a + c)(b + d)

and X

 ‹  a + c 4/3

b + d 4/3 a4/3 = a4/3 + c 4/3 + b4/3 + d 4/3 ≥ 2 +2 , 2 2

it suffices to show that  3 9 (a + c)4/3 + (b + d)4/3 ≥ 8[(a + c)2 + (b + d)2 + (a + c)(b + d)]2 . Due to homogeneity, we may assume that b + d = 1. Putting a + c = t 3 , t > 0, the inequality becomes 9(t 4 + 1)3 ≥ 8(t 6 + 1 + t 3 )2 ,  ‹  ‹2 1 3 1 9 t2 + 2 ≥ 8 t3 + 3 + 1 . t t

216

Vasile Cîrtoaje

1 Setting x = t + , x ≥ 2, the inequality turns into t 9(x 2 − 2)3 ≥ 8(x 3 − 3x + 1)2 , which is equivalent to (x − 2)2 (x 4 + 4x 3 + 6x 2 − 8x − 20) ≥ 0. This is true since x 4 + 4x 3 + 6x 2 − 8x − 20 = x 4 + 4x 2 (x − 2) + 4x(x − 2) + 10(x 2 − 2) > 0., Thus, the proof is completed. The equality holds for a = b = c = d. Second Solution. Due to homogeneity, we may assume that a + b + c + d = 1. In this case, we write the inequality as  a 2  b ‹2  c 2  d ‹2 4 + + + ≥ . 1−d 1−a 1− b 1−c 9 Let (x, y, z, t) be a permutation of (a, b, c, d) such that x ≥ y ≥ z ≥ t. Since

1 1 1 1 ≤ ≤ ≤ , (1 − t)2 (1 − z)2 (1 − y)2 (1 − x)2

by the rearrangement inequality, we have  x 2  y 2  z ‹2  t 2 + + + ≤ 1− t 1−z 1− y 1− x  a 2  b ‹2  c 2  d ‹2 + + + . ≤ 1−d 1−a 1− b 1−c Therefore, it suffices to show that x ≥ y ≥ z ≥ t > 0 and x + y + z + t = 1 yield U +V ≥ where

4 , 9

 x 2  t 2 U= + , 1− t 1− x

Cyclic Inequalities

217

V=

 y 2  z ‹2 + . 1−z 1− y

Let s = x + t,

p = x t, s ∈ (0, 1),

Since x 2 + t 2 = s2 − 2p,

x 3 + t 3 = s3 − 3ps,

x 4 + t 4 = s4 − 4ps2 + 2p2 ,

we get x 2 + t 2 − 2(x 3 + t 3 ) + x 4 + t 4 (1 − s + p)2 2 2p − 2(1 − s)(1 − 2s)p + s2 (1 − s)2 , = p2 + 2(1 − s)p + (1 − s)2

U=

(2 − U)p2 − 2(1 − s)(1 − 2s + U)p + (1 − s)2 (s2 − U) = 0. The quadratic in p has the discriminant D = (1 − s)2 [(1 − 2s + U)2 − (2 − U)(s2 − U)]. From the necessary condition D ≥ 0, we get U≥

4s − 1 − 2s2 . (2 − s)2

V≥

4r − 1 − 2r 2 , (2 − r)2

Analogously,

where r = y + z. Taking into account that s + r = 1 and s, r ∈ (0, 1), we get U +V ≥

5(s2 + r 2 ) − 2(s4 + r 4 ) 4s − 1 − 2s2 4r − 1 − 2r 2 + = , (2 − s)2 (2 − r)2 (2 + sr)2

hence U +V − ≥

4 45(s2 + r 2 ) − 18(s2 + r 2 )2 − 18 + 2(1 − 4sr)2 ≥ 9 9(2 + sr)2

5(s2 + r 2 ) − 2(s2 + r 2 )2 − 2 (2 − s2 − r 2 )(2s2 + 2r 2 − 1) = ≥ 0, (2 + sr)2 (2 + sr)2

since 2 − s2 − r 2 > 2 − (s + r)2 = 1, 2s2 + 2r 2 − 1 ≥ (s + r)2 − 1 = 0.

218

Vasile Cîrtoaje

P 1.164. If a, b, c, d are positive real numbers such that a + b + c + d = 3, then a b(b + c) + bc(c + d) + cd(d + a) + d a(a + b) ≤ 4. (Pham Kim Hung, 2007) Solution. Write the inequality as X

a b2 +

X

a bc ≤ 4,

(a b2 + cd 2 + bcd + d a b) + (bc 2 + d a2 + a bc + cd a) ≤ 4, (b + d)(a b + cd) + (a + c)(bc + d a) ≤ 4. Without loss of generality, assume that a + c ≤ b + d. Since (a b + cd) + (bc + d a) = (a + c)(b + d), we can rewrite the inequality as (a + c)(b + d)2 + (a + c − b − d)(bc + d a) ≤ 4. Since a + c − b − d ≤ 0, it suffices to show that (a + c)(b + d)2 ≤ 4. Indeed, by the AM-GM inequality, we have (a + c)



b+d 2

‹

b+d 2

‹

≤

 ‹ 1 b+d b+d 3 a+c+ + = 1. 27 2 2

The equality holds for a = b = 0, c = 1 and d = 2 (or any cyclic permutation).

P 1.165. If a ≥ b ≥ c ≥ d ≥ 0 and a + b + c + d = 2, then a b(b + c) + bc(c + d) + cd(d + a) + d a(a + b) ≤ 1. (Vasile Cîrtoaje, 2007) Solution. Write the inequality as X

a b2 +

X

a bc ≤ 1.

Since X X a b2 − a2 b = (a b2 + bc 2 +ca2 −a2 b− b2 c−c 2 a)+(cd 2 +d a2 +ac 2 −c 2 d −d 2 a−a2 c)

Cyclic Inequalities

219

= (a − b)(b − c)(c − a) + (c − d)(d − a)(a − c) ≤ 0, it suffices to show that X

a b2 +

X

a2 b + 2

X

a bc ≤ 2.

Indeed, X X X X X a b2 + a2 b + 2 a bc = (a b2 + a2 b + a bc + a bd) = (a + b + c + d) ab (a + c) + (b + d) = 2(a + c)(b + d) ≤ 2 2 •

˜2

= 2. ˜ • 1 ,1 . The equality holds for a = b = t and c = d = 1 − t, where t ∈ 2

P 1.166. Let a, b, c, d be nonnegative real numbers such that a + b + c + d = 4. If k ≥ then

37 , 27

a b(b + kc) + bc(c + kd) + cd(d + ka) + d a(a + k b) ≤ 4(1 + k). (Vasile Cîrtoaje, 2007) Solution. Write the inequality in the homogeneous form a b(b + kc) + bc(c + kd) + cd(d + ka) + d a(a + k b) ≤

(1 + k)(a + b + c + d)3 . 16

Assume that d = min{a, b, c, d} and use the substitution a = d + x,

b = d + y,

c = d + z,

where x, y, z ≥ 0. The inequality can be restated as 4Ad + B ≥ 0, where A = (3k − 1)(x 2 + y 2 + z 2 ) − 2(k + 1) y(x + z) + (6 − 2k)xz, B = (1 + k)(x + y + z)3 − 16(x y 2 + yz 2 + k x yz). It suffices to show that A ≥ 0 and B ≥ 0. We have A = (3k − 1) y 2 + (3k − 1)(x + z)2 − 2(k + 1) y(x + z) − 8(k − 1)xz ≥ (3k − 1) y 2 + (3k − 1)(x + z)2 − 2(k + 1) y(x + z) − 2(k − 1)(x + z)2 = (3k − 1) y 2 + (k + 1)(x + z)2 − 2(k + 1) y(x + z) Æ ≥ 2 (3k − 1)(k + 1) y(x + z) − 2(k + 1) y(x + z) €p Š p p =2 k+1 3k − 1 − k + 1 y(x + z) ≥ 0.

220

Vasile Cîrtoaje

Since (x + y + z)3 − 16x yz ≥ 0, the inequality B ≥ 0 holds for all k ≥ the inequality B ≥ 0 can be written as 4

37 37 if it holds for k = . In this particular case, 27 27

 x + y + z 3 37 x yz. ≥ x y 2 + yz 2 + 3 27

Actually, the following sharper inequality holds (see P 2.27) 4

 x + y + z 3 3 ≥ x y 2 + yz 2 + x yz. 3 2

Thus, the proof is completed. The equality holds for a = b = c = d = 1. If k = the equality also holds for a =

37 , then 27

8 4 , b = and c = d = 0 (or any cyclic permutation). 3 3

P 1.167. Let a, b, c, d be positive real numbers such that a ≤ b ≤ c ≤ d. Prove that  ‹ a b c d a c b d 2 + + + ≥4+ + + + . b c d a c a d b (Vasile Cîrtoaje, 2012) First Solution. Let 

‹ a b c d a c b d E(a, b, c, d) = 2 + + + −4− − − − . b c d a c a d b We show that E(a, b, c, d) ≥ E(b, b, c, d) ≥ E(b, b, c, c). We have ‹ 1 2d 2 c + − − ≥ 0, E(a, b, c, d) − E(b, b, c, d) = (b − a) c ab b ab 

since

1 2d 2 c 1 2c 2 c + − − ≥ + − − c ab b ab c ab b ab =

c 2 1 c 2 (b − c)2 1 + − ≥ + 2− = ≥ 0. c ab b c b b b2 c

Cyclic Inequalities

221

Also, E(b, b, c, d) − E(b, b, c, c) = (d − c) since



‹ 1 2c − b − ≥ 0, b cd

1 2c − b (b − c)2 1 2c − b − ≥ − = ≥ 0. b cd b c2 bc 2

Because E(b, b, c, c) = 0, the proof is completed. The equality holds for a = b and c = d. Second Solution. Using the substitution x=

a , b

y=

b c , z= , c d

0 < x, y, z ≤ 1,

the inequality becomes as follows  ‹ 1 1 1 2 x + y +z+ ≥4+ xy + + yz + , x yz xy yz  ‹ 1 2 1 1 y(2 − x − z) + − − − 2(2 − x − z) ≥ 0, y xz x z  ‹ 1 (2 − x − z) y + − 2 ≥ 0. x yz The last inequality is true since 2 − x − y ≥ 0 and y+

1 1 − 2 ≥ y + − 2 ≥ 0. x yz y

P 1.168. Let a, b, c, d be positive real numbers such that a ≤ b ≤ c ≤ d and a bcd = 1. Prove that a b c d + + + ≥ a b + bc + cd + d a. b c d a (Vasile Cîrtoaje, 2012) Solution. Write the inequality as follows a2 cd + b2 d a + c 2 a b + d 2 bc ≥ a b + bc + cd + d a, ac(ad + bc) + bd(a b + cd) ≥ (ad + bc) + (a b + cd), (ac − 1)(ad + bc) + (bd − 1)(a b + cd) ≥ 0.

222

Vasile Cîrtoaje

Since ac − 1 =

1 − 1 ≥ 1 − bd bd

and bd ≥

p

a bcd = 1,

we have (ac − 1)(ad + bc) + (bd − 1)(a b + cd) ≥ (1 − bd)(ad + bc) + (bd − 1)(a b + cd) = (bd − 1)(a − c)(b − d) ≥ 0. The equality holds for a = b = 1/c = 1/d ≥ 1.

P 1.169. Let a, b, c, d be positive real numbers such that a ≤ b ≤ c ≤ d and a bcd = 1. Prove that a b c d 4 + + + + ≥ 2(a + b + c + d). b c d a (Vasile Cîrtoaje, 2012) Solution. Making the substitution x=

s 4

a , b

y=

v tb c

, z=

s 4

c , d

0 < x, y, z ≤ 1,

we need to show that E(x, y, z) ≥ 0, where  ‹ yz 1 z 1 3 E(x, y, z) = 4 + x + z + y + 4 2 4 − 2 x yz + + + . x y z x x y x yz 3 4

4

2

We will show that E(x, y, z) ≥ E(x, 1, z) ≥ E(x, 1, 1) ≥ 0. The left inequality is equivalent to (1 − y)E1 (x, y, z) ≥ 0, where

 ‹  1+ y z 2 z 1 3 E1 (x, y, z) = −1 − y + 4 2 4 + 2 x z + − + . x y z x y x xz 3

To prove it, we show that E1 (x, y, z) ≥ E1 (x, 1, z) ≥ 0.

(*)

Cyclic Inequalities We have

223

‹ 1 E1 (x, 1, z) = 2(1 − x z) − 1 ≥ 0. x 4z4 3



Since E1 (x, y, z − E1 (x, 1, z) = (1 − y)E2 (x, y, z), where E2 (x, y, z) = 1 +

1 + 2y 2 − x 4 y 2z4 y



‹ z 1 + 3 , x xz

we need to show E2 (x, y, z) ≥ 0. Indeed,  ‹ 1 2 z 1 1 E2 (x, y, z) = 1 + 4 2 4 − + − x y z y x xz 3 x 4 z 4  ‹  ‹ 2 2 z 1 1 2 1 1 1 ≥ 2 2− + 3− 4 4 = − z2 − 2 + 3 3 x yz y x xz x z x yz xz z x z    ‹ 1 − z3 1 − z 1 1 1 2 2 − z2 − 2 + 3 = + 3 ≥ ≥ 0. x yz z z z x yz z z The middle inequality in (*) is equivalent to (1 − z)F (x, z) ≥ 0, where F (x, z) = (1 + z + z 2 + z 3 )



‹  ‹ 1 2 1 + z + z2 3 − 1 + 2 x + − . x 4z4 x xz

It is true since F (x, z) >

1 3 1 + z + z2 1 3 1 + z + z2 − 1 + − ≥ − 1 + − x 4z4 x xz xz x xz

2− x −z ≥ 0. x The right inequality in (*) is also true since =

x 4 E(x, 1, 1) = x 8 − 2x 7 + 6x 4 − 6x 3 + 1 = (x − 1)2 (x 6 − x 4 − 2x 3 + 3x 2 + 2x + 1) ≥ (x − 1)2 (x 6 − x 4 − 2x 3 + 2x 2 ) = x 2 (x − 1)4 (x 2 + 2x + 2) ≥ 0. The proof is completed. The equality holds for a = b = c = d = 1.

224

Vasile Cîrtoaje

P 1.170. Let A = {a1 , a2 , a3 , a4 } be a set of real numbers such that a1 + a2 + a3 + a4 = 0. Prove that there exists a permutation B = {a, b, c, d} of A such that a2 + b2 + c 2 + d 2 + 3(a b + bc + cd + d a) ≥ 0. Solution. Write the desired inequality as a2 + b2 + c 2 + d 2 + 3(a b + bc + cd + d a) ≥ (a + b + c + d)2 , a b + bc + cd + d a ≥ 2(ac + bd), (a b + cd − ac − bd) + (bc + d a − ac − bd) ≥ 0. (a − d)(b − c) + (a − b)(d − c) ≥ 0. Clearly, this inequality is true for a ≤ b ≤ d ≤ c. The equality occurs when A has three equal elements.

P 1.171. Let a, b, c, d, e be positive real numbers such that a + b + c + d + e = 5. Prove that 4 a b c d e 1+ ≥ + + + + . a bcd e b c d e a Solution. Let (x, y, z, t, u) be a permutation of (a, b, c, d, e) such that x ≥ y ≥ z ≥ t ≥ u. By the rearrangement inequality, we have y z a b c d e x t u + + + + ≤ + + + + b c d e a u t z y x ‹ x u  y t = + +2 + + + 2 − 3 = 4(p + q) − 3, u x t y where

 1 x u p= + + 2 ≥ 1, 4 u x

1 q= 4



‹ y t + + 2 ≥ 1. t y

Since p + q ≤ 1 + pq,

4(p + q) − 3 ≤ 1 + 4pq,

it suffices to show that x yz tupq ≤ 1.

Cyclic Inequalities

225

This inequality is equivalent to z

 x + u 2  y + t 2 2

2

≤ 1.

Indeed, by the AM-GM inequality, we get

z

 x + u 2  y + t 2 2

2



z+

≤

y + t 5 x +u x +u y + t + + + 2 2 2 2  = 1. 5

The equality holds for a = b = c = d = e = 1. Remark. Similarly, we can prove the following generalization (Michael Rozenberg). • If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an = n, then an a1 a2 4 ≥ + + ··· + . a1 a2 · · · an a2 a3 a1

n−4+

P 1.172. If a, b, c, d, e are real numbers such that a + b + c + d + e = 0, then p p 5+1 a b + bc + cd + d e + ea 5−1 ≤ 2 ≤ . − 2 2 2 2 4 a +b +c +d +e 4 Solution. From (a + b + c + d + e)2 = 0, we get X

a2 + 2

X

ab + 2

X

ac = 0.

Therefore, for any real k, we have X X X a2 + (2k + 2) ab = 2a(k b − c), and, by the AM-GM inequality, we get X X X X X a2 + (2k + 2) ab ≤ [a2 + (k b − c)2 ] = (k2 + 2) a2 − 2k a b, hence X

a2 ≥

2(2k + 1) X a b. k2 + 1

226

Vasile Cîrtoaje

p p −1 − 5 −1 + 5 Choosing k = and k = , we get the left and the right inequality, 2 2 respectively. The equality in both inequalities occurs when a = k b − c, b = kc − d, c = kd − e, d = ke − a, e = ka − b; that is, when a = x,

b = y,

d = −k(x + y),

c = −x + k y, e = k x − y,

where x and y are real numbers.

P 1.173. Let a, b, c, d, e be positive real numbers such that a2 + b2 + c 2 + d 2 + e2 = 5. Prove that a2 b2 c2 d2 e2 5 + + + + ≥ . b+c+d c+d+e d+e+a e+a+b a+b+c 3 (Pham Van Thuan, 2005) Solution. By the AM-GM Inequality, we get 2b + 2c + 2d ≤ (b2 + 1) + (c 2 + 1) + (d 2 + 1) = 8 − a2 − e2 . Therefore, it suffices to show that X

5 a2 ≥ . 8 − a2 − e2 6

By the Cauchy-Schwarz Inequality, we have X

€X Š2 a2 a2 25 X X ≥X = 2 2 8−a −e a2 (8 − a2 − e2 ) 40 − a4 − a2 e2 =

50 50 5 X ≥ ”X —2 = 6 , 2 2 2 1 80 − (a + e ) 80 − (a2 + e2 ) 5

and the proof is completed. The equality holds for a = b = c = d = e = 1.

Cyclic Inequalities

227

P 1.174. Let a, b, c, d, e be nonnegative real numbers such that a + b + c + d + e = 5. Prove that 729 (a2 + b2 )(b2 + c 2 )(c 2 + d 2 )(d 2 + e2 )(e2 + a2 ) ≤ . 2 (Vasile Cîrtoaje, 2007) Solution. Let E(a, b, c, d, e) be the left hand side of the inequality. Without loss of generality, assume that e = min{a, b, c, d, e}. We will show first that  e e  E(a, b, c, d, e) ≤ E a + , b, c, d + , 0 , 2 2

(*)

which is equivalent to (a2 + b2 )(c 2 + d 2 )(d 2 + e2 )(e2 + a2 ) ≤ ˜• ˜ •  e 2  e 2  e 2 e 2 + b2 c 2 + d + d+ a+ . ≤ a+ 2 2 2 2 This is true since

 e 2 + b2 , a2 + b2 ≤ a + 2  e 2 c2 + d 2 ≤ c2 + d + , 2  e 2 d 2 + e2 ≤ d 2 + d e ≤ d + , 2  e 2 . e2 + a2 ≤ ae + a2 ≤ a + 2 According to (*), it suffices to prove the desired inequality for e = 0; that is, to show that a + b + c + d = 5 involves F (a, b, c, d) ≤

729 , 2

where F (a, b, c, d) = a2 d 2 (a2 + b2 )(b2 + c 2 )(c 2 + d 2 ). Without loss of generality, assume that c = min{b, c}. We will show that  c c , F (a, b, c, d) ≤ E a, b + , 0, d + 2 2

(**)

228

Vasile Cîrtoaje

which is equivalent to • ˜ c 2  c 2 c 2 2  c 2  b+ d+ . d (a + b )(b + c )(c + d ) ≤ d + a + b+ 2 2 2 2 2

2

2

2

This is true since

2

2

2



 c 2 d 2 (c 2 + d 2 ) ≤ d + , 2  c 2 a2 + b2 ≤ a2 + b + , 2  c 2 . b2 + c 2 ≤ b2 + bc ≤ b + 2

According to (**), it suffices to prove the inequality F (a, b, c, d) ≤ is, to show that a + b + d = 5 involves G(a, b, d) ≤

729 for c = 0; that 2

729 , 2

where G(a, b, d) = a2 b2 d 4 (a2 + b2 ). We will show that ‹ a+b a+b G(a, b, d) ≤ G , ,d . 2 2 

This is true if 32a2 b2 (a2 + b2 ) ≤ (a + b)6 . Indeed, (a + b)6 − 32a2 b2 (a2 + b2 ) ≥ 4a b(a + b)4 − 32a2 b2 (a2 + b2 ) = 4a b(a − b)4 ≥ 0. To end the proof, we only need to prove that  ‹ a+b a+b 729 G , ,d ≤ 2 2 2 for a + b + d = 5. Putting u =

a+b , we need to show that 2u + d = 5 involves 2 G(u, u, d) ≤

that is, u6 d 4 ≤

729 ; 2

729 . 4

Cyclic Inequalities

229

This is true if u3 d 2 ≤

27 . 2

By the AM-GM inequality, we have 5=

2u 2u 2u d d + + + + ≥5 3 3 3 2 2

v u ‹3  2 t 5 2u t 3

2

,

from which the conclusion follows. The equality holds for a = b = e = 0 (or any cyclic permutation).

3 , c = 0, d = 2 and 2

P 1.175. If a, b, c, d, e ∈ [1, 5], then a−b b−c c−d d−e e−a + + + + ≥ 0. b+c c+d d+e e+a a+b (Vasile Cîrtoaje, 2002) Solution. Write the inequality as Xa − b

‹ 2 10 + ≥ , b+c 3 3

X 3a − b + 2c b+c

≥ 10.

Since 3a − b + 2c ≥ 3 − 5 + 2 = 0, we may apply the Cauchy-Schwarz inequality to get P P X 3a − b + 2c [ (3a − b + 2c)]2 16( a)2 P P . ≥P =P b+c (b + c)(3a − b + 2c) a2 + 4 a b + 3 ac Therefore, it suffices to show that X X X X 8( a)2 ≥ 5 a2 + 20 a b + 15 ac. Since

X X X X ( a)2 = a2 + 2 ab + 2 ac,

this inequality is equivalent to 3

X

a2 +

X

ac ≥ 4

X

a b.

230

Vasile Cîrtoaje

Indeed, 3

X

a2 +

X

ac − 4

X

ab =

1X (a − 2b + c)2 ≥ 0. 2

The equality holds for a = b = c = d = e. Conjecture. If a, b, c, d, e are positive real numbers, then a−b b−c c−d d−e e−a + + + + ≥ 0. b+c c+d d+e e+a a+b

P 1.176. If a, b, c, d, e, f ∈ [1, 3], then f −a a−b b−c c−d d−e e− f + + + + + ≥ 0. b+c c+d d+e e+ f f +a a+b (Vasile Cîrtoaje, 2002) Solution. Write the inequality as Xa − b b+c

+

‹ 1 ≥ 3, 2

X 2a − b + c b+c

≥ 6.

Since 2a − b + c ≥ 2 − 3 + 1 = 0, we may apply the Cauchy-Schwarz inequality to get P P X 2a − b + c 2( a)2 [ (2a − b + c)]2 P . =P ≥P b+c (b + c)(2a − b + c) a b + ac Thus, we still have to show that X X X ( a)2 ≥ 3( ab + ac). Let x = a + d, Since

X

ab +

y = b + e, z = c + f .

X

ac = x y + yz + z x,

we have X X X ( a)2 − 3( ab + ac) = (x + y + z)2 − 3(x y + yz + z x) ≥ 0. The equality holds for a = c = e and b = d = f .

Cyclic Inequalities

231

P 1.177. If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers, then n X i=1

ai n ≤ , ai−1 + 2ai + ai+1 4

where a0 = an and an+1 = a1 . (Vasile Cîrtoaje, 2008) Solution. Applying the Cauchy-Schwarz inequality, we have n X i=1

n

X ai ai = ai−1 + 2ai + ai+1 (ai−1 + ai ) + (ai + ai+1 ) i=1  ‹ n 1X 1 1 ≤ ai + 4 i=1 ai−1 + ai ai + ai+1 ‚ n Œ n X ai ai 1 X = + 4 i=1 ai−1 + ai i=1 ai + ai+1 ‚ n Œ n X ai n 1 X ai+1 + = . = 4 i=1 ai + ai+1 i=1 ai + ai+1 4

The equality holds for a1 = a2 = · · · = an .

P 1.178. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. Prove that 1 1 1 + + ··· + ≤ 1. n − 2 + a1 + a2 n − 2 + a2 + a3 n − 2 + an + a1 (Vasile Cîrtoaje, 2008) First Solution. Let r = lowing inequalities

n−2 . We can get the desired inequality by summing the foln a3r + a4r + · · · + anr n−2 ≤ r , n − 2 + a1 + a2 a1 + a2r + · · · + anr a1r + a4r + · · · + anr n−2 ≤ r , n − 2 + a2 + a3 a1 + a2r + · · · + anr .................................... r a2r + a3r + · · · + an−1 n−2 ≤ r . n − 2 + an + a1 a1 + a2r + · · · + anr

232

Vasile Cîrtoaje

The first inequality is equivalent to (a1 + a2 )(a3r + a4r + · · · + anr ) ≥ (n − 2)(a1r + a2r ). By the AM-GM inequality, we have n−2

r

a3r + a4r + · · · + anr ≥ (n − 2)(a3 a4 · · · an ) n−2 =

r

(a1 a2 ) n−2

.

Therefore, it suffices to show that r

a1 + a2 ≥ (a1 a2 ) n−2 (a1r + a2r ), or, equivalently,  n−2 n−2  1 a1 + a2 ≥ (a1 a2 ) n a1 n + a2 n . This is equivalent to the obvious inequality 

n−1 n

a1

n−1 n

− a2



1 n

1 n

a1 − a2



≥ 0.

The equality holds for a1 = a2 = · · · = an . Second Solution. Since a1 + a2 n−2 =1− , n − 2 + a1 + a2 n − 2 + a1 + a2 we can write the desired inequality as n X i=1

ai + ai+1 ≥ 2, ai + ai+1 + n − 2

where an+1 = a1 . Using now the Cauchy-Schwarz inequality, we get ‚ n X i=1

Œ2

n X p

ai + ai+1 ai + ai+1 i=1 ≥ n ai + ai+1 + n − 2 X (ai + ai+1 + n − 2) i=1 n X

2 =

i=1

ai + 2

X Æ 1≤i< j≤n n X

(ai + ai+1 )(a j + a j+1 )

ai + n(n − 2)

2

i=1

.

Cyclic Inequalities

233

Therefore, it suffices to prove that X q

(ai + ai+1 )(a j + a j+1 ) ≥

1≤i< j≤n

n X

ai + n(n − 2).

i=1

Setting an+2 = a2 , by the Cauchy-Schwarz inequality and the AM-GM inequality, we have X q (ai + ai+1 )(a j + a j+1 ) = 1≤i< j≤n

=

n Æ X

(ai + ai+1 )(ai+1 + ai+2 ) +

≥

(ai + ai+1 )(a j + a j+1 )

1≤i< j≤n j6=i+1

i=1 n X

X q

ai+1 +

p


p ai ai+2 + n(n − 3) n a1 a2 · · · an

i=1

=

n X

ai + n(n − 3) +

i=1

≥

n X

n X p

ai ai+2

i=1 n

X p ai + n(n − 3) + n n a1 a2 · · · an = ai + n(n − 2).

i=1

i=1

P 1.179. If a1 , a2 , ..., an ≥ 1, then Y

a1 +

‹ ‹  1 1 1 1 + n − 2 ≥ nn−2 (a1 + a2 + ... + an ) + + ... + . a2 a1 a2 an (Vasile Cîrtoaje, 2011)

Solution. Write the inequality as E(a1 , a2 , ..., an ) ≥ 0, and denote   ‹ ‹ 1 1 1 A = a2 + + n − 2 a3 + + n − 2 ... an−1 + +n−2 . a3 a4 an 

We will prove that E(a1 , a2 , ..., an ) ≥ E(1, a2 , ..., an ). If this is true, then E(a1 , a2 , ..., an ) ≥ E(1, a2 , ..., an ) ≥ ... ≥ E(1, 1, ..., 1, an ) = 0.

234

Vasile Cîrtoaje

We have



‹ C E(a1 , a2 , ..., an ) − E(1, a2 , ..., an ) = (a1 − 1) B − , a1

where B = A(an + n − 2) − nn−2 C =A





‹ 1 1 1 , + + ... + a2 a3 an

‹ 1 + n − 2 − nn−2 (a2 + a3 + ... + an ). a2

Since a1 − 1 ≥ 0, we need to show that B−

C ≥ 0. a1

According to the AM-GM inequality, we have  v  v   v  v t an−1 t a2 t t a3 n−2 n a2 n n n A≥ n n ... n =n , a3 a4 an an p an + n − 2 ≥ (n − 1) n−1 an , r

A(an + n − 2) ≥ (n − 1)nn−2 and hence

n

1

a2 ann−1 ≥ (n − 1)nn−2 ,

‹  1 1 1 − − ... − ≥ 0. B ≥ nn−2 n − 1 − a2 a3 an

If C ≤ 0, then B−

C (−C) =B+ ≥ 0, a1 a1

and if C ≥ 0, then  ‹  ‹  ‹ C 1 1 1 n−2 n−2 B− ≥ B − C = A an − +n a2 − + ... + n an − ≥ 0. a1 a2 a2 an The equality holds when n − 1 numbers of a1 , a2 , ..., an are equal to 1.

P 1.180. If a1 , a2 , ..., an ≥ 1, then  ‹ ‹  ‹  ‹ ‹  ‹ an a1 a2 1 1 1 n a1 + a2 + ... an + +2 ≥2 1+ 1+ ... 1 + . a1 a2 an a2 a3 a1 (Vasile Cîrtoaje, 2011)

Cyclic Inequalities

235

Solution. Write the inequality as E(a1 , a2 , ..., an ) ≥ 0, and denote  ‹  ‹ 1 1 A = a2 + ... an + , a2 an ‹  ‹  an−1 a . B = 1 + 2 ... 1 + a3 an We will prove that E(a1 , a2 , ..., an ) ≥ E(1, a2 , ..., an ). If this is true, then E(a1 , a2 , ..., an ) ≥ E(1, a2 , ..., an ) ≥ ... ≥ E(1, 1, ..., 1, an ) = 0. We have



‹ D E(a1 , a2 , ..., an ) − E(1, a2 , ..., an ) = (a1 − 1) C − , a1

where C = A−

2B , a2

D = A − 2Ban . Since a1 − 1 ≥ 0, we need to show that C−

D ≥ 0. a1

First, we prove that C ≥ 0; that is, (a22 + 1)...(an2 + 1) ≥ 2(a2 + a3 )...(an−1 + an ). By squaring, this inequality becomes 2 + 1)(an2 + 1)](an2 + 1) ≥ (a22 + 1)[(a22 + 1)(a32 + 1)]...[(an−1

≥ 4(a2 + a3 )2 ...(an−1 + an )2 . By the Cauchy-Schwarz inequality, we have 2 (a22 + 1)(a32 + 1) ≥ (a2 + a3 )2 , ... , (an−1 + 1)(an2 + 1) ≥ (an−1 + an )2 .

Therefore, we still have to show that (a22 + 1)(an2 + 1) ≥ 4, which is clearly true for a2 ≥ 1 and an ≥ 1. Now, let us prove that C − D/a1 ≥ 0. If D ≤ 0, then D (−D) C− =C+ ≥ 0, a1 a1

236

Vasile Cîrtoaje

and if D ≥ 0, then

 ‹ D 1 C− ≥ C − D = 2B an − ≥ 0. a1 a2

The equality holds when n − 1 of a1 , a2 , ..., an are equal to 1.

P 1.181. Let k and n be positive integers, and let a1 , a2 , ..., an be real numbers such that a1 ≤ a2 ≤ · · · ≤ an . Consider the inequality (a1 + a2 + · · · + an )2 ≥ n(a1 ak+1 + a2 ak+2 + · · · + an an+k ), where an+i = ai for any positive integer i. Prove that this inequality holds (a) for n = 2k; (b) for n = 4k. (Vasile Cîrtoaje, 2004) Solution. (a) We need to prove that (a1 + a2 + · · · + a2k )2 ≥ 4k(a1 ak+1 + a2 ak+2 + · · · + ak a2k ). If x is a real number such that ak ≤ x ≤ ak+1 , then (x − a1 )(ak+1 − x) + (x − a2 )(ak+2 − x) + · · · + (x − ak )(a2k − x) ≥ 0. Expanding and multiplying by 4k, we get 4k x(a1 + a2 + · · · + a2k ) ≥ 4k2 x 2 + 4k(a1 ak+1 + a2 ak+2 + · · · + ak a2k ). On the other hand, by the AM-GM inequality, we have (a1 + a2 + · · · + a2k )2 + 4k2 x 2 ≥ 4k x(a1 + a2 + · · · + a2k ). Adding these inequalities yields the desired inequality. The equality holds for a j+1 = a j+2 = · · · = a j+k = where j ∈ {1, 2, · · · , k − 1}.

a1 + a2 + · · · a2k , 2k

Cyclic Inequalities

237

(b) We need to show that (a1 + a2 + · · · + a4k )2 ≥ 4k(a1 ak+1 + a2 ak+2 + · · · + a4k ak ). Using the substitution bi = ai + a2k+i ,

i = 1, 2, · · · , 2k,

this inequality becomes (b1 + b2 + · · · + b2k )2 ≥ 4k(b1 bk+1 + b2 bk+2 + · · · + bk b2k ), which is just the inequality in (a). The equality holds for  a j+1 = a j+2 = · · · = a j+k = a      a j+2k+1 = a j+2k+2 = · · · = a j+3k = b ,      a1 + a2 + · · · + a4k = 2k(a + b) where a ≤ b are real numbers, and j ∈ {1, 2, · · · , k − 1}. Remark. Actually, the inequality holds for any integer k satisfying

n n l ek ≤ . 4 2

P 1.182. If a1 , a2 , . . . , an are real numbers, then a1 (a1 + a2 ) + a2 (a2 + a3 ) · · · + an (an + a1 ) ≥

2 (a1 + a2 + · · · + an )2 . n

Solution. Making the substitution a=

1 (a1 + a2 + · · · + an ) n

and x i = ai − a,

i = 1, 2, · · · , n,

we have x1 + x2 + · · · + x n = 0 and X 2 a1 (a1 + a2 ) − (a1 + a2 + · · · + an )2 = (x 1 + a)(x 1 + x 2 + 2a) − 2na2 n X 1X = x 1 (x 1 + x 2 ) = (x 1 − x 2 )2 ≥ 0. 2 The equality holds for a1 = a2 = · · · = an - if n is odd, and for a1 = a3 = · · · = an−1 and a2 = a4 = · · · = an - if n is even. X

238

Vasile Cîrtoaje

P 1.183. If f is a convex function on (0, ∞) and a1 , a2 , . . . , an are positive numbers, then ‹ X  ‹  n n X 1 1 ≥ f ai + , f ai + ai+1 ai i=1 i=1 where an+1 = a1 . (Vasile Cîrtoaje, 2009) Solution. Use the induction method. For n = 2, the inequality has the form  ‹  ‹  ‹  ‹ 1 1 1 1 f a1 + + f a2 + ≥ f a1 + + f a2 + . a2 a1 a1 a2 Assume that a1 ≥ a2 . Since a1 + and



ª § 1 1 1 1 1 = max a1 + , a2 + , a1 + , a2 + a2 a2 a1 a1 a2

‹  ‹  ‹  ‹ 1 1 1 1 a1 + + a2 + = a1 + + a2 + , a2 a1 a1 a2

the inequality follows from Lemma below (which is a particular case of Karamata’s inequality). Suppose now that the inequality is valid for n, and prove it for n + 1. Assume that an+1 = min{a1 , a2 , · · · , an+1 }. Using the induction hypothesis, it suffices to show that  ‹  ‹  ‹  ‹ 1 1 1 1 + f an+1 + ≥ f an + + f an+1 + . f an + an+1 a1 a1 an+1 Since an + and



1 an+1

an +

1 1 1 = max an + , an+1 + , an + , an+1 + an+1 a1 a1 an+1 §

1 an+1

‹

1

1 + an+1 + a1 

‹

ª

‹  ‹ 1 1 = an + + an+1 + , a1 an+1 

the inequality follows also from Lemma below. Lemma. Let f be a convex function on a real interval I. If a, b, c, d ∈ I such that a = max{a, b, c, d},

a + b = c + d,

then f (a) + f (b) ≥ f (c) + f (d).

Cyclic Inequalities

239

Proof. Since a = max{a, b, c, d},

b = min{a, b, c, d},

there are p, q ∈ [0, 1] such that c = pa + (1 − p)b,

d = qa + (1 − q)b.

For the non-trivial case a > b, the relation a + b = c + d involves p + q = 1. By Jensen’s inequality, we have f (c) + f (d) ≤ p f (a) + (1 − p) f (b) + q f (a) + (1 − q) f (b) = f (a) + f (b).

P 1.184. If f is a convex function on a real interval I, a1 , a2 , . . . , an ∈ I and a1 + a2 + · · · + an , a= n then n ai − ai+1  n(n − 2) nX  f (a1 ) + f (a2 ) + · · · + f (an ) + f (a) ≥ f a+ , 2 2 i=1 n where an+1 = a1 . (Vasile Cîrtoaje and Darij Grinberg, 2007) Solution (by Michel Bataille). Let bj =

1 X ak , n − 1 k6= j

j = 1, 2, · · · , n.

By Jensen’s inequality, we have  ‹  ‹  ai − ai+1  ai + (n − 1)bi+1 1 1 f a+ =f ≤ f (ai ) + 1 − f (bi+1 ), n n n n hence n n n  X X ai − ai+1  X n f a+ ≤ f (ai ) + (n − 1) f (bi ). n i=1 i=1 i=1 Thus, it suffices to show that n n n X X X 2 f (ai ) + n(n − 2) f (a) ≥ f (ai ) + (n − 1) f (bi ), i=1

i=1

i=1

which is just Popoviciu’s inequality n X i=1

f (ai ) + n(n − 2) f (a) ≥ (n − 1)

n X i=1

f (bi ).

240

Vasile Cîrtoaje

Chapter 2

Noncyclic Inequalities 2.1

Applications

2.1. If a, b, c are nonnegative real numbers such that a + b + c = 3, then (a b + c)(ac + b) ≤ 4.

2.2. If a, b, c are nonnegative real numbers, then a3 + b3 + c 3 − 3a bc ≥

1 (b + c − 2a)3 . 4

2.3. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ c. Prove that (a)

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

(b)

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

2.4. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ c. Prove that (a)

a3 + b3 + c 3 − 3a bc ≥ 3(a2 − b2 )(b − c);

(b)

a3 + b3 + c 3 − 3a bc ≥

241

9 (a − b)(b2 − c 2 ). 2

242

Vasile Cîrtoaje

2.5. Let a, b, c be nonnegative real numbers such that a = max{a, b, c}. Prove that a6 + b6 + c 6 − 3a2 b2 c 2 ≥ 2(b4 + c 4 + 4b2 c 2 )(b − c)2 .

2.6. Let a, b, c be nonnegative real numbers such that a = max{a, b, c}. Prove that a2 + b2 + c 2 ≥

9a bc 5 + (b − c)2 . a+b+c 3

2.7. If a, b, c are nonnegative real numbers, no two of which are zero, then 1 16 6 1 + + ≥ . 2 2 2 (a + b) (a + c) (b + c) a b + bc + ca

2.8. If a, b, c are nonnegative real numbers, no two of which are zero, then 1 1 2 5 + + ≥ . (a + b)2 (a + c)2 (b + c)2 2(a b + bc + ca)

2.9. If a, b, c are positive real numbers, then (a + b)3 (a + c)3 ≥ 4a2 bc(2a + b + c)2 .

2.10. If a, b, c are positive real numbers such that a bc = 1, then (a)

a b 1 + + ≥ a + b + 1; b c a

(b)

a b 1 p + + ≥ 3(a2 + b2 + 1). b c a

2.11. If a, b, c are positive real numbers such that a bc ≥ 1, then a

b

a b b c c c ≥ 1.

Noncyclic Inequalities

243

2.12. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a b + bc + ca = 3. Prove that a b2 c 3 < 4.

2.13. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a b + bc + ca = Prove that a b2 c 2 ≤

5 . 3

1 . 3

2.14. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a b + bc + ca = 3. Prove that 9 ; 8

(a)

a b2 c ≤

(b)

a b4 c ≤ 2;

(c)

a2 b3 c ≤ 2.

2.15. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a+b+c =

1 1 1 + + . a b c

Prove that a b2 c 3 ≥ 1.

2.16. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a + b + c = a bc + 2. Prove that (1 − b)(1 − a b3 c) ≥ 0.

244

Vasile Cîrtoaje

2.17. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a+b+c = Prove that b≥

1 1 1 + + . a b c

1 . a+c−1

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

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

(b)

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

2.19. Let a, b, c be real numbers, no two of which are zero. If bc ≥ 0, then (a)

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

(b)

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

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

2.21. Let a, b, c be positive real numbers, b 6= c. Prove that ab ac (b + c)2 + ≤ . (a + b)2 (a + c)2 4(b − c)2

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

Noncyclic Inequalities

245

2.23. Let a, b, c be nonnegative real numbers such that a + b > 0. Prove that a bc ≥ (b + c − a)(c + a − b)(a + b − c) +

a b(a − b)2 . a+b

2.24. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ c. Prove that (a)

a bc ≥ (b + c − a)(c + a − b)(a + b − c) +

2a b(a − b)2 ; a+b

(b)

a bc ≥ (b + c − a)(c + a − b)(a + b − c) +

27b(a − b)4 . 4a2

2.25. Let a, b, c be nonnegative real numbers such that a + b > 0. Prove that X

2

2 2

a (a − b)(a − c) ≥ a b



a−b a+b

‹2

.

2.26. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that a b2 + bc 2 + 2ca2 ≤ 8.

2.27. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 3 a b2 + bc 2 + a bc ≤ 4. 2

2.28. Let a, b, c be nonnegative real numbers such that a + b + c = 5. Prove that a b2 + bc 2 + 2a bc ≤ 20.

2.29. If a, b, c are nonnegative real numbers, then a3 + b3 + c 3 − a2 b − b2 c − c 2 a ≥

8 (a − b)(b − c)2 . 9

246

Vasile Cîrtoaje

2.30. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ c. Prove that ‹  P 2 a−b 2 (a) a (a − b)(a − c) ≥ 4a2 b2 ; a+b (b)

P

a2 (a − b)(a − c) ≥

27b(a − b)4 . 4a

2.31. If a, b, c are positive real numbers, then a b c 2(a − c)2 + + ≥3+ . b c a (a + c)2

2.32. If a, b, c are positive real numbers, then (a − c)2 a b c + + ≥3+ . b c a a b + bc + ca

2.33. If a, b, c are positive real numbers, then a b c 4(a − c)2 + + ≥3+ . b c a (a + b + c)2

2.34. If a ≥ b ≥ c > 0, then a b c 3(b − c)2 + + ≥3+ . b c a a b + bc + ca

2.35. If a, b, c are positive real numbers, then a2 b2 c 2 4(a − c)2 + + ≥a+b+c+ . b c a a+b+c 2.36. If a ≥ b ≥ c > 0, then a2 b2 c 2 6(b − c)2 + + ≥a+b+c+ . b c a a+b+c

Noncyclic Inequalities

247

2.37. If a ≥ b ≥ c > 0, then a2 b2 c 2 + + > 5(a − b). b c a

2.38. If a, b, c are positive real numbers, then b c 3 27(b − c)2 a + + ≥ + . b+c c+a a+b 2 16(a + b + c)2

2.39. Let a, b, c be positive real numbers such that a = min{a, b, c}. Prove that a b c 3 9(b − c)2 + + ≥ + . b+c c+a a+b 2 4(a + b + c)2

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

2.41. Let a, b, c be positive real numbers such that a = min{a, b, c}. Prove that b c 3 (b − c)2 a + + ≥ + . b+c c+a a+b 2 4bc 2.42. Let a, b, c be positive real numbers such that a = min{a, b, c}. Prove that (a)

a b + bc + ca 2(b − c)2 + ≤ 1; a2 + b2 + c 2 3(b2 + c 2 )

(b)

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

2.43. Let a, b, c be nonnegative real numbers such that a = max{a, b, c}. Prove that (a)

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

(b)

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

248

Vasile Cîrtoaje

2.44. Let a, b, c be positive real numbers such that a = min{a, b, c}. Prove that (a)

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

(b)

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

2.45. If a, b, c are positive real numbers, then a2 + b2 + c 2 9(a − c)2 ≥1+ . a b + bc + ca 4(a + b + c)2

2.46. Let a, b, c be nonnegative real numbers, no two of which are zero. If a = min{a, b, c}, then 1 1 1 6 . +p +p ≥ p b+c a2 − a b + b2 b2 − bc + c 2 c 2 − ca + a2

2.47. If a ≥ 1 ≥ b ≥ c ≥ 0 such that a b + bc + ca = a bc + 2, then

p ac ≤ 4 − 2 2.

2.48. If 0 < a ≤ b ≤ c such that a + b + c = 3, then a4 (b4 + c 4 ) ≤ 2.

2.49. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then a2 + b2 + c 2 − a − b − c ≥

5 (a − c)2 . 8

2.50. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then a3 + b3 + c 3 5 ≥ 1 + (a − c)2 . a+b+c 9

Noncyclic Inequalities

249

2.51. If a, b, c are nonnegative real numbers such that a ≥ b ≥ c,

a b + bc + ca = 3,

then (a)

7 a3 + b3 + c 3 ≥ 1 + (a − b)2 ; a+b+c 9

(b)

a3 + b3 + c 3 2 ≥ 1 + (b − c)2 . a+b+c 3

2.52. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then a4 + b4 + c 4 − a2 − b2 − c 2 ≥

11 (a − c)2 . 4

2.53. If a, b, c are nonnegative real numbers such that a ≥ b ≥ c,

a b + bc + ca = 3,

then (a) (b)

11 (a − b)2 ; 3 10 (b − c)2 . a4 + b4 + c 4 − a2 − b2 − c 2 ≥ 3

a4 + b4 + c 4 − a2 − b2 − c 2 ≥

2.54. Let a, b, c be nonnegative real numbers such that a ≤ b ≤ c,

a + b + c = 3.

Find the greatest real number k such that Æ (56b2 + 25)(56c 2 + 25) + k(b − c)2 ≤ 14(b + c)2 + 25.

2.55. If a ≥ b ≥ c > 0 such that a bc = 1, then a 3(a + b + c) ≤ 8 + . c

250

Vasile Cîrtoaje

2.56. If a ≥ b ≥ c > 0, then (a + b − c)(a2 b − b2 c + c 2 a) ≥ (a b − bc + ca)2 .

2.57. If a ≥ b ≥ c ≥ 0, then p 2(a − c)2 (a − c)2 3 ≤ a + b + c − 3 a bc ≤ . 2(a + c) a + 5c

2.58. If a ≥ b ≥ c ≥ d ≥ 0, then p 3(a − d)2 (a − d)2 4 ≤ a + b + c + d − 4 a bcd ≤ . a + 3d a + 5d

2.59. If a ≥ b ≥ c > 0, then (a) (b)

p 3(a − b)2 3 a + b + c − 3 a bc ≥ ; 5a + 4b p 64(a − b)2 3 . a + b + c − 3 a bc ≥ 7(11a + 24b)

2.60. If a ≥ b ≥ c > 0, then (a) (b)

p 3(b − c)2 3 a + b + c − 3 a bc ≥ ; 4b + 5c p 25(b − c)2 3 a + b + c − 3 a bc ≥ . 7(3b + 11c)

2.61. If a ≥ b ≥ c > 0, then a+ b+c−3

p 3

a bc ≥

3(a − c)2 . 4(a + b + c)

2.62. If a ≥ b ≥ c > 0, then (a)

a6 + b6 + c 6 − 3a2 b2 c 2 ≥ 12a2 c 2 (b − c)2 ;

(b)

a6 + b6 + c 6 − 3a2 b2 c 2 ≥ 10a3 c(b − c)2 .

Noncyclic Inequalities

251

2.63. Let k and a, b, c be positive real numbers, and let  ‹  ‹ 1 1 k 1 1 k 2 2 2 E = (ka + b + c) + + , F = (ka + b + c ) 2 + 2 + 2 . a b c a b c (a) If k ≥ 1, then v t F − (k − 2)2 2k (b) If 0 < k ≤ 1, then

v t F − k2 k+1

+2≥

E − (k − 2)2 ; 2k

+2≥

E − k2 . k+1

2.64. If a, b, c are positive real numbers, then a b 25c + + > 1. 2b + 6c 7c + a 9a + 8b

2.65. If a, b, c are positive real numbers such that 1 1 1 ≥ + , a b c then

1 1 55 1 + + ≥ . a+b b+c c+a 12(a + b + c)

2.66. If a, b, c are positive real numbers such that 1 1 1 ≥ + , a b c then a2

1 1 1 189 + 2 + 2 ≥ . 2 2 2 2 +b b +c c +a 40(a + b2 + c 2 )

2.67. If a, b, c are the lengths of the sides of a triangle, then a3 (b + c) + bc(b2 + c 2 ) ≥ a(b3 + c 3 ).

252

Vasile Cîrtoaje

2.68. If a, b, c are the lengths of the sides of a triangle, then (a + c)2 (b + c)2 (a + b)2 + ≥ . 2a b + c 2 2ac + b2 2bc + a2

2.69. If a, b, c are the lengths of the sides of a triangle, then a+b a+c b+c + ≥ . a b + c 2 ac + b2 bc + a2

2.70. If a, b, c are the lengths of the sides of a triangle, then b(a + c) c(a + b) a(b + c) + ≥ . ac + b2 a b + c2 bc + a2

2.71. Let a, b, c, d be nonnegative real numbers such that a2 − a b + b2 = c 2 − cd + d 2 . Prove that (a + b)(c + d) ≥ 2(a b + cd).

2.72. Let a, b ∈ (0, 1], a ≤ b. (a) If a ≤

(b) If b ≥

1 , then e

2a a ≥ a b + b a ;

1 , then e

2.73. If 0 ≤ a ≤ b and b ≥

2b b ≥ a b + b a .

1 , then 2 2b2b ≥ a2b + b2a .

Noncyclic Inequalities

253

2.74. If a ≥ b ≥ 0, then a−b a b−a ≤ 1 + p ; a

(a)

a a−b ≥ 1 −

(b)

3(a − b) . p 4 a

2.75. Let a, b, c and x, y, z be positive real numbers such that x + y + z = a + b + c. Prove that a x 2 + b y 2 + cz 2 + x yz ≥ 4a bc.

2.76. Let a, b, c and x, y, z be positive real numbers such that x + y + z = a + b + c. Prove that

x(3x + a) y(3 y + b) z(3z + c) + + ≥ 12. bc ca ab

2.77. Let a, b, c be given positive numbers. Find the minimum value F (a, b, c) of E(x, y, z) =

by ax cz + + y +z z+ x x + y

for any nonnegative numbers x, y, z, no two of which are zero.

2.78. Let a, b, c and x, y, z be real numbers. (a) If a b + bc + ca > 0, then [(b + c)x + (c + a) y + (a + b)z]2 ≥ 4(a b + bc + ca)(x y + yz + z x); (b) If a, b, c ≥ 0, then [(b + c)x + (c + a) y + (a + b)z]2 ≥ 4(a + b + c)(a yz + bz x + c x y).

254

Vasile Cîrtoaje

2.79. Find the best real numbers x, y, z such that p p p p ( a + b + c) a + b + c ≥ x a + y b + zc for all a ≥ b ≥ c ≥ 0.

2.80. Let a, b, c and x, y, z be positive real numbers such that a b c + + = 1. yz z x x y Prove that (a)

x + y +z ≥

Æ

4(a + b + c +

x + y +z >

(b)

p

p

a+b+

ab +

p

p

bc +

b+c+

p

p

p 3 ca ) + 3 a bc;

c + a.

2.81. Let a, b, c be the lengths of the sides of a triangle. If x, y, z are real numbers, then ( y a2 + z b2 + x c 2 )(za2 + x b2 + y c 2 ) ≥ (x y + yz + z x)(a2 b2 + b2 c 2 + c 2 a2 ).

2.82. If a, b, c, d are real numbers, then 6(a2 + b2 + c 2 + d 2 ) + (a + b + c + d)2 ≥ 12(a b + bc + cd).

2.83. If a, b, c, d are positive real numbers, then a2

1 1 1 4 1 + 2 + 2 + 2 ≥ . + a b b + bc c + cd d + d a ac + bd

2.84. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and a b + bc + cd + d a = 3. Prove that a3 bcd < 4.

Noncyclic Inequalities

255

2.85. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and a b + bc + cd + d a = 6. Prove that acd ≤ 2.

2.86. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and a b + bc + cd + d a = 9. Prove that a bd ≤ 4.

2.87. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and a2 + b2 + c 2 + d 2 = 10. Prove that 2b + 4d ≤ 3c + 5.

2.88. If a, b, c, d are positive real numbers such that a ≥ b ≥ c ≥ d, then (a + b + c + d)2 ≥ 8(ac + bd).

2.89. Let a, b, c, d be positive real numbers such that a ≤ b ≤ c ≤ d and a bcd = 1. Prove that a b c d 4 + + + + ≥ 2(ac + c b + bd + d a). b c d a 2.90. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and 3(a2 + b2 + c 2 + d 2 ) = (a + b + c + d)2 . Prove that (a) (b) (c)

a+d ≤ 2; b+c

p a+c 7+2 6 ≤ ; b+d 5 p a+c 3+ 5 ≤ . c+d 2

256

Vasile Cîrtoaje

2.91. Let a, b, c, d be nonnegative real numbers such that a ≥ b ≥ c ≥ d and 2(a2 + b2 + c 2 + d 2 ) = (a + b + c + d)2 . Prove that

p a ≥ b + 3c + (2 3 − 1)d.

2.92. If a ≥ b ≥ c ≥ d ≥ 0, then a+ b+c+d −4

p 4

a bcd ≥

p p 3 p ( b − 2 c + d)2 . 2

2.93. If a ≥ b ≥ c ≥ d ≥ 0, then (a) (b) (c) (d) (e) (f)

p p p 2 p 4 a + b + c + d − 4 a bcd ≥ (3 b − 2 c − d)2 ; 9 p p p 1 p 4 a + b + c + d − 4 a bcd ≥ (3 b − c − 2 d)2 ; 5 p p p 3 p 4 a + b + c + d − 4 a bcd ≥ ( b − 3 c + 2 d)2 ; 8 p p p 1 p 4 a + b + c + d − 4 a bcd ≥ (2 b − 3 c + d)2 ; 2 p p p p 1 4 a + b + c + d − 4 a bcd ≥ (2 b + c − 3 d)2 ; 6 p p 4 p 4 a + b + c + d − 4 a bcd ≥ ( b − d)2 . 3

2.94. If a, b, c, d, e are real numbers, then p 3 a b + bc + cd + d e ≤ . a2 + b2 + c 2 + d 2 + e2 2

2.95. If a, b, c, d, e, f are nonnegative real numbers such that a ≥ b ≥ c ≥ d ≥ e ≥ f, then (a + b + c + d + e + f )2 ≥ 8(ac + bd + ce + d f ).

Noncyclic Inequalities

257

2.96. If a1 ≥ a2 ≥ · · · ≥ a8 ≥ 0, then p p p a1 + a2 + · · · + a8 − 8 8 a1 a2 · · · a8 ≥ 3( a6 − a7 )2 .

2.97. Let a1 , a2 , . . . , an and b1 , b2 , · · · , bn be real numbers. Prove that a1 b1 + · · · + an bn +

q

(a12 + · · · + an2 )(b12 + · · · + bn2 ) ≥

2 (a1 + · · · + an )(b1 + · · · + bn ). n

2.98. Let a1 , a2 , . . . , an be positive real numbers such that a1 ≥ 2a2 . Prove that (5n − 1)(a12 + a22 + · · · + an2 ) ≥ 5(a1 + a2 + · · · + an )2 .

2.99. If a1 ≥ a2 ≥ · · · ≥ an > 0 such that a1 + a2 + · · · + an = n, then 1 1 1 4(n − 1)2 (a1 − a2 )2 . + + ··· + −n≥ a1 a2 an n3

2.100. If a1 ≥ a2 ≥ · · · ≥ an ≥ 0, then (a) (b)

1 p p p p a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ ( a1 + a2 − 2 an )2 ; 3 1 p p p p a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ (2 a1 − an−1 − an )2 . 4

2.101. If a1 ≥ a2 ≥ · · · ≥ an ≥ 0, n ≥ 3, then p p p n−1 p a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ ( an−2 + an−1 − 2 an )2 . 2n

2.102. If a1 ≥ a2 ≥ · · · ≥ an ≥ 0, n ≥ 3, then  ‹ p p p p 2 n (2 an−2 − an−1 − an )2 . a1 + a2 + · · · + an − n a1 a2 · · · an ≥ 1 − n

258

Vasile Cîrtoaje

2.103. Let a1 ≥ a2 ≥ · · · ≥ an ≥ 0. If n ≤ k ≤ n − 1, 2 then

p p 2k(n − k) p ( ak − ak+1 )2 . a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ n

2.104. If a1 ≥ a2 ≥ · · · ≥ an ≥ 0, n ≥ 4, then  ‹ 2 p p p p n (a) a1 + a2 + · · · + an − n a1 a2 · · · an ≥ 1 − (2 an−2 − 3 an−1 + an )2 ; n  ‹ 1 p 1 p p p 1− ( an−2 − 3 an−1 + 2 an )2 . (b) a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ 2 n

2.105. If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that a1 ≤ a2 ≤ · · · ≤ a n ,

a1 + a2 + · · · + an = 0,

then a12 + a22 + · · · + an2 + na1 an ≤ 0.

2.106. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = n. (a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then a13 + a23 + · · · + an3 + 2n ≥ 3(a12 + a22 + · · · + an2 ); (b) If a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then a13 + a23 + · · · + an3 + 2n ≤ 3(a12 + a22 + · · · + an2 ).

2.107. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. (a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then a14 + a24 + · · · + an4 + 5n ≥ 6(a12 + a22 + · · · + an2 ); (b) If a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then a14 + a24 + · · · + an4 + 6n ≤ 7(a12 + a22 + · · · + an2 ).

Noncyclic Inequalities

259

2.108. If a1 , a2 , . . . , an are positive real numbers such that 1 1 1 + + ··· + = n, a1 a2 an

a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

a12 + a22 + · · · + an2 + 2n ≥ 3(a1 + a2 + · · · + an ). 2.109. If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n,

a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then (a)

(b)

a1 + 1 a12

+1 1

a12

+3

+ +

a2 + 1 a22

+1 1

a22

+3

+ ··· + + ··· +

an + 1 ≤ n; an2 + 1 1 a12

+3

≤

n . 4

2.110. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n,

a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

a12 − 1 (a1 + 3)2

+

a22 − 1 (a2 + 3)2

+ ··· +

an2 − 1 (an + 3)2

≥ 0.

2.111. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

1 3a13

+4

+

1 3a23

+4

a1 + a2 + · · · + an = n, + ··· +

1 n ≥ . +4 7

3an3

2.112. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

a1 + a2 + · · · + an = n,

v v v t 3a1 t 3a2 t 3an + + ··· + ≤ n. 4 − a1 4 − a2 4 − an

260

Vasile Cîrtoaje

2.113. If a1 , a2 , . . . , an are nonnegative real numbers such that a12 + a22 + · · · + an2 = n,

a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

1 1 1 n + + ··· + ≤ . 3 − a1 3 − a2 3 − an 2

2.114. If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n,

a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

(1 + a12 )(1 + a22 ) · · · (1 + an2 ) ≥ 2n .

2.115. If a1 , a2 , . . . , an are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

a1 a2 · · · an = 1,

1 1 1 n + + ··· + ≥ . 2 2 2 (a1 + 1) (a2 + 1) (an + 1) 4

2.116. If a1 , a2 , . . . , an are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

a1 a2 · · · an = 1,

1 1 1 n + + ··· + ≥ . 2 2 2 (a1 + 2) (a2 + 2) (an + 2) 9

2.117. If a1 , a2 , . . . , an are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then a1n + a2n + · · · + ann − n ≥ n2



a1 a2 · · · an = 1,

‹ 1 1 1 + + ··· + −n . a1 a2 an

Noncyclic Inequalities

261

2.118. If a, b, c are positive real numbers such that a ≥ 1 ≥ b ≥ c, then

a bc = 1,

1− b 1−c 1−a + + ≥ 0. 2 2 3+a 3+ b 3 + c2

2.119. If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n, then a14 + a24 + · · · + an4 − n ≥

a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an , 14 2 (a + a22 + · · · + an2 − n). 3 1

262

Vasile Cîrtoaje

Noncyclic Inequalities

2.2

263

Solutions

P 2.1. If a, b, c are nonnegative real numbers such that a + b + c = 3, then (a b + c)(ac + b) ≤ 4. Solution. By the AM-GM inequality, we have (a b + c) + (ac + b) (a b + c)(ac + b) ≤ 2 •

˜2

=

(a + 1)2 (b + c)2 . 4

Therefore, it suffices to show that (a + 1)(b + c) ≤ 4. Indeed, (a + 1) + (b + c) (a + 1)(b + c) ≤ 2 •

˜2

= 4.

The equality holds for a = b = c = 1, for a = 1, b = 0, c = 2, and for a = 1, b = 2, c = 0.

P 2.2. If a, b, c are nonnegative real numbers, then a3 + b3 + c 3 − 3a bc ≥

1 (b + c − 2a)3 . 4

Solution. Write the inequality as 2(a + b + c)[(a − b)2 + (b − c)2 + (c − a)2 ] ≥ (b + c − 2a)3 . Consider the non-trivial case b + c − 2a ≥ 0. Since a + b + c ≥ b + c − a, it suffices to show that 2(a − b)2 + 2(c − a)2 ≥ (b + c − 2a)2 . Indeed, we have 2(a − b)2 + 2(c − a)2 − (b + c − 2a)2 = (b − c)2 ≥ 0. The equality holds for a = b = c, and also for a = 0 and b = c.

264

Vasile Cîrtoaje

P 2.3. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ c. Prove that (a)

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

(b)

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

Solution. (a) Write the inequality as (a + b + c)(a2 + b2 + c 2 − a b − bc − ca) ≥ 2(2b − a − c)3 . For the non-trivial case 2b − a − c ≥ 0, since a + b + c ≥ 2(2b − a − c), it suffices to show that a2 + b2 + c 2 − a b − bc − ca ≥ (2b − a − c)2 . This is equivalent to the obvious inequality 3(a − b)(b − c) ≥ 0. The equality holds for a = b = c, and also for a = b and c = 0. (b) Write the inequality as (a + b + c)(a2 + b2 + c 2 − a b − bc − ca) ≥ (a − 2b + c)3 . For the non-trivial case a − 2b + c ≥ 0, since a + b + c ≥ a − 2b + c, it suffices to show that a2 + b2 + c 2 − a b − bc − ca ≥ (a − 2b + c)2 . This is equivalent to the obvious inequality 3(a − b)(b − c) ≥ 0. The equality holds for a = b = c, and also for b = c = 0.

P 2.4. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ c. Prove that (a)

a3 + b3 + c 3 − 3a bc ≥ 3(a2 − b2 )(b − c);

(b)

a3 + b3 + c 3 − 3a bc ≥

9 (a − b)(b2 − c 2 ). 2

Noncyclic Inequalities

265

Solution. (a) Write the inequality as (a + b + c)(a2 + b2 + c 2 − a b − bc − ca) ≥ 3(a + b)(a − b)(b − c)). Since a + b + c ≥ a + b, it suffices to show that a2 + b2 + c 2 − a b − bc − ca ≥ 3(a − b)(b − c). Indeed, a2 + b2 + c 2 − a b − bc − ca − 3(a − b)(b − c) = (a − 2b + c)2 ≥ 0. The equality holds for a = b = c, and also for a = 2b and c = 0. (b) Write the inequality as (a + b + c)(a2 + b2 + c 2 − a b − bc − ca) ≥ Since a + b + c ≥

9 (a − b)(b − c)(b + c). 2

3 (b + c), it suffices to show that 2 a2 + b2 + c 2 − a b − bc − ca ≥ 3(a − b)(b − c).

This is equivalent to the obvious inequality (a − 2b + c)2 ≥ 0. The equality holds for a = b = c.

P 2.5. Let a, b, c be nonnegative real numbers such that a = max{a, b, c}. Prove that a6 + b6 + c 6 − 3a2 b2 c 2 ≥ 2(b4 + c 4 + 4b2 c 2 )(b − c)2 .

Solution. Because the inequality is symmetric in b and c, we may assume that b ≥ c; that is, a ≥ b ≥ c. We will show that a6 + b6 + c 6 − 3a2 b2 c 2 ≥ 2b6 + c 6 − 3b4 c 2 ≥ 2(b4 + c 4 + 4b2 c 2 )(b − c)2 . The left inequality is equivalent to the obvious inequality (a2 − b2 )(a4 + a2 b2 + b4 − 3b2 c 2 ) ≥ 0.

266

Vasile Cîrtoaje

The right inequality is equivalent to (b2 − c 2 )2 (2b2 + c 2 ) ≥ 2(b4 + c 4 + 4b2 c 2 )(b − c)2 , (b − c)2 [(b + c)2 (2b2 + c 2 ) − 2(b4 + c 4 + 4b2 c 2 )] ≥ 0, c(b − c)3 (4b2 − bc + c 2 ) ≥ 0. Clearly, the last inequality is true. The equality holds for a = b = c, for a = b and c = 0, and for a = c and b = 0.

P 2.6. Let a, b, c be nonnegative real numbers such that a = max{a, b, c}. Prove that a2 + b2 + c 2 ≥

9a bc 5 + (b − c)2 . a+b+c 3

Solution. Because the inequality is symmetric in b and c, we may assume that b ≥ c. Write the inequality as follows: (a2 + b2 + c 2 )(a + b + c) − 9a bc ≥

5 (a + b + c)(b − c)2 ; 3

5 (a + b + c)(b − c)2 ; 3 X X 10 (a + b + c) (b − c)2 + 2 a(b − c)2 ≥ (a + b + c)(b − c)2 . 3 It suffices to show that a3 + b3 + c 3 − 3a bc +

X

a(b − c)2 ≥

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

10 (a + b + c)(b − c)2 . 3

Moreover, it suffices to prove that (a + b + c)[(b − c)2 + (b − c)2 ] + 2a(b − c)2 + 2b(b − c)2 ≥

10 (a + b + c)(b − c)2 . 3

Clearly, this is true if 2(a + b + c) + 2a + 2b ≥

10 (a + b + c). 3

This inequality reduces to a + b − 2c ≥ 0. The equality holds for a = b = c.

Noncyclic Inequalities

267

P 2.7. If a, b, c are nonnegative real numbers, no two of which are zero, then 1 16 6 1 + + ≥ . 2 2 2 (a + b) (a + c) (b + c) a b + bc + ca (Vasile Cîrtoaje, 2014) Solution (by Nguyen Van Quy). Since the equality holds for a = 0 and b = c, we write the desired inequality in the form  ‹2 16 1 1 6 2 + + ≥ + , 2 (b + c) a+b a+c a b + bc + ca (a + b)(a + c) and apply then the AM-GM inequality  ‹2  ‹ 16 1 1 8 1 1 + + ≥ + . (b + c)2 a+b a+c b+c a+b a+c Therefore, it suffices to show that  ‹ 8 1 1 6 2 + ≥ + . b+c a+b a+c a b + bc + ca (a + b)(a + c) Since (a + b)(a + c) ≥ a b + bc + ca, it is enough to show that  ‹ 8 1 1 8 + ≥ . b+c a+b a+c a b + bc + ca Write this inequality as (2a + b + c)(ab + bc + ca) ≥ (a + b)(b + c)(c + a). Indeed, (2a + b + c)(a b + bc + ca) ≥ (a + b + c)(a b + bc + ca) ≥ (a + b)(b + c)(c + a). This completes the proof. The equality holds for a = 0 and b = c.

P 2.8. If a, b, c are nonnegative real numbers, no two of which are zero, then 1 1 2 5 + + ≥ . 2 2 2 (a + b) (a + c) (b + c) 2(a b + bc + ca)

268

Vasile Cîrtoaje

Solution. This inequality follows from Iran 1996 inequality (see P 1.71 in Volume 2 for k = 2), namely 1 1 1 9 , + + ≥ 2 2 2 (a + b) (a + c) (b + c) 4(a b + bc + ca) and the inequality in P 2.7, namely 1 1 16 6 + + ≥ . (a + b)2 (a + c)2 (b + c)2 a b + bc + ca Indeed, summing the first inequality multiplied by 14 and the second inequality, we get the desired inequality. The equality holds for a = 0 and b = c.

P 2.9. If a, b, c are positive real numbers, then (a + b)3 (a + c)3 ≥ 4a2 bc(2a + b + c)2 . (XZLBQ, 2014) Solution (by Nguyen Van Quy). Write the inequality as (a + b)2 (a + c)2 (2a + b + c)2 ≥ . 4a2 bc (a + b)(a + c) Since (a + b)2 (a + c)2 = [(a − b)2 + 4a b][(a − c)2 + 4ac] ≥ 4ac(a − b)2 + 4a b(a − c)2 + 16a2 bc, it suffices to show that (a − b)2 (a − c)2 (2a + b + c)2 + +4≥ , ab ac (a + b)(a + c) which is equivalent to (a − b)2 (a − c)2 (b − c)2 + ≥ . ab ac (a + b)(a + c) Indeed, by the Cauchy-Schwarz inequality, we have (a − b)2 (a − c)2 (a − b − a + c)2 (b − c)2 + ≥ ≥ . ab ac a b + ac (a + b)(a + c) The equality holds for a = b = c.

Noncyclic Inequalities

269

P 2.10. If a, b, c are positive real numbers such that a bc = 1, then (a)

a b 1 + + ≥ a + b + 1; b c a

(b)

a b 1 p + + ≥ 3(a2 + b2 + 1). b c a (Vasile Cîrtoaje, 2007)

Solution. (a) First Solution. Write the inequality as  ‹  ‹  ‹ a b b 1 1 2 + + + + + a ≥ 3a + 2b + 2. b c c a a By the AM-GM inequality, we have v v ‹  ‹  ‹  t t b 2 b 1 1 a b 3 a + + + +a ≥3 +2 + 2 ≥ 3a + 2b + 2. 2 + b c c a a bc ca The equality holds for a = b = c = 1. Second Solution. Since c =

1 , the inequality becomes as follows: ab 1 a + a b2 + ≥ a + b + 1, b a

1 1 b 1 + b2 + 2 ≥ 1 + + , b a a a 1 1 1 − (b + 1) + b2 + − 1 ≥ 0, a2 a b  ‹2 1 b+1 (b − 1)2 (3b + 4) − + ≥ 0. a 2 4b (b) Write the inequality as  ‹ 1 1 Æ 2 a + b + ≥ 3(a2 + b2 + 1). b a By squaring, this inequality becomes  ‹ 1 1 2 2 4 a b + 2b − 3 + 2 + 2 ≥ b2 + 3 − . b a b Since b4 + 2b − 3 +

1 1 (b − 1)2 (2b + 1) > 2b − 3 + = ≥ 0, b2 b2 b2

270

Vasile Cîrtoaje

by the AM-GM inequality, we have a

2



v ‹ t 1 1 1 b + 2b − 3 + 2 + 2 ≥ 2 b4 + 2b − 3 + 2 . b a b 4

Thus, it suffices to prove that 2

v t

b4 + 2b − 3 +

1 2 ≥ b2 + 3 − . b2 b

Squaring again, we get the inequality b5 − 2b3 + 4b2 − 7b + 4 ≥ 0, which is equivalent to the obvious inequality b(b2 − 1)2 + 4(b − 1)2 ≥ 0. The equality holds for a = b = c = 1.

P 2.11. If a, b, c are positive real numbers such that a bc ≥ 1, then a

b

a b b c c c ≥ 1. (Vasile Cîrtoaje, 2011) Solution. Write the inequality as a b ln a + ln b + c ln c ≥ 0. b c Since f (x) = x ln x is a convex function on (0, ∞), by Jensen’s inequality, we have ‹  ‹ pa + q b + r c pa + q b + r c pa ln a + q b ln b + r c ln c ≥ (p + q + r) ln p+q+r p+q+r 

pa + q b + r c = (pa + q b + r c) ln p+q+r 

for all p, q, r > 0. Choosing p=

1 1 , q= , b c

r = 1,

‹

Noncyclic Inequalities

271

we get  a b + +c b a b a b c  ln a + ln b + c ln c ≥ ( + + c) ln  . 1 1 b c b c + +1 b c Thus, it suffices to show that 

1 1 a b + + c ≥ + + 1. b c b c Since a ≥

1 , we need to show that bc 1 b 1 1 + + c ≥ + + 1. b2 c c b c

This is equivalent to 1 c + b + c 2 ≥ + 1 + c, 2 b b  ‹ 1 1 c2 − 1 + + b − 1 + 2 ≥ 0, b b ‹2  2 (b − 1) (4b + 3) b+1 + ≥ 0. c− 2b 4b2 The equality holds for a = b = c = 1.

P 2.12. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a b + bc + ca = 3. Prove that a b2 c 3 < 4. (Vasile Cîrtoaje, 2012) Solution. From a b + bc + ca = 3, it is clear that c
4(a + b)3 − 27a b2 = 4a3 + 12a2 b − 15a b2 + 4b3 = (a + 4b)(2a − b)2 ≥ 0.

272

Vasile Cîrtoaje

P 2.13. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a b + bc + ca = Prove that a b2 c 2 ≤

5 . 3

1 . 3 (Vasile Cîrtoaje, 2012)

Solution. By the AM-GM inequality, we have p

bc.

bc + bc ≤

5 . 3

a b + ca ≥ 2a Thus, from a b + bc + ca =

5 , we get 3 2a

p

Therefore, it suffices to show that (5 − 3bc)b2 c 2 1 ≤ . p 3 6 bc Setting

p

bc = t, this inequality becomes 3t 5 − 5t 3 + 2 ≥ 0.

Indeed, be the AM-GM inequality, we have 3t 5 + 2 = t 5 + t 5 + t 5 + 1 + 1 ≥ 5 The equality holds for a =

p 5

t 5 · t 5 · t 5 · 1 · 1 = 5t 3 .

1 and b = c = 1. 3

P 2.14. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a b + bc + ca = 3. Prove that 9 ; 8

(a)

a b2 c ≤

(b)

a b4 c ≤ 2;

(c)

a b3 c 2 ≤ 2. (Vasile Cîrtoaje, 2012)

Noncyclic Inequalities

273

Solution. From (b − a)(b − c) ≤ 0, we get b2 + ac ≤ b(a + c), b2 + ac ≤ 3 − ac, b2 + 2ac ≤ 3. (a) We have 9 − 8a b2 c ≥ 9 − 4b2 (3 − b2 ) = (2b2 − 3)2 ≥ 0. s s 1 3 3 The equality holds for a = and b = c = . 2 2 2 (b) We have 4 − 2a b4 c ≥ 4 − b4 (3 − b2 ) = (b2 − 2)2 (b2 + 1) ≥ 0. p p 2 The equality holds for a = and b = c = 2. 4 (c) Write the desired inequality as follows: 2(a b + bc + ca)3 ≥ 27a b3 c 2 ,  ca 3 2 a+c+ ≥ 27ac 2 . b Since ca/b ≥ a, it suffices to show that 2(2a + c)3 ≥ 27ac 2 , which is equivalent to the obvious inequality (a + 2c)(4a − c)2 ≥ 0. p p 2 and b = c = 2. The equality holds for a = 4

P 2.15. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a+b+c =

1 1 1 + + . a b c

Prove that a b2 c 3 ≥ 1. (Vasile Cîrtoaje, 1998)

274

Vasile Cîrtoaje

First Solution. Write the inequality in the homogeneous form ˜ • a bc(a + b + c) 3 2 3 ab c ≥ . a b + bc + ca This is true if (a b + bc + ca)3 ≥ a2 b(a + b + c)3 . Since (a b + bc + ca)2 ≥ 3a bc(a + b + c), it suffices to show that 3c(a b + bc + ca) ≥ a(a + b + c)2 . Indeed, 3c(a b + bc + ca) − a(a + b + c)2 ≥ (a + b + c)(a b + bc + ca) − a(a + b + c)2 = (a + b + c)(bc − a2 ) ≥ 0. The equality holds for a = b = c = 1. Second Solution. Let us show that a ≤ 1 and bc ≥ 1. Indeed, if a > 1, then 1 < a ≤ b ≤ c and 1 1 1 1 − a2 1 − b2 1 − c 2 + + < 0, a+b+c− − − = a b c a b c which is false. On the other hand, from a ≤ 1 and  ‹ 1 1 a − = (b + c) −1 , a bc we get bc ≥ 1. Similarly, we can prove that c ≥ 1 and a b ≤ 1. Since bc ≥ 1, it suffices to show that a bc 2 ≥ 1. Taking account of a b ≤ 1, we have  ‹ ‹  p  1 p ‹ 1 1 1 c − = (a + b) − 1 ≥ 2 ab − 1 = 2 p − ab , c ab ab ab p 1 1 c − ≥ p − a b, c ab p   ‹ 1 ab c−p 1+ ≥ 0. c ab Clearly, the last inequality yields a bc 2 ≥ 0.

Noncyclic Inequalities

275

P 2.16. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a + b + c = a bc + 2. Prove that (1 − b)(1 − a b3 c) ≥ 0. (Vasile Cîrtoaje, 1999) Solution. Let us show that a ≤ 1 and c ≥ 1. To do this, we write the hypothesis a + b + c = a bc + 2 as (1 − a)(1 − c) + (1 − b)(1 − ac) = 0,

(*)

If a > 1, then 1 < a ≤ b ≤ c, which contradicts (*). Similarly, if c < 1, then a ≤ b ≤ c < 1, which also contradicts (*). Therefore, we have a ≤ 1,

c ≥ 1.

According to (*), we get (1 − b)(1 − ac) = (1 − a)(c − 1) ≥ 0.

(**)

There are two cases to consider. Case 1: b ≥ 1. According to (**), we have ac ≥ 1. Therefore, a b3 c = ac · b3 ≥ 1, hence (1 − b)(1 − a b3 c) ≥ 0. Case 2: b ≤ 1. According to (**), we have ac ≤ 1. Therefore, a b3 c = ac · b3 ≤ 1, and hence (1 − b)(1 − a b3 c) ≥ 0. This completes the proof. The equality holds for a = b = 1 or b = c = 1.

P 2.17. Let a, b, c be positive real numbers such that a ≤ b ≤ c and a+b+c = Prove that b≥

1 1 1 + + . a b c

1 . a+c−1 (Vasile Cîrtoaje, 2007)

276

Vasile Cîrtoaje

Solution. Let us show that a ≤ 1, From a + b + c =

c ≥ 1.

1 1 1 + + and a b c

a+b+c+

(a − 1)2 (b − 1)2 (c − 1)2 1 1 1 + + −6= + + ≥ 0, a b c a b c

we get a+b+c =

1 1 1 + + ≥ 3. a b c

Then, 1 1 ≥ a 3



‹ 1 1 1 + + ≥ 1, a b c

c≥

a+b+c ≥ 1. 3

Further, consider the following two cases. Case 1: a bc ≥ 1. Write the desired inequality as a+c−1− We have a+c−1−

1 ≥ 0. b

1 a bc − 1 = (1 − a)(c − 1) + ≥ 0. b b

Case 2: a bc ≤ 1. Since a+c−1−

1 1 1 = + − 1 − b, b a c

the desired inequality is equivalent to 1 1 + − 1 − b ≥ 0. a c We have

 ‹ ‹ 1 1 1 − a bc 1 1 + −1− b = −1 1− + ≥ 0. a c a c ac

This completes the proof. The equality holds for a = b = c = 1.

P 2.18. Let a, b, c be real numbers, no two of which are zero. Prove that (a)

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

(b)

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

Noncyclic Inequalities

277

Solution. (a) Consider two cases. Case 1: 2a2 ≤ b2 + c 2 . By the Cauchy-Schwarz inequality, we have [(b − a) + (a − c)]2 (b − c)2 (a − b)2 (a − c)2 ) + ≥ = . a2 + b2 a2 + c 2 (a2 + b2 ) + (a2 + c 2 ) 2a2 + b2 + c 2 Thus, it suffices to show that 1 1 ≥ , 2a2 + b2 + c 2 2(b2 + c 2 ) which reduces to b2 + c 2 ≥ 2a2 . Case 2: 2a2 ≥ b2 + c 2 . By the Cauchy-Schwarz inequality, we have (a − b)2 (a − c)2 [c(b − a) + b(a − c)]2 a2 (b − c)2 + ≥ = . a2 + b2 a2 + c 2 c 2 (a2 + b2 ) + b2 (a2 + c 2 ) a2 (b2 + c 2 ) + 2b2 c 2 Therefore, it suffices to prove that a2 1 ≥ , a2 (b2 + c 2 ) + 2b2 c 2 2(b2 + c 2 ) which reduces to a2 (b2 + c 2 ) ≥ 2b2 c 2 . This is true since 2a2 (b2 + c 2 ) − 4b2 c 2 ≥ (b2 + c 2 )2 − 4b2 c 2 = (b2 − c 2 )2 . The equality holds for a = b = c. (b) The inequality follows from the inequality in (a) by replacing a with −a. The equality holds for −a = b = c.

P 2.19. Let a, b, c be real numbers, no two of which are zero. If bc ≥ 0, then (a)

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

(b)

(a + b)2 (a + c)2 (b − c)2 + ≥ . a2 + b2 a2 + c 2 (b + c)2 (Vasile Cîrtoaje, 2011)

278

Vasile Cîrtoaje

Solution. (a) Consider two cases: a2 ≤ bc and a2 ≥ bc. Case 1: a2 ≤ bc. By the Cauchy-Schwarz inequality, we have (a − b)2 (a − c)2 ) [(b − a) + (a − c)]2 (b − c)2 + ≥ = . a2 + b2 a2 + c 2 (a2 + b2 ) + (a2 + c 2 ) 2a2 + b2 + c 2 Thus, it suffices to show that 1 1 ≥ , 2a2 + b2 + c 2 (b + c)2 which is equivalent to a2 ≤ bc. Case 2: a2 ≥ bc. By the Cauchy-Schwarz inequality, we have (a − b)2 (a − c)2 [c(b − a) + b(a − c)]2 a2 (b − c)2 + ≥ = . a2 + b2 a2 + c 2 c 2 (a2 + b2 ) + b2 (a2 + c 2 ) a2 (b2 + c 2 ) + 2b2 c 2 Therefore, it suffices to prove that 1 a2 ≥ , 2 2 2 2 2 a (b + c ) + 2b c (b + c)2 which reduces to bc(a2 − bc) ≥ 0. The equality holds for a = b = c, for b = 0 and a = c, and for c = 0 and a = b. (b) The inequality follows from the inequality in (a) by replacing a with −a. The equality holds for −a = b = c, for b = 0 and a + c = 0, and for c = 0 and a + b = 0.

P 2.20. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that |a − b|3 |a − c|3 |b − c|3 + ≥ . a3 + b3 a3 + c 3 (b + c)3 (Vasile Cîrtoaje, 2013) Solution. Without loss of generality, assume that b ≥ c. Thus, we have three cases to consider: a ≥ b ≥ c, b ≥ c ≥ a and b ≥ a ≥ c. Case 1: a ≥ b ≥ c. It suffices to show that |a − c|3 |b − c|)3 ≥ , (a + c)3 (b + c)3 which is equivalent to a−c b−c ≥ . a+c b+c

Noncyclic Inequalities

279

Indeed, a−c b−c 2c(a − b) − = ≥ 0. a+c b+c (a + c)(b + c) Case 2: b ≥ c ≥ a. It suffices to show that (b − c)3 (b − a)3 ≥ . a3 + b3 (b + c)3 Indeed, (b − a)3 (b − c)3 (b − c)3 b − c)3 ≥ ≥ ≥ . a3 + b3 a3 + b3 b3 + c 3 (b + c)3 Case 3: b ≥ a ≥ c. We need to prove that (b − c)3 (b − a)3 (a − c)3 + ≥ . a3 + b3 a3 + c 3 (b + c)3 Using the substitution x=

b−a , a+b

y=

we have b=

a−c , 0 ≤ x < 1, 0 ≤ y ≤ 1, a+c

1+ x a, 1− x

c=

1− y a, 1+ y

(b − a)3 =

8x 3 a3 , (1 − x)3

(a − c)3 =

8 y3 a3 , (1 + y)3

a3 + b3 =

2(1 + 3x 3 ) , (1 − x)3

a3 + c 3 =

2(1 + 3 y 2 ) , (1 + y)3

x+y b−c = . b+c 1+ xy Thus, the desired inequality becomes 4 y3 (x + y)3 4x 3 + ≥ , 1 + 3x 2 1 + 3 y 2 (1 + x y)3 x 2 + y 2 − x y + 3x 2 y 2 (x + y)2 ≥ , (1 + 3x 2 )(1 + 3 y 2 ) 4(1 + x y)3 s − p + 3p2 s + 2p ≥ , 1 + 3s + 9p2 4(1 + p)3 where s = x 2 + y 2,

p = x y, 0 ≤ p < 1, 2p ≤ s ≤ 1 + p2 .

280

Vasile Cîrtoaje

Therefore, we need to show that f (s) ≥ 0, where f (s) = 4(1 + p)3 (s − p + 3p2 ) − (s + 2p)(3s + 1 + 9p2 ). Since f is a concave function, it suffices to show that f (2p) ≥ 0 and f (1 + p2 ) ≥ 0. Indeed, we have f (2p) = 4p3 (3p + 1)(p + 3) ≥ 0, f (1 + p2 ) = 16p3 (p + 1)2 ≥ 0. Thus, the proof is completed. The equality holds for a = b = c, for b = 0 and a = c, and for c = 0 and a = b.

P 2.21. Let a, b, c be positive real numbers, b 6= c. Prove that ab ac (b + c)2 + ≤ . (a + b)2 (a + c)2 4(b − c)2 (Vasile Cîrtoaje, 2010) Solution. Write the inequality as (a − b)2 (a − c)2 (b + c)2 + + ≥ 2. (a + b)2 (a + c)2 (b − c)2 Making the substitution x=

a−b , a+b

y=

a−c b+c , z= , a+c b−c

we can write the inequality as x 2 + y 2 + z 2 ≥ 2. From 1+ x =

2a 2a 2b , 1+ y = , 1+z = a+b a+c b−c

1− x =

2b 2c −2c , 1− y = , 1−z = , a+b a+c b−c

and

we get (1 + x)(1 − y)(1 + z) + (1 − x)(1 + y)(1 − z) = 0, x y + yz − z x = 1. Therefore, we have x 2 + y 2 + z 2 − 2 = x 2 + y 2 + z 2 − 2(x y + yz − z x) = (x − y + z)2 ≥ 0.

Noncyclic Inequalities

281

The equality holds for x + z = y; that is, (b + c)a2 − (b2 + c 2 − 6bc)a + bc(b + c) = 0.

P 2.22. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 3bc + a2 3a b − c 2 3ac − b2 ≥ + 2 . b2 + c 2 a2 + b2 a + c2 (Vasile Cîrtoaje, 2014) Solution (by Nguyen Van Quy). Write the inequality as b2 c2 3bc 3a b 3ac a2 + + + ≥ 2 + . b2 + c 2 a2 + c 2 a2 + b2 b2 + c 2 a + b2 a2 + c 2 By the Cauchy-Schwarz inequality, we have b2 c2 (b2 + c 2 )2 (b2 + c 2 )2 + ≥ = a2 + c 2 a2 + b2 b2 (a2 + c 2 ) + c 2 (a2 + b2 ) a2 (b2 + c 2 ) + 2b2 c 2 a2 2b2 c 2 a2 (b2 + c 2 ) + 2b2 c 2 = 2 − − , ≥2− (b2 + c 2 )2 b2 + c 2 (b2 + c 2 )2 hence

a2 b2 c2 2b2 c 2 + + ≥ 2 − . b2 + c 2 a2 + c 2 a2 + b2 (b2 + c 2 )2

Therefore, it suffices to show that 2−

2b2 c 2 3bc 3a b 3ac + 2 ≥ 2 + 2 . 2 2 2 2 2 (b + c ) b +c a +b a + c2

This inequality is equivalent to    ‹  ‹  ‹ 1 2b2 c 2 3 3a b 3 3ac 3 3bc − 2 + − + − ≥ − , 2 (b + c 2 )2 2 a2 + b2 2 a2 + c 2 2 b2 + c 2 (b2 − c 2 )2 (a − b)2 (a − c)2 (b − c)2 + + ≥ . 3(b2 + c 2 )2 a2 + b2 a2 + c 2 b2 + c 2 Using the inequality in P 2.19-(a), namely (a − b)2 (a − c)2 (b − c)2 + ≥ , a2 + b2 a2 + c 2 (b + c)2

282

Vasile Cîrtoaje

it is enough to prove that 1 1 (b + c)2 + ≥ 2 , 3(b2 + c 2 )2 (b + c)2 b + c2 which is equivalent to 1 2(b2 − bc + c 2 ) ≥ . (b + c)2 3(b2 + c 2 )2 We have 3(b2 + c 2 )2 − 2(b + c)2 (b2 − bc + c 2 ) = 3(b2 + c 2 )2 − 2(b + c)(b3 + c 3 ) = b4 + c 4 + 6b2 c 2 − 2bc(b2 + c 2 ) ≥ (b2 + c 2 )2 − 2bc(b2 + c 2 ) = (b2 + c 2 )(b − c)2 ≥ 0. The equality holds for a = b = c.

P 2.23. Let a, b, c be nonnegative real numbers such that a + b > 0. Prove that a bc ≥ (b + c − a)(c + a − b)(a + b − c) +

a b(a − b)2 . a+b (Vasile Cîrtoaje, 2011)

Solution. Since (b + c − a)(c + a − b)(a + b − c) =

2(a2 b2 + b2 c 2 + c 2 a2 ) − a4 − b4 − c 4 , a+b+c

we can rewrite the inequality as a4 + b4 + c 4 + a bc(a + b + c) ≥ 2(a2 b2 + b2 c 2 + c 2 a2 ) +

a b(a + b + c)(a − b)2 . a+b

By Schur’s inequality of fourth degree, we have a4 + b4 + c 4 + a bc(a + b + c) ≥

X

a b(a2 + b2 ).

Therefore, it suffices to prove that X

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

a b(a + b + c)(a − b)2 , a+b

Noncyclic Inequalities

283

which is equivalent to X

a b(a − b)2 ≥

a b(a + b + c)(a − b)2 , a+b

or

a bc(a − b)2 . a+b This inequality follows immediately from the Cauchy-Schwarz inequality p p (a + b)[bc(b − c)2 + ca(c − a)2 ] ≥ [ a bc(b − c) + a bc(c − a)]2 . bc(b − c)2 + ca(c − a)2 ≥

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

P 2.24. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ c. Prove that (a)

a bc ≥ (b + c − a)(c + a − b)(a + b − c) +

(b)

a bc ≥ (b + c − a)(c + a − b)(a + b − c) +

2a b(a − b)2 ; a+b 27b(a − b)4 . 4a2 (Vasile Cîrtoaje, 2011)

Solution. (a) Write the inequality as X

a(a − b)(a − c) ≥

2a b(a − b)2 . a+b

Since c(c − a)(c − b) ≥ 0, it suffices to show that a(a − b)(a − c) + b(b − c)(c − a) ≥

2a b(a − b)2 . a+b

Since a(a − b)(a − c) = a(a − b)[(a − b) + (b − c)] = a(a − b)2 + a(a − b)(b − c) ≥

2a b(a − b)2 + a(a − b)(b − c), a+b

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

284

Vasile Cîrtoaje

The equality holds for a = b = c, and for a = b and c = 0. (b) Write the inequality as X

a(a − b)(a − c) ≥

27b(a − b)4 . 4a2

Since c(c − a)(c − b) ≥ 0, it suffices to show that a(a − b)(a − c) + b(b − c)(c − a) ≥

27b(a − b)4 , 4a2

or a(a − b)2 + a(a − b)(b − c) + b(b − c)(c − a) ≥ Since a(a − b)2 −

27b(a − b)4 . 4a2

(a − b)2 (a − 3b)2 27b(a − b)4 = , 4a2 4a2

it suffices to show that a(a − b)(b − c) + b(b − c)(b − a) ≥ 0. This inequality is equivalent to (a − b)2 (b − c) ≥ 0. The equality holds for a = b = c, and for a/3 = b = c.

P 2.25. Let a, b, c be nonnegative real numbers such that a + b > 0. Prove that X

2

2 2

a (a − b)(a − c) ≥ a b



a−b a+b

‹2

. (Vasile Cîrtoaje, 2011)

Solution. Without loss of generality, assume that a ≥ b. There three cases to consider. Case 1. c ≥ a ≥ b. Since a2 (a − b)(a − c) + c 2 (c − a)(c − b) ≥ a2 (a − b)(a − c) + c 2 (c − a)(a − b) = (a − b)(c − a)2 (c + a) ≥ 0, it suffices to show that b2 (a − b)(c − b) ≥ a2 b2



a−b a+b

‹2

.

Noncyclic Inequalities

285

Since c − b ≥ a − b, this is true if 1≥

 a 2 , a+b

which is clearly true. Case 2. a ≥ b ≥ c. Since c 2 (c − a)(c − b) ≥ 0 and a2 (a − b)(a − c) + b2 (b − c)(b − a) = (a − b)2 [a2 + a b + b2 − c(a + b)] ≥ (a − b)2 [a2 + a b + b2 − b(a + b)] = a2 (a − b)2 , it suffices to show that 

b 1≥ a+b

‹2

,

which is clarly true. Case 3. a ≥ c ≥ b. Since b2 (b − c)(b − a) ≥ b2 (c − b)2 and a2 (a − b)(a − c) + c 2 (c − a)(c − b) = (a − c)2 [a2 + ac + c 2 − b(a + c)] ≥ (a − c)2 [a2 + ac + c 2 − c(a + c)] = a2 (a − c)2 , it suffices to show that 2

2

2

2

2 2

b (c − b) + a (a − c) ≥ a b



a−b a+b

‹2

.

By the Cauchy-Schwarz inequality, we have  ‹  1 1  2 2 2 2 + b (c − b) + a (a − c) ≥ [(c − b) + (a − c)]2 = (a − b)2 . b2 a2 Therefore, it suffices to prove that  ‹2 a2 b2 (a − b)2 2 2 a−b ≥a b , a2 + b2 a+b which is clearly true. This completes the proof. The equality holds for a = b = c, and for a = 0 and b = c (or any cyclic permutation). Remark. Similarly, we can prove the following generalization. • Let a, b, c be nonnegative real numbers such that a + b > 0. If k is a positive natural number, then  ‹ X ab k k a (a − b)(a − c) ≥ (a − b)2 . a+b

286

Vasile Cîrtoaje

P 2.26. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that a b2 + bc 2 + 2ca2 ≤ 8. Solution. By the AM-GM inequality, we get 1  a a 3 1 1 ca ≤ c+ + = (c + a)3 ≤ (a + b + c)3 = 1. 27 2 2 27 27 2

Therefore, it suffices to show that a b2 + bc 2 + ca2 ≤ 4, which is the inequality in P 1.2. The equality occurs for a = 2, b = 0, c = 1.

P 2.27. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 3 a b2 + bc 2 + a bc ≤ 4. 2 (Vasile Cîrtoaje and Vo Quoc Ba Can, 2007) Solution. Consider two cases. Case 1: c ≥ 2b. We have  3 c a b2 + bc 2 + a bc = b(a + c)2 − a b a − b + ≤ b(a + c)2 2 2

= 4b

a + ca + c 2

2

 ≤ 4

b+

a + c a + c 3 + 2 2  = 4. 3

Case 2: 2b > c. Write the desired inequality as f (a) ≥ 0, where ‹ a+b+c 3 3 − a b2 − bc 2 − a bc, f (a) = 4 3 2  ‹2 a+b+c 3 f 0 (a) = 4 − b2 − bc. 3 2 

The equation f 0 (a) = 0 has the positive root v 3 t b(2b + 3c) (2b − c)(5b + 8c) a1 = −b−c = p . 2 2 6 2b(2b + c) + 8(b + c)

Noncyclic Inequalities

287

Since f 0 (a) < 0 for 0 ≤ a < a1 and f 0 (a) > 0 for a > a1 , f (a) is decreasing on [0, a1 ] and increasing on [a1 , ∞); consequently, f (a) ≥ f (a1 ). To complete the proof, it suffices to show that f (a1 ) ≥ 0. Indeed, since a1 + b + c 4 3 

‹2

= b2 +

3 bc, 2

we have ‹ ‹  a1 + b + c 3 3 − a1 b2 + bc − bc 2 3 2  ‹  ‹ a1 + b + c 3 3 = b2 + bc − a1 b2 + bc − bc 2 3 2 2  ‹ b + c − 2a1 3 = b2 + bc − bc 2 3 2 v ‚ Œ ‹ t 2b2 + 3bc  3 2 b + bc − bc 2 = b+c− 2 2 ” — Æ b 4b2 + 10bc + 2c 2 − (2b + 3c) 2b(2b + 3c) = 4 bc(2b − c)2 (b + 2c) = ≥ 0. p 2[4b2 + 10bc + 2c 2 + (2b + 3c) 2b(2b + 3c)]

f (a1 ) = 4



Thus, the proof is completed. The equality holds for a = 0, b = 1, c = 2, and for a = 1, b = 2, c = 0.

P 2.28. Let a, b, c be nonnegative real numbers such that a + b + c = 5. Prove that a b2 + bc 2 + 2a bc ≤ 20. (Vo Quoc Ba Can, 2011) Solution. Write the inequality as b(a b + c 2 + 2ac) ≤ 20. We see that the equality holds for a = 1 and b = c = 2. From (a − b/2)2 ≥ 0, it is clear that b2 a b ≤ a2 + . 4

288

Vasile Cîrtoaje

Therefore, for b ≤ 4, we have    b2 b2 2 2 b(a b + c + 2ac) − 20 ≤ b a + + c + 2ac − 20 = b (a + c) + − 20 4 4   b2 5 2 = b (5 − b) + − 20 = (b − 4)(b − 2)2 ≤ 0. 4 4 

2

2

Consider now that b > 4. We have a b2 + bc 2 + 2a bc − 20 = a b2 + b(5 − a − b)2 + 2a b(5 − a − b) − 20 = b3 + a b2 − 10b2 − a2 b + 25b − 20 < b3 + (5 − b)b2 − 10b2 + 25b − 20 = −5(b − 4)(b − 1) < 0.

P 2.29. If a, b, c are nonnegative real numbers, then a3 + b3 + c 3 − a2 b − b2 c − c 2 a ≥

8 (a − b)(b − c)2 . 9

Solution. Since 3(a3 + b3 + c 3 − a2 b − b2 c − c 2 a) =

X X (2a3 − 3a2 b + b3 ) = (2a + b)(a − b)2 ,

we can write the inequality as (2a + b)(a − b)2 + (2b + c)(b − c)2 + (2c + a)(c − a)2 ≥

8 (a − b)(b − c)2 . 3

If a ≤ b, then (2a + b)(a − b)2 + (2b + c)(b − c)2 + (2c + a)(c − a)2 ≥ 0 ≥

8 (a − b)(b − c)2 . 3

If a ≥ b, then there are two cases to consider: b ≥ c and b ≤ c. Case 1: a ≥ b ≥ c. It suffices to show that (2c + a)(a − c)2 ≥

8 (a − b)(b − c)2 . 3

By the AM-GM inequality, we have (a − b)(b − c)2 = 4(a − b)



b−c 2

‹

b−c 2

‹

≤4

•

(a − b) + (b − c)/2 + (b − c)/2 3

˜3

Noncyclic Inequalities

289 =

4 (a − c)3 . 27

Therefore, it suffices to show that (2c + a)(a − c)2 ≥

32 (a − c)3 , 81

which is obvious. Case 2: a ≥ b, c ≥ b. Making the substitution a = b + p and c = b + q, where p ≥ 0 and q ≥ 0, the inequality becomes (3b + 2p)p2 + (3b + q)q2 + (3b + p + 2q)(p − q)2 ≥

8 2 pq , 3

or 3[p2 + q2 + (p − q)2 ]b + 2p3 + q3 + (p + 2q)(p − q)2 ≥

8 2 pq . 3

It suffices to show that 2p3 + q3 + (p + 2q)(p − q)2 ≥

8 2 pq , 3

which is equivalent to 2p3 + 2q3 ≥

34 2 pq . 9

By the AM-GM inequality, we have 2p3 + 2q3 = 2p3 + q3 + q3 ≥ 3 because

Æ 3

2p3 q6 ≥

34 2 pq , 9

p 34 3 . 3 2> 9

The equality holds for a = b = c.

P 2.30. Let a, b, c be nonnegative real numbers such that a ≥ b ≥ c. Prove that ‹  P 2 a−b 2 (a) a (a − b)(a − c) ≥ 4a2 b2 ; a+b (b)

P

a2 (a − b)(a − c) ≥

27b(a − b)4 . 4a (Vasile Cîrtoaje, 2011)

290

Vasile Cîrtoaje

Solution. (a) Since c 2 (c − a)(c − b) ≥ 0, it suffices to show that  ‹2 2 2 2 2 a−b a (a − b)(a − c) + b (b − c)(b − a) ≥ 4a b . a+b Since a2 (a − b)(a − c) = a2 (a − b)[(a − b) + (b − c)]  ‹2 2 2 2 2 2 a−b = a (a − b) + a (a − b)(b − c) ≥ 4a b + a2 (a − b)(b − c), a+b it suffices to show that a2 (a − b)(b − c) + b2 (b − c)(b − a) ≥ 0. This inequality is equivalent to (a − b)2 (a + b)(b − c) ≥ 0. The equality holds for a = b = c, and for a = b and c = 0. (b) Since c 2 (c − a)(c − b) ≥ 0, it suffices to show that a2 (a − b)(a − c) + b2 (b − c)(c − a) ≥

27b(a − b)4 , 4a

or a2 (a − b)2 + a(a − b)(b − c) + b2 (b − c)(c − a) ≥ Since a2 (a − b)2 −

27b(a − b)4 . 4a

27b(a − b)4 (a − b)2 (a − 3b)2 (4a − 3b) = ≥ 0, 4a 4a

it suffices to show that a2 (a − b)(b − c) + b2 (b − c)(b − a) ≥ 0. This inequality is equivalent to (a − b)2 (a + b)(b − c) ≥ 0. The equality holds for a = b = c, and for a/3 = b = 2.

P 2.31. If a, b, c are positive real numbers, then a b c 2(a − c)2 + + ≥3+ . b c a (a + c)2

Noncyclic Inequalities

291

Solution. Since a b + ≥2 b c it suffices to show that c +2 a Using the substitution x =

s

s

s

a , c

2(a − c)2 a ≥3+ . c (a + c)2

a , this inequality becomes as follows: c 1 2(1 − x 2 )2 + 2x ≥ 3 + , x2 (1 + x 2 )2

(1 − x)2 (1 + 2x) 2(1 − x 2 )2 ≥ . x2 (1 + x 2 )2 We need to show that

1 + 2x 2(1 + x)2 ≥ , x2 (1 + x 2 )2

which is equivalent to 2x 5 − 3x 4 + 2x + 1 ≥ 0. For 0 < x ≤ 1, we have 2x 5 − 3x 4 + 2x + 1 > −3x 4 + 2x + 1 ≥ −3x + 2x + 1 ≥ 0. Also, for x ≥ 1, we have 2x 5 − 3x 4 + 2x + 1 > 2x 5 − 3x 4 + 2x − 1 = (x − 1)2 (2x 3 + x 2 − 1) ≥ 0. The equality holds for a = b = c.

P 2.32. If a, b, c are positive real numbers, then a b c (a − c)2 + + ≥3+ . b c a a b + bc + ca (Vasile Cîrtoaje, Vo Quoc Ba Can, 2008) First Solution. By expanding, the inequality can be written as b2 +

bc 2 ca2 a b2 + + ≥ 2a b + 2bc. a b c

292

Vasile Cîrtoaje

We can get this inequality by summing up the AM-GM inequalities ab +

b2 +

bc 2 ≥ 2bc, a

ca2 a b2 + ≥ 3a b. b c

The equality holds for a = b = c. Second Solution. From (a + b + c)

=

X  a2 b



‹ X 2 X X a b c a bc + + −3 = + −2 a b c a b a



− 2a + b +

X  bc a

‹

−a =

X (a − b)2 b

+

1 X a(b − c)2 , 2 bc

we get ‹ a b c (a − b)2 (b − c)2 (c − a)2 (a + b + c) + + −3 ≥ + + . b c a b c a 

By the Cauchy-Schwarz inequality, we have (a − c)2 (a − b)2 (b − c)2 + ≥ . b c b+c Therefore, ‹ a b c (a − c)2 (c − a)2 (a + b + c) + + −3 ≥ + , b c a b+c a 

which is equivalent to a b c (a − c)2 + + −3≥ . b c a a(b + c) From this result, the desired inequality follows immediately.

P 2.33. If a, b, c are positive real numbers, then a b c 4(a − c)2 + + ≥3+ . b c a (a + b + c)2

Noncyclic Inequalities

293

Solution. As we have shown at the second solution of the preceding problem P 2.32, the following inequality holds: (a − c)2 a b c + + −3≥ . b c a a(b + c) Therefore, it suffices to show that 4 1 ≥ . a(b + c) (a + b + c)2 Indeed,

1 4 (a − b − c)2 − = ≥ 0. a(b + c) (a + b + c)2 a(b + c)(a + b + c)2

The equality holds for a = b = c.

P 2.34. If a ≥ b ≥ c > 0, then a b c 3(b − c)2 + + ≥3+ . b c a a b + bc + ca Solution. Since

c (a − b)(a − c) a c + −1− = ≥ 0, b a b ab

it suffices to show that

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

Indeed, we have b c 3(b − c)2 (b − c)2 (a b + ac − 2bc) + −2− = . c b a b + bc + ca bc(a b + bc + ca) The equality holds for a = b = c.

P 2.35. If a, b, c are positive real numbers, then a2 b2 c 2 4(a − c)2 + + ≥a+b+c+ . b c a a+b+c (Balkan MO, 2005, 2008)

294

Vasile Cîrtoaje

Solution. Write the inequality as follows:  2   2   2  4(a − c)2 a b c + b − 2a + + c − 2b + + a − 2c ≥ , b c a a+b+c (a − b)2 (b − c)2 (a − c)2 4(a − c)2 + + ≥ . b c a a+b+c By the Cauchy-Schwarz inequality, we have [(a − b) + (b − c) + (a − c)]2 4(a − c)2 (a − b)2 (b − c)2 (a − c)2 + + ≥ = . b c a b+c+a a+b+c p b 1+ 5 The equality holds for a = b = c, and also for a = b + c and = . c 2

P 2.36. If a ≥ b ≥ c > 0, then 6(b − c)2 a2 b2 c 2 + + ≥a+b+c+ . b c a a+b+c (Vasile Cîrtoaje, 2014) Solution. Write the inequality as follows:  2   2   2  a b c 6(b − c)2 + b − 2a + + c − 2b + + a − 2c ≥ , b c a a+b+c 6(b − c)2 (a − b)2 (b − c)2 (a − c)2 + + ≥ , b c a a+b+c (a − b)2 (a − c)2 (a + b − 5c)(b − c)2 + + ≥ 0. b a c(a + b + c) Since (a − c)2 = [(a − b) + (b − c)]2 = (a − b)2 + 2(a − b)(b − c) + (b − c)2 , we have

(a − b)2 (a − c)2 (a − c)2 2(a − b)(b − c) + (b − c)2 + ≥ ≥ . b a a a Therefore, it suffices to show that 2(a − b)(b − c) + (b − c)2 (a + b − 5c)(b − c)2 + ≥ 0, a c(a + b + c)

Noncyclic Inequalities

295

which can be written as 2(a − b)(b − c) (a − c)2 + a b + bc − 2ca + (b − c)2 ≥ 0. a ac(a + b + c) Since (a − c)2 + a b + bc − 2ca = (a − c)2 + a(b − c) − c(a − b) ≥ −c(a − b), it is enough to prove that a−b 2(a − b)(b − c) − (b − c)2 ≥ 0. a a(a + b + c) Indeed,  ‹ 2(a − b)(b − c) a−b (a − b)(b − c) b−c 2 − (b − c) = 2− ≥ 0. a a(a + b + c) a a+b+c The equality holds for a = b = c.

P 2.37. If a ≥ b ≥ c > 0, then a2 b2 c 2 + + > 5(a − b). b c a (Vasile Cîrtoaje, 2014) Solution. Consider two cases: a ≤ 2b and a ≥ 2b. Case 1: a ≤ 2b. It suffices to show that a2 b2 + ≥ 5(a − b), b b which is equivalent to the obvious inequality (2b − a)(3b − a) ≥ 0. Case 2: a ≥ 2b. Since  ‹ b2 c 2 b2 b b+c + −b− = (b − c) − c a a c a  ‹ (b − c)2 (2b + c) b b+c ≥ (b − c) − = ≥ 0, c 2b 2bc

296

Vasile Cîrtoaje

it suffices to show that

b2 a2 +b+ ≥ 5(a − b), b a

which is equivalent to x(x − 2)(3 − x) < 1, where x = a/b ≥ 2. For the non-trivial case 2 ≤ x ≤ 3, we have x(x − 2)(3 − x) ≥ x

•

(x − 2) + (3 − x) 2

˜2

=

x < 1. 4

P 2.38. If a, b, c are positive real numbers, then a b c 3 27(b − c)2 + + ≥ + . b+c c+a a+b 2 16(a + b + c)2 (Vasile Cîrtoaje, 2014) Solution. Write the inequality as follows:  9 X a 27(b − c)2 +1 ≥ + , b+c 2 16(a + b + c)2 ”X — X (b + c)

1 b+c

‹

27(b − c)2 ≥ 9 + P 2 . 2 (b + c)

Replacing b + c, c + a, a + b by a, b, c, respectively, we need to show that (a + b + c)



‹ 1 1 1 27(b − c)2 + + ≥9+ , a b c 2(a + b + c)2

where a, b, c are the side-lengths of a non-degenerate triangle. Write this inequality in the form  ‹ a+b+c 1 1 54bc 27(b + c)2 + (a + b + c) + + ≥ 9 + . a b c (a + b + c)2 2(a + b + c)2 Applying the AM-GM inequality gives v ‹ t 6(b + c) 1 1 54bc (a + b + c) + + ≥ 6 . b c (a + b + c)2 a+b+c 

Noncyclic Inequalities

297

Therefore, it suffices to show that v t 6(b + c) a+b+c 27(b + c)2 +6 ≥9+ , a a+b+c 2(a + b + c)2 which can be rewritten as v t 6(b + c) 27(b + c)2 1 +6 ≥9+ . b+c a+b+c 2(a + b + c)2 1− a+b+c Using the substitution p p 2 b+c 3 < 2t < 6, = t 2, a+b+c 3 this inequality becomes 1 + 4t ≥ 3 + 2t 4 , 3 − 2t 2 2t 6 − 3t 4 − 4t 3 + 3t 2 + 6t − 4 ≥ 0, (t − 1)2 (2t 4 + 4t 3 + 3t 2 − 2t − 4) ≥ 0,   (t − 1)2 (4t 2 − 3)(t 2 + 2t + 2) + t 2 + 2t − 2 ≥ 0. p 3 . The original inequality is an equality for Clearly, the last inequality is true for t > 2 a = b = c.

P 2.39. Let a, b, c be positive real numbers such that a = min{a, b, c}. Prove that a b c 3 9(b − c)2 + + ≥ + . b+c c+a a+b 2 4(a + b + c)2 (Vasile Cîrtoaje, 2014) Solution. Write the inequality as  9 X a 9(b − c)2 +1 ≥ + , b+c 2 4(a + b + c)2 ”X — X (b + c)

1 b+c

‹

≥9+

18(b − c)2 . [(b + c) + (c + a) + (a + b)]2

298

Vasile Cîrtoaje

Replacing b + c, c + a, a + b by a, b, c, respectively, we need to show that (a + b + c)



‹ 1 1 1 18(b − c)2 + + ≥9+ , a b c (a + b + c)2

where a, b, c are the side-lengths of a non-degenerate triangle, a = max{a, b, c}. Since (a + b + c)2 ≥

9 (b + c)2 ≥ 9bc, 4

it suffices to show that ‹ 1 1 1 2(b − c)2 + + ≥9+ . (a + b + c) a b c bc 

Write the inequality as follows: (a − b)2 (a − c)2 (b − c)2 2(b − c)2 + + ≥ , ab ac bc bc c(a − b)2 + b(a − c)2 ≥ a(b − c)2 , (b + c)a2 − (b + c)2 a + bc(b + c) ≥ 0, (b + c)(a − b)(a − c) ≥ 0. Clearly, the last inequality is true. The original inequality is an equality for a = b = c.

P 2.40. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a b c 3 (b − c)2 + + ≥ + . b+c c+a a+b 2 2(b + c)2 (Vasile Cîrtoaje, 2014) First Solution. Write the inequality as follows: 2bc a b c + + + ≥ 2, 2 (b + c) b+c c+a a+b a(b + c) + 2bc b c + + ≥ 2, (b + c)2 c+a a+b By the Cauchy-Schwarz inequality, we have b c (b + c)2 (b + c)2 + ≥ = . c+a a+b b(c + a) + c(a + b) a(b + c) + 2bc

Noncyclic Inequalities

299

Therefore, it suffices to prove that (b + c)2 a(b + c) + 2bc + ≥ 2, (b + c)2 a(b + c) + 2bc which is obvious. The original inequality is an equality for a = b = c, for a = b and c = 0, and for a = c and b = 0. Second Solution. Write the inequality as follows:  9 X a (b − c)2 +1 ≥ + , b+c 2 2(b + c)2 ”X — X (b + c)

1 b+c

‹

≥9+

(b − c)2 . (b + c)2

Replacing b + c, c + a, a + b by a, b, c, respectively, we need to show that (a + b + c)



‹ 1 1 1 (b − c)2 + + ≥9+ , a b c a2

where a, b, c are the lengths of the sides of a triangle. Write this inequality as (b − c)2 (a − b)2 (a − c)2 (b − c)2 + + ≥ , ab ac bc a2 a[c(a − b)2 + b(a − c)2 ] ≥ (bc − a2 )(b − c)2 . Without loss of generality, assume that b ≥ c. Since a ≥ b − c, it suffices to show that c(a − b)2 + b(a − c)2 ≥ (bc − a2 )(b − c). Indeed, we have c(a − b)2 + b(a − c)2 − (bc − a2 )(b − c) = 2b(a − c)2 ≥ 0.

P 2.41. Let a, b, c be positive real numbers such that a = min{a, b, c}. Prove that a b c 3 (b − c)2 + + ≥ + . b+c c+a a+b 2 4bc (Vasile Cîrtoaje, 2014)

300

Vasile Cîrtoaje

First Solution (by Nguyen Van Quy). Notice that for a = min{a, b, c}, we have 4bc ≥ (a + b)(a + c) ≥ 2a(b + c). Hence, a 2a2 ≥ , b+c (a + b)(a + c)

(b − c)2 (b − c)2 ≤ . 4bc (a + b)(a + c)

So, it suffices to show that 2a2 b c 3 (b − c)2 + + ≥ + , (a + b)(a + c) c + a a + b 2 (a + b)(a + c) which is equivalent to the obvious inequality (a − b)(a − c) ≥ 0. The proof is completed. The original inequality is an equality for a = b = c. Second Solution. Let E(a, b, c) =

b c a + + . b+c c+a a+b

Without loss of generality, assume that b ≤ c, hence a ≤ b ≤ c. We will show that E(a, b, c) ≥ E(b, b, c) ≥

3 (b − c)2 + . 2 4bc

We have a−b b(b − a) c(b − a) + + b + c (a + c)(b + c) 2b(a + b) ˜ • c (b − a) − c + = (b − a) (a + c)(b + c) 2b(a + b) (b − a)[2b(b2 − a2 ) + c(c − b)(a + 2b + c)] = ≥0 2b(a + b)(a + c)(b + c)

E(a, b, c) − E(b, b, c) =

and 3 (b − c)2 E(b, b, c) − − = 2 4bc

‹ 2b c 3 (b − c)2 + − − b + c 2b 2 4bc 2 2 (b − c) (b − c) = − 2b(b + c) 4bc 3 (c − b) = ≥ 0. 4bc(b + c) 

Noncyclic Inequalities

301

Remark. From the second solution, we get E(a, b, c) −

3 (b − c)2 (b − c)2 3 ≥ E(b, b, c) − = = . 2 2 2b(b + c) 2(b + c) min{b, c}

Therefore, the following sharper inequality holds for a = min{a, b, c}: b c 3 (b − c)2 a + + ≥ + . b+c c+a a+b 2 2(b + c) min{b, c} Because this inequality is an equality for a = min{b, c}, it follows that is the best function f (b, c) such that

(b − c)2 2(b + c) min{b, c}

a b c 3 + + ≥ + f (b, c) b+c c+a a+b 2 for all positive a, b, c with a = min{a, b, c}. Also, from the first solution, we get b c 3 (b − c)2 3 (b − c)2 a + + ≥ + ≥ + . b+c c+a a+b 2 (a + b)(a + c) 2 2(b + c) min{b, c}

P 2.42. Let a, b, c be positive real numbers such that a = min{a, b, c}. Prove that (a)

a b + bc + ca 2(b − c)2 + ≤ 1; a2 + b2 + c 2 3(b2 + c 2 )

(b)

a b + bc + ca (a − b)2 + ≤ 1. a2 + b2 + c 2 2(a2 + b2 ) (Vasile Cîrtoaje, 2014)

Solution. (a) First Solution. Since 3(b2 + c 2 ) ≥ 2(a2 + b2 + c 2 ), it suffices to show that a b + bc + ca (b − c)2 + ≤ 1. a2 + b2 + c 2 a2 + b2 + c 2 This inequality is equivalent to (a − b)(a − c) ≥ 0, which is clearly true. The equality holds for a = b = c.

302

Vasile Cîrtoaje

Second Solution. Write the inequality as follows: (b − c)2 + (a − b)2 + (a − c)2 4(b − c)2 ≤ , 3(b2 + c 2 ) a2 + b2 + c 2 3(b2 + c 2 )[(a − b)2 + (a − c)2 ] ≥ (b − c)2 (4a2 + b2 + c 2 ), 3(b2 + c 2 )[(b − c)2 + 2(a − b)(a − c)] ≥ (b − c)2 (4a2 + b2 + c 2 ), 6(b2 + c 2 )(a − b)(a − c) + 2(b − c)2 (b2 + c 2 − 2a2 ) ≥ 0. The last inequality is true because (a − b)(a − c) ≥ 0 and b2 + c 2 − 2a2 ≥ 0. (b) Write the inequality as follows: (a + b)2 a b + (a + b)c ≤ , a2 + b2 + c 2 2(a2 + b2 ) (a + b)2 c 2 − 2(a + b)(a2 + b2 )c + (a2 + b2 )2 ≥ 0, [(a + b)c − (a2 + b2 )]2 ≥ 0. The equality holds for c =

a2 + b2 . a+b

P 2.43. Let a, b, c be nonnegative real numbers such that a = max{a, b, c}. Prove that (a)

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

(b)

a b + bc + ca 2(b − c)2 + ≤ 1. a2 + b2 + c 2 (a + b + c)2 (Vasile Cîrtoaje, 2014)

Solution. Without loss of generality, assume that a ≥ b ≥ c. (a) Write the inequality as follows: (b − c)2 + (a − b)2 + (a − c)2 (b − c)2 ≤ , a b + bc + ca a2 + b2 + c 2 (a b + bc + ca)[(a − b)2 + (a − c)2 ] ≥ (b − c)2 (a2 + b2 + c 2 − a b − bc − ca). Since a b + bc + ca ≥ a b ≥ b2 ≥ (b − c)2 , it suffices to show that (a − b)2 + (a − c)2 ≥ a2 + b2 + c 2 − a b − bc − ca.

Noncyclic Inequalities

303

Indeed, (a − b)2 + (a − c)2 − (a2 + b2 + c 2 − a b − bc − ca) = (a − b)(a − c) ≥ 0. The equality holds for a = b = c, for a = b and c = 0, and for a = c and b = 0. (b) Write the inequality as follows: 4(b − c)2 (b − c)2 + (a − b)2 + (a − c)2 ≤ , (a + b + c)2 a2 + b2 + c 2 (a + b + c)2 [(a − b)2 + (a − c)2 ] ≥ (b − c)2 [3(a2 + b2 + c 2 ) − 2(a b + bc + ca)], (a + b + c)2 [(b − c)2 + 2(a − b)(a − c)] ≥ (b − c)2 [3(a2 + b2 + c 2 ) − 2(a b + bc + ca)], (a + b + c)2 (a − b)(a − c) ≥ (b − c)2 [a2 + b2 + c 2 − 2(a b + bc + ca)]. Since a2 + b2 + c 2 − 2(a b + bc + ca) = (a − b)2 − c(2a + 2b − c) ≤ (a − b)2 , it suffices to show that (a + b + c)2 (a − c) ≥ (b − c)2 (a − b). This inequality is true because (a + b + c)2 ≥ (b − c)2 and a − c ≥ a − b. The equality holds for a = b = c, for a = b and c = 0, and for a = c and b = 0.

P 2.44. Let a, b, c be positive real numbers such that a = min{a, b, c}. Prove that (a)

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

(b)

(a − b)2 a2 + b2 + c 2 ≥1+ . a b + bc + ca (a + b)2 (Vasile Cîrtoaje, 2014)

304

Vasile Cîrtoaje

Solution. (a) First Solution. Since 3(b + c)2 ≥ 12bc ≥ 4(a b + bc + ca), it suffices to prove that (b − c)2 a2 + b2 + c 2 ≥1+ , a b + bc + ca a b + bc + ca which is equivalent to the obvious inequality (a − b)(a − c) ≥ 0. The equality holds for a = b = c. Second Solution. Since (b + c)2 ≥ 4bc, it suffices to prove that a2 + b2 + c 2 (b − c)2 ≥1+ . a b + bc + ca 3bc Write this inequality as follows: (a − b)2 + (a − c)2 + (b − c)2 2(b − c)2 ≥ , a b + bc + ca 3bc 3bc[(a − b)2 + (a − c)2 ] ≥ (b − c)2 (2a b + 2ac − bc), 3bc[(b − c)2 + 2(a − b)(a − c)] ≥ (b − c)2 (2a b + 2ac − bc), 6bc(a − b)(a − c) + 2(b − c)2 (2bc − a b − ac) ≥ 0. The last inequality is true because (a − b)(a − c) ≥ 0 and 2bc − a b − ac = (a − b)(a − c) + (bc − a2 ) ≥ 0.

(b) Write the inequality as follows: a2 + b2 + c 2 2(a2 + b2 ) ≥ , a b + (a + b)c (a + b)2 (a + b)2 c 2 − 2(a + b)(a2 + b2 )c + (a2 + b2 )2 ≥ 0, [(a + b)c − (a2 + b2 )]2 ≥ 0. The equality holds for c =

a2 + b2 . a+b

Noncyclic Inequalities

305

P 2.45. If a, b, c are positive real numbers, then a2 + b2 + c 2 9(a − c)2 ≥1+ . a b + bc + ca 4(a + b + c)2 (Vasile Cîrtoaje, 2014) Solution. Clearly, it is enough to consider that a ≥ b ≥ c. Write the inequality as follows: (b − c)2 + (a − b)2 + (a − c)2 9(a − c)2 ≥ , a b + bc + ca 2(a + b + c)2 2(a + b + c)2 [(b − c)2 + (a − b)2 ] ≥ (a − c)2 [5(a b + bc + ca) − 2(a2 + b2 + c 2 )], 2(a + b + c)2 [(a − c)2 − 2(a − b)(b − c)] ≥ (a − c)2 [5(a b + bc + ca) − 2(a2 + b2 + c 2 )], (a − c)2 [4(a2 + b2 + c 2 ) − (a b + bc + ca)] ≥ 4(a + b + c)2 (a − b)(a − c), Since (a − c)2 = [(a − b) + (b − c)]2 ≥ 4(a − b)(b − c), it suffices to show that 4(a2 + b2 + c 2 ) − (a b + bc + ca) ≥ (a + b + c)2 . Indeed, 4(a2 + b2 + c 2 ) − (a b + bc + ca) − (a + b + c)2 = 3(a2 + b2 + c 2 − a b − bc − ca) ≥ 0. The equality holds for a = b = c.

P 2.46. Let a, b, c be nonnegative real numbers, no two of which are zero. If a = min{a, b, c}, then 1 1 1 6 +p +p ≥ . p 2 2 2 2 2 2 b+c a − ab + b b − bc + c c − ca + a Solution. Since X it suffices to show that

p

1 a2 − a b + b2

≥

1 1 1 + , +p b b2 − bc + c 2 c

1 1 1 6 +p + ≥ . 2 2 b c b + c b − bc + c

Write this inequality as v b c t b2 + c 2 + 2bc + + ≥ 4, c b b2 + c 2 − bc

306

Vasile Cîrtoaje

which is equivalent to v tx +2 x −1 where x = is true if

≥ 4 − x,

b c + , x ≥ 2. Consider the non-trivial case 2 ≤ x ≤ 4. Clearly, the inequality c b x +2 ≥ (4 − x)2 , x −1

which is equivalent to (x − 2)(x 2 − 7x + 9) ≤ 0. This inequality holds because x 2 − 7x + 9 < x 2 − 7x + 10 = (x − 2)(x − 5) ≤ 0. The equality holds for a = b = c, and also a = 0 and b = c (or any cyclic permutation).

P 2.47. If a ≥ 1 ≥ b ≥ c ≥ 0 such that a b + bc + ca = a bc + 2, then

p ac ≤ 4 − 2 2. (Vasile Cîrtoaje, 2012)

Solution. By hypothesis, we have a= Then, ac ≤

2 − bc . b + c − bc

1 (2 − bc)(b + c) 2 − bc 2 − bc a(b + c) = = ≤ p , 2bc 2 2(b + c − bc) 2 − bc 2 − b+c

and it suffices to show that

p 2 − bc p ≤ 4 − 2 2. 2 − bc This is true since is equivalent to the obvious inequality p p ( bc − 2 + 2)2 ≥ 0. p The equality holds for a = 2 and b = c = 2 − 2.

Noncyclic Inequalities

307

P 2.48. If 0 < a ≤ b ≤ c such that a + b + c = 3, then a4 (b4 + c 4 ) ≤ 2. (Vasile Cîrtoaje, 2012) Solution. Since x 4 − y 4 − 4 y 3 (x − y) = (x − y)(x 3 + x 2 y + x y 2 − 3 y 3 ) = (x − y)[(x 3 − y 3 ) + y(x 2 − y 2 ) + y 2 (x − y)], we have x 4 − y 4 ≥ 4 y 3 (x − y) for any positive numbers x, y. Using this inequality, we will show that b4 + c 4 ≤ a4 + (b + c − a)4 . Indeed, (a4 − b4 ) + (b + c − a)4 − c 4 ≥ 4b3 (a − b) + 4c 3 (b + c − a − c) = 4(a − b)(a3 − b3 ) ≥ 0. Therefore, it suffices to show that a4 [a4 + (b + c − a)4 ] ≤ 2, which is equivalent to f (a) ≤ 2, where f (a) = a8 + a4 (3 − 2a)4 , 0 < a ≤ 1. If f 0 (a) ≥ 0 for 0 < a ≤ 1, then f (a) is increasing, and hence f (a) ≤ f (1) = 2. Since f 0 (a) = 4a3 [2a4 − (4a − 3)(3 − 2a)3 ], we have f 0 (a) ≥ 0 if

2a4 ≥ (4a − 3)(3 − 2a)3 .

Clearly, this is true for 0 < a ≤ 3/4. It is also true for 3/4 < a ≤ 1 if h(a) ≥ 0, where h(a) = ln 2 + 4 ln a − ln(4a − 3) + 3 ln(3 − 2a), 3/4 < a ≤ 1. From h0 (a) =

3(7a − 6) , a(4a − 3)(3 − 2a)

it follows that h(a) is decreasing on (3/4, 6/7] and increasing on [6/7, 1]. Thus, 6 7 3 9 32 h(a) ≥ h( ) = ln 2 + 4 ln − ln − 3 ln = ln > 0. 7 7 7 7 27 This completes the proof. The equality holds for a = b = c = 1.

308

Vasile Cîrtoaje

P 2.49. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then a2 + b2 + c 2 − a − b − c ≥

5 (a − c)2 . 8 (Vasile Cîrtoaje, 2014)

Solution. Denote q = a b + bc + ca,

S = (a − b)2 + (b − c)2 + (c − a)2 .

Summing the identities 1 S a2 + b2 + c 2 − (a + b + c)2 = 3 3 and s p q 1 1 2 (a + b + c) − (a + b + c) = (a + b + c)(a + b + c − 3q ) 3 3 3 (a + b + c)S = , p 6(a + b + c + 3q ) we get   1 a+b+c 5S a +b +c −a−b−c = 2+ S≥ . p 6 12 a + b + c + 3q 2

2

2

Therefore, it suffices to show that 5 5S ≥ (a − c)2 , 12 8 which is equivalent to 2(a − b)2 + 2(b − c)2 ≥ (a − c)2 . Indeed, 2(a − b)2 + 2(b − c)2 − (a − c)2 = 2(a − b)2 + 2(b − c)2 )2 − [(a − b) + (b − c)]2 = [(a − b) − (b − c)]2 = (a − 2b + c)2 ≥ 0. This completes the proof. The equality holds for a = b = c = 1.

P 2.50. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then 5 a3 + b3 + c 3 ≥ 1 + (a − c)2 . a+b+c 9 (Vasile Cîrtoaje, 2014)

Noncyclic Inequalities

309

Solution. It suffices to consider the case a ≥ b ≥ c. Denote p = a + b + c. Summing the identities a3 + b3 + c 3 −

(a + b)(a − b)2 + (b + c)(b − c)2 + (c + a)(c − a)2 1 p(a2 + b2 + c 2 ) = 3 3

and 1 (a − b)2 + (b − c)2 + (c − a)2 1 p(a2 + b2 + c 2 ) − p(a b + bc + ca) = p · , 3 3 6 we get a3 + b3 + c 3 − a − b − c =

(a + b)(a − b)2 + (b + c)(b − c)2 + (c + a)(c − a)2 3 (a − b)2 + (b − c)2 + (c − a)2 . +p· 6

Therefore, we can write the desired inequality in the homogeneous form (a + b)(a − b)2 + (b + c)(b − c)2 + (c + a)(c − a)2 (a − b)2 + (b − c)2 + (c − a)2 +p· ≥ 3 6 ≥

5 p(a − c)2 , 9

which is equivalent to (9a + 9b + 3c)(a − b)2 + (3a + 9b + 9c)(b − c)2 ≥ (a + 7b + c)(a − c)2 . Using the substitution b = c + x,

a = c + x + y,

x, y ≥ 0,

the inequality becomes 3(7c + 6x + 3 y) y 2 + 3(7c + 4x + y)x 2 ≥ (9c + 8x + y)(x + y)2 , which is equivalent to the obvious inequality 6c(2x 2 + 2 y 2 − 3x y) + 2(x − 2 y)2 (2x + y) ≥ 0. p Thisp completes the proof. The equality holds for a = b = c = 1, and also for a = 3/ 2, b = 2, c = 0.

310

Vasile Cîrtoaje

P 2.51. If a, b, c are nonnegative real numbers such that a ≥ b ≥ c,

a b + bc + ca = 3,

then (a)

7 a3 + b3 + c 3 ≥ 1 + (a − b)2 ; a+b+c 9

(b)

a3 + b3 + c 3 2 ≥ 1 + (b − c)2 . a+b+c 3 (Vasile Cîrtoaje, 2014)

Solution. As we have shown in the proof of the preceding problem, a3 + b3 + c 3 − a − b − c =

(a + b)(a − b)2 + (b + c)(b − c)2 + (c + a)(c − a)2 3 (a − b)2 + (b − c)2 + (c − a)2 +p· , 6

where p = a + b + c. (a) Write the desired inequality in the homogeneous form (a + b)(a − b)2 + (b + c)(b − c)2 + (c + a)(c − a)2 (a − b)2 + (b − c)2 + (c − a)2 7 +p· ≥ p(a−b)2 . 3 6 9 Since (a + b)(a − b)2 + (b + c)(b − c)2 + (c + a)(c − a)2 ≥ (a + b)(a − b)2 + (c + a)(c − a)2 ≥ (2a + b + c)(a − b)2 and (a − b)2 + (b − c)2 + (c − a)2 ≥ (a − b)2 + (c − a)2 ≥ 2(a − b)2 , it suffices to show that

7p 2a + b + c p + ≥ , 3 3 9

which is equivalent to the obvious inequality 2a − b − c ≥ 0. The equality holds for a = b = c = 1. (b) Write the desired inequality in the homogeneous form (a + b)(a − b)2 + (b + c)(b − c)2 + (c + a)(c − a)2 (a − b)2 + (b − c)2 + (c − a)2 2 +p· ≥ p(a−b)2 . 3 6 3

Noncyclic Inequalities

311

Since (a + b)(a − b)2 + (b + c)(b − c)2 + (c + a)(c − a)2 ≥ (b + c)(b − c)2 + (c + a)(c − a)2 ≥ (a + b + 2c)(b − c)2 and (a − b)2 + (b − c)2 + (c − a)2 ≥ (b − c)2 + (c − a)2 ≥ 2(b − c)2 , it suffices to show that

2p a + b + 2c p + ≥ , 3 3 3 which is equivalent topthe obvious inequality c ≥ 0. The equality holds for a = b = c = 1, and also for a = b = 3, c = 0.

P 2.52. If a, b, c are nonnegative real numbers such that a b + bc + ca = 3, then a4 + b4 + c 4 − a2 − b2 − c 2 ≥

11 (a − c)2 . 4 (Vasile Cîrtoaje, 2014)

Solution. It suffices to consider the case a ≥ b ≥ c. Denote S = a2 + b2 + c 2 ,

q = a b + bc + ca.

Summing the identities (a2 − b2 )2 + (b2 − c 2 )2 + (c 2 − a2 )2 1 a4 + b4 + c 4 − S 2 = 3 3 and

1 2 1 (a − b)2 + (b − c)2 + (c − a)2 S − Sq = S · , 3 3 6

we get a4 + b4 + c 4 − a2 − b2 − c 2 =

(a2 − b2 )2 + (b2 − c 2 )2 + (c 2 − a2 )2 3 (a − b)2 + (b − c)2 + (c − a)2 +S· . 6

Therefore, we can write the desired inequality in the homogeneous form (a2 − b2 )2 + (b2 − c 2 )2 + (c 2 − a2 )2 (a − b)2 + (b − c)2 + (c − a)2 11 +S· ≥ q(a − c)2 . 3 6 12

312

Vasile Cîrtoaje

Since (a − b)2 + (b − c)2 ≥

1 1 [(a − b) + (b − c)]2 = (a − c)2 , 2 2

it suffices to prove that 11 (a2 − b2 )2 + (b2 − c 2 )2 + (c 2 − a2 )2 S(a − c)2 + ≥ q(a − c)2 , 3 4 12 which is equivalent to 4(a + b)2 (a − b)2 + 4(b + c)2 (b − c)2 + E(a − c)2 ≥ 0, where E = 4(a + c)2 + 3S − 11q. Using the substitution b = c + x,

a = c + x + y,

x, y ≥ 0,

the inequality becomes 4(2c + 2x + y)2 y 2 + 4(2c + x)2 x 2 + E(x + y)2 ≥ 0, where E = −8c 2 − 16x c − x 2 + 7 y 2 + 3x y. Write this inequality as Ac 2 + D ≥ 2Bc, where A = 8(x − y)2 , B = 8 y(x − y)(2x + y), D = 3x 4 + 11 y 4 + 28x 2 y 2 + 33x y 3 + x 3 y. p Since Ac 2 + D ≥ 2c AD, it suffices to show that AD ≥ B 2 . Indeed, AD − B 2 = 8(x − y)2 [3x 4 + 11 y 4 + 28x 2 y 2 + 33x y 3 + x 3 y − 8 y 2 (2x + y)2 ] = 8(x − y)2 [x 4 + y 4 + 2(x 2 − y 2 )2 + x y(x 2 + y 2 )] ≥ 0. This completes the proof. The equality holds for a = b = c = 1.

P 2.53. If a, b, c are nonnegative real numbers such that a ≥ b ≥ c,

a b + bc + ca = 3,

then (a) (b)

11 (a − b)2 ; 3 10 a4 + b4 + c 4 − a2 − b2 − c 2 ≥ (b − c)2 . 3 a4 + b4 + c 4 − a2 − b2 − c 2 ≥

(Vasile Cîrtoaje, 2014)

Noncyclic Inequalities

313

Solution. Denote S = a2 + b2 + c 2 ,

q = a b + bc + ca.

As we have shown in the proof of the preceding problem, a4 + b4 + c 4 − a2 − b2 − c 2 =

(a2 − b2 )2 + (b2 − c 2 )2 + (c 2 − a2 )2 3 2 (a − b) + (b − c)2 + (c − a)2 . +S· 6

(a) Write the desired inequality in the homogeneous form (a2 − b2 )2 + (b2 − c 2 )2 + (c 2 − a2 )2 + S ·

11 (a − b)2 + (b − c)2 + (c − a)2 ≥ q(a − b)2 . 2 3

Since (a2 − b2 )2 + (b2 − c 2 )2 + (c 2 − a2 )2 ≥ (a2 − b2 )2 + (a2 − c 2 )2 ≥ 2(a2 − b2 )2 and (a − b)2 + (b − c)2 + (c − a)2 ≥ (a − b)2 + (a − c)2 ≥ 2(a − b)2 , it suffices to prove that 2(a + b)2 + a2 + b2 + c 2 ≥

11 (a b + bc + ca); 3

that is, 9(a2 + b2 ) + a b + 3c 2 ≥ 11c(a + b). Since 9(a2 + b2 ) + a b −

19 17 (a + b)2 = (a − b)2 ≥ 0, 4 4

we have 19 (a + b)2 + 3c 2 − 11c(a + b) 4 (a + b − 2c)(19a + 19b − 6c) = ≥ 0. 4

9(a2 + b2 ) + a b + 3c 2 − 11c(a + b) ≥

The equality holds for a = b = c = 1. (b) Write the desired inequality in the homogeneous form (a2 − b2 )2 + (b2 − c 2 )2 + (c 2 − a2 )2 + S ·

(a − b)2 + (b − c)2 + (c − a)2 10 ≥ q(b − c)2 . 2 3

Since (a2 − b2 )2 + (b2 − c 2 )2 + (c 2 − a2 )2 ≥ (b2 − c 2 )2 + (a2 − c 2 )2

314

Vasile Cîrtoaje ≥ (b + c)2 (b − c)2 + (a + c)2 (b − c)2

and (a − b)2 + (b − c)2 + (c − a)2 ≥ (b − c)2 + (a − c)2 ≥ 2(b − c)2 , it suffices to prove that (b + c)2 + (a + c)2 + a2 + b2 + c 2 ≥

10 (a b + bc + ca); 3

that is, 6(a2 + b2 ) − 10a b + 9c 2 ≥ 4c(a + b). Since

1 11 6(a2 + b2 ) − 10a b − (a + b)2 = (a − b)2 ≥ 0, 2 2

we have 1 (a + b)2 + 9c 2 − 4c(a + b) 2 v t9 ≥2 c(a + b) − 4c(a + b) 2 p = (3 2 − 4)c(a + b) ≥ 0.

6(a2 + b2 ) − 10a b + 9c 2 − 4c(a + b) ≥

The equality holds for a = b = c = 1. Remark. Similarly, we can prove the following refinement of the inequality in (b): p 33 (b − c)2 , a +b +c −a −b −c ≥ 2 p 3 + 33 with equality for a = b = c = 1, and also for a = b = c. 4 4

4

4

2

2

2

1+

P 2.54. Let a, b, c be nonnegative real numbers such that a ≤ b ≤ c,

a + b + c = 3.

Find the greatest real number k such that Æ

(56b2 + 25)(56c 2 + 25) + k(b − c)2 ≤ 14(b + c)2 + 25. (Vasile Cîrtoaje, 2014)

Noncyclic Inequalities

315

Solution. For a = b = 0 and c = 3, the inequality becomes 115 + 9k ≤ 126 + 25,

k ≤ 4.

To show that 4 is the greatest possible value of k, we need to prove the inequality Æ

(56b2 + 25)(56c 2 + 25) + 4(b − c)2 ≤ 14(b + c)2 + 25,

which is equivalent to Æ

(56b2 + 25)(56c 2 + 25) ≤ 10(b2 + c 2 ) + 36bc + 25.

By squaring, the inequality becomes as follows (10b2 + 10c 2 + 36bc)2 − 562 b2 c 2 ≥ 50[28(b2 + c 2 ) − (10b2 + 10c 2 + 36bc)], 20(b − c)2 (5b2 + 5c 2 + 46bc) ≥ 900(b − c)2 , 20(b − c)2 (5b2 + 5c 2 + 46bc − 45) ≥ 0. Therefore, we need to show that 5(b + c)2 + 36bc − 45 ≥ 0. From (a − b)(a − c) ≥ 0, we get bc ≥ a(b + c) − a2 = a(3 − a) − a2 = 3a − 2a2 . Thus, 5(b + c)2 + 36bc − 45 ≥ 5(3 − a)2 + 36(3a − 2a2 ) − 45 = a(78 − 67a) ≥ 0. The proof is completed. If k = 4, then the equality holds for a = b = c = 1 and also for a = b = 0 and c = 3.

P 2.55. If a ≥ b ≥ c > 0 such that a bc = 1, then a 3(a + b + c) ≤ 8 + . c (Vasile Cîrtoaje, 2009)

316

Vasile Cîrtoaje

Solution. Write the inequality in the homogeneous form a 3(a + b + c) ≤8+ , p 3 c a bc which is equivalent to 3(x 3 + y 3 + z 3 ) x3 ≤8+ 3, x yz z We show that

x ≥ y ≥ z > 0.

  x 3 + y 3 + z3 x 3 + 2z 3 1 x3 ≤ ≤ 8+ 3 . x yz xz 2 3 z

Write the left inequality as ( y − z)[x 3 + z 3 − yz( y + z)] ≥ 0. This is true since x 3 + z 3 − yz( y + z) ≥ y 3 + z 3 − yz( y + z) = ( y + z)( y − z)2 ≥ 0. Write the left inequality as (x − z)(x 3 − 2x 2 z − 2xz 2 + 6z 3 ) ≥ 0. This is also true since x 3 − 2x 2 z − 2xz 2 + 6z 3 = (x − z)3 + z(x 2 − 5xz + 7z 2 ) ≥ 0. The equality holds for a = b = c = 1.

P 2.56. If a ≥ b ≥ c > 0, then (a + b − c)(a2 b − b2 c + c 2 a) ≥ (a b − bc + ca)2 . Solution. Making the substitution a = (p + 1)c,

b = (q + 1)c,

p ≥ q ≥ 0,

we get a + b − c = (p + q + 1)c, a2 b − b2 c + c 2 a = (p2 q + p2 + 2pq − q2 + 3p − q + 1)c 3 ,

Noncyclic Inequalities

317 a b − bc + ca = (pq + 2p + 1)c.

Thus, the inequality becomes (p + q + 1)(p2 q + p2 + 2pq − q2 + 3p − q + 1) ≥ (pq + 2p + 1)2 , which is equivalent to the obvious inequality p3 (q + 1) + q2 (p − q) + 2q(p − q) ≥ 0. The equality holds for a = b = c.

P 2.57. If a ≥ b ≥ c ≥ 0, then p (a − c)2 2(a − c)2 3 ≤ a + b + c − 3 a bc ≤ . 2(a + c) a + 5c (Vasile Cîrtoaje, 2007) Solution. To prove the inequality a+ b+c−3

p 3

a bc ≥

(a − c)2 , 2(a + c)

we will show that a+ b+c−3

p 3

p (a − c)2 a bc ≥ a + c − 2 ac ≥ . 2(a + c)

(*)

The left inequality is equivalent to p p 3 b + 2 ac ≥ 3 a bc, which is a consequence of the AM-GM inequality. The right inequality in (*) can be written as follows: p a2 + c 2 + 6ac ≥ 4(a + c) ac, p p ( a − b)4 ≥ 0. The equality holds for a = b = c. To prove the inequality a+ b+c−3

p 3

a bc ≤

2(a − c)2 , a + 5c

318

Vasile Cîrtoaje

we will show that a+ b+c−3

p 3

a bc ≤ 2a + c − 3

p 3

a2 c ≤

2(a − c)2 . a + 5c

(**)

Write the left inequality as p p p 3 3 3 a − b − 3 ac ( a − b) ≥ 0, p p p p p p 3 3 3 3 3 3 ( a − b)( a2 + a b + b2 − 3 ac) ≥ 0. The last inequality is true since p p p p p 3 3 3 3 3 a2 + a b + b2 ≥ 3 a b ≥ 3 ac. The right inequality in (**) is an equality for c = 0. For c > 0, due to homogeneity, we may assume that c = 1. In addition, making the substitution a = x 3 , x ≥ 1, the right inequality in (**) becomes in succession (x 3 + 5)(2x 3 − 3x 2 + 1) ≤ 2(x 3 − 1)2 , (x − 1)2 (x 3 + 2x 2 − 2x − 1) ≥ 0, (x − 1)3 (x 2 + 3x + 1) ≥ 0. The equality holds for a = b = c, and also for a = b and c = 0.

P 2.58. If a ≥ b ≥ c ≥ d ≥ 0, then p (a − d)2 3(a − d)2 4 ≤ a + b + c + d − 4 a bcd ≤ . a + 3d a + 5d (Vasile Cîrtoaje, 2009) Solution. To prove the inequality a+ b+c+d −4

p 4

a bcd ≥

(a − d)2 , a + 3d

we will show that a+ b+c+d −4

p 4

a bcd ≥ a + d − 2

p

ad ≤

The left inequality is equivalent to b+c+2

p

ad ≥ 4

p 4

a bcd,

(a − d)2 . a + 3d

(*)

Noncyclic Inequalities

319

which is a consequence of the AM-GM inequality. The right inequality in (*) can be written as follows: p p (a − d)2 ≥ (a + 3d)( a − d)2 , p p p 2 d ( a − d)3 ≥ 0. The equality holds for a = b = c = d, and also for b = c = d = 0. To prove the inequality a+ b+c+d −4

p 4

a bcd ≤

3(a − d)2 , a + 5d

we will show that a+ b+c+d −4

p 4

a bcd ≤ 2a + c + d − 4

p 4

a2 cd ≤

3(a − d)2 . a + 5d

(**)

Write the left inequality as p p p 4 4 4 a − b − 4 acd ( a − b) ≥ 0, p p p p p p p 4 4 4 4 4 4 4 ( a − b)( a3 + a2 b + a b2 + b3 − 4 acd) ≥ 0. The last inequality is true since p 4

a3 +

p 4

a2 b +

p 4

p p p p p p 4 4 4 4 4 4 a b2 + b3 − 4 acd ≥ a3 + a2 b + b3 − 3 a b2 p p p 4 4 4 ≥ a3 + 2 b3 − 3 a b2 ≥ 0.

Write the right inequality in (**) as F (c) ≥ 0, where F (c) = 3(a − d)2 − (a + 5d)(2a + c + d − 4

p 4

a2 cd).

Since F is a concave function and d ≤ c ≤ a, it suffices to show that F (d) ≥ 0 and F (a) ≥ 0. We have p p p p p p F (d) = 3(a − d)2 − 2(a + 5d)( a − d)2 = ( a − d)3 ( a + 7 d) ≥ 0 and F (a) = 3(a − d)2 − (a + 5d)(3a + d − 4

p 4

a3 d).

Setting a = 1 (due to homogeneity) and substituting d = x 4 , 0 ≤ x ≤ 1, the inequality F (a) ≥ 0 becomes 3(1 − x 4 )2 − (1 + 5x 4 )(3 + x 4 − 4x) ≥ 0.

320

Vasile Cîrtoaje

Since 3 + x 4 − 4x = (1 − x)2 (3 + 2x + x 2 ), we need to show that 3(1 + x + x 2 + x 3 )2 − (1 + 5x 4 )(3 + 2x + x 2 ) ≥ 0, which is equivalent to x(2 + 4x + 6x 2 − 3x 3 − 2x 4 − x 5 ) ≥ 0. This inequality is true since 2 + 4x + 6x 2 − 3x 3 − 2x 4 − x 5 > 6x 2 − 3x 3 − 2x 4 − x 5 ≥ 0. The equality holds for a = b = c = d, and also for a = b = c and d = 0. Remark. The following generalization holds. • If a1 ≥ a2 ≥ · · · ≥ an ≥ 0, then p (n − 1)(a1 − an )2 , a1 + a2 + · · · + an − n n a1 a2 · · · an ≤ a1 + k n a n where

8 , n od d n+1 kn = .  7 − 8,  n even  n     7−

P 2.59. If a ≥ b ≥ c > 0, then p 3(a − b)2 3 a + b + c − 3 a bc ≥ ; 5a + 4b p 64(a − b)2 3 a + b + c − 3 a bc ≥ . 7(11a + 24b)

(a) (b)

(Vasile Cîrtoaje, 2009) Solution. We use the inequality a+ b+c−3 which is equivalent to

p 3

a bc ≥ a + 2b − 3

p 3

a b2 ,

p p 3 3 a b( b − c) ≥ b − c, p p p p p p 3 3 3 3 3 3 ( b − c)(3 a b − b2 − bc − c 2 ) ≥ 0. 3

p 3

Noncyclic Inequalities

321

Since a ≥ b ≥ c, the last inequality is obvious. (a) It suffices to show that a + 2b − 3

p 3

a b2 ≥

3(a − b)2 . 5a + 4b

Setting b = 1 and a = x 3 , x ≥ 1, this inequality becomes as follows: (5x 3 + 4)(x 3 − 3x + 2) ≥ 3(x 3 − 1)2 , (x − 1)2 (2x 4 + 4x 3 − 9x 2 − 2x + 5) ≥ 0, (x − 1)4 (2x 2 + 8x + 5) ≥ 0. The equality holds for a = b = c. (b) It suffices to show that a + 2b − 3

p 3

a b2 ≥

64(a − b)2 . 7(11a + 24b)

Setting b = 1 and a = x 3 , x ≥ 1, this inequality becomes in succession: 7(11x 3 + 24)(x 3 − 3x + 2) ≥ 64(x 3 − 1)2 , (x − 1)2 (13x 4 + 26x 3 − 192x 2 + 40x + 272) ≥ 0, (x − 1)2 (x − 2)2 (13x 3 + 78x + 68) ≥ 0. a The equality holds for a = b = c, and for = b = c. 8

P 2.60. If a ≥ b ≥ c > 0, then (a) (b)

p 3(b − c)2 3 a + b + c − 3 a bc ≥ ; 4b + 5c p 25(b − c)2 3 a + b + c − 3 a bc ≥ . 7(3b + 11c) (Vasile Cîrtoaje, 2009)

Solution. We use the inequality a+ b+c−3

p 3

a bc ≥ 2b + c − 3

which is equivalent to a−b≥3

p 3

p 3

p p 3 3 bc( a − b),

b2 c,

322

Vasile Cîrtoaje p p p p p p 3 3 3 3 3 3 ( a − b)( a2 + a b + b2 − 3 bc) ≥ 0.

Since a ≥ b ≥ c, the last inequality is obvious. (a) It suffices to show that 2b + c − 3

p 3

b2 c ≥

3(b − c)2 . 4b + 5c

Setting c = 1 and b = x 3 , x ≥ 1, this inequality becomes as follows: (4x 3 + 5)(2x 3 − 3x 2 + 1) ≥ 3(x 3 − 1)2 , (x − 1)2 (5x 4 − 2x 3 − 9x 2 + 4x + 2) ≥ 0, (x − 1)4 (5x 2 + 8x + 2) ≥ 0. The equality holds for a = b = c. (b) It suffices to show that 2b + c − 3

p 3

b2 c ≥

25(b − c)2 . 7(3b + 11c)

Setting c = 1 and b = x 3 , x ≥ 1, this inequality becomes in succession: 7(3x 3 + 11)(2x 3 − 3x 2 + 1) ≥ 25(x 3 − 1)2 , (x − 1)2 (17x 4 − 29x 3 − 75x 2 + 104x + 52) ≥ 0, (x − 1)2 (x − 2)2 (17x 3 + 39x + 13) ≥ 0. The equality holds for a = b = c, and for a = b = 8c. Remark. The following generalization holds. • If a1 ≥ a2 ≥ · · · ≥ an ≥ 0, then p a1 + a2 + · · · + an − n n a1 a2 · · · an ≥

3i(n − j + 1)(ai − a j )2 2(2n + i − 2 j + 2)ai + 2(n + 2i − j + 1)a j

for all i < j.

P 2.61. If a ≥ b ≥ c > 0, then a+ b+c−3

p 3

a bc ≥

3(a − c)2 . 4(a + b + c) (Vasile Cîrtoaje, 2009)

Noncyclic Inequalities

323

Solution. Due to homogeneity, assume that a + b + c = 3. Let x=

 a + c 2 2

y = ac,

,

x ≥ y.

We have

‹  a − c 2 3− b 2 . x= , x−y= 2 2 The desired inequality is equivalent to p 3 3 − 3 b y ≥ x − y. 

There are two cases to consider. Case 1: b ≤ 1. By the AM-GM inequality, we have p p 3 y + 2 b ≥ 3 b y. Thus, it suffices to show that 3−2

p

b ≥ x.

Indeed, 3−2

p

b− x =3−2

p

b−



3− b 2

‹2

=

p p 1 (1 − b )3 (3 + b ) ≥ 0. 4

Case 2: b ≥ 1. From v  a + c 2 t a+c a+c 3 , a+b+c = b+ + ≥3 b 2 2 2 we get 3≥3

p 3

b x.

Therefore, it suffices to prove that p p 3 3 3 b x − 3 b y ≥ x − y, which is equivalent to Š p Æ p 3 3 p € p p 3 3 ( x − 3 y) 3 b − x 2 − 3 x y − y 2 ≥ 0. Since 3− b y≤x= 2 

‹2

≤ 1 ≤ b,

the last inequality is clearly true. The equality holds for a = b = c

324

Vasile Cîrtoaje

P 2.62. If a ≥ b ≥ c > 0, then (a)

a6 + b6 + c 6 − 3a2 b2 c 2 ≥ 12a2 c 2 (b − c)2 ;

(b)

a6 + b6 + c 6 − 3a2 b2 c 2 ≥ 10a3 c(b − c)2 . (Vasile Cîrtoaje, 2014)

Solution. (a) Let us denote E(a, b, c) = a6 + b6 + c 6 − 3a2 b2 c 2 − 12a2 c 2 (b − c)2 . We will show that E(a, b, c) ≥ E(b, b, c) ≥ 0. We have E(a, b, c) − E(b, b, c) = (a2 − b2 )[a4 + a2 b2 + b4 − 3b2 c 2 − 12c 2 (b − c)2 ] ≥ (a2 − b2 )[3b2 (b2 − c 2 ) − 12c 2 (b − c)2 ] = 3(a2 − b2 )(b − c)[b3 + c(b − 2c)2 ] ≥ 0. Also, E(b, b, c) = 2b6 + c 6 − 3b4 c 2 − 12b2 c 2 (b − c)2 = (b2 − c 2 )2 (2b2 + c 2 ) − 12b2 c 2 (b − c)2 = (b − c)2 (2b4 + 4b3 c − 9b2 c 2 + 2bc 3 + c 4 ) = (b − c)3 (2b3 + 6b2 c 2 − 3bc 2 − c 3 ) ≥ 0. The equality holds for a = b = c. (b) Let E(a, b, c) = a6 + b6 + c 6 − 3a2 b2 c 2 − 12a2 c 2 (b − c)2 . We will show that E(a, b, c) ≥ E(b, b, c) ≥ 0. To prove the left inequality, it suffices to show that for fixed b and c, the function f (a) = E(a, b, c) is increasing on [b, ∞); that is, f 0 a) ≥ 0. Indeed, we have f 0 (a) = 6a[a4 − b2 c 2 − 5ac(b − c)2 ] ≥ 6a[a4 − a2 c 2 − 5ac(a − c)2 ] = 6a2 (a − c)[a(a + c) − 5c(a − c)] = 6a2 (a − c)[(a − 2c)2 + c 2 ] ≥ 0. With regard to the right inequality, we have E(b, b, c) = 2b6 + c 6 − 3b4 c 2 − 10b3 c(b − c)2 = (b2 − c 2 )2 (2b2 + c 2 ) − 10b3 c(b − c)2 = (b − c)2 g(b, c),

Noncyclic Inequalities

325

where g(b, c) = 2b4 − 6b3 c + 3b2 c 2 + 2bc 3 + c 4 . Since g(b, c) = 2b(b − c)(b − 2c)2 + c · h(b, c),

h(b, c) = 4b3 − 13b2 c + 10bc 2 + c 3 ,

it suffices to show that h(b, c) ≥ 0. For b ≥ 2c, we have h(b, c) = b(b − 2c)(4b − 5c) + c 3 > 0. Also, for c ≤ b ≤ 2c, we have 2h(b, c) = −(b − 2c)(b − c)2 + b(3b − 5c)2 ≥ 0. Thus, the proof is completed. The equality holds for a = b = c.

P 2.63. Let k and a, b, c be positive real numbers, and let  ‹  ‹ k 1 1 k 1 1 2 2 2 E = (ka + b + c) + + , F = (ka + b + c ) 2 + 2 + 2 . a b c a b c

(a) If k ≥ 1, then v t F − (k − 2)2 2k

+2≥

E − (k − 2)2 ; 2k

+2≥

E − k2 . k+1

(b) If 0 < k ≤ 1, then v t F − k2 k+1

(Vasile Cîrtoaje, 2007) Solution. Due to homogeneity, we may assume that bc = 1. Under this assumption, if we denote 1 1 1 x =a+ , y = b+ =c+ a b c (x ≥ 2, y ≥ 2), then  ‹ ‹  ‹ 1 k 1 k E = ka + b + +b+ = (ka + y) + y = k2 + k x y + y 2 b a b a

326

Vasile Cîrtoaje

and ‹ ‹   ‹ 1 k 1 k 2 2 2 2 = (ka + y − 2) F = ka2 + b2 + 2 + b + + y − 2 b a2 b2 a2 = k2 + k(x 2 − 2)( y 2 − 2) + ( y 2 − 2)2 . (a) Write the inequality as 2kF − 2k(k − 2)2 ≥ (E − k2 − 4)2 . We have E − k2 − 4 = k x y + y 2 − 4 > 0, (E − k2 − 4)2 = k2 x 2 y 2 + 2k x y( y 2 − 4) + ( y 2 − 4)2 , and F − (k − 2)2 = 4k + k(x 2 − 2)( y 2 − 2) + y 2 ( y 2 − 4), 2kF − 2k(k − 2)2 = 8k2 + 2k2 (x 2 − 2)( y 2 − 2) + 2k y 2 ( y 2 − 4). Therefore, 2kF − 2k(k − 2)2 − (E − k2 − 4)2 = ( y 2 − 4)[k2 (x 2 − 4) − 2k y(x − y) − ( y 2 − 4)]. Since y 2 − 4 ≥ 0, we still need to show that k2 (x 2 − 4) − 2k y(x − y) ≥ y 2 − 4. We will show that k2 (x 2 − 4) − 2k y(x − y) ≥ (x 2 − 4) − 2 y(x − y) ≥ y 2 − 4. The right inequality reduces to (x − y)2 ≥ 0, while the left inequality is equivalent to (k − 1)[(k + 1)(x 2 − 4) − 2 y(x − y)] ≥ 0. This is true because (k + 1)(x 2 − 4) − 2 y(x − y) ≥ 2(x 2 − 4) − 2 y(x − y) = 2(x − y)2 + 2(x y − 4) ≥ 0. The equality holds for b = c. If k = 1, then the equality holds also for b = c or c = a. (b) Write the inequality as (k + 1)(F − k2 ) ≥ (E − k2 − 2k − 2)2 . We have E − k2 − 2k − 2 = k(x y − 2) + y 2 − 2 > 0,

Noncyclic Inequalities

327

(E − k2 − 2k − 2)2 = k2 (x y − 2)2 + 2k(x y − 2)( y 2 − 2) + ( y 2 − 2)2 , and (k + 1)(F − k2 ) = k2 (x 2 − 2)( y 2 − 2) + k( y 2 − 2)(x 2 + y 2 − 4) + ( y 2 − 2)2 . Thus, (k + 1)(F − k2 ) − (E − k2 − 2k − 2)2 = k(x − y)2 ( y 2 − 2k − 2) ≥ k(x − y)2 ( y 2 − 4) ≥ 0. If 0 < k < 1, then the equality holds for a = b or a = c.

P 2.64. If a, b, c are positive real numbers, then a b 25c + + > 1. 2b + 6c 7c + a 9a + 8b Solution. By the Cauchy-Schwarz inequality, we have a b 25c (a + b + 5c)2 + + ≥ . 2b + 6c 7c + a 9a + 8b a(2b + 6c) + b(7c + a) + c(9a + 8b) Therefore, it suffices to show that (a + b + 5c)2 ≥ 3a b + 15bc + 15ca, which is equivalent to E ≥ 0, where E = a2 + b2 + 25c 2 − a b − 5bc − 5ca. Indeed, we have 2E = (a − b)2 + a2 + b2 + 50c 2 − 10bc − 10ca = (a − b)2 + (a − 5c)2 + (b − 5c)2 ≥ 0.

P 2.65. If a, b, c are positive real numbers such that 1 1 1 ≥ + , a b c then

1 1 1 55 + + ≥ . a+b b+c c+a 12(a + b + c) (Vasile Cîrtoaje, 2014)

328

Vasile Cîrtoaje

Solution. Denote x=

bc , b+c

a ≤ x,

and write the desired inequality as X a+b+c b+c

≥

55 , 12

b c 19 a + + ≥ . b+c c+a a+b 12 Using the Cauchy-Schwarz inequality b c (b + c)2 + ≥ , c+a a+b b(c + a) + c(a + b) it suffices to show that F (a, b, c) ≥ where F (a, b, c) =

19 , 12

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

We will show that F (a, b, c) ≥ F (x, b, c) ≥

19 . 12

Since  1 (b + c)3 F (a, b, c) − F (x, b, c) = (x − a) − + , b + c (a(b + c) + 2bc)(x(b + c) + 2bc) 

we need to prove that (b + c)4 ≥ [a(b + c) + 2bc][(x(b + c) + 2bc]. Since a(b + c) + 2bc ≤ x(b + c) + 2bc, it is enough to show that (b + c)2 ≥ x(b + c) + 2bc, which is equivalent to the obvious inequality (b + c)2 ≥ 3bc. Also, we have F (x, b, c) −

bc 19 (b + c)2 19 (b − c)2 (4b2 + 5bc + 4c 2 ) = + − = ≥ 0. 12 (b + c)2 3bc 12 12bc(b + c)2

The equality occurs for 2a = b = c.

Noncyclic Inequalities

329

P 2.66. If a, b, c are positive real numbers such that 1 1 1 ≥ + , a b c then

1 1 1 189 + + ≥ . a2 + b2 b2 + c 2 c 2 + a2 40(a2 + b2 + c 2 ) (Vasile Cîrtoaje, 2014)

Solution. Denote x=

bc , b+c

a ≤ x,

and write the desired inequality as X a2 + b2 + c 2 b2

+ c2

≥

189 , 40

a2 b2 c2 69 + + ≥ . b2 + c 2 c 2 + a2 a2 + b2 40 Using the Cauchy-Schwarz inequality c2 (b2 + c 2 )2 b2 + ≥ , c 2 + a2 a2 + b2 b2 (c 2 + a2 ) + c 2 (a2 + b2 ) it suffices to show that F (a, b, c) ≥ where F (a, b, c) =

69 , 40

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

We will show that F (a, b, c) ≥ F (x, b, c) ≥

69 . 40

Since  1 (b2 + c 2 )3 F (a, b, c)−F (x, b, c) = (x −a ) − 2 + , b + c 2 (a2 (b2 + c 2 ) + 2b2 c 2 ) (x 2 (b2 + c 2 ) + 2b2 c 2 ) 2

2



we need to prove that (b2 + c 2 )4 ≥ [a2 (b2 + c 2 ) + 2b2 c 2 ][x 2 (b2 + c 2 ) + 2b2 c 2 ]. Since a2 (b2 + c 2 ) + 2b2 c 2 ≤ x 2 (b2 + c 2 ) + 2b2 c 2 , it is enough to show that (b2 + c 2 )2 ≥ x 2 (b2 + c 2 ) + 2b2 c 2 ,

330

Vasile Cîrtoaje

which is equivalent to (b4 + c 4 )(b + c)2 ≥ b2 c 2 (b2 + c 2 ). This inequality follows from b4 + c 4 > b2 c 2 and (b + c)2 > b2 + c 2 . Also, we have F (x, b, c) =

x2 (b2 + c 2 )2 + . b2 + c 2 x 2 (b2 + c 2 ) + 2b2 c 2

Since 2b2 c 2 ≤ 4x 2 (b2 + c 2 ), we have F (x, b, c) ≥

(b2 + c 2 )2 1 t x2 + = + , 2 2 2 2 2 b +c 5x (b + c ) t 5

where t=

b2 + c 2 ≥ 8. x2

Therefore,

69 1 t 69 (t − 8)(8t − 5) ≥ + − = ≥ 0. 40 t 5 40 40t The equality occurs for 2a = b = c. F (x, b, c) −

P 2.67. If a, b, c are the lengths of the sides of a triangle, then a3 (b + c) + bc(b2 + c 2 ) ≥ a(b3 + c 3 ). (Vasile Cîrtoaje, 2010) First Solution. Because the inequality is symmetric in b and c, we may assume that b ≥ c. Consider the following two cases. Case 1: a ≥ b. It suffices to show that a3 (b + c) ≥ a(b3 + c 3 ). We have a3 (b + c) − a(b3 + c 3 ) ≥ a b2 (b + c) − a(b3 + c 3 ) = ac(b2 − c 2 ) ≥ 0. Case 2: a ≤ b. Write the inequality as c(a3 + b3 ) − c 3 (a − b) + a b(a2 − b2 ) ≥ 0. It suffices to show that c(a3 + b3 ) + a b(a2 − b2 ) ≥ 0.

Noncyclic Inequalities

331

We have c(a3 + b3 ) + a b(a2 − b2 ) ≥ c(a3 + b3 ) − a bc(a + b) = c(a + b)(a − b)2 ≥ 0. The equality holds for a degenerate triangle with a = b and c = 0, or a = c and b = 0. Second Solution. Consider two cases. Case 1: b2 + c 2 ≥ a(b + c). Write the inequality as bc(b2 + c 2 ) ≥ a(b + c)(b2 + c 2 − bc − a2 ). It suffices to show that bc ≥ b2 + c 2 − bc − a2 , which is equivalent to the obvious inequality a2 ≥ (b − c)2 . Case 2: a(b + c) ≥ b2 + c 2 . Write the inequality as a(b + c)(a2 + bc) ≥ (b2 + c 2 )(a b + ac − bc). It suffices to show that a2 + bc ≥ a b + ac − bc, which is equivalent to the obvious inequality bc ≥ (a − c)(b − a).

P 2.68. If a, b, c are the lengths of the sides of a triangle, then (a + b)2 (a + c)2 (b + c)2 + ≥ . 2a b + c 2 2ac + b2 2bc + a2 (Vasile Cîrtoaje, 2010) Solution. By the Cauchy-Schwarz inequality, we have (a + b)2 (a + c)2 (2a + b + c)2 + ≥ . 2a b + c 2 2ac + b2 2a(b + c) + b2 + c 2 Therefore, it suffices to show that (2a + b + c)2 (b + c)2 ≥ . 2a(b + c) + b2 + c 2 2bc + a2

332

Vasile Cîrtoaje

We will show that

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

The left inequality reduces to 4a2 ≥ (b − c)2 , and the right inequality reduces to 2a2 ≥ (b − c)2 . These are true because a2 ≥ (b − c)2 . The equality holds for a degenerate triangle with a = 0 and b = c.

P 2.69. If a, b, c are the lengths of the sides of a triangle, then a+c b+c a+b + ≥ . 2 2 ab + c ac + b bc + a2 (Vasile Cîrtoaje, 2010) Solution. Without loss of generality, assume that b ≥ c. Since a + b ≥ a + c and a b + c 2 − (ac + b2 ) = (b − c)(a − b − c) ≤ 0, by Chebyshev’s inequality, we have  ‹ a+b a+c 1 1 1 + ≥ [(a + b) + (a + c)] + a b + c 2 ac + b2 2 a b + c 2 ac + b2 ≥

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

On the other hand,

b+c b+c 2(b + c) ≤ = 2 . 1 bc + a2 b + c2 2 (b − c) + bc 2 Therefore, it suffices to show that 2(2a + b + c) 2(b + c) ≥ 2 , 2 2 a(b + c) + b + c b + c2 which is equivalent to a(b − c)2 ≥ 0. The equality holds for a degenerate triangle with a = 0 and b = c.

P 2.70. If a, b, c are the lengths of the sides of a triangle, then b(a + c) c(a + b) a(b + c) + ≥ . ac + b2 a b + c2 bc + a2 (Vo Quoc Ba Can, 2010)

Noncyclic Inequalities

333

Solution. Without loss of generality, assume that b ≥ c. Since a b + c 2 − (ac + b2 ) = (b − c)(a − b − c) ≤ 0, it suffices to prove that b(a + c) c(a + b) a(b + c) + ≥ , 2 2 ac + b ac + b bc + a2 which is equivalent to

2bc + a(b + c) a(b + c) ≥ , ac + b2 bc + a2  ‹ 1 1 2bc ≥ a(b + c) − , ac + b2 bc + a2 ac + b2 2bc(bc + a2 ) ≥ a(b + c)(b − a)(a + b − c).

Consider the non-trivial case b ≥ a. Since c ≥ b − a, it suffices to show that 2b(bc + a2 ) ≥ a(b + c)(a + b − c). We have 2b(bc + a2 ) − a(b + c)(a + b − c) = a b(a − b) + c(2b2 − a2 + ac) ≥ −a bc + c(2b2 − a2 + ac) = ac(b + c − a) + 2bc(b − a) ≥ 0. The equality holds for a degenerate triangle with a = b and c = 0, or a = c and b = 0.

P 2.71. Let a, b, c, d be nonnegative real numbers such that a2 − a b + b2 = c 2 − cd + d 2 . Prove that (a + b)(c + d) ≥ 2(a b + cd). (Vasile Cîrtoaje, 2000) Solution. Let x = a2 − a b + b2 = c 2 − cd + d 2 . Without loss of generality, assume that a b ≥ cd. Then, x ≥ a b ≥ cd, (a + b)2 = x + 3a b, (c + d)2 = x + 3cd.

334

Vasile Cîrtoaje

By squaring, the desired inequality can be restated as (x + 3a b)(x + 3cd) ≥ 4(a b + cd)2 . It is true since (x + 3a b)(x + 3cd) − 4(a b + cd)2 ≥ (a b + 3a b)(a b + 3cd) − 4(a b + cd)2 = 4cd(a b − cd) ≥ 0. The equality occurs for a = b = c = d, and for a = b = c and d = 0 (or any cyclic permutation).

P 2.72. Let a, b ∈ (0, 1], a ≤ b. (a) If a ≤

(b) If b ≥

1 , then e

2a a ≥ a b + b a ;

1 , then e

2b b ≥ a b + b a . (Vasile Cîrtoaje, 2012)

Solution. (a) We need to show that f (a) ≥ f (b), where f (x) = a x + x a ,

x ∈ [a, b].

This is true if f (x) is decreasing; that is, if f 0 (x) ≤ 0 on [a, b]. Since f 0 (x) = a(x a−1 + a x−1 ln a) ≤ a(x a−1 − a x−1 ), it suffices to show that x a−1 ≤ a x−1 for 0 < a ≤ x ≤ 1. Consider the non-trivial case 0 < a ≤ x < 1, and write the inequality as g(x) ≥ g(a), where ln x g(x) = . 1− x Clearly, it suffices to show that g 0 (x) ≥ 0 for 0 < x < 1. We have g 0 (x) =

h(x) 1 , h(x) = − 1 + ln x. 2 (1 − x) x

Noncyclic Inequalities

335

Since

x −1 < 0, x2 h(x) is strictly decreasing, h(x) > h(1) = 0, g 0 (x) > 0. This completes the proof. The equality holds for a = b. h0 (x) =

(b) We need to show that f (b) ≥ f (a), where f (x) = x b + b x ,

x ∈ [a, b].

This is true if f (x) is increasing; that is, if f 0 (x) ≥ 0 on [a, b]. Since f 0 (x) = b(x b−1 + b x−1 ln b) ≥ b(x b−1 − b x−1 ), it suffices to show that x b−1 ≥ b x−1 for 0 < x ≤ b ≤ 1. As we shown at (a), this inequality is true. The equality holds for a = b.

P 2.73. If 0 ≤ a ≤ b and b ≥

1 , then 2 2b2b ≥ a2b + b2a . (Vasile Cîrtoaje, 2012)

Solution. We need to show that f (a) ≤ f (b), where f (x) = x 2b + b2x ,

x ∈ [0, b].

From   f 00 (x) = 2b 2b2x−1 ln2 b + (2b − 1)x 2b−2 > 0,

x ∈ (0, b],

it follows that f (x) is convex on [0, b]. Therefore, we have f (a) ≤ max{ f (0), f (b)}. To prove that f (a) ≤ f (b), it suffices to show that f (0) ≤ f (b). Using Bernoulli’s inequality, we get f (b) − f (0) = 2b2b − 1 = 2[1 + (b − 1)]2b − 1 ≥ 2[1 + 2b(b − 1)] − 1 = (2b − 1)2 ≥ 0. 1 1 The equality holds for a = b ≥ , and also for a = 0 and b = . 2 2

336

Vasile Cîrtoaje

P 2.74. If a ≥ b ≥ 0, then (a)

a−b a b−a ≤ 1 + p ; a

(b)

a a−b ≥ 1 −

3(a − b) . p 4 a (Vasile Cîrtoaje, 2010)

Solution. (a) We write the inequality as  ‹ a−b (a − b) ln a + ln 1 + p ≥ 0, a and prove it by adding the inequalities  ‹ a−b a − b (a − b)2 ln 1 + p − p + ≥ 0, 2a a a a − b (a − b)2 (a − b) ln a + p − ≥ 0. 2a a Denoting a−b x= p , a we can write the first inequality as f (x) ≥ 0 for x ≥ 0, where f (x) = ln(1 + x) − x + From f 0 (x) =

x2 . 2

x2 ≥ 0, 1+ x

it follows that f is increasing, and hence f (x) ≥ f (0) = 0. The second inequality is true if 1 a−b ln a + p − ≥ 0. 2a a It suffices to prove that g(a) ≥ 0, where 1 1 g(a) = ln a + p − . a 2 From

p 2 a−1 g (a) = p , 2a a 0

Noncyclic Inequalities

337

it follows that

 ‹ 1 3 g(a) ≥ g = − ln 4 > 0. 4 2

The equality holds for a = b. 3(a − b) > 0, write the inequality as p 4 a  ‹ 3a − 3b (a − b) ln a ≥ ln 1 − , p 4 a

(b) Consider the non-trivial case 1 −

and prove it by adding the inequalities  ‹ 3a − 3b 3(a − b) 0 ≥ ln 1 − + , p p 4 a 4 a (a − b) ln a +

3(a − b) ≥ 0. p 4 a

Denoting x=

3(a − b) , p 4 a

we can write the first inequality as f (x) ≤ 0 for 0 ≤ x < 0, where f (x) = ln(1 − x) + x. From f 0 (x) =

−x ≤ 0, 1− x

it follows that f is decreasing, and hence f (x) ≤ f (0) = 0. The second inequality is true if g(a) ≥ 0, where 3 g(a) = ln a + p . 4 a From g 0 (a) = it follows that

p 8 a−3 p , 8a a

‹ 9 3e g(a) ≥ g = 2 ln > 0. 64 8

The equality holds for a = b.



338

Vasile Cîrtoaje

P 2.75. Let a, b, c and x, y, z be positive real numbers such that x + y + z = a + b + c. Prove that a x 2 + b y 2 + cz 2 + x yz ≥ 4a bc. (Vasile Cîrtoaje, 1989) First Solution. Among the numbers a−

y +z , 2

b−

z+x , 2

c−

x+y , 2

there are two of them with the same sign; let pq ≥ 0, where p= b−

x+y z+x , q=c− . 2 2

We have b= p+

x +z , 2

c =q+

x+y , 2

a = x + y +z− b−c =

y +z − p − q. 2

Then, a x 2 + b y 2 + cz 2 + x yz − 4a bc =   y +z x + y 2 x +z 2  − p − q x2 + p + y + q+ z = 2 2 2 y +z  x + y x + z +x yz − 4 −p−q p+ q+ 2 2 2 = 4pq(p + q) + 2p2 (x + y) + 2q2 (x + z) + 4pq x   x + y x +z 2 2 = 4q p + + 4p q + + 4pq x 2 2 = 4(q2 b + p2 c + pq x) ≥ 0. x+y y +z z+x The equality holds for a = , b= ,c= . 2 2 2 Second Solution. Consider the following two cases. Case 1: x 2 ≥ 4bc. We have a x 2 + b y 2 + cz 2 + x yz − 4a bc > a x 2 − 4a bc ≥ 0. Case 2: x 2 ≤ 4bc. Let

u = x + y + z = a + b + c.

Substituting z = u − x − y,

a = u − b − c,

Noncyclic Inequalities

339

the inequality can be restated as Au2 + Bu + C ≥ 0, where A = c, B = (x 2 − 4bc) − 2c(x + y) + x y, C = −(b + c)(x 2 − 4bc) + b y 2 + c(x + y)2 − x y(x + y). Since the quadratic in u has the discriminant D = (x 2 − bc)(2c − x − y)2 ≤ 0, the conclusion follows.

P 2.76. Let a, b, c and x, y, z be positive real numbers such that x + y + z = a + b + c. Prove that

x(3x + a) y(3 y + b) z(3z + c) + + ≥ 12. bc ca ab (Vasile Cîrtoaje, 1990)

Solution. Write the inequality as 1 a x 2 + b y 2 + cz 2 + (a2 x + b2 y + c 2 z) ≥ 4a bc. 3 Applying the Cauchy-Schwarz inequality, we have a2 x + b2 y + c 2 z ≥

x yz(x + y + z)2 a+b+c = ≥ 3x yz. 1 1 1 x y + yz + z x + + x y z

Therefore, it suffices to show that a x 2 + b y 2 + cz 2 + x yz ≥ 4a bc, which is just the inequality in the preceding P 2.75. The equality holds for x = y = z = a = b = c.

340

Vasile Cîrtoaje

P 2.77. Let a, b, c be given positive numbers. Find the minimum value F (a, b, c) of E(x, y, z) =

by ax cz + + y +z z+ x x + y

for any nonnegative numbers x, y, z, no two of which are zero. (Vasile Cîrtoaje, 2006) Solution. Assume that a = max{a, b, c}. There are two cases to consider. p p p Case 1: a < b + c. Using the Cauchy-Schwarz inequality, we get X a(x + y + z) − a( y + z) X a X = (x + y + z) − a y +z y +z Pp 2 X p p ( a) p a+b+c ≥ (x + y + z) P − a = a b + bc + ca − . 2 ( y + z) E=

The equality holds for

y +z x+y z+x p = p = p ; c a b

that is, for

y x z p =p p p =p p p p . b+ c− a c+ a− b a+ b− c p p p Case 2: a ≥ b + c. Let us denote p p x+y y +z z+x A = ( b + c)2 , , Y= , Z= . X= 2 2 2 p

We have E≥

by Ax cz + + y +z z+ x x + y

A(Y + Z − X ) b(Z + X − Y ) c(X + Y − Z) + + 2X 2Y 2Z  ‹  ‹  ‹ p 1 Y X 1 Z Y 1 X Z A +b + b +c + c +A − b − c − bc 2 X Y 2 Y Z 2 Z X p p p p p ≥ Ab + bc + cA − b − c − bc = 2 bc. s y c The equality holds for x = 0 and = . Therefore, for a = max{a, b, c}, we have z b  p p p p p a+b+c p  , a< b+ c  a b + bc + ca − 2 F (a, b, c) = p p p p .  a≥ b+ c  2 bc, =

Noncyclic Inequalities

341

P 2.78. Let a, b, c and x, y, z be real numbers. (a) If a b + bc + ca > 0, then [(b + c)x + (c + a) y + (a + b)z]2 ≥ 4(a b + bc + ca)(x y + yz + z x); (b) If a, b, c ≥ 0, then [(b + c)x + (c + a) y + (a + b)z]2 ≥ 4(a + b + c)(a yz + bz x + c x y). (Vasile Cîrtoaje, 1995) Solution. (a) First Solution. The condition a b + bc + ca > 0 yields b + c 6= 0. Indeed, if b + c = 0, then a b + bc + ca = −b2 ≤ 0, which is false. The desired inequality is equivalent to D ≥ 0, where D is the discriminant of the quadratic function f (t) = (at − x)(bt − y) + (bt − y)(c t − z) + (c t − z)(at − x). For the sake of contradiction, assume that D < 0 for some real numbers a, b, c and x, y, z. Since the coefficient of t 2 is positive, we have f (t) > 0 for all real t. This is not true, because for (bt − y) + (c t − z) = 0, we get f

 y +z b+c

bz − c y =− b+c 

‹2

≤ 0.

For pqr 6= 0, the equality holds when y x z = = . a b c Second Solution. If x y + yz + z x ≤ 0, then the inequality is obviously true. Otherwise, due to homogeneity, we may assume that x + y + z = a + b + c. Then, by the AM-GM inequality, we have Æ 2 (a b + bc + ca)(x y + yz + z x) ≤ (ab + bc + ca) + (x y + yz + z x) =

(a + b + c)2 − a2 − b2 − c 2 (x + y + z)2 − x 2 − y 2 − z 2 + 2 2 = (a + b + c)(x + y + z) −

a2 + x 2 b2 + y 2 c 2 + z 2 − − 2 2 2

≤ (a + b + c)(x + y + z) − ax − b y − cz = (b + c)x + (c + a) y + (a + b)z. (b) Assume that x is between y and z, that is, (x − y)(x − z) ≤ 0. Consider the non-trivial case a + b + c 6= 0, hence a + b + c > 0. The desired inequality is equivalent to D ≥ 0, where D is the discriminant of the quadratic function f (t) = a(t − y)(t − z) + b(t − z)(t − x) + c(t − x)(t − y).

342

Vasile Cîrtoaje

For the sake of contradiction, assume that D < 0 for some non-positive numbers a, b, c and real numbers x, y, z. Since the coefficient of t 2 is positive, we have f (t) > 0 for all real t. This is false, because f (x) = a(x − y)(x − z) ≤ 0. The equality holds for x = y = z, and also for a = 0 and x = or c = 0 and z =

bx + a y . b+a

c y + bz az + c x , or b = 0 and y = , c+b a+c

Remark 1. For x = b, y = c, z = a, from the inequality in (b), we get the following cyclic inequality: (a2 + b2 + c 2 + a b + bc + ca)2 ≥ 4(a + b + c)(ab2 + bc 2 + ca2 ), p b 5−1 where a, b, c ≥ 0. The equality holds for a = b = c, and also for a = 0 and = (or any c 2 cyclic permutation). Notice that this inequality is equivalent to a4 + b4 + c 4 − a2 b2 − b2 c 2 − c 2 a2 ≥ 2(ab3 + bc 3 + ca3 − a3 b − b3 c − c 3 a), which is similar to the inequality in P 3.93 from Volume 1. Remark 2. For x = 1/b, y = 1/c, z = 1/a, from the inequality in (b), we get the following cyclic inequality:  ‹2  ‹ b c a 1 1 1 + + + 3 ≥ 4(a + b + c) + + , a b c a b c where a, b, c > 0. The equality holds for a = b = c. Remark 3. For a = x(x − y + z), b = y( y − z + x), c = z(z − x + y), the inequality in (b) turns into (x 2 y + y 2 z + z 2 x)2 ≥ x yz(x + y + z)(x 2 + y 2 + z 2 ). where x, y, z are the lengths of the sides of a triangle. The equality holds for an equilateral triangle, and for a degenerate triangle with y = x + z and x 3 = z 2 (x + z) (or any cyclic permutation).

P 2.79. Find the best real numbers x, y, z such that p p p p ( a + b + c) a + b + c ≥ x a + y b + zc for all a ≥ b ≥ c ≥ 0. Solution. For a = 1 and b = c = 0, for a = b = 1 and c = 0, and for a = b = c = 1, we get respectively p p x ≤ 1, x + y ≤ 2 2, x + y + z ≤ 3 3,

Noncyclic Inequalities

343

which yield x a + y b + zc = x(a − b) + (x + y)(b − c) + (x + y + z)c p p p p p ≤ a − b + 2 2 (b − c) + 3 3 c = a + (2 2 − 1)b + (3 3 − 2 2)c. Therefore, if the following inequality holds p p p p p p p ( a + b + c) a + b + c ≥ a + (2 2 − 1)b + (3 3 − 2 2)c, then x = 1,

p y = 2 2 − 1,

p p z =3 3−2 2

are the best values of x, y, z. Since p p p p p p ( a + b + c )2 = a + (2 a b + b) + (2 ac + 2 bc + c) ≥ a + 3b + 5c, it suffices to show that p p p (a + 3b + 5c)(a + b + c) ≥ [a + (2 2 − 1)b + (3 3 − 2 2)c]2 , which is equivalent to the obvious inequality p p p p (3 − 2 2)b(a − b) + (3 + 2 2 − 3 3)c(a − b) + 3(5 − 2 6)c(b − c) ≥ 0. p p p For x = 1, y = 2 2 − 1, z = 3 3 − 2 2, the equality holds when a = b = c, when a = b and c = 0, and when b = c = 0.

P 2.80. Let a, b, c and x, y, z be positive real numbers such that a b c + + = 1. yz z x x y Prove that (a) (b)

x + y +z ≥

Æ

4(a + b + c +

x + y +z >

p

p

a+b+

ab +

p

p

bc +

b+c+

p

p

p 3 ca ) + 3 a bc;

c + a.

344

Vasile Cîrtoaje

Solution. (a) We will show that  ‹ p p p p b c a 3 + + (x + y + z)2 ≥ 4(a + b + c + a b + bc + ca ) + 3 a bc. yz z x x y Since 

‹ X a x 2 X a( y 2 + z 2 ) b c a + + (x + y + z)2 = + , yz z x x y yz yz

by the AM-GM inequality, we get ‹  p b c a 3 + + (x 2 + y 2 + z 2 ) ≥ 3 a bc + 2(a + b + c). yz z x x y On the other hand, the Cauchy-Schwarz inequality yields  ‹ p p p a b c 2 + + ( yz + z x + x y) ≥ 2( a + a + c)2 . yz z x x y Adding these inequalities, the desired inequality follows. The equality holds for p p p x = y = z = 3a = 3b = 3c. (b) According to the inequality in (a), it suffices to show that p p p p p p 4(a + b + c + a b + bc + ca ) ≥ ( a + b + b + c + c + a)2 . This inequality is equivalent to XÆ p p p ( a + b + c)2 ≥ (a + b)(a + c), which follows immediately from the inequality in P 2.22 from Volume 2.

P 2.81. Let a, b, c be the lengths of the sides of a triangle. If x, y, z are real numbers, then ( y a2 + z b2 + x c 2 )(za2 + x b2 + y c 2 ) ≥ (x y + yz + z x)(a2 b2 + b2 c 2 + c 2 a2 ). (Vasile Cîrtoaje, 2001) First Solution. Write the inequality as follows: x 2 b2 c 2 + y 2 c 2 a2 + z 2 a2 b2 ≥ ≥ yza2 (b2 + c 2 − a2 ) + z x b2 (c 2 + a2 − b2 ) + x y c 2 (a2 + b2 − c 2 ), x 2 b2 c 2 + y 2 c 2 a2 + z 2 a2 b2 ≥ a bc(2 yza cos A + 2z x b cos B + 2x y c cos C),

Noncyclic Inequalities

345

x2 + a2 x − a The equality holds for

y 2 z2 2 yz cos A 2z x cos B 2x y cos C + 2 ≥ + + , 2 b c bc ca ab 2  y 2 y z z cos C − cos B + sin C − sin B ≥ 0. b c b c

y x z = 2 = 2. 2 a b c Second Solution. Write the inequality as b2 c 2 x 2 − B x + C ≥ 0, where B = c 2 (a2 + b2 − c 2 ) y + b2 (a2 − b2 + c 2 )z, C = a2 [c 2 y 2 − (b2 + c 2 − a2 ) yz + b2 z 2 ]. It suffices to show that 4b2 c 2 C ≥ B 2 , which is equivalent to A(c 2 y − b2 z)2 ≥ 0, where A = 2a2 b2 + 2b2 c 2 + 2c 2 a2 − a4 − b4 − c 4 . This inequality is true since A = (a + b + c)(a + b − c)(b + c − a)(c + a − b) ≥ 0. Remark 1. For x = 1/b, y = 1/c and z = 1/a, we get the well-known inequality a3 b + b3 c + c 3 a ≥ a2 b2 + b2 c 2 + c 2 a2 . Remark 2. For x = 1/c 2 , y = 1/a2 and z = 1/b2 , we get the elegant cyclic inequality of Walker  2   ‹ a b2 c 2 1 1 1 2 2 2 3 + + ≥ (a + b + c ) 2 + 2 + 2 . b2 c 2 a2 a b c

P 2.82. If a, b, c, d are real numbers, then 6(a2 + b2 + c 2 + d 2 ) + (a + b + c + d)2 ≥ 12(a b + bc + cd). (Vasile Cîrtoaje, 2005)

346

Vasile Cîrtoaje

Solution. Let E(a, b, c, d) = 6(a2 + b2 + c 2 + d 2 ) + (a + b + c + d)2 − 12(a b + bc + cd). First Solution. We have E(x + a, x + b, x + c, x + d) = = 4x 2 + 4(2a − b − c + 2d)x + 7(a2 + b2 + c 2 + d 2 ) + 2(ac + ad + bd) − 10(a b + bc + cd) = (2x + 2a − b − c + 2d)2 + 3(a2 + 2b2 + 2c 2 + d 2 − 2a b + 2ac − 2ad − 4bc + 2bd − 2cd) = (2x + 2a − b − c + 2d)2 + 3(b − c)2 + 3(a − b + c − d)2 . For x = 0, we get E(a, b, c, d) = (2a − b − c + 2d)2 + 3(b − c)2 + 3(a − b + c − d)2 ≥ 0. The equality holds for 2a = b = c = 2d. Second Solution. Let x = a − b and y = d − c. We have E = 6(x 2 + y 2 ) + [x + y + 2(b + c)]2 − 12bc = 3(x − y)2 + 4(x + y)2 + 4(x + y)(b + c) + (b + c)2 + 3(b − c)2 = 3(x − y)2 + (2x + 2 y + b + c)2 + 3(b − c)2 ≥ 0.

P 2.83. If a, b, c, d are positive real numbers, then a2

1 1 1 1 4 + 2 + 2 + 2 ≥ . + a b b + bc c + cd d + d a ac + bd

Solution. Write the inequality as follows: ‹ X  ac + bd + 1 ≥ 8, a2 + a b X a(c + a) + b(d + a) ≥ 8, a(a + b) X c + a X b(d + a) + ≥ 8. a+b a(a + b) By the AM-GM inequality, we have v X b(d + a) Y b(d + a) t 4 ≥4 = 4. a(a + b) a(a + b)

Noncyclic Inequalities

347

Therefore, it suffices to prove the inequality X c+a a+b

≥ 4,

which is equivalent to 1 1 (a + c) + a+b c+d 

‹

‹ 1 1 + (b + d) + ≥ 4. b+c d+a 

This inequality follows immediately from 1 1 4 + ≥ a+b c+d (a + b) + (c + d) and

1 4 1 + ≥ . b+c d+a (b + c) + (d + a)

The equality occurs for a = b = c = d.

P 2.84. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and a b + bc + cd + d a = 3. Prove that a3 bcd < 4. (Vasile Cîrtoaje, 2012) Solution. Write the desired inequality as 4(a b + bc + cd + d a)3 > 27a3 bcd,  ‹ bc + cd 3 4 b+d+ > 27bcd. a It suffices to show that 4(b + d)3 ≥ 27bcd. In addition, this inequality is true if 4(b + d)3 ≥ 27b2 d. Indeed, 4(b + d)3 − 27b2 d = (4b + d)(b − 2d)2 ≥ 0.

348

Vasile Cîrtoaje

P 2.85. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and a b + bc + cd + d a = 6. Prove that acd ≤ 2. (Vasile Cîrtoaje, 2012) Solution. Write the desired inequality in the homogeneous form (a + c)3 (b + d)3 ≥ 54a2 c 2 d 2 . Since b ≥ c, it remains to show that (a + c)3 (c + d)3 ≥ 54a2 c 2 d 2 . On the other hand, by the AM-GM inequality, we have 3 aa a 27 2 (a + c) = + + c ≥ 27 a c. c= 2 2 2 2 4 3

a

Therefore, it suffices to show that (c + d)3 ≥ 8cd 2 . Indeed, (c + d)3 − 8cd 2 = (c − d)(c 2 + 4cd − d 2 ) ≥ 0. The equality holds for a = 2 and b = c = d = 1.

P 2.86. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and a b + bc + cd + d a = 9. Prove that a bd ≤ 4. (Vasile Cîrtoaje, 2012) Solution. Write the desired inequality in the homogeneous form (a + c)3 (b + d)3 ≥

729 2 2 2 a b d . 16

Noncyclic Inequalities

349

Since c ≥ d, we only need to show that (a + d)3 (b + d)3 ≥

729 2 2 2 a b d . 16

On the other hand, by the AM-GM inequality, we have (a + d)3 =

a 2

+

3 aa a 27 2 + d ≥ 27 d= a d 2 2 2 4

and, similarly, (b + d)3 ≥

27 2 b d 4

Multiplying these inequalities, the desired inequality holds. The equality holds for a = b = 2 and c = d = 1.

P 2.87. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and a2 + b2 + c 2 + d 2 = 10. Prove that 2b + 4d ≤ 3c + 5. (Vasile Cîrtoaje, 2012) Solution. Write the desired inequality in the homogeneous form v t5 2

(a2 + b2 + c 2 + d 2 ) ≥ 2b − 3c + 4d.

It is true if 5(a2 + b2 + c 2 + d 2 ) ≥ 2(2b − 3c + 4d)2 . Since a ≥ b, it remains to show that 5(2b2 + c 2 + d 2 ) ≥ 2(2b − 3c + 4d)2 , which is equivalent to 2b2 + 24bc + 48cd ≥ 13c 2 + 27d 2 + 32bd. Since cd ≥ d 2 , it suffices to prove that 2b2 + 24bc + 48cd ≥ 13c 2 + 27cd + 32bd,

350

Vasile Cîrtoaje

which is equivalent to 2b2 + 24bc ≥ 13c 2 + (32b − 21c)d. Since 32b − 21c > 0 and c ≥ d, it is enough to show that 2b2 + 24bc ≥ 13c 2 + (32b − 21c)c. This reduces to the obvious inequality 2(b − 2c)2 ≥ 0. The equality holds for a = b = 2 and c = d = 1.

P 2.88. If a, b, c, d are positive real numbers such that a ≥ b ≥ c ≥ d, then (a + b + c + d)2 ≥ 8(ac + bd). (Vasile Cîrtoaje, 2005) Solution. We have (a + b + c + d)2 − 8(ac + bd) = a2 + 2(b + d − 3c)a + (b + c + d)2 − 8bd = (a + b + d − 3c)2 − (b + d − 3c)2 + (b + d + c)2 − 8bd = (a + b + d − 3c)2 + 8(b − c)(c − d) ≥ 0. The equality holds for b = c =

a+d . 2

P 2.89. Let a, b, c, d be positive real numbers such that a ≤ b ≤ c ≤ d and a bcd = 1. Prove that a b c d 4 + + + + ≥ 2(ac + c b + bd + d a). b c d a (Vasile Cîrtoaje, 2012) Solution. Since

b d b d (d − b)(c − a) + − − = ≥ 0, c a a c ca we only need to prove that 4+

a b c d + + + ≥ 2(ac + bd) + 2(bc + ad), b a d c

Noncyclic Inequalities

351

which is equivalent to (a + b)2 (c + d)2 + ≥ 2(a + b)(c + d), ab cd 

a+b c+d − p p cd ab

‹2

≥ 0.

The proof is completed. The equality holds for a = b = c = d = 1.

P 2.90. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and 3(a2 + b2 + c 2 + d 2 ) = (a + b + c + d)2 . Prove that a+d ≤ 2; b+c

(a)

p 7+2 6 a+c ≤ ; b+d 5 p a+c 3+ 5 ≤ . c+d 2

(b) (c)

(Vasile Cîrtoaje, 2010) Solution. (a) Since (a + d)(b + c) − 2(ad + bc) = (a − b)(c − d) + (a − c)(b − d) ≥ 0, we have a2 + b2 + c 2 + d 2 = (a + d)2 + (b + c)2 − 2(ad + bc) ≥ (a + d)2 + (b + c)2 − (a + d)(b + c). Thus, according to the hypothesis 3(a2 + b2 + c 2 + d 2 ) = (a + b + c + d)2 , we get 1 (a + b + c + d)2 ≥ (a + d)2 + (b + c)2 − (a + d)(b + c), 3 ‹ ‹  a+d 1 a+d −2 − ≤ 0, b+c b+c 2 from where the desired result follows. The equality holds for a/3 = b = c = d.

352

Vasile Cîrtoaje (b) From (a − d)(b − c) ≥ 0 end the AM-GM inequality, we have 2(ac + bc) ≤ (a + d)(b + c) ≤

(a + b + c + d)2 , 4

hence a2 + b2 + c 2 + d 2 = (a + c)2 + (b + d)2 − 2(ac + bd) ≥ (a + c)2 + (b + d)2 −

(a + b + c + d)2 . 4

Thus, using the hypothesis 3(a2 + b2 + c 2 + d 2 ) = (a + b + c + d)2 , we get (a + b + c + d)2 1 (a + b + c + d)2 ≥ (a + c)2 + (b + d)2 − , 3 4  p  p  a+c 7+2 6 a+c 7−2 6 − − ≤ 0, b+d 2 b+d 2 from where the desired result follows. The equality holds for p p (3 − 6)a = b = c = (3 + 6)d. (c) Writing the hypothesis 3(a2 + b2 + c 2 + d 2 ) = (a + b + c + d)2 as (2b − a − c − d)2 = 3(2ac + 2cd + 2d a − a2 − c 2 − d 2 ), we get 2(ac + cd + d a) ≥ a2 + c 2 + d 2 , hence

p

a≤

p

c+

p

d.

Thus, it suffices to prove that p p p ( c + d)2 + c 3+ 5 ≤ , c+d 2 which is equivalent to p p p p 1+ 5 3+ 5 ( c + d)2 ≤ c+ d. 2 2 This inequality follows from the Cauchy-Schwarz inequality, as follows:  p p  ‹ p p 1+ 5 3+ 5 2 2 2 c+ d ( c + d) ≤ p + p 2 2 1+ 5 3+ 5 p p 1+ 5 3+ 5 = c+ d. 2 2

Noncyclic Inequalities

353

The equality holds for b c d a p = = = p . 4 2 3− 5 3+ 5

P 2.91. Let a, b, c, d be nonnegative real numbers such that a ≥ b ≥ c ≥ d and 2(a2 + b2 + c 2 + d 2 ) = (a + b + c + d)2 . Prove that

p a ≥ b + 3c + (2 3 − 1)d. (Vasile Cîrtoaje, 2010)

First Solution. From the hypothesis, we get p a = b + c + d + 2 bc + cd + d a. Notice that it is not possible to have p a = b + c + d − 2 bc + cd + d a, bc + cd + d a 6= 0, since a ≥ b involves

c+d ≥2

p

bc + cd + d a,

2

(c − d) ≥ 4b(c + d), (c − d)2 ≥ 4c(c + d), d 2 ≥ 3c(c + 2d), which is not true. Thus, we can rewrite the desired inequality as p p b + c + d − 2 bc + cd + d a ≥ b + 3c + (2 3 − 1)d, Æ p b(c + d) + cd ≥ c + ( 3 − 1)d. Since b ≥ c, it suffices to show that Æ p c(c + d) + cd ≥ c + ( 3 − 1)d. By squaring, we get the obvious inequality d(c − d) ≥ 0. The equality holds for a = b a a and c = d = 0, for = b = c and d = 0, and for p = b = c = d. 4 3+2 2 Second Solution (by Vo Quoc Ba Can). Write the hypothesis as (a − b)2 + (c − d)2 ≥ 2(a + b)(c + d).

354

Vasile Cîrtoaje

Since a + b ≥ (a − b) + 2c, we get (a − b)2 + (c − d)2 ≥ 2[(a − b) + 2c](c + d), which is equivalent to (a − b)2 − 2(c + d)(a − b) − 3c 2 − 6cd + d 2 . From this, we get a− b ≥ c+d +2

p

c 2 + 2cd.

Thus, it suffices to show that p p c + d + 2 c 2 + 2cd ≥ 3c + (2 3 − 1)d, that is, p

p c 2 + 2cd ≥ c + ( 3 − 1)d.

By squaring, we get the obvious inequality d(c − d) ≥ 0.

P 2.92. If a ≥ b ≥ c ≥ d ≥ 0, then a+ b+c+d −4

p 4

a bcd ≥

p p 3 p ( b − 2 c + d)2 . 2 (Vasile Cîrtoaje, 2010)

Solution. First, we show that a−4

p 4

a bcd ≥ b − 4

Write this inequality as a−b≥4

p 4

p 4

b2 cd.

p p 4 4 bcd( a − b)

and prove that the following sharper inequality holds a−b≥4

p 4

p p 4 4 b3 ( a − b).

Indeed, a− b−4

p 4

p p p p p p p p 4 4 4 4 4 4 4 4 b3 ( a − b) = ( a − b)( a3 + a2 b + a b2 − 3 b3 ) ≥ 0.

Noncyclic Inequalities

355

Thus, it is enough to show that 2b + c + d − 4

p 4

b2 cd ≥

p p 3 p ( b − 2 c + d)2 , 2

which is equivalent to yz ≥

3 (x − 2 y + z)2 , 2

yz)2 + ( y − z)2 ≥

3 (x − 2 y + z)2 , 2

2x 2 + y 2 + z 2 − 4x 2(x −

p

p

where x ≥ y ≥ z ≥ 0. Since x−

p

yz ≥ x −

y +z ≥ 0, 2

it suffices to show that (2x − y − z)2 + 2( y − z)2 ≥ 3(x − 2 y + z)2 , which is equivalent to the obvious inequality (x − y)(x + 9 y − 10z) ≥ 0. The equality holds for a = b = c = d.

P 2.93. If a ≥ b ≥ c ≥ d ≥ 0, then (a) (b) (c) (d) (e) (f)

p p p 2 p 4 a + b + c + d − 4 a bcd ≥ (3 b − 2 c − d)2 ; 9 p p p 1 p 4 a + b + c + d − 4 a bcd ≥ (3 b − c − 2 d)2 ; 5 p p p 3 p 4 a + b + c + d − 4 a bcd ≥ ( b − 3 c + 2 d)2 ; 8 p p p 1 p 4 a + b + c + d − 4 a bcd ≥ (2 b − 3 c + d)2 ; 2 p p p p 1 4 a + b + c + d − 4 a bcd ≥ (2 b + c − 3 d)2 ; 6 p p 4 p 4 a + b + c + d − 4 a bcd ≥ ( b − d)2 . 3 (Vasile Cîrtoaje, 2010)

356

Vasile Cîrtoaje

Solution. (a) First Solution. Let p p x = b − d, Since a+ b+c+d −4

p 4

y=

p

b−

p

c,

x ≥ y ≥ 0.

p p p p a bcd ≥ ( a − d)2 + ( b − c)2 ≥ x 2 + y 2 ,

it suffices to show that

2 (x + 2 y)2 . 9

x2 + y2 ≥ Indeed,

9(x 2 + y 2 ) − 2(x + 2 y)2 = (x − y)(7x − y) ≥ 0. The equality holds for a = b = c = d, and also for a = b and c = d. Second Solution. We show that a−4

p 4

a bcd ≥ b − 4

Write this inequality as a−b≥4

p 4

p 4

b2 cd.

p p 4 4 bcd( a − b),

and prove that the following sharper inequality holds a−b≥4

p 4

p p 4 4 b3 ( a − b).

Indeed, a− b−4

p 4

p p p p p p p p 4 4 4 4 4 4 4 4 b3 ( a − b) = ( a − b)( a3 + a2 b + a b2 − 3 b3 ) ≥ 0.

Thus, it is enough to show that 2b + c + d − 4

p 4

b2 cd ≥

p p 2 p (3 b − 2 b − c)2 , 9

which is equivalent to 2x 2 + y 2 + z 2 − 4x where x ≥ y ≥ z ≥ 0. Since

p

yz ≥

2 (3x − 2 y − z)2 , 9

p 2 x y ≤ x + y,

it suffices to show that 2x 2 + y 2 + z 2 − 2x( y + z) ≥

2 (3x − 2 y − z)2 , 9

Noncyclic Inequalities

357

which is equivalent to the obvious inequality ( y − z)(6x + y − 7z) ≥ 0. Third Solution. Let A=

p

b−

p

B=

c,

p

c−

p

d,

x ≥ y ≥ 0.

We have a+ b+c+d −4

p 4

p p p p p p 4 4 a bcd = ( a − c)2 + ( b − d)2 + 2( ac − bd)2 p p p p p p p 4 4 ≥ ( b − c)2 + ( b − d)2 + 2 b( c − d)2 p 2 bB 2 2 2 = A + (A + B) + p p 4 ( 4 c + d)2 1 3 ≥ A2 + (A + B)2 + B 2 = 2A2 + 2AB + B 2 . 2 2

Therefore, it suffices to show that 3 2 2A2 + 2AB + B 2 ≥ (3A + B)2 . 2 9 This inequality reduces to  ‹ 23 B 6A + B ≥ 0. 2 (b) First Solution. Let p p x = b − d, Since a+ b+c+d −4

p 4

y=

p

b−

p

c,

x ≥ y ≥ 0.

p p p p a bcd ≥ ( a − d)2 + ( b − c)2 ≥ x 2 + y 2 ,

it suffices to show that x2 + y2 ≥

1 (2x + y)2 . 5

Indeed, 5(x 2 + y 2 ) − (2x + y)2 = (x − 2 y)2 ≥ 0. The equality holds for a = b = c = d. Second Solution. As it is shown at the second solution of (a), it suffices to prove that 2b + c + d − 4

p 4

b2 cd ≥

p p 1 p (3 b − b − 2 c)2 , 5

358

Vasile Cîrtoaje

which is equivalent to 2x 2 + y 2 + z 2 − 4x where x ≥ y ≥ z ≥ 0. Since

p

yz ≥

1 (3x − 2 y − z)2 , 5

p 2 x y ≤ x + y,

it suffices to show that 2x 2 + y 2 + z 2 − 2x( y + z) ≥

1 (3x − 2 y − z)2 , 5

which is equivalent to the obvious inequality (x − 2 y + z)2 ≥ 0. (c) As it is shown at the second solution of (a), we have p p 4 4 a + b + c + d − 4 a bcd ≥ 2b + c + d − 4 b2 cd. Since 2b + c + d − 4

p 4

p p p p 4 b2 cd = 2( b − cd)2 + ( c − d)2  p 2 p p p p c+ d + ( c − d)2 ≥2 b− 2 1 = (2A + B)2 + B 2 , 2

where A=

p

b−

p

c,

B=

p

c−

p

d,

it suffices to prove that 1 3 (2A + B)2 + B 2 ≥ (A − 2B)2 , 2 8 which is equivalent to the obvious inequality A(13A + 28B) ≥ 0. The equality holds for a = b = c = d. (d) First Solution. Let p p x = b − d, Since a+ b+c+d −4

p 4

y=

p

b−

p

c,

x ≥ y ≥ 0.

p p p p a bcd ≥ ( a − d)2 + ( b − c)2 ≥ x 2 + y 2 ,

Noncyclic Inequalities

359

it suffices to show that x2 + y2 ≥

1 (x − 3 y)2 . 2

Indeed, 2(x 2 + y 2 ) − (x − 3 y)2 = (x − y)(x + 7 y) ≥ 0. The equality holds for a = b = c = d, and also for a = b and c = d. Second Solution. As it is shown at the second solution of (a), it suffices to prove that 2b + c + d − 4

p 4

b2 cd ≥

p p 1 p (2 b − 3 c + d)2 , 2

which is equivalent to 2x 2 + y 2 + z 2 − 4x where x ≥ y ≥ z ≥ 0. Since

p

yz ≥

1 (2x − 3 y + z)2 , 2

p 2 x y ≤ x + y,

it suffices to show that 2x 2 + y 2 + z 2 − 2x( y + z) ≥

1 (2x − 3 y + z)2 . 2

We can write this inequality in the obvious form ( y − z)(8x − 7 y − z) ≥ 0. (e) As it is shown at (c), we have a+ b+c+d −4 where A=

p

p 4

b−

p

a bcd ≥

c,

B=

1 (2A + B)2 + B 2 , 2 p

c−

p

d.

Thus, it suffices to show that 1 1 (2A + B)2 + B 2 ≥ (2A + 3B)2 . 2 6 This inequality is equivalent to A2 ≥ 0. The equality holds for a = b = c = d. (f) As it is shown at (c), we have a+ b+c+d −4 where A=

p

p 4

b−

p

a bcd ≥

c,

B=

1 (2A + B)2 + B 2 , 2 p

c−

p

d.

360

Vasile Cîrtoaje

Thus, it suffices to show that 4 1 (2A + B)2 + B 2 ≥ (A + B)2 . 2 3 This inequality is equivalent to (2A − B)2 ≥ 0. The equality holds for a = b = c = d.

P 2.94. If a, b, c, d, e are real numbers, then p a b + bc + cd + d e 3 ≤ . 2 2 2 2 2 a +b +c +d +e 2 Solution. Using the AM-GM inequality, we have ‹  ‹  ‹  ‹  2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 a +b +c +d +e = a + b + b + c + c + d + d +e 3 3 2 2 3 3 v v v v t t2 t1 t1 1 1 2 ≥ 2 a2 · b2 + 2 b2 · c 2 + 2 c2 · d 2 + 2 d 2 · e2 3 3 2 2 3 3 2 ≥ p (a b + bc + cd + d a). 3 The equality holds for b c d a = p = = p = e. 3 2 3 Remark. The following more general inequality holds a1 a2 + a2 a3 + · · · + an−1 an a12 with equality for

+ · · · + an2

≤ cos

π , n+1

an a1 a2 = ··· = π = nπ . 2π sin n+1 sin n+1 sin n+1

Denoting ci = we have

+ a22

(i+1)π n+1 iπ 2 sin n+1

sin

,

i = 1, 2, · · · , n − 1,

1 π + ci+1 = cos 4ci n+1

Noncyclic Inequalities

361

and (a12 + a22 + · · · + an2 ) cos

π = n+1 ‹ 2 + cn−1 an−1 +

‹  1 1 1 2 = + + c2 a2 + · · · + an2 4c1 4cn−2 4cn−1 ‹  ‹  ‹  1 2 1 1 2 2 2 2 2 a + c2 a2 + a + · · · + cn−1 an−1 + a = c1 a1 + 4c1 2 4c2 3 4cn−1 n v v v t t t 1 1 1 2 ≥ 2 c1 a12 · a22 + 2 c2 a22 · a32 + · · · + 2 cn−1 an−1 · a2 4c1 4c2 4cn−1 n c1 a12



≥ a1 a2 + a2 a3 + · · · + an−1 an .

P 2.95. If a, b, c, d, e, f are nonnegative real numbers such that a ≥ b ≥ c ≥ d ≥ e ≥ f, then (a + b + c + d + e + f )2 ≥ 8(ac + bd + ce + d f ). (Vasile Cîrtoaje, 2005) First Solution. Let us denote x = b + c + d + e + f , and write the inequality as follows: (a + x)2 − 8(ac + bd + ce + d f ) ≥ 0, (a + x − 4c)2 + 8(a + x)c − 16c 2 − 8(ac + bd + ce + d f ) ≥ 0, (a + x − 4c)2 − 8[c 2 − (b + d + f )c + d(b + f )] ≥ 0, (a + x − 4c)2 − 8(c − d)(c − b − f ) ≥ 0, (a + x − 4c)2 + 8(c − d)(b − c + f ) ≥ 0. The last inequality is clearly true. The equality holds for c = d = (a + b + e + f )/2, and for c = b + f = (a + d + e)/2; that is, for a = b = c = d and e = f = 0, and for b = c = (a + d + e)/2 and f = 0.

P 2.96. If a1 ≥ a2 ≥ · · · ≥ a8 ≥ 0, then p p p a1 + a2 + · · · + a8 − 8 8 a1 a2 · · · a8 ≥ 3( a6 − a7 )2 .

362

Vasile Cîrtoaje

Solution. Let us denote x=

p 6

a1 a2 · · · a6 , y =

p

x ≥ a6 ≥ a7 ≥ y.

a7 a8 ,

By the AM-GM inequality, we have a1 + a2 + · · · + a6 ≥ 6x, Also, we have p

a6 −

p

a7 ≤

p

a7 + a8 ≥ 2 y.

x−

p

y.

Thus, it suffices to show that 6x + 2 y − 8

Æ 8

p p x 6 y 2 ≥ 3( x − y)2 .

For the nontrivial case y > 0, we can set y = 1 (due to homogeneity). In addition, setting x = t 4 , t ≥ 1, the inequality can be restated as 6t 4 + 2 − 8t 3 ≥ 3(t 2 − 1)2 , which is equivalent to (t − 1)3 (3t + 1) ≥ 0. The equality holds for a1 = a2 = · · · = a8 .

P 2.97. Let a1 , a2 , . . . , an and b1 , b2 , · · · , bn be real numbers. Prove that a1 b1 + · · · + an bn +

q

(a12 + · · · + an2 )(b12 + · · · + bn2 ) ≥

2 (a1 + · · · + an )(b1 + · · · + bn ). n (Vasile Cîrtoaje, 1989)

First Solution. Write the inequality as q where b =

(a12 + · · · + an2 )(b12 + · · · + bn2 ) ≥ a1 (2b − b1 ) + · · · + an (2b − bn ),

1 (b1 + · · · + bn ). Using the substitution n x i = 2b − bi ,

we have

n X i=1

x i = 2nb −

n X i=1

i = 1, 2, · · · , n,

bi = 2nb − nb = nb,

Noncyclic Inequalities n X

bi2 =

363 n n n n X X X X (2b − x i )2 = 4nb2 − 4b xi + x i2 = x i2 .

i=1

i=1

i=1

i=1

i=1

Therefore, the desired inequality can be restated as q (a12 + · · · + an2 )(x 12 + · · · + x n2 ) ≥ a1 x 1 + · · · + an x n , which is just the Cauchy-Schwarz inequality. In the case a1 a2 · · · an 6= 0, the equality holds for 2b − bn 2b − b1 2b − b2 = = ··· = ≥ 0. a1 a2 an Second Solution. Consider the nontrivial case where a12 +· · ·+an2 > 0 and b12 +· · ·+ bn2 > 0, denote v u 2 t b1 + · · · + bn2 p= , a12 + · · · + an2 and use the substitution bi = p x i for all i to have a12 + · · · + an2 = x 12 + · · · + x n2 . The desired inequality becomes (a1 x 1 + · · · + an x n ) + (a12 + · · · + an2 ) ≥ (a1 + x 1 )2 + · · · + (an + x n )2 ≥

2 (a1 + · · · + an )(x 1 + · · · + x n ), n

4 (a1 + · · · + an )(x 1 + · · · + x n ). n

Since 4(a1 + · · · + an )(x 1 + · · · + x n ) ≤ [(a1 + · · · + an ) + (x 1 + · · · + x n )]2 , it suffices to show that (a1 + x 1 )2 + · · · + (an + x n )2 ≥

1 [(a1 + x 1 ) + · · · + (an + x n )]2 . n

This follows immediately from the Cauchy-Schwarz inequality. Remark. Substituting bi = 1/ai for all i, we get the following inequality v    ‹ u 1 1 1 1 t 2 2 2 n + n (a1 + · · · + an ) 2 + · · · + 2 ≥ 2(a1 + · · · + an ) + ··· + . an a1 an a1 For a1 ≤ a2 ≤ · · · ≤ an and even n = 2k, the equality holds if a1 = a2 = · · · = ak ,

ak+1 = ak+2 = · · · = a2k .

364

Vasile Cîrtoaje

For odd n, the equality holds only if a1 = a2 = · · · = an . Conjecture. If a1 , a2 , . . . , an are positive real numbers and n is odd, then v   u 1 1 t 2 2 2 2 n + 1 + (n − 1)(a1 + · · · + an ) 2 + · · · + 2 − n2 + 1 ≥ an a1 ‹ 1 1 ≥ 2(a1 + · · · + an ) + ··· + . a1 an 

For a1 ≤ a2 ≤ · · · ≤ an and n = 2k + 1, the equality holds when a1 = a2 = · · · = ak ,

ak+1 = ak+2 = · · · = a2k+1

or a1 = a2 = · · · = ak+1 ,

ak+2 = ak+3 = · · · = a2k+1 .

P 2.98. Let a1 , a2 , . . . , an be positive real numbers such that a1 ≥ 2a2 . Prove that (5n − 1)(a12 + a22 + · · · + an2 ) ≥ 5(a1 + a2 + · · · + an )2 . (Vasile Cîrtoaje, 2009) Solution. Let a1 = ka2 , k ≥ 2. By the Cauchy-Schwarz inequality, we have a12 + a22 + · · · + an2 = (k2 + 1)a22 + a32 + · · · + an2 ≥

[(k + 1)a2 + a3 + · · · + an ]2

(k + 1)2 +n−2 k2 + 1 Therefore, it suffices to show that

=

(a1 + a2 + · · · + an )2 . 2k +n−1 k2 + 1

5n − 1 2k ≥ 2 + n − 1. 5 k +1 This inequality is equivalent to (k − 2)(2k − 1) ≥ 0, which is obviously true for k ≥ 2. The equality holds if and only if k = 2 and 5a22 + a32 + · · · + an2 =

(3a2 + a3 + · · · + an )2 9 5

+n−2

that is, if and only if 5a1 5a = 2 = a3 = · · · = an . 6 3

;

Noncyclic Inequalities

365

P 2.99. If a1 ≥ a2 ≥ · · · ≥ an > 0 such that a1 + a2 + · · · + an = n, then 1 1 4(n − 1)2 1 + + ··· + −n≥ (a1 − a2 )2 . a1 a2 an n3 (Vasile Cîrtoaje, 2009) Solution. Since 1 1 1 (n − 1)2 (n − 1)2 + + ··· + ≥ = a2 a3 an a2 + a3 + · · · + an n − a1 and a1 − a2 ≤ a1 − it suffices to show that

a2 + a3 + · · · + an n − a1 n(a1 − 1) = a1 − = , n−1 n−1 n−1

1 (n − 1)2 4 + − n ≥ (a1 − 1)2 . a1 n − a1 n

This is equivalent to the obvious inequality (a − 1)2 (2a − n)2 ≥ 0. n The equality holds if a1 = a2 = · · · = an = 1, and also for a1 = and a2 = a3 = · · · = 2 n . an = 2(n − 1)

P 2.100. If a1 ≥ a2 ≥ · · · ≥ an ≥ 0, then (a) (b)

1 p p p p a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ ( a1 + a2 − 2 an )2 ; 3 1 p p p p a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ (2 a1 − an−1 − an )2 . 4 (Vasile Cîrtoaje, 2010)

p p Solution. (a) For n = 2, the inequality is equivalent to ( a1 − a2 )2 ≥ 0. Consider further n ≥ 3. By the AM-GM inequality, we have p p a3 + · · · + an−1 + 3 3 a1 a2 an ≥ n n a1 a2 · · · an . Therefore, it suffices to prove that p p 1 p p a1 + a2 + an − 3 3 a1 a2 an ≥ ( a1 + a2 − 2 an )2 . 3

366

Vasile Cîrtoaje

Setting x=

p

a1 +

p

a2

2

‹2

, x ≥ an ,

since a1 + a2 ≥ 2x and a1 a2 ≤ x 2 , it suffices to show that 2x + an − 3

Æ 3

x 2 an ≥

p 4 p ( x − an )2 . 3

For the nontrivial case an > 0, we may consider an = 1 (due to homogeneity). In addition, substituting x = y 6 , y ≥ 1, the inequality can be restated as 2 y6 + 1 − 3 y4 ≥

4 3 ( y − 1)2 , 3

( y − 1)2 [3( y + 1)2 (2 y 2 + 1) − 4( y 2 + y + 1)2 ] ≥ 0. This inequality is true if Æ ( y + 1) 3(2 y 2 + 1) ≥ 2( y 2 + y + 1). Since Æ

3(2 y 2 + 1) ≥ 2 y + 1,

we have Æ ( y + 1) 3(2 y 2 + 1) − 2( y 2 + y + 1) ≥ ( y + 1)(2 y + 1) − 2( y 2 + y + 1) = y − 1 ≥ 0. This complete the proof. The equality holds if a1 = a2 = · · · = an . p p (b) For n = 2, the inequality is equivalent to ( a1 − a2 )2 ≥ 0. Consider further n ≥ 3. By the AM-GM inequality, we have p p a2 + a3 + · · · + an−2 + 3 3 a1 an−1 an ≥ n n a1 a2 · · · an . Therefore, it suffices to prove that p p p 1 p a1 + an−1 + an − 3 3 a1 an−1 an ≥ (2 a1 − an−1 − an )2 . 4 Setting p x = an−1 an , x ≤ a1 , p p p since an−1 + an ≥ 2x and an−1 + an ≥ 2 x, it suffices to show that Æ p p 3 a1 + 2x − 3 a1 x 2 ≥ ( a1 − x)2 . Due to homogeneity, we may consider a1 = 1. In addition, substituting x = y 6 , y ≤ 1, the inequality becomes 1 + 2 y 6 − 3 y 4 ≥ (1 − y 3 )2 ,

Noncyclic Inequalities

367

which is equivalent to the obvious inequality y 3 ( y − 1)2 ( y + 2) ≥ 0. This complete the proof. The equality holds if a1 = a2 = · · · = an .

P 2.101. If a1 ≥ a2 ≥ · · · ≥ an ≥ 0, n ≥ 3, then p p p n−1 p ( an−2 + an−1 − 2 an )2 . a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ 2n (Vasile Cîrtoaje, 2010) Solution. Let us denote x=

a1 + a2 + · · · + an−1 , n−1

x ≥ an .

By the AM-GM inequality, we have a1 a2 · · · an−1 ≤ x n−1 . Also,

p

an−2 + 2

p

an−1

≤

v ta

n−2

v + an−1 t a1 + a2 + · · · + an−1 p ≤ = x. 2 n−1

Then, it suffices to show that (n − 1)x + an − n

Æ n

x n−1 an ≥

p 2(n − 1) p ( x − an )2 . n

For the nontrivial case an > 0, we may consider an = 1 (due to homogeneity). In addition, substituting x = t 2n , t ≥ 1, the inequality becomes g(t) ≥ 0, where g(t) = (n − 1)t 2n + 1 − nt 2n−2 −

2(n − 1) n (t − 1)2 . n

We have g 0 (t) = 2(n − 1)t n−1 h(t), where h(t) = n(t n − t n−2 ) − 2(t n − 1). Since h0 (t) = n(n − 2)t n−3 (t 2 − 1) ≥ 0, h(t) is increasing, h(t) ≥ h(1) = 0, g 0 (t) ≥ 0, g(t) is increasing, hence g(t) ≥ g(1) = 0. This completes the proof. The equality holds for a1 = a2 = · · · = an .

368

Vasile Cîrtoaje

P 2.102. If a1 ≥ a2 ≥ · · · ≥ an ≥ 0, n ≥ 3, then  ‹ p p p p 2 a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ 1 − (2 an−2 − an−1 − an )2 . n (Vasile Cîrtoaje, 2010) Solution. Let us denote x=

p

n−2

a1 a2 · · · an−2 ,

y=

p

an−1 an ,

x ≥ an−2 ≥ y.

By the AM-GM inequality, we have a1 + a2 + · · · + an−2 ≥ (n − 2)x, an−1 + an ≥ 2 y, and p

an−1 +

p

p an ≥ 2 y.

Then, it suffices to show that (n − 2)x + 2 y − n

Æ n

x n−2 y 2 ≥

4(n − 2) p p ( x − y)2 . n

For the nontrivial case y > 0, we may consider y = 1 (due to homogeneity). In addition, substituting x = t 2n , t ≥ 1, the inequality becomes g(t) ≥ 0, where g(t) = (n − 2)t 2n + 2 − nt 2n−4 −

4(n − 2) n (t − 1)2 . n

We have g 0 (t) = 2(n − 2)t n−1 h(t), where h(t) = (n − 4)t n − nt n−4 + 4. Since h0 (t) = n(n − 4)t n−5 (t 4 − 1) ≥ 0, h(t) is increasing, h(t) ≥ h(1) = 0, g 0 (t) ≥ 0, g(t) is increasing, hence g(t) ≥ g(1) = 0. This completes the proof. The equality holds for a1 = a2 = · · · = an . If n = 4, then the equality holds also for a1 = a2 and a3 = a4 .

Noncyclic Inequalities

369

P 2.103. Let a1 ≥ a2 ≥ · · · ≥ an ≥ 0. If n ≤ k ≤ n − 1, 2 then

p p 2k(n − k) p a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ ( ak − ak+1 )2 . n (Vasile Cîrtoaje, 2010)

Solution. Let us denote p p x = k a1 a2 · · · ak , y = n−k ak+1 ak+2 · · · an ,

x ≥ ak ≥ ak+1 ≥ y.

By the AM-GM inequality, we have a1 + a2 + · · · + ak ≥ k x, Also, we have

p

ak −

p

ak+1 + ak+2 + · · · + an ≥ (n − k) y. ak+1 ≤

p

x−

p

y.

Thus, it suffices to show that 2k(n − k) p p ( x − y)2 . n For the nontrivial case y > 0, we can set y = 1 (due to homogeneity). In addition, setting x = t 2n , t ≥ 1, the inequality becomes f (t) ≥ 0, where k x + (n − k) y − n

Æ n

x k y n−k ≥

f (t) = kt 2n + n − k − nt 2k −

2k(n − k) n (t − 1)2 . n

We have f 0 (t) = 2kt n−1 h(t), where h(t) = n(t n − t 2k−n ) − 2(n − k)(t n − 1). Since h0 (t) = n(2k − n)(t n−1 − t 2k−n−1 ) ≥ 0, h(t) is increasing for t ≥ 1, h(t) ≥ h(1) = 0, f 0 (t) ≥ 0, f (t) is increasing, f (t) ≥ f (1) = 0. This completes the proof. The equality holds for a1 = a2 = · · · = an . If n is even and 2k = n, then the equality holds for a1 = a2 = · · · = ak and ak+1 = ak+2 = · · · = an . Remark. A more general inequality is given by the following statement (Vasile Cîrtoaje, 2010). • Let a1 ≥ a2 ≥ · · · ≥ an ≥ 0. If 1 ≤ k ≤ j ≤ n and k + j ≥ n, then p 2k(n − j + 1) p p ( ak − a j )2 . a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ n+k− j+1

370

Vasile Cîrtoaje

P 2.104. If a1 ≥ a2 ≥ · · · ≥ an ≥ 0, n ≥ 4, then  ‹ 2 p p p p (2 an−2 − 3 an−1 + an )2 ; (a) a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ 1 − n  ‹ 1 1 p p p p n (b) a1 + a2 + · · · + an − n a1 a2 · · · an ≥ 1− ( an−2 − 3 an−1 + 2 an )2 . 2 n (Vasile Cîrtoaje, 2010) Solution. Let x=

p

an−2 −

p

an−1 ≥ 0,

y=

p

an−1 −

p

an ≥ 0.

For k = n−2 and k = n−1, the inequality in the preceding P 2.103 becomes respectively p 4(n − 2)x 2 a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ n and

p 2(n − 1) y 2 a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ . n

Therefore, p 2 a1 + a2 + · · · + an − n n a1 a2 · · · an ≥ max{2(n − 2)x 2 , (n − 1) y 2 }. n (a) It suffices to show that max{4(n − 2)x 2 , 2(n − 1) y 2 } ≥ (n − 2)(2x − y)2 . This is true since 4(n − 2)x 2 ≥ (n − 2)(2x − y)2 for 2x − y ≥ 0, and 2(n − 1) y 2 ≥ (n − 2) y 2 ≥ (n − 2)( y − 2x)2 for y − 2x ≥ 0. The equality holds for a1 = a2 = · · · = an . If n = 4, then the equality holds for a1 = a2 and a3 = a4 . (b) It suffices to show that max{8(n − 2)x 2 , 4(n − 1) y 2 } ≥ (n − 1)(x − 2 y)2 . This is true since 8(n − 2)x 2 ≥ (n − 1)x 2 ≥ (n − 1)(x − 2 y)2 for x − 2 y ≥ 0, and

4(n − 1) y 2 ≥ (n − 1)(2 y − x)2

for 2 y − x ≥ 0. The equality holds for a1 = a2 = · · · = an .

Noncyclic Inequalities

371

P 2.105. If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that a1 ≤ a2 ≤ · · · ≤ a n ,

a1 + a2 + · · · + an = 0,

then a12 + a22 + · · · + an2 + na1 an ≤ 0. (Vasile Cîrtoaje, 2009) Solution. Let a1 = a < 0 and an = b > 0 be fixed. We claim that for a ≤ a2 · · · ≤ an−1 ≤ b,

a2 + · · · + an−1 = −a − b,

2 the sum S = a22 +· · ·+ an−1 is maximal when at most one of a2 , . . . , an−1 is different from a or b. Indeed, if a < ai ≤ a j < b, then ai2 + a2j < ci2 + c 2j for all ci and c j such that

a ≤ ci < ai ≤ a j < c j ≤ b,

ci + c j = ai + a j ,

because ai2 + a2j − ci2 − c 2j = (ai − ci )(ai + ci ) + (a j − c j )(a j + c j ) = (ai − ci )(ai + ci − a j − c j ) < 0. Therefore, it suffices to consider the case where n − 3 of the numbers a2 , . . . , an−1 are equal to a or b. More precisely, assume that k of a2 , . . . , an−1 are equal to a and m of a2 , . . . , an−1 are equal to b, where k + m = n − 3, k ≥ 0, m ≥ 0. Therefore, it suffices to show that (k + 1)a2 + c 2 + (m + 1)b2 + (k + m + 3)a b ≤ 0, where a ≤ c ≤ b,

(k + 1)a + c + (m + 1)b = 0.

We have (k + 1)a2 + c 2 + (m + 1)b2 + (k + m + 3)a b = c 2 + (a + b)[(k + 1)a + (m + 1)b] + a b = c 2 − (a + b)c + a b = (c − a)(c − b) ≤ 0. The equality holds if and only if a2 , . . . , an−1 ∈ {a1 , an }.

P 2.106. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = n. (a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then a13 + a23 + · · · + an3 + 2n ≥ 3(a12 + a22 + · · · + an2 ); (b) If a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then a13 + a23 + · · · + an3 + 2n ≤ 3(a12 + a22 + · · · + an2 ). (Vasile Cîrtoaje, 2007)

372

Vasile Cîrtoaje

Solution. (a) Write the inequality as X (a13 − 3a12 + 3a1 − 1) ≥ 0, X (a1 − 1)3 ≥ 0, (a1 − 1)3 ≥ (1 − a2 )3 + · · · + (1 − an )3 , [(1 − a2 ) + · · · + (1 − an )]3 ≥ (1 − a2 )3 + · · · + (1 − an )3 . Clearly, the last form of the desired inequality is true. The equality holds for a1 = a2 = · · · = an = 1, and also for a1 = 2, a2 = · · · = an−1 = 1, an = 0. (b) Similarly, write the inequality as X (a13 − 3a12 + 3a1 − 1) ≤ 0, X (1 − a1 )3 ≥ 0, (1 − a1 )3 ≥ (a2 − 1)3 + · · · + (an − 1)3 , [(a2 − 1) + · · · + (an − 1)]3 ≥ (a2 − 1)3 + · · · + (an − 1)3 . The last form of the desired inequality is obvious. The equality holds for a1 = a2 = · · · = an = 1, and also for a1 = 0, a2 = · · · = an−1 = 1, an = 2.

P 2.107. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. (a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then a14 + a24 + · · · + an4 + 5n ≥ 6(a12 + a22 + · · · + an2 ); (b) If a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then a14 + a24 + · · · + an4 + 6n ≤ 7(a12 + a22 + · · · + an2 ). (Vasile Cîrtoaje, 2007) Solution. (a) Write the inequality as X (a14 − 6a12 + 8a1 − 3) ≥ 0, X (a1 − 1)3 (a1 + 3) ≥ 0, (a1 − 1)3 (a1 + 3) ≥ (1 − a2 )3 (a2 + 3) + · · · + (1 − an )3 (an + 3).

Noncyclic Inequalities

373

Since (a1 − 1)3 = [(1 − a2 ) + · · · + (1 − an )]3 ≥ (1 − a2 )3 + · · · + (1 − an )3 , it suffices to show that [(1 − a2 )3 + · · · + (1 − an )3 ](a1 + 3) ≥ (1 − a2 )3 (a2 + 3) + · · · + (1 − an )3 (an + 3), which is equivalent to the obvious inequality (1 − a2 )3 (a1 − a2 ) + · · · + (1 − an )3 (a1 − an ) ≥ 0. The equality holds for a1 = a2 = · · · = an = 1. (b) Write the inequality as X (a14 − 7a12 + 10a1 − 4) ≤ 0, X (a1 − 1)2 (a12 + 2a1 − 4) ≤ 0, (a2 − 1)2 (a22 + 2a2 − 4) + · · · + (an − 1)2 (an2 + 2an − 4) ≤ (1 − a1 )2 (4 − 2a1 − a12 ). Since (1 − a1 )2 = [(a2 − 1) + · · · + (an − 1)]2 ≥ (a2 − 1)2 + · · · + (an − 1)2 , it suffices to show that (a2 −1)2 (a22 +2a2 −4)+· · ·+(an −1)2 (an2 +2an −4) ≤ [(a2 −1)2 +· · ·+(an −1)2 ](4−2a1 −a12 ), which is equivalent to (a2 − 1)2 (a12 + a22 + 2a1 + 2a2 − 8) + · · · + (an − 1)2 (a12 + an2 + 2a1 + 2an − 8) ≤ 0. This inequality is true if a12 + an2 + 2a1 + 2an − 8 ≤ 0. Since a1 + an = n − (a2 + · · · + an−1 ) = 2 + (1 − a2 ) + · · · + (1 − an ) ≤ 2, we have a12 + an2 + 2a1 + 2an − 8 = (a1 + an + 1)2 − 9 − 2a1 an ≤ (a1 + an + 1)2 − 9 ≤ 0. The equality holds for a1 = a2 = · · · = an = 1, and also for a1 = 0, a2 = · · · = an−1 = 1, an = 2.

374

Vasile Cîrtoaje

P 2.108. If a1 , a2 , . . . , an are positive real numbers such that 1 1 1 + + ··· + = n, a1 a2 an

a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

a12 + a22 + · · · + an2 + 2n ≥ 3(a1 + a2 + · · · + an ). (Vasile Cîrtoaje, 2008) Solution. Write the inequality as follows: (a1 − 1)(a1 − 2) + (a2 − 1)(a2 − 2) + · · · + (an − 1)(an − 2) ≥ 0, (a1 − 1)(a1 − 2) ≥ (1 − a2 )(a2 − 2) + · · · + (1 − an )(an − 2),   ‹  ‹ ‹ 1 1 1 2 2 1− (a1 − 2a1 ) ≥ − 1 (a2 − 2a2 ) + · · · + − 1 (an2 − 2an ), a1 a2 an • ‹  ‹˜  ‹  ‹ 1 1 1 1 2 2 − 1 + ··· + − 1 (a1 −2a1 ) ≥ − 1 (a2 −2a2 )+· · ·+ − 1 (an2 −2an ), a2 an a2 an   ‹ ‹ 1 1 2 2 − 1 (a1 − 2a1 − a2 + 2a2 ) + · · · + − 1 (a12 − 2a1 − an2 + 2an ) ≥ 0, a2 an  ‹  ‹ 1 1 − 1 (a1 − a2 )(a1 + a2 − 2) + · · · + − 1 (a1 − an )(a1 + an − 2) ≥ 0. a2 an Clearly, it suffices to prove that a1 + an − 2 ≥ 0. Indeed, a1 + an − 2 = n − 2 − (a2 + · · · + an−1 ) = (1 − a2 ) + · · · + (1 − an−1 ) ≥ 0. The equality holds for a1 = a2 = · · · = an = 1.

P 2.109. If a1 , a2 , . . . , an are real numbers such that a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,

a1 + a2 + · · · + an = n,

then (a)

(b)

a1 + 1 a12

+1 1

a12

+3

+ +

a2 + 1 a22

+1 1

a22

+3

+ ··· + + ··· +

an + 1 ≤ n; an2 + 1 1 a12

+3

≤

n . 4 (Vasile Cîrtoaje, 2009)

Noncyclic Inequalities

375

Solution. (a) Write the inequality as       an + 1 a1 + 1 a2 + 1 + 1− 2 + ··· + 1 − 2 ≥ 0, 1− 2 an + 1 a1 + 1 a2 + 1 a1 (a1 − 1) a12

+1

+

a2 (a2 − 1) a22

+1

a2 (a2 − 1) a22

+1

+ ··· +

+ ··· +

an (an − 1) ≥ 0, an2 + 1

an (an − 1) a1 (1 − a1 ) ≥ , 2 an + 1 a12 + 1

an (an − 1) a1 [(a2 − 1) + · · · + (an − 1)] ≥ , an2 + 1 +1 a12 + 1     an a1 a1 a2 (a2 − 1) 2 − + · · · + (an − 1) 2 − ≥ 0, an + 1 a12 + 1 a2 + 1 a12 + 1 a2 (a2 − 1) a22

+ ··· +

(a2 − 1)(a2 − a1 )(1 − a1 a2 ) a22

+1

+ ··· +

(an − 1)(an − a1 )(1 − a1 an ) ≥ 0. an2 + 1

For a1 ≥ 0, it suffices to show that 1 − a1 an ≥ 0. Indeed, p 2 a1 an ≤ a1 + an = 2 + (1 − a2 ) + · · · + (1 − an−1 ) ≤ 2. For a1 ≤ 0, the inequality is also true because 1 − a1 a2 > 0, · · · , 1 − a1 an > 0. The equality holds for a1 = a2 = · · · = an = 1. (b) As in case (a), we write the inequality as (a2 − 1)(a2 − a1 )(3 − a1 a2 − a1 − a2 ) a22

+3

+···+

(an − 1)(an − a1 )(3 − a1 an − a1 − an ) a22 + 3

≥ 0.

For a1 ≥ 0, it suffices to show that 3 − a1 an − a1 − an ≥ 0. From (1 − a1 )(an − 1) ≥ 0, we get 3 − a1 an ≥ 4 − a1 − an , hence 1 (3 − a1 an − a1 − an ) ≥ 2 − a1 − an = (a2 − 1) + · · · + (an−1 − 1) ≥ 0. 2 For a1 ≤ 0, the inequality is also true because 3 − a1 a2 − a1 − a2 > 2 − a1 − a2 = (a3 − 1) + · · · + (an − 1) ≥ 0, ··· 3 − a1 an − a1 − an > 2 − a1 − an = (a2 − 1) + · · · + (an−1 − 1) ≥ 0. The equality holds for a1 = a2 = · · · = an = 1.

376

Vasile Cîrtoaje

P 2.110. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n,

a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

a12 − 1 (a1 + 3)2

a22 − 1

+

(a2 + 3)2

+ ··· +

an2 − 1 (an + 3)2

≥ 0. (Vasile Cirtoaje, 2009)

Solution. Write the inequality as follows: a22 − 1 (a2 + 3)2 a22 − 1

+ ··· +

an2 − 1

an2 − 1 (an + 3)2

≥

1 − a12 (a1 + 3)2

,

[(a2 − 1) + · · · + (an − 1)](1 + a1 ) , (a2 (an (a1 + 3)2 • ˜ • ˜ an + 1 a1 + 1 a1 + 1 a2 + 1 (a2 − 1) − + · · · + (an − 1) − ≥ 0, (a2 + 3)2 (a1 + 3)2 (an + 3)2 (a1 + 3)2 + 3)2

+ ··· +

+ 3)2

≥

(an − 1)(an − a1 )(3 − a1 − an − a1 an ) (a2 − 1)(a2 − a1 )(3 − a1 − a2 − a1 a2 ) + · · · + ≥ 0. (a1 + 3)2 (a2 + 3)2 (a1 + 3)2 (an + 3)2 It suffices to show that 3 − a1 − an − a1 an ≥ 0. Since 1 1 3 − a1 − an − a1 an ≥ 3 − a1 − an − (a1 + an )2 = (2 − a1 − an )(6 + a1 + an ) ≥ 0, 4 4 we only need to show that a1 + an ≤ 2. Indeed, we have a1 + an = 2 + (1 − a2 ) + · · · + (1 − an−1 ) ≤ 2. The equality holds for a1 = a2 = · · · = an = 1.

P 2.111. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

1 3a13

+4

+

1 3a23

+4

a1 + a2 + · · · + an = n,

+ ··· +

1 n ≥ . 3an3 + 4 7 (Vasile Cirtoaje, 2009)

Noncyclic Inequalities

377

Solution. Write the inequality as follows:     1 1 1 1 1 1 − − , + ··· + ≥ − 3 3 3 3an + 4 7 7 3a1 + 4 3a2 + 4 7 1 − a23 3a23 + 4 1 − a23

+ ··· +

1 − an3

+ ··· + ≥

1 − an3 3an3 + 4

≥

a13 − 1 3a13 + 4

,

[(1 − a2 ) + · · · + (1 − an )](1 + a1 + a12 )

, 3a13 + 4     1 + an + an2 1 + a1 + a12 1 + a2 + a22 1 + a1 + a12 (1 − a2 ) − + · · · + (1 − an ) ≥ 0. − 3an3 + 4 3a23 + 4 3a13 + 4 3a13 + 4 3a23 + 4

It suffices to show that

3an3 + 4

1 + ai + ai2 3ai3 + 4

−

1 + a1 + a12 3a13 + 4

≥0

for i = 2, . . . , n. Write these inequalities as (a1 − ai )Ei ≥ 0, where Ei = 3a12 ai2 + 3a1 ai (a1 + ai ) + 3(a12 + a1 ai + ai2 ) − 4(a1 + ai ) − 4 = (a1 + ai )(3a1 + 3ai − 4 + 3a1 ai ) + 3a12 ai2 − 3a1 ai − 4. Since a1 + ai ≥ a1 + an = 2 + (1 − a2 ) + · · · + (1 − an−1 ) ≥ 2, we have Ei ≥ 2(6 − 4 + 3a1 ai ) + 3a12 ai2 − 3a1 ai − 4 = 3a1 ai + 3a12 ai2 ≥ 0. The equality holds for a1 = a2 = · · · = an = 1, and also for a1 = 2, a2 = · · · = an−1 = 1, an = 0.

P 2.112. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

a1 + a2 + · · · + an = n,

v v v t 3a1 t 3a2 t 3an + + ··· + ≤ n. 4 − a1 4 − a2 4 − an (Vasile Cirtoaje, 2009)

378

Vasile Cîrtoaje

Solution. Write the inequality as follows: v  v  v  t 3a1 t 3a2 t 3an −1 + − 1 + ··· + − 1 ≤ 0. 4 − a1 4 − a2 4 − an an − 1 a1 − 1 a2 − 1 + + ··· + ≤ 0, p p p 4 − an + 3an (4 − an ) 4 − a1 + 3a1 (4 − a1 ) 4 − a2 + 3a2 (4 − a2 ) an − 1 (a2 − 1) + · · · + (an − 1) a2 − 1 , + ··· + ≤ p p p 4 − an + 3an (4 − an ) 4 − a1 + 3a1 (4 − a1 ) 4 − a2 + 3a2 (4 − a2 ) (a2 − 1)E2 + · · · + (an − 1)En ≥ 0, where Ej =

1 1 − , p Æ 4 − a1 + 3a1 (4 − a1 ) 4 − a j + 3a j (4 − a j )

j = 2, . . . , n.

It suffices to show that all E j ≥ 0. The inequality E j ≥ 0 is equivalent to q

Æ

3a j (4 − a j ) −

Æ

3a1 (4 − a1 ) ≥ a j − a1 ,

3(a j − a1 )(4 − a1 − a j ) ≥ a j − a1 . p 3a j (4 − a j ) + 3a1 (4 − a1 )

This is true if Æ

3a1 (4 − a1 ) +

q

3a j (4 − a j ) ≤ 3(4 − a1 − a j ).

Denote x = a1 + a j . Since Æ

3a1 (4 − a1 ) +

q

3a j (4 − a j ) ≤

q

2[3a1 (4 − a1 ) + 3a j (4 − a j )] ≤

p

24x − 3x 2 ,

it suffices to show that p

24x − 3x 2 ≤ 3(4 − x),

which is equivalent to (2 − x)(6 − x) ≥ 0. To end the proof, we need to show that x ≤ 2. Indeed x − 2 = a1 + ai − 2 ≤ a1 + an − 2 = (1 − a2 ) + · · · + (1 − an−1 ) ≤ 0. The equality holds for a1 = a2 = · · · = an = 1.

Noncyclic Inequalities

379

P 2.113. If a1 , a2 , . . . , an are nonnegative real numbers such that a12 + a22 + · · · + an2 = n,

a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

1 1 n 1 + + ··· + ≤ . 3 − a1 3 − a2 3 − an 2 (Vasile Cirtoaje, 2009)

Solution. Write the inequality as follows: 

‹  ‹  ‹ 2 2 2 −1 + − 1 + ··· + − 1 ≤ 0. 3 − a1 3 − a1 3 − a1 an − 1 a1 − 1 a2 − 1 + + ··· + ≤ 0, 3 − a1 3 − a2 3 − an an − 1 1 − a1 a2 − 1 + ··· + ≤ , 3 − a2 3 − an 3 − a1 a22 − 1

(1 + a2 )(3 − a2 ) a22 − 1 (1 + a2 )(3 − a2 )

+ ··· +

+ ··· +

an2 − 1 (1 + an )(3 − an ) an2 − 1

(1 + an )(3 − an )

≤

≤

1 − a12 (1 + a1 )(3 − a1 )

(a22 − 1) + · · · + (an2 − 1) (1 + a1 )(3 − a1 )

(a22 − 1)E2 + · · · + (an2 − 1)En ≤ 0, where Ej =

1 1 − , (1 + a j )(3 − a j ) (1 + a1 )(3 − a1 )

j = 2, . . . , n.

It suffices to show that all E j ≤ 0. The inequality E j ≤ 0 is equivalent to (a j − a1 )(a1 + a j − 2) ≤ 0, which is true because a1 + ai − 2 ≤ a1 + an − 2 = (1 − a2 ) + · · · + (1 − an−1 ) ≤ 0. The equality holds for a1 = a2 = · · · = an = 1.

,

,

380

Vasile Cîrtoaje

P 2.114. If a1 , a2 , . . . , an are real numbers such that a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,

a1 + a2 + · · · + an = n,

then (1 + a12 )(1 + a22 ) · · · (1 + an2 ) ≥ 2n . (Vasile Cirtoaje, 2009) Solution. Use the substitutions a1 = 1 − S and a2 = b2 + 1, . . . , an = bn + 1, where S and b2 , . . . , bn are nonnegative real numbers such that S = b2 + · · · + bn . We have

1 1 (1 + a12 ) = 1 − S + S 2 , 2 2

and, by Lemma below, 1

2

(1 + a22 ) · · · (1 + an2 ) n−1

 ‹  ‹ 1 2 1 2 1 = 1 + b2 + b2 · · · 1 + b n + b n ≥ 1 + S + S 2 . 2 2 2

Therefore, it suffices to show that  ‹ ‹ 1 2 1 2 1−S+ S 1 + S + S ≥ 1, 2 2 which is equivalent to the obvious inequality S 4 ≥ 0. The equality holds for a1 = a2 = · · · = an = 1. Lemma. If c1 , c2 , . . . , ck are nonnegative real numbers such that c1 + c2 + · · · + ck = S, then  ‹ ‹  ‹ 1 2 1 2 1 2 1 1 + c1 + c1 1 + c2 + c2 · · · 1 + ck + ck ≥ 1 + S + S 2 . 2 2 2 2 Proof. We have ‹ Y  X  1 2 1 + ci + ci ≥ 1 + ci + 2 1≤i≤k 1≤i≤k X  ≥1+ ci + 1≤i≤k

1 = 1 + S + S2. 2

‹ ‹ ‹ X  1 2 1 2 1 2 c + ci + ci cj + cj 2 i 2 2 1≤i< j≤k ‹ X 1 2 c + ci c j 2 i 1≤i< j≤k

Noncyclic Inequalities

381

P 2.115. If a1 , a2 , . . . , an are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

a1 a2 · · · an = 1,

1 1 1 n + + ··· + ≥ . 2 2 2 (a1 + 1) (a2 + 1) (an + 1) 4 (Vasile Cirtoaje, 2009)

Solution (by Arthurzyz). We use the induction method. For n = 2, the desired inequality is equivalent to the obvious inequality 2(a12 + 1) ≥ (a1 + 1)2 . Let us denote E(a1 , a2 , . . . , an ) =

1 1 1 + + ··· + . (a1 + 1)2 (a2 + 1)2 (an + 1)2

To end the proof, it suffices to show that E(a1 , a2 , a3 , . . . , an ) ≥ E(a1 a2 , 1, a3 , . . . , an ) ≥ 0. Since a1 a2 =

1 ≥1 a3 . . . an

and (a1 a2 )a3 . . . an = 1, the right inequality follows by the induction hypothesis. The left inequality is equivalent to 1 1 1 1 + ≥ + , 2 2 2 (a1 + 1) (a2 + 1) (a1 a2 + 1) 4 a1 (1 − a2 )(a1 a2 + a1 + 2) (1 − a2 )(a2 + 3) ≥ , 2 4(a2 + 1) (a1 + 1)2 (a1 a2 + 1)2 which is true if

a1 (a1 a2 + a1 + 2) a2 + 3 ≥ . 4(a2 + 1)2 (a1 + 1)2 (a1 a2 + 1)2

We can write this inequality in the obvious form 2a1 + 2 (a1 a2 + 1)2 a2 + 3 a2 + a1 a2 · · · ≥ 1. 4a1 a2 2a2 + 2 a2 + 1 a1 + a1 a2 + 2 The equality holds for a1 = a2 = · · · = an = 1.

382

Vasile Cîrtoaje

P 2.116. If a1 , a2 , . . . , an are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

a1 a2 · · · an = 1,

1 1 n 1 + + · · · + ≥ . (a1 + 2)2 (a2 + 2)2 (an + 2)2 9 (Vasile Cirtoaje, 2009)

Solution. We use the induction method. For n = 2, the desired inequality is equivalent to (a1 − 1)4 ≥ 0. Let us denote E(a1 , a2 , . . . , an ) =

1 1 1 + + ··· + . 2 2 (a1 + 2) (a2 + 2) (an + 2)2

To end the proof, it suffices to show that E(a1 , a2 , a3 , . . . , an−1 , an ) ≥ E(a1 , 1, a3 , . . . , an−1 , a2 an ) ≥ 0. Since a2 an ≤ an ≤ an−1 and a1 a3 . . . an−1 (a2 an ) = 1, the right inequality follows by the induction hypothesis. The left inequality is equivalent to 1 1 1 1 + ≥ + . 2 2 2 (a2 + 2) (an + 2) (a2 an + 2) 9 Denoting s = a2 + an and p = a2 an , where s ≤ 2 and p ≤ 1, the inequality becomes s2 + 4s + 8 − 2p p2 + 4p + 13 ≥ , (2s + 4 + p)2 9(p + 2)2 (1 + p − s)(As + B) ≥ 0, where A = 16 − 20p − 5p2 ,

B = 80 − 32p − 29p2 − p3 > 0.

Since 1 + p − s = (1 − a2 )(1 − an ) ≥ 0, we only need to show that As + B ≥ 0. For the nontrivial case A < 0, we get As + B ≥ 2A + B = 112 − 72p − 39p2 − p3 = (1 − p)(112 + 40p + p2 ) ≥ 0. This completes the proof. The equality holds for a1 = a2 = · · · = an = 1.

Noncyclic Inequalities

383

P 2.117. If a1 , a2 , . . . , an are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then a1n

+ a2n

+ · · · + ann

2

−n≥n



a1 a2 · · · an = 1,

‹ 1 1 1 + + ··· + −n . a1 a2 an (Vasile Cirtoaje, 2009)

Solution. We use the induction method. For n = 2, the desired inequality is equivalent to (a1 − 1)4 ≥ 0. Let us denote E(a1 , a2 , . . . , an ) =

a1n

+ a2n

+ · · · + ann

2

−n



‹ 1 1 1 . + + ··· + a1 a2 an

To end the proof, it suffices to show that E(a1 , a2 , a3 , . . . , an−1 , an ) ≥ E(a1 , 1, a3 , . . . , an−1 , a2 an ) ≥ 0. Since a2 an ≤ an ≤ an−1 and a1 a3 . . . an−1 (a2 an ) = 1, the right inequality follows by the induction hypothesis. The left inequality is equivalent to  ‹ 1 1 1 2 n n n n a2 + an − 1 − a2 an ≥ n + −1− , a2 a n a2 a n ‹ ‹  1 1 (1 − a2n )(1 − ann ) ≤ n2 −1 −1 , a2 an which is true if (1 + a2 + · · · + a2n−1 )(1 + an + · · · + ann−1 ) ≤

n2 . a2 an

Since a2 ≤ 1 and an ≤ 1, this inequality is clearly true. This completes the proof. The equality holds for a1 = a2 = · · · = an = 1.

384

Vasile Cîrtoaje

P 2.118. If a, b, c are positive real numbers such that a ≥ 1 ≥ b ≥ c, then

a bc = 1,

1−a 1− b 1−c + + ≥ 0. 3 + a2 3 + b2 3 + c 2 (Vasile Cirtoaje, 2009)

Solution. Denote the left side of the inequality by E(a, b, c). We will show that E(a, b, c) ≥ E(a b, 1, c) ≥ 0. Let a + b = s, We have

a b = p.

p p ≥ a bc = 1, s ≥ 2 p ≥ 2.

Therefore, 1− b ab − 1 1−a + + 2 2 3+a 3+ b 3 + a2 b2 2 s − (3 + p)s + 2(3 − p) p−1 = + 2 2 3s + (p − 3) 3 + p2 (3 + p)(s − p − 1)(ps + p − 3) ≥ 0, = (3 + p2 )[3s2 + (p − 3)2 ]

E(a, b, c) − E(a b, 1, c) =

since s − p − 1 = (a − 1)(1 − b) ≥ 0, Also, we have E(a b, 1, c) = E(1/c, 1, c) =

ps + p − 3 ≥ 2p + p − 3 ≥ 0. (1 − c)4 ≥ 0. (3c 2 + 1)(3 + c 2 )

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

P 2.119. If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n, then a14 + a24 + · · · + an4 − n ≥

a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an , 14 2 (a + a22 + · · · + an2 − n). 3 1 (Vasile Cirtoaje, 2009)

Noncyclic Inequalities

385

Solution (by Linqaszayi). Using the substitution ai = 1 + x i , i = 1, 2, . . . , n, which implies x1 ≥ x2 ≥ 0 ≥ x3 ≥ · · · ≥ x n,

x 1 + x 2 + · · · + x n = 0,

we need to show that E(x 1 , x 2 , . . . , x n = 3

n X

x i 4 + 12

i=1

n X

xi3 + 4

i=1

n X

x i 2.

i=1

We will prove that E(x 1 , x 2 , . . . , x n ) − E(

x1 + x2 x1 + x2 , , x 3 , . . . , x n ) ≥ 0. 2 2

The left inequality is true because x 14 + x 24 ≥ 2

 x + x 4 1 2 , 2

x 13 + x 23 ≥ 2

 x + x 3 1 2 , 2

x 12 + x 22 ≥ 2

 x + x 2 1 2 . 2

Replacing x 3 , · · · , x n with −x 3 , · · · , −x n , the left inequality becomes 6A4 + 24A3 + 8A2 + 3(x 34 + · · · + x n4 ) − 12(x 33 + · · · + x n3 ) + 4(x 32 + · · · + x n2 ) ≥ 0, where A= Since 4

A ≥

x3 + · · · + x n , 2

x 34 + · · · + x n4

3

0 ≤ x3 ≤ · · · ≤ x n.

x 33 + · · · + x n3

2

x 32 + · · · + x n2

, A ≥ , A ≥ , 16 8 4 it suffices to show that ‹  3 + 3 (x 34 + · · · + x n4 ) + (3 − 12)(x 33 + · · · + x n3 ) + (2 + 4)(x 32 + · · · + x n2 ) ≥ 0, 8 which is equivalent to the obvious inequality x 32 (3x 3 − 4)2 + · · · + x n2 (3x n2 − 4)2 ≥ 0. This completes the proof. The equality holds for a1 = a2 = · · · = an = 1, and also for a1 = a2 =

5 , 3

a3 = · · · = an−1 = 1,

an =

−1 . 3

386

Vasile Cîrtoaje

Chapter 3

Bibliography

387