Extensions and refinements of Jensen’s inequalities [Volume 4]

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Vasile Cîrtoaje C j

MATHEM MATICAL INEQUA Q ALITIES Volume 4

EXTENSIONS AND REFINEMENTS OF JENSEN’SS INEQUALITY OF JENSEN

UNIVERSITY OF PLOIESTI, ROMANIA 2015

Contents 1 Half Convex Function Method 1 1.1 Theoretical Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2 Half Convex Function Method for Ordered Variables 97 2.1 Theoretical Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 2.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 2.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 3 Partially Convex Function Method 141 3.1 Theoretical Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 3.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 3.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 4 Partially Convex Function Method for Ordered Variables 193 4.1 Theoretical Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 4.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 4.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 5 Bibliography

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Chapter 1

Half Convex Function Method 1.1

Theoretical Basis

Half Convex Function Theorem (Vasile Cirtoaje, 2004). Let f (u) be a function defined on a real interval I and convex on Iu≥s or Iu≤s , where s ∈ I. The inequality f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f

a + a + ··· + a  1 2 n n

holds for all a1 , a2 , . . . , an ∈ I satisfying a1 + a2 + · · · + an = ns if and only if f (x) + (n − 1) f ( y) ≥ n f (s) for all x, y ∈ I such that x + (n − 1) y = ns. Proof (for the right convexity). The necessity is obvious. Without loss of generality, assume that a1 ≤ a2 ≤ · · · ≤ an . If a1 ≥ s, then the required inequality follows from Jensen’s inequality for convex functions. Otherwise, there exists k ∈ {1, 2, . . . , n − 1} such that a1 ≤ · · · ≤ ak < s ≤ ak+1 ≤ · · · ≤ an . Since f is convex on Iu≥s , we can apply Jensen’s inequality to get f (ak+1 ) + · · · + f (an ) ≥ (n − k) f (z),

z=

ak+1 + · · · + an . n−k

Thus, it suffices to show that f (a1 ) + · · · + f (ak ) + (n − k) f (z) ≥ n f (s). Let b1 , . . . , bk ∈ I defined by ai + (n − 1)bi = ns, 1

i = 1, . . . , k.

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We claim that z ≥ b1 ≥ · · · ≥ bk > s. Indeed, we have b1 ≥ · · · ≥ bk , bk − s =

s − ak > 0, n−1

(n − 1)b1 = ns − a1 = (a2 + · · · + ak ) + ak+1 + · · · + an ≤ (k − 1)s + ak+1 + · · · + an = (k − 1)s + (n − k)z ≤ (n − 1)z. By hypothesis, we have f (a1 ) + (n − 1) f (b1 ) ≥ n f (s), ··· f (ak ) + (n − 1) f (bk ) ≥ n f (s), hence f (a1 ) + · · · + f (ak ) + (n − 1)[ f (b1 ) + · · · + f (bk )] ≥ kn f (s). Consequently, it suffices to show that kn f (s) − (n − 1)[ f (b1 ) + · · · + f (bk )] + (n − k) f (z) ≥ n f (s), which is equivalent to p f (z) + (k − p) f (s) ≥ f (b1 ) + · · · + f (bk ),

p=

n−k ≤ 1. n−1

By Jensen’s inequality, we have p f (z) + (1 − p) f (s) ≥ f (w),

w = pz + (1 − p)s ≥ s.

Thus, we only need to show that f (w) + (k − 1) f (s) ≥ f (b1 ) + · · · + f (bk ). Since the decreasingly ordered vector A~k = (w, s, . . . , s) majorizes the decreasingly ordered vector B~k = (b1 , b2 , . . . , bk ), this inequality follows from Karamata’s inequality for convex functions. The proof for the left convexity is similar. Remark 1. Let us denote g(u) =

f (u) − f (s) , u−s

h(x, y) =

g(x) − g( y) . x−y

Half Convex Function Method

3

In many applications, it is useful to replace the hypothesis f (x) + (n − 1) f ( y) ≥ n f (s) in Half Convex Function Theorem (HCF Theorem) by the equivalent condition h(x, y) ≥ 0 for all x, y ∈ I such that x + (n − 1) y = ns. This equivalence is true since f (x) + (n − 1) f ( y) − n f (s) = [ f (x) − f (s)] + (n − 1)[ f ( y) − f (s)] = (x − s)g(x) + (n − 1)( y − s)g( y) n−1 (x − y)[g(x) − g( y)] = n n−1 = (x − y)2 h(x, y). n Remark 2. Assume that f is differentiable on I, and let H(x, y) =

f 0 (x) − f 0 ( y) . x−y

Then, the desired inequality of Jensen’s type in HCF Theorem holds true by replacing the hypothesis f (x) + (n − 1) f ( y) ≥ n f (s) with the more restrictive condition H(x, y) ≥ 0 for all x, y ∈ I such that x + (n − 1) y = ns. To prove this claim, we will show that the new condition implies f (x)+(n−1) f ( y) ≥ n f (s) for all x, y ∈ I such that x + (n − 1) y = ns. Write this inequality as f1 (x) ≥ n f (s), where  ns − x  f1 (x) = f (x) + (n − 1) f ( y) = f (x) + (n − 1) f . n−1 From  ns − x  n = f 0 (x) − f 0 ( y) = (x − s)H(x, y), f10 (x) = f 0 (x) − f 0 n−1 n−1 it follows that f1 is decreasing for x ≤ s and increasing for x ≥ s; therefore, f1 (x) ≥ f1 (s) = n f (s). Remark 3. The inequality in HCF Theorem becomes an equality for a1 = a2 = · · · = an = s and also for a1 = x,

a2 = · · · = an = y

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(or any cyclic permutation), where x, y ∈ I satisfy the equations x + (n − 1) y = ns,

f (x) + (n − 1) f ( y) = n f (s).

For x 6= y, these equations are equivalent to x + (n − 1) y = ns, h(x, y) = 0. Remark 4. HCF Theorem is also valid in the case when I = [a, b] \ {u0 } or I = (a, b) \ {u0 }, where a, b, u0 are real numbers such that a < u0 < b. Clearly, two cases are possible: (1) u0 < s - when f is right convex, i.e. convex on Iu≥s ; (2) u0 > s - when f is left convex, i.e. convex on Iu≤s . Remark 5. In HCF Theorem for the right convexity (when HCF Theorem is called RHCF Theorem), it suffices to consider x ≤ s ≤ y. This claim is true because in the proof of RHCF Theorem, the hypothesis f (x) + (n − 1) f ( y) ≥ n f (s) is used to get the inequalities f (ai ) + (n − 1) f (bi ) ≥ n f (s),

i = 1, 2, · · · , k,

where ai ≤ s ≤ bi . Similarly, in HCF Theorem for the left convexity (when HCF Theorem is called LHCF Theorem), it suffices to consider x ≥ s ≥ y. The following theorem (LCRCF-Theorem) is also useful to prove some symmetric inequalities. Left Convex-Right Concave Function Theorem (Vasile Cirtoaje, 2004). Let a ≤ c be real numbers, let f be a continuous function on I = [a, ∞), strictly convex on [a, c] and strictly concave on [c, ∞), and let E(a1 , a2 , . . . , an ) = f (a1 ) + f (a2 ) + · · · + f (an ). If a1 , a2 , . . . , an ∈ I such that a1 + a2 + · · · + an = S = const ant, then (a) E is minimum for a1 = a2 = · · · = an−1 ≤ an ;

Half Convex Function Method

5

(b) E is maximum for either a1 = a or a < a1 ≤ a2 = · · · = an . Proof. Without loss of generality, assume that a1 ≤ a2 ≤ . . . ≤ an . Since E(a1 , a2 , . . . , an ) is a continuous function on the compact set Λ = {(a1 , a2 , . . . , an ) : a1 + a2 + · · · + an = S, a1 , a2 , . . . , an ∈ I}, E attains its minimum and maximum values. (a) For the sake of contradiction, suppose that E is minimum at a point (b1 , b2 , . . . , bn ) with b1 ≤ b2 ≤ . . . ≤ bn and b1 < bn−1 . For bn−1 ≤ c, by Jensen’s inequality for strictly convex functions, we have ‹  b1 + bn−1 , f (b1 ) + f (bn−1 ) > 2 f 2 while for bn−1 > c, by Karamata’s inequality for strictly concave functions, we have f (bn−1 ) + f (bn ) > f (c) + f (bn−1 + bn − c). The both results contradict the assumption that E is minimum at (b1 , b2 , . . . , bn ). (b) For the sake of contradiction, suppose that E is maximum at a point (b1 , b2 , . . . , bn ) with a < b1 ≤ b2 ≤ . . . ≤ bn and b2 < bn . There are three cases to consider. Case 1: b2 ≥ c. By Jensen’s inequality for strictly concave functions, we have  ‹ b2 + bn f (b2 ) + f (bn ) < 2 f . 2 Case 2: b2 < c and b1 + b2 −a ≤ c. By Karamata’s inequality for strictly convex functions, we have f (b1 ) + f (b2 ) < f (a) + f (b1 + b2 − a). Case 3: b2 < c and b1 + b2 −c ≥ a. By Karamata’s inequality for strictly convex functions, we have f (b1 ) + f (b2 ) < f (b1 + b2 − c) + f (c). Clearly, the result of each case contradicts the assumption that E is msximum at (b1 , b2 , . . . , bn ). Remark 6. The part (a) in LCRCF-Theorem is also true in the case where I = (a, ∞) and lim x→a f (x) = ∞.

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Half Convex Function Method

1.2

7

Applications

1.1. If a, b, c are real numbers such that a + b + c = 3, then 3(a4 + b4 + c 4 ) + a2 + b2 + c 2 + 6 ≥ 6(a3 + b3 + c 3 ).

1.2. If a1 , a2 , . . . , an ≥

1 − 2n such that a1 + a2 + · · · + an = n, then n−2 a13 + a23 + · · · + an3 ≥ n.

1.3. If a1 , a2 , . . . , an ≥

−n such that a1 + a2 + · · · + an = n, then n−2 a13 + a23 + · · · + an3 ≥ a12 + a22 + · · · + an2 .

1.4. If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n, then (n2 − 3n + 3)(a14 + a24 + · · · + an4 − n) ≥ 2(n2 − n + 1)(a12 + a22 + · · · + an2 − n).

1.5. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then (n2 + n + 1)(a13 + a23 + · · · + an3 − n) ≥ (n + 1)(a14 + a24 + · · · + an4 − n).

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

p a4 + b4 + c 4 − 3 + 2(7 + 3 7)(a3 + b3 + c 3 − 3) ≥ 0; p a4 + b4 + c 4 − 3 + 2(7 − 3 7)(a3 + b3 + c 3 − 3) ≥ 0.

1.7. Let k ≥ 3 and n ≥ 3 be integer numbers such that k ≤ n + 1. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then a1k + a2k + · · · + ank − n a12 + a22 + · · · + an2 − n

≥ (n − 1)

•

˜ n k−1 −1 . n−1

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1.8. Let k ≥ 3 and n ≥ 3 be integer numbers. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then a1k + a2k + · · · + ank − n a12 + a22 + · · · + an2 − n

≤

nk−1 − 1 . n−1

1.9. If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an = n, then  ‹ 1 1 1 2 n + + ··· + − n ≥ 4(n − 1)(a12 + a22 + · · · + an2 − n). a1 a2 an

1.10. If a1 , a2 , . . . , a8 are positive real numbers such that a1 + a2 + · · · + a8 = n, then 1 a12

+

1 a22

+ ··· +

1 a82

≥ a12 + a22 + · · · + a82 .

1.11. If a1 , a2 , . . . , an are positive real numbers such that a12

+ a22

+ · · · + an2



p

−n≥2 1+

1 1 1 + + ··· + = n, then a1 a2 an

 n−1 (a1 + a2 + · · · + an − n). n

1.12. If a, b, c are nonnegative real numbers, no two of which are zero, then  ‹ 1 1 2 1 1 1 1 + + ≤ + + . 3a + b + c 3b + c + a 3c + a + b 5 b+c c+a a+b

1.13. If a, b, c, d ≥ 3 −

p 7 such that a + b + c + d = 4, then 1 1 1 1 4 + + + ≥ . 2 2 2 2 2+a 2+ b 2+c 2+d 3

1.14. If a1 , a2 , . . . , an ∈ [0, n − 2] such that a1 + a2 + · · · + an = n, then 1 n + a12

+

1 n + a22

+ ··· +

1 n ≤ . 2 n + an n+1

Half Convex Function Method

9

1.15. If a, b, c are nonnegative real numbers such that a + b + c = 3, then 3− b 3−c 3 3−a + + ≥ . 2 2 2 9+a 9+ b 9+c 5 1.16. If a, b, c are nonnegative real numbers such that a + b + c = 3, then 1 1 1 3 + + ≥ . 2 2 2 5+a+a 5+ b+ b 5+c+c 7 1.17. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then 1 1 1 1 1 + + + ≤ . 10 + a + a2 10 + b + b2 10 + c + c 2 10 + d + d 2 3 1.18. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If 1 k ≥ 1 − , then n 1 n 1 1 ≥ . + + ··· + 1 + kan2 1+k 1 + ka12 1 + ka22 1.19. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = n. If 0 < k ≤ n−1 , then 2 n −n+1 1 n 1 1 ≤ . + + ··· + 2 2 2 1 + kan 1+k 1 + ka1 1 + ka2 1.20. Let a1 , a2 , . . . , an be nonnegative numbers such that a1 + a2 + · · · + an = n. If n2 k≥ , then 4(n − 1) a1 (a1 − 1) a12

+k

+

a2 (a2 − 1) a22

+k

+ ··· +

an (an − 1) ≥ 0. an2 + k

1.21. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then 1 − an 1 − a1 1 − a2 + + ··· + ≤ 0. 2 2 (n − 2a1 ) (n − 2a2 ) (n − 2an )2

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1.22. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If n , then k ≥1+ p n−1 a12 − 1 (a1 + k)2

+

a22 − 1 (a2 + k)2

+ ··· +

an2 − 1 (an + k)2

≥ 0.

1.23. Let a1s , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If 2n − 1 0< k ≤1+ , then n−1 a12 − 1 (a1 + k)2

+

a22 − 1 (a2 + k)2

+ ··· +

an2 − 1 (an + k)2

≤ 0.

1.24. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If (n − 1)(2n − 1) k≥ , then n2 1 1 + ka13

+

1 1 + ka23

+ ··· +

1 n ≥ . 3 1 + kan 1+k

1.25. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If n−1 , then 0 0 for u ≥ 1, hence f is convex for u ≥ 1. By HCF Theorem, it suffices to show that f (x) + 2 f ( y) ≥ 3 f (1) for all real x, y such that x + 2 y = 3. Using Remark 1, we only need to show that h(x, y) ≥ 0, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

We have g(u) = u3 + u2 + u + 1 + p(u2 + u + 1) + u + 1 = u3 + (p + 1)(u2 + u + 1), h(x, y) = x 2 + x y + y 2 ) + (p + 1)(x + y + 1) = 3 y 2 − (10 + p) y + 13 + 4p  ‹ 10 + p 2 =3 y− ≥ 0. 6 In accordance with Remark 3, the equality holds for a = b = c = 1, and also for a = −(1 + p)/3 and b = c = (10 + p)/6 (or any cyclic permutation). p (a) Since p = 2(7 + 3 7),p the equality holds for a = b = c = 1, and also for p a = −5 − 2 7 and b = c = 4 + 7 (or any cyclic permutation). p (b) Since p p = 2(7 − 3 7), p the equality holds for a = b = c = 1, and also for a = −5 + 2 7 and b = c = 4 − 7 (or any cyclic permutation). Remark. Similarly, we can prove the following generalization:

22

Vasile Cîrtoaje • If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n, then a14 + a24 + · · · + an4 − n + p(a12 + a22 + · · · + an2 − n) ≥ 0,

where p=

2(n2 − n + 1) ± 2

p

3(n2 − n + 1)(n2 − 3n + 3) . (n − 2)2

The equality holds for a1 = a2 = · · · = an = 1, and also for a1 =

2(n2 − n − 1) + (n − 2)p −2(n2 − 3n + 1) − (n − 1)(n − 2)p , a = a = · · · = a = 2 3 n 2(n2 − 3n + 3) 2(n2 − 3n + 3)

(or any cyclic permutation).

P 1.7. Let k ≥ 3 and (n ≥ 3) be integer numbers such that k ≤ n + 1. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then a1k + a2k + · · · + ank − n a12 + a22 + · · · + an2 − n

≥ (n − 1)

•

˜ n k−1 −1 . n−1 (Vasile Cirtoaje, 2012)

Solution. Denote •

m = (n − 1)

˜  n k−2  n k−3 n k−1 −1 = + + · · · + 1, n−1 n−1 n−1

and write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

where f (u) = uk − mu2 ,

u ∈ [0, n].

We will show that f (u) is convex for u ∈ [1, n]. Since f 00 (u) = k(k − 1)uk−2 − 2m ≥ k(k − 1) − 2m, we need to show that k(k − 1)  n k−2  n k−3 ≥ + + · · · + 1. 2 n−1 n−1

Half Convex Function Method

23

Since n ≥ k − 1, this inequality is true if  ‹  ‹ k(k − 1) k − 1 k−2 k − 1 k−3 ≥ + + · · · + 1. 2 k−2 k−2 By Bernoulli’s inequality, we have 

k−1 k−2

‹j

=

1 1−

1 k−1


j ≤

1 1−

j k−1

=

k−1 , k− j−1

0 ≤ j ≤ k − 2.

Therefore, it suffices to show that  ‹ k(k − 1) 1 1 ≥ (k − 1) 1 + + · · · + . 2 2 k−1 This is true if

k 1 1 ≥ 1 + + ··· + , 2 2 k−1 which can be easily proved by induction. According to HCF Theorem and Remark 1, we only need to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

We have (uk − 1) − m(u2 − 1) = (uk−1 + uk−2 + · · · + 1) − m(u + 1), u−1  k−1  x − y k−1 x k−2 − y k−1 h(x, y) = + + ··· + 1 − m x−y x−y  k−1   k−2   2  x − y k−2  n k−3 x − y2 x − y k−1  n k−2 n + +· · ·+ = − − − . x−y n−1 x−y n−1 x−y n−1 g(u) =

It suffices to show that

x j+1 − y j+1  n  j ≥ x−y n−1

n i for x = 6 y and j = 1, 2, · · · , k − 2. This is true if f j ( y) ≥ 0 for y ∈ 0, , where n−1  n j f j ( y) = x j + x j−1 y + · · · + x y j−1 + y j − , x = n − (n − 1) y. n−1 h

Since x 0 = −(n − 1) and n − 1 ≥ k − 2 ≥ j, we get f j0 ( y) = −(n − 1)[ j x j−1 + ( j − 1)x j−2 y + · · · + y j−1 ] + x j−1 + 2x j−2 y + · · · + j y j−1

24

Vasile Cîrtoaje ≤ − j[ j x j−1 + ( j − 1)x j−2 y + · · · + y j−1 ] + x j−1 + 2x j−2 y + · · · + j y j−1

= −( j · j − 1)x j−1 − [ j · ( j − 1) − 2]x j−2 y − · · · − ( j · 2 − j + 1)x y j−2 ≤ 0. n Therefore, f j ( y) is decreasing, hence f j ( y) is minimal for y = , when x = 0. Thus, n−1  n  = 0. f j ( y) ≥ f j n−1 This completes the proof. From x + (n − 1) y = n and h(x, y) = 0, we get x = 0, n y= . Therefore, in accordance with Remark 3, the equality holds for a1 = 0 and n−1 n a2 = a3 = · · · = an = (or any cyclic permutation). n−1 Remark. For k = 3 and k = 4, we get the following statements (Vasile Cirtoaje, 2002): • If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then (n − 1)(a13 + a23 + · · · + an3 − n) ≥ (2n − 1)(a12 + a22 + · · · + an2 − n), with equality for a1 = a2 = · · · = an = 1, and also for a1 = 0 and a2 = a3 = · · · = an = n (or any cyclic permutation). n−1 • If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then (n − 1)2 (a14 + a24 + · · · + an4 − n) ≥ (3n2 − 3n + 1)(a12 + a22 + · · · + an2 − n), with equality for a1 = a2 = · · · = an = 1, and also for a1 = 0 and a2 = a3 = · · · = an = n (or any cyclic permutation). n−1

P 1.8. Let k ≥ 3 and n ≥ 3 be integer numbers. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then a1k + a2k + · · · + ank − n a12 + a22 + · · · + an2 − n

≤

nk−1 − 1 . n−1 (Vasile Cirtoaje, 2012)

Solution. Denote m=

nk−1 − 1 = nk−2 + nk−3 + · · · + 1, n−1

and write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

Half Convex Function Method

25

where f (u) = mu2 − uk ,

u ∈ [0, n].

We will show that f (u) is convex for u ∈ [0, 1]. Since f 00 (u) = 2m − k(k − 1)uk−2 ≥ 2m − k(k − 1), we need to show that nk−2 + nk−3 + · · · + 1 ≥

k(k − 1) . 2

3k−2 + 3k−3 + · · · + 1 ≥

k(k − 1) . 2

This is true if

By Bernoulli’s inequality, we have 3k−2 + 3k−3 + · · · + 1 ≥ [1 + 2(k − 2)] + [1 + 2(k − 3)] + · · · + [1 + 2 · 0] k(k − 1) . 2 According to HCF Theorem and Remark 1, we only need to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where = (k − 1) + (k − 2)(k − 1) = (k − 1)2 >

h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

We have g(u) =

m(u2 − 1) − (uk − 1) = m(u + 1) − (uk−1 + uk−2 + · · · + 1), u−1

x k−1 − y k−1 x k−2 − y k−1 − − ··· − 1 x−y x−y       x k−1 − y k−1 x k−2 − y k−2 x2 − y2 k−2 k−3 = n − + n − + ··· + n − . x−y x−y x−y h(x, y) = m −

It suffices to show that nj ≥

x j+1 − y j+1 x−y

for x 6= y and j = 1, 2, · · · , k − 2. We will show that n j ≥ (x + y) j ≥

x j+1 − y j+1 . x−y

The left inequality is true since n − (x + y) = x + (n − 1) y − (x + y) = (n − 2) y ≥ 0.

26

Vasile Cîrtoaje

The right inequality is also true since  ‹  ‹ j j−1 j j j (x + y) = x + x y + ··· + x y j−1 + y j 1 j−1 and

x j+1 − y j+1 = x j + x j−1 y + · · · + x y j−1 + y j . x−y

In accordance with Remark 3, the equality holds for a1 = n and a2 = a3 = · · · = an = 0 (or any cyclic permutation). Remark. For k = 3 and k = 4, we get the following statements (Vasile Cirtoaje, 2002): • If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then a13 + a23 + · · · + an3 − n ≤ (n + 1)(a12 + a22 + · · · + an2 − n), with equality for a1 = a2 = · · · = an = 1, and also for a1 = n and a2 = a3 = · · · = an = 0 (or any cyclic permutation). • If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then a14 + a24 + · · · + an4 − n ≤ (n2 + n + 1)(a12 + a22 + · · · + an2 − n), with equality for a1 = a2 = · · · = an = 1, and also for a1 = n and a2 = a3 = · · · = an = 0 (or any cyclic permutation).

P 1.9. If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an = n, then  ‹ 1 1 1 n2 + + ··· + − n ≥ 4(n − 1)(a12 + a22 + · · · + an2 − n). a1 a2 an (Vasile Cirtoaje, 2004) Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

s=

a1 + a2 + · · · + an = 1, n

n2 − 4(n − 1)u2 , u ∈ (0, n). u

For u ∈ (0, 1], we have f 00 (u) =

2n2 − 8(n − 1) ≥ 2n2 − 8(n − 1) = 2(n − 2)2 ≥ 0. u3

Half Convex Function Method

27

Thus, f is convex on (0, 1]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y > 0 such that x + (n − 1) y = n, where h(x, y) =

g(x) − g( y) , x−y

We have g(u) = and h(x, y) =

g(u) =

f (u) − f (1) . u−1

−n2 − 4(n − 1)(u + 1) u

[n − (2n − 2) y]2 n2 − 4(n − 1) = ≥ 0. xy xy

In accordance with Remark 3, the equality holds for a1 = a2 = · · · = an = 1, and also for a1 = n/2 and a2 = a3 = · · · = an = n/(2n − 2) (or any cyclic permutation).

P 1.10. If a1 , a2 , . . . , a8 are positive real numbers such that a1 + a2 + · · · + a8 = n, then 1 a12

+

1 a22

+ ··· +

1 a82

≥ a12 + a22 + · · · + a82 . (Vasile Cirtoaje, 2007)

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (a8 ) ≥ 8 f (s), where f (u) =

s=

a1 + a2 + · · · + a8 = 1, 8

1 − u2 , u ∈ (0, n). u2

For u ∈ (0, 1], we have f 00 (u) =

6 − 2 ≥ 6 − 2 > 0. u4

Thus, f is convex on (0, 1]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y > 0 such that x + 7 y = 8, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

We have g(u) = −u − 1 −

f (u) − f (1) . u−1

1 1 − 2 u u

28

Vasile Cîrtoaje

and h(x, y) = −1 + From 8 = x + 7 y ≥ 2

p

x+y 1 + 2 2. xy x y

7x y, we get x y ≤ 16/7. Therefore, 7(x + y) 112 y 2 − 170 y + 72 1 + = xy 16x y 16x y 2 2 112 y − 176 y + 72 14 y − 22 y + 9 > = > 0. 16x y 2x y

h(x, y) ≥ −1 +

The equality holds for a1 = a2 = · · · = a8 = 1. Remark. Similarly, we can prove the following generalization. • If a1 , a2 , . . . , an (n ≥ 4) are positive real numbers such that a1 + a2 + · · · + an = n, then
1 1 1 8 2 a1 + a22 + · · · + an2 . + 2 + ··· + 2 + 8 − n ≥ 2 an n a1 a2

P 1.11. If a1 , a2 , . . . , an are positive real numbers such that a12

+ a22

+ · · · + an2



p

−n≥2 1+

1 1 1 + + ··· + = n, then a1 a2 an

 n−1 (a1 + a2 + · · · + an − n). n (Vasile Cirtoaje, 2006)

Solution. Replacing each ai by 1/ai , we need to prove that f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where

s=

a1 + a2 + · · · + an = 1, n

  p 1 n−1 1 f (u) = 2 − 2 1 + , u ∈ (0, n). u n u

For u ∈ (0, 1], we have € Š € p 6 − 4 1 + n−1 u 6−4 1+ n f 00 (u) = ≥ u4 u4

p

n−1 n

Š

p 2( n − 1 − 1)2 = ≥ 0. nu4

Thus, f is convex on (0, 1]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y > 0 such that x + (n − 1) y = n, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

Half Convex Function Method We have

29

  p −1 2 n−1 1 g(u) = 2 + 1 + u n u

and 1 h(x, y) = xy We only need to show that



 p 2 n−1 1 1 + −1− . x y n

p 2 n−1 1 1 + ≥1+ . x y n

Indeed, using the Cauchy-Schwarz inequality, we get p p p (1 + n − 1)2 (1 + n − 1)2 2 n−1 1 1 + ≥ = =1+ . x y x + (n − 1) y n n In accordance p with Remark 3, the equality holds forpa1 = a2 = · · · = an = 1, and also for 1+ n−1 n−1+ n−1 a1 = and a2 = a3 = · · · = an = (or any cyclic permutation). n n

P 1.12. If a, b, c are nonnegative real numbers, no two of which are zero, then  ‹ 1 1 1 2 1 1 1 + + ≤ + + . 3a + b + c 3b + c + a 3c + a + b 5 b+c c+a a+b (Vasile Cirtoaje, 2006) Solution. Due to homogeneity, we may assume that a + b + c = 3. So, we need to show that a+b+c f (a) + f (b) + f (c) ≥ 3 f (s), s = = 1, 3 where 2 5 − , u ∈ [0, 3). f (u) = 3 − u 2u + 3 For u ∈ [1, 3), we have f 00 (u) =

4 40 36[2u3 + 3u2 + 9(u − 1)(3 − u)] − = > 0; (3 − u)3 (2u + 3)3 (3 − u)3 (2u + 3)3

therefore, f is convex on [1, 3). By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + 2 y = 3, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

30

Vasile Cîrtoaje

We have g(u) =

2 1 + 3 − u 2u + 3

and 9(2x + 2 y − 3) 1 2 − = (3 − x)(3 − y) (2x + 3)(2 y + 3) (3 − x)(3 − y)(2x + 3)(2 y + 3) 9x ≥ 0. = (3 − x)(3 − y)(2x + 3)(2 y + 3)

h(x, y) =

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

P 1.13. If a, b, c, d ≥ 3 −

p 7 such that a + b + c + d = 4, then

1 1 1 1 4 + + + ≥ . 2 2 2 2 2+a 2+ b 2+c 2+d 3 (Vasile Cirtoaje, 2008) Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) ≥ 4 f (s), where f (u) =

s=

a+b+c+d = 1, 4

p 1 , u ≥ 3 − 7. 2 2+u

For u ≥ 1, we have f 00 (u) =

3(3u2 − 2) > 0. (2 + u2 )3

Thus, f is convex for u ≥ 1. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + 3 y = 4. We have g(u) = and h(x, y) = since

f (u) − f (1) −1 − u = u−1 3(2 + u2 )

xy + x + y −2 g(x) − g( y) = ≥0 x−y 3(2 + x 2 )(2 + y 2 )

−x 2 + 6x − 2 (3 + xy + x + y −2= = 3

p p 7 − x)(x − 3 + 7) ≥ 0. 3

Half Convex Function Method

31

In accordance with Remark 3, the p equality holds for a = b = c = d = 1, and also for p 1+ 7 (or any cyclic permutation). a = 3 − 7 and b = c = d = 3 Remark. Similarly, we can prove the following generalization. p • If a1 , a2 , . . . , an ≥ n − 1 − n2 − 3n + 3 such that a1 + a2 + · · · + an = n, then 1

+

2 + a12

1 2 + a22

+ ··· +

n 1 ≥ , 2 + an2 3

with equality for a1 = a2 p= · · · = an = 1, and also for a1 = n − 1 − 1 + n2 − 3n + 3 a2 = a3 = · · · = an = (or any cyclic permutation). n−1

p

n2 − 3n + 3 and

P 1.14. If a1 , a2 , . . . , an ∈ [0, n − 2] such that a1 + a2 + · · · + an = n, then 1 n + a12

+

1 n + a22

+ ··· +

n 1 . ≤ 2 n + an n+1 (Vasile Cirtoaje, 2008)

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

s=

a1 + a2 + · · · + an = 1, n

−1 , u ∈ [0, n − 2], n ≥ 3. n + u2

For u ∈ [0, 1], we have f 00 (u) =

2(n − 3u2 ) ≥ 0. (n + u2 )3

Thus, f is convex on [0, 1]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. We have g(u) = and h(x, y) =

f (u) − f (1) u+1 = u−1 (n + 1)(n + u2 )

g(x) − g( y) n− x − y − xy = . x−y (n + 1)(n + x 2 )(n + y 2 )

We need to show that n − x − y − x y ≥ 0.

32

Vasile Cîrtoaje

Indeed, we have n− x − y − xy =

(n − x)(n − 2 − x) ≥ 0. n−1

The equality holds for a1 = a2 = · · · = an = 1, and also for a1 = n − 2 and a2 = a3 = 2 · · · = an = (or any cyclic permutation). n−1

P 1.15. If a, b, c are nonnegative real numbers such that a + b + c = 3, then 3−a 3− b 3−c 3 + + ≥ . 9 + a2 9 + b2 9 + c 2 5 (Vasile Cirtoaje, 2013) Solution. Write the inequality as f (a) + f (b) + f (c) ≥ 3 f (s), where f (u) =

s=

a+b+c = 1, 3

3−u , u ∈ [0, 3]. 9 + u2

For u ∈ [1, 3], we have 1 00 u2 (9 − u) + 27(u − 1) f (u) = > 0. 2 (9 + u2 )3 Thus, f is convex on [1, 3]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + 2 y = 3, where h(x, y) =

g(x) − g( y) , x−y

We have g(u) = and h(x, y) =

g(u) =

f (u) − f (1) . u−1

−(6 + u) 5(9 + u2 )

x y + 6x + 6 y − 9 x(9 − x) = ≥ 0. 5(9 + x 2 )(9 + y 2 ) 10(9 + x 2 )(9 + y 2 )

The equality holds for a = b = c = 1, and also for a = 0 and b = c = 3/2 (or any cyclic permutation). Remark. Similarly, we can prove the following generalization.

Half Convex Function Method

33

• If a1 , a2 , . . . , an (n ≥ 3) are nonnegative real numbers such that a1 + a2 + · · · + an = n, then n − a1 n2

+ (n2

− 3n + 1)a12

+

n − a2 n2

+ (n2

− 3n + 1)a22

+ ··· +

n2

n − an n ≥ , 2 − 3n + 1)an 2n − 1

+ (n2

with equality for a1 = a2 = · · · = an = 1, and also for a1 = 0 and a2 = a3 = · · · = an = n (or any cyclic permutation). n−1

P 1.16. If a, b, c are nonnegative real numbers such that a + b + c = 3, then 1 1 3 1 + + ≥ . 2 2 2 5+a+a 5+ b+ b 5+c+c 7 (Vasile Cirtoaje, 2008) Solution. Write the inequality as f (a) + f (b) + f (c) ≥ 3 f (s), where f (u) =

s=

a+b+c = 1, 3

1 , u ∈ [0, 3]. 5 + u + u2

For u ≥ 1, we have f 00 (u) =

2(3u2 + 3u − 4) > 0. (5 + u + u2 )3

Thus, f is convex on [1,3]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + 2 y = 3. We have g(u) =

f (u) − f (1) −2 − u = u−1 7(5 + u + u2 )

and h(x, y) =

g(x) − g( y) x y + 2(x + y) − 3 x(5 − x) = = ≥ 0. 2 2 x−y 7(5 + x + x )(5 + y + y ) 14(5 + x + x 2 )(5 + y + y 2 )

In accordance with Remark 3, the equality holds for a = b = c = 1, and also for a = 0 and b = c = 3/2 (or any cyclic permutation). Remark. Similarly, we can prove the following generalization.

34

Vasile Cîrtoaje

• Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If 2(2n − 1) 0 0. (10 + u + u2 )3

Thus, f is convex on [0,1]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + 3 y = 4. We have g(u) = and h(x, y) =

f (u) − f (1) 2+u = u−1 12(10 + u + u2 )

g(x) − g( y) 8 − 2(x + y) − x y = . x−y 12(10 + x + x 2 )(10 + y + y 2 )

We need to show that 8 − 2(x + y) − x y ≥ 0. Indeed, we have 8 − 2(x + y) − x y = 3 y 2 ≥ 0.

Half Convex Function Method

35

The equality holds for a = b = c = d = 1, and also for a = 4 and b = c = d = 0 (or any cyclic permutation). Remark. Similarly, we can prove the following generalization. • Let a1 , a2 , . . . , an (n ≥ 4) be nonnegative real numbers such that a1 + a2 +· · ·+ an = n. If k ≥ 2n + 2, then 1 k + a1 + a12

1

+

k + a2 + a22

+ ··· +

1 n ≤ , 2 k + an + an k+2

with equality for a1 = a2 = · · · = an = 1. If k = 2n + 2, then the equality holds also for a1 = n and a2 = a3 = · · · = an = 0 (or any cyclic permutation).

P 1.18. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If 1 k ≥ 1 − , then n 1 n 1 1 ≥ . + + ··· + 2 2 2 1 + ka 1 + k 1 + ka1 1 + ka2 n (Vasile Cirtoaje, 2005) Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

a1 + a2 + · · · + an = 1, n

1 , u ∈ [0, n]. 1 + ku2

For u ∈ [1, n], we have f 00 (u) = since

s=

2k(3ku2 − 1) > 0, (1 + ku2 )3

 ‹ 1 3 3ku − 1 ≥ 3k − 1 ≥ 3 1 − − 1 = 2 − > 0. n n 2

Thus, f is convex on [1, n]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. We have g(u) = and h(x, y) =

f (u) − f (1) −k(u + 1) = u−1 (1 + k)(1 + ku2 )

g(x) − g( y) k2 (x + y + x y) − k = . x−y (1 + k)(1 + k x 2 )(1 + k y 2 )

36

Vasile Cîrtoaje

We need to show that k(x + y + x y) − 1 ≥ 0. Indeed, we have  ‹ 1 x(2n − 2 − x) ≥ 0. k(x + y + x y) − 1 ≥ 1 − (x + y + x y) − 1 = n n 1 The equality holds for a1 = a2 = · · · = an = 1. If k = 1 − , then the equality holds also n n for a1 = 0 and a2 = a3 = · · · = an = (or any cyclic permutation). n−1

P 1.19. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = n. If 0 < k ≤ n−1 , then n2 − n + 1 n 1 1 1 ≤ . + + ··· + 2 2 2 1 + ka 1 + k 1 + ka1 1 + ka2 n (Vasile Cirtoaje, 2005) Solution. Replacing all negative numbers ai by −ai , we need to show the same inequality for a1 , a2 , . . . , an ≥ 0 such that a1 + a2 + · · · + an ≥ n. Since the left side of the desired inequality is decreasing with respect to each ai , is sufficient to consider that a1 + a2 + · · · + an = n. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

s=

a1 + a2 + · · · + an = 1, n

−1 , u ∈ [0, n]. 1 + ku2

For u ∈ [0, 1], we have f 00 (u) =

2k(1 − 3ku2 ) ≥ 0, (1 + ku2 )3

since 1 − 3ku2 ≥ 1 − 3k ≥ 1 −

3(n − 1) (n − 2)2 = ≥ 0. n2 − n + 1 n2 − n + 1

Thus, f is convex on [0, 1]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. We have g(u) =

f (u) − f (1) k(u + 1) = u−1 (1 + k)(1 + ku2 )

Half Convex Function Method and h(x, y) =

37

k − k2 (x + y + x y) g(x) − g( y) = . x−y (1 + k)(1 + k x 2 )(1 + k y 2 )

We need to show that 1 − k(x + y + x y) ≥ 0. Indeed, we have 1 − k(x + y + x y) ≥ 1 −

n−1 (x − n + 1)2 (x + y + x y) = ≥ 0. n2 − n + 1 n2 − n + 1

The equality holds for a1 = a2 = · · · = an = 1. If k = also for a1 = n − 1 and a2 = a3 = · · · = an =

n2

n−1 , then the equality holds −n+1

1 (or any cyclic permutation). n−1

P 1.20. Let a1 , a2 , . . . , an be nonnegative numbers such that a1 + a2 + · · · + an = n. If n2 , then k≥ 4(n − 1) a1 (a1 − 1) a12

+k

+

a2 (a2 − 1) a22

+k

+ ··· +

an (an − 1) ≥ 0. an2 + k (Vasile Cîrtoaje, 2012)

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) = From f 0 (u) =

u2 + 2ku − k , (u2 + k)2

s=

a1 + a2 + · · · + an = 1, n

u(u − 1) , u ∈ [0, n]. u2 + k f 00 (u) =

2(k2 − u3 ) + 6ku(1 − u) , (u2 + k)3

it follow that f is convex on [0, 1]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

We have g(u) =

u , 2 u +k

h(x, y) =

k−xy n2 − 4(n − 1)x y ≥ . (x 2 + k)( y 2 + k) 4(n − 1)(x 2 + k)( y 2 + k)

38

Vasile Cîrtoaje

We only need to show that n2 ≥ 4(n − 1)x y. Indeed, this follows by the AM-GM inequality, as follows: ” Æ —2 n2 = [x + (n − 1) y]2 ≥ 2 (n − 1)x y = 4(n − 1)x y. The equality holds for a1 = a2 = · · · = an = 1, and also for a1 = n/2 and a2 = a3 = · · · = an = n/(2n − 2) (or any cyclic permutation).

P 1.21. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then 1 − an 1 − a1 1 − a2 + + ··· + ≤ 0. 2 2 (n − 2a1 ) (n − 2a2 ) (n − 2an )2 (Vasile Cîrtoaje, 2012) Solution. For n = 2, the inequality is an identity. Consider further n ≥ 3 and write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) = From f 0 (u) =

s=

a1 + a2 + · · · + an = 1, n

u−1 , u ∈ I = [0, n] \ {n/2}. (n − 2u)2 2u + n − 4 , (n − 2u)3

f 00 (u) =

8(u + n − 3) , (n − 2u)4

it follow that f is convex on [0, 1]. By HCF Theorem, Remark 1 and Remark 4, it suffices to show that h(x, y) ≥ 0 for x, y ∈ I such that x + (n − 1) y = n, where h(x, y) =

g(x) − g( y) , x−y

We have g(u) = and h(x, y) =

g(u) =

f (u) − f (1) . u−1

1 (n − 2u)2

4(n − x − y) 4(n − 2) y = ≥ 0. 2 2 (n − 2x) (n − 2 y) (n − 2x)2 (n − 2 y)2

In accordance with Remark 3, the equality holds for a1 = a2 = · · · = an = 1, and also for a1 = n and a2 = a3 = · · · = an = 0 (or any cyclic permutation).

Half Convex Function Method

39

P 1.22. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If n , then k ≥1+ p n−1 a12 − 1 (a1 + k)2

+

a22 − 1 (a2 + k)2

+ ··· +

an2 − 1 (an + k)2

≥ 0. (Vasile Cirtoaje, 2008)

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

s=

a1 + a2 + · · · + an = 1, n

u2 − 1 , u ∈ [0, n]. (u + k)2

For u ∈ [0, 1], we have f 00 (u) =

2(k2 − 3 − 2ku) 2(k2 − 2k − 3) 2(k + 1)(k − 3) ≥ = ≥ 0. (u + k)4 (u + k)4 (u + k)4

Thus, f is convex on [0, 1]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. We have g(u) = and h(x, y) =

f (u) − f (1) u+1 = u−1 (u + k)2

g(x) − g( y) (k − 1)2 − (1 + x)(1 + y) = . x−y (x + k)2 ( y + k)2

Since (k − 1)2 ≥

n2 , n−1

we need to show that n2 ≥ (n − 1)(1 + x)(1 + y). Indeed, n2 − (n − 1)(1 + x)(1 + y) = n2 − (1 + x)(2n − 1 − x) = (x − n + 1)2 ≥ 0. Thus, the proof is completed. The equality holds for a1 = a2 = · · · = an = 1. If k = n 1 1+ p , then the equality holds also for a1 = n − 1 and a2 = a3 = · · · = an = n−1 n−1 (or any cyclic permutation).

40

Vasile Cîrtoaje

P 1.23. Let as 1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If 2n − 1 0< k ≤1+ , then n−1 a12 − 1 (a1 + k)2

+

a22 − 1 (a2 + k)2

+ ··· +

an2 − 1 (an + k)2

≤ 0. (Vasile Cirtoaje, 2008)

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

s=

a1 + a2 + · · · + an = 1, n

1 − u2 , u ∈ [0, n]. (u + k)2

For u ≥ 1, we have f 00 (u) =

2(2ku − k2 + 3) 2(2k − k2 + 3) 2(1 + k)(3 − k) ≥ = > 0. (u + k)4 (u + k)4 (u + k)4

Thus, f is convex on [1, n]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. We have g(u) = and h(x, y) =

f (u) − f (1) −u − 1 = u−1 (u + k)2

2k − k2 + x + y g(x) − g( y) 2k − k2 + x + y + x y = ≥ . x−y (x + k)2 ( y + k)2 (x + k)2 ( y + k)2

Since x+y≥

x + (n − 1) y n = , n−1 n−1

we get 2k − k2 + x + y ≥ 2k − k2 +

n 2n − 1 = −(k − 1)2 + ≥ 0, n−1 n−1

hence h(x, y) ≥ 0.sThus, the proof is completed. The equality holds for a1 = a2 = · · · = 2n − 1 an = 1. If k = 1 + , then the equality holds also for a1 = 0 and a2 = a3 = · · · = n−1 n an = (or any cyclic permutation). n−1

Half Convex Function Method

41

P 1.24. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If (n − 1)(2n − 1) k≥ , then n2 1 1 n 1 + + ··· + ≥ . 3 3 3 1 + kan 1+k 1 + ka1 1 + ka2 (Vasile Cirtoaje, 2008) Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) = For u ∈ [1, n], we have f 00 (u) =

s=

a1 + a2 + · · · + an = 1, n

1 , u ∈ [0, n]. 1 + ku3

6ku(2ku3 − 1) 6ku(2k − 1) ≥ > 0. (1 + ku3 )3 (1 + ku3 )3

Thus, f is convex on [1, n]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where h(x, y) =

g(x) − g( y) , x−y

We have g(u) = and

g(u) =

f (u) − f (1) . u−1

−k(u2 + u + 1) (1 + k)(1 + ku3 )

x 2 y 2 + x y(x + y − 1) + (x + y)2 − (x + y + 1)/k 1 h(x, y) = . k2 (1 + k)(1 + k x 3 )(1 + k y 3 )

Since x+y≥

x + (n − 1) y n = > 1, n−1 n−1

it suffices to show that k(x + y)2 ≥ x + y + 1. From x + y ≥ k(x + y) ≥

n , we get n−1

2n − 1 n

and

• ˜ n 2n − 1 − 1 = 1. n−1 n (n − 1)(2n − 1) The equality holds for a1 = a2 = · · · = an = 1. If k = , then the equality n2 n holds also for a1 = 0 and a2 = a3 = · · · = an = (or any cyclic permutation). n−1 k(x + y)2 − x − y ≥ (x + y)[k(x + y) − 1] ≥

42

Vasile Cîrtoaje

P 1.25. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If n−1 , then 0 0. (1 + ku3 )3 (1 + ku3 )3

Thus, f is convex on [0, 1]. By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where h(x, y) =

g(x) − g( y) , x−y

We have g(u) =

g(u) =

f (u) − f (1) . u−1

k(u2 + u + 1) (1 + k)(1 + ku3 )

and

(x + y + 1)/k − x 2 y 2 − x y(x + y − 1) − (x + y)2 1 h(x, y) = . k2 (1 + k)(1 + k x 3 )(1 + k y 3 ) It suffices to show that (n2 − 2n + 2)(x + y + 1) − x 2 y 2 − x y(x + y − 1) − (x + y)2 ≥ 0, n−1 which is equivalent to [2 + n y − (n − 1) y 2 ][1 − (n − 1) y]2 ≥ 0.

This is true since 2 + n y − (n − 1) y 2 = 2 + y[n − (n − 1) y] = 2 + x y > 0. In accord with Remark 3, the equality holds for a1 = a2 = · · · = an = 1. If k = n−1 1 , then the equality holds also for a1 = n − 1 and a2 = a3 = · · · = an = 2 n − 2n + 2 n−1 (or any cyclic permutation).

Half Convex Function Method

43

P 1.26. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If n2 k≥ , then n−1 v v v t a1 t a2 t an n + + ··· + ≤p . k − a1 k − a2 k − an k−1 (Vasile Cirtoaje, 2008) Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) = −

s

s=

a1 + a2 + · · · + an = 1, n

u , u ∈ [0, n]. k−u

For u ∈ [0, 1], we have f 00 (u) =

k(k − 4u) 4u3/2 (k − u)5/2

≥

k(k − 4) 4u3/2 (k − u)5/2

≥ 0.

Thus, f is convex on [0, 1]. By HCF Theorem, it suffices to prove that f (x) + (n − 1) f ( y) ≥ n f (1) for all x, y ≥ 0 such that x + (n − 1) y = n. We write the inequality as v v t (k − 1)x t (k − 1) y + (n − 1) ≤ n, k−x k− y – ™ v v t (k − 1) y t (n − 1)k(1 − y) 1+ ≤ 1 + (n − 1) 1 − . (n − 1) y + k − n k− y Let z=

v t (k − 1) y k− y

,

which yields y= 1− y =

(k − 1)(1 − z 2 ) , z2 + k − 1

kz 2 , z2 + k − 1

(n − 1) y + k − n =

(k − 1)(nz 2 + k − n) , z2 + k − 1

hence k(1 − y) k(1 − z 2 ) 1 = = 2 (n − 1) y + k − n k − n(1 − z ) 1/(1 − z 2 ) − n/k 1 n(1 − z 2 ) ≤ = . 1/(1 − z 2 ) − (n − 1)/n (n − 1)z 2 + 1

44

Vasile Cîrtoaje

Therefore, it suffices to show that v t n(n − 1)(1 − z 2 ) 1+ ≤ 1 + (n − 1)(1 − z). (n − 1)z 2 + 1 By squaring, we get the obvious inequality (z − 1)2 [(n − 1)z − 1]2 ≥ 0. So, we only need to show that 1 + (n − 1)(1 − z) ≥ 0, that is, z ≤ n/(n − 1). Since y= and

we have

n− x n ≤ n−1 n−1

1 n−1 1 ≤ 2 = 2 , k−1 n /(n − 1) − 1 n −n+1 v v v t t k−1 t k−1 n z= ≤ = k/ y − 1 k(n − 1)/n − 1 n − 1 − 1/(k − 1) p v t n n n2 − n + 1 ≤ = < . 2 n − 1 − (n − 1)/(n − n + 1) n−1 n−1

The proof is completed. The equality holds for a1 = a2 = · · · = an = 1. If k = n2 n(n − 1)2 , then the equality holds also for a1 = 2 and a2 = a3 = · · · = an = n−1 n − 2n + 2 n (or any cyclic permutation). 2 (n − 1)(n − 2n + 2)

P 1.27. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then 2 2 2 n−a1 + n−a2 + · · · + n−an ≥ 1. (Vasile Cîrtoaje, 2006) Solution. Let k = ln n. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where

s=

a1 + a2 + · · · + an = 1, n

2

f (u) = n−u , u ∈ [0, n]. For u ≥ 1, we have 2

2

2

f 00 (u) = 2kn−u (2ku2 − 1) ≥ 2kn−u (2k − 1) ≥ 2kn−u (2 ln 2 − 1) > 0;

Half Convex Function Method

45

therefore, f is convex on [1, n]. By HCF Theorem and Remark 5, it suffices to show that f (x) + (n − 1) f ( y) ≥ n f (1) for 0 ≤ x ≤ 1 ≤ y and x + (n − 1) y = n. The desired inequality is equivalent to g(x) ≥ 0, where 2

2

g(x) = n−x + (n − 1)n− y ,

y=

n− x , 0 ≤ x ≤ 1. n−1

Since y 0 = −1/(n − 1), we get 2

2

2

2

g 0 (x) = −2x kn−x − 2(n − 1)k y y 0 n− y = 2k( y n− y − x n−x ). The derivative g 0 (x) has the same sign as g1 (x), where 2

2

g1 (x) = ln( y n− y ) − ln(x n−x ) = ln y − ln x + k(x 2 − y 2 ). From g10 (x) =

• ˜ y0 1 −1 2k + 2(n − 2)k x − + 2k(x − y y 0 ) = n + , y x x(n − x) (n − 1)2

we see that g10 (x) has for 0 < x ≤ 1 the same sign as h(x) =

−(n − 1)2 + x(n − x)(1 + nx − 2x). 2k

Since h0 (x) = n + 2(n2 − 2n − 1)x − 3(n − 2)x 2 ≥ nx + 2(n2 − 2n − 1)x − 3(n − 2)x = 2(n − 1)(n − 2)x ≥ 0, h is strictly increasing on [0, 1]. From −(n − 1)2 h(0) = < 0, 2k

 ‹ 1 h(1) = (n − 1) 1 − > 0, 2k 2

it follows that there is x 1 ∈ (0, 1) such that h(x) < 0 for x ∈ [0, x 1 ), h(x 1 ) = 0 and h(x) > 0 for x ∈ (x 1 , 1]. Therefore, g1 is strictly decreasing on (0, x 1 ] and strictly increasing on [x 1 , 1]. Since lim x→0 g1 (x) = ∞ and g1 (1) = 0, there is x 2 ∈ (0, x 1 ) such that g1 (x) > 0 for x ∈ (0, x 2 ), g1 (x 2 ) = 0 and g1 (x) < 0 for x ∈ (x 2 , 1). Consequently, g is strictly increasing on [0, x 2 ] and strictly decreasing on [x 2 , 1]. From g(0) > 0 and g(1) = 0, it follows that g(x) ≥ 0 for x ∈ [0, 1]. The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

P 1.28. If a, b, c, d, e are nonnegative real numbers such that a + b + c + d + e = 5, then then (a2 + 1)(b2 + 1)(c 2 + 1)(d 2 + 1)(e2 + 1) ≥ (a + 1)(b + 1)(c + 1)(d + 1)(e + 1).

46

Vasile Cîrtoaje

Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) + f (e) ≥ n f (s),

s=

a+b+c+d+e = 1, 5

where f (u) = ln(u2 + 1) − ln(u + 1),

u ∈ [0, 5].

For u ∈ [0, 1], we have f 00 (u) =

2(1 − u2 ) 1 (u2 − u4 ) + 4u(1 − u2 ) + u2 + 3 + = > 0. (u2 + 1)2 (u + 1)2 (u2 + 1)2 (u + 1)2

Therefore, f is convex on [0, 1]. By HCF Theorem and Remark 2, we only need to show that H(x, y) ≥ 0 for x, y ≥ 0 such that x + 4 y = 5, where H(x, y) =

2(1 − x y) f 0 (x) − f 0 ( y) 1 = 2 + ; x−y (x + 1)( y 2 + 1) (x + 1)( y + 1)

that is, (x 2 + 1)( y 2 + 1)H(x, y) = 2(1 − x y) + Since

x2 + 1 x +1 ≥ , x +1 2

(x 2 + 1)( y 2 + 1) . (x + 1)( y + 1)

y2 + 1 y +1 ≥ , y +1 2

it suffices to prove that 2(1 − x y) +

(x + 1)( y + 1) ≥ 0, 4

which is equivalent to x + y + 9 − 7x y ≥ 0. Indeed, x + y + 9 − 7x y = 28x 2 − 38x + 14 = 28(x − 19/28)2 + 31/28 > 0. The proof is completed. The equality holds for a = b = c = d = e = 1.

P 1.29. If a1 , a2 , · · · , a10 are nonnegative real numbers such that a1 + a2 + · · · + a10 = 10, then 2 (1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a10 + a10 ) ≥ 1. (Vasile Cîrtoaje, 2006)

Half Convex Function Method

47

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (a10 ) ≥ 10 f (s),

s=

a1 + a2 + · · · + a10 = 1, 10

where f (u) = ln(1 − u + u2 ), From f 00 (u) =

u ∈ [0, 10].

1 + 2u(1 − u) , (1 − u + u2 )2

it follows that f 00 (u) > 0 for u ∈ [0, 1], hence f is convex on [0, 1]. According to HCF Theorem, we only need to show that f (x) + 9 f ( y) ≥ 10 f (1) for all x, y ≥ 0 such that x + 9 y = 10. Using Remark 2, it suffices to prove that H(x, y) ≥ 0, where H(x, y) = Since H(x, y) =

f 0 (x) − f 0 ( y) . x−y

1 + x + y − 2x y , (1 − x + x 2 )(1 − y + y 2 )

we need to show that 1 + x + y − 2x y ≥ 0. Indeed, 7 1 + x + y − 2x y = 18 y − 28 y + 11 = 18 y − 9 

2

‹2

+

1 > 0. 9

The proof is completed. The equality holds for a1 = a2 = · · · = a10 = 1.

P 1.30. If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers such that a1 + a2 + · · · + an = 1, then       ‹ p p p 1 1 1 1 n p n− p . p − a2 · · · p − an ≥ p − a1 a1 a2 an n (Vasile Cîrtoaje, 2006) Solution. Apply HCF Theorem to the function  ‹ p 1 1 f (u) = ln p − u = ln(1 − u) − ln u, 2 u

u ∈ (0, 1),

for s = 1/n. From f 0 (u) =

−1 1 − , 1 − u 2u

f 00 (u) =

1 − 2u − u2 , 2u2 (1 − u)2

48

Vasile Cîrtoaje

p it follows that f is convex on (0, 2 − 1]. Since s=

1 1 p ≤ < 2 − 1, n 3

f is also convex on (0, s]. First Solution. By HCF Theorem, it suffices to show that f (x) + (n − 1) f ( y) ≥ n f (1/n) for all x, y > 0 such that x + (n − 1) y = 1; that is, to show that 

p 1 p − x x

‹

1 p p − y y

n−1 ≥

 p

1 n− p n

‹n

.

Write this inequality as nn/2 (1 − y)n−1 ≥ (n − 1)n−1 x 1/2 y (n−3)/2 . By squaring, this inequality becomes as follows nn (1 − y)2n−2 ≥ (n − 1)2n−2 x y n−3 , (2 − 2 y)2n−2 ≥ •

1 n · + x + (n − 3) y n

˜2n−2

(2n − 2)2n−2 x y n−3 , nn

≥ [n + 1 + (n − 3)]n+1+(n−3) ·

1 · x · y n−3 . nn

Clearly, the last inequality follows from the AM-GM inequality. The proof is completed. The equality holds for a1 = a2 = · · · = an = 1/n. Second Solution. By HCF Theorem and Remark 2, it suffices to prove that H(x, y) ≥ 0 for x, y > 0 such that x + (n − 1) y = 1, where H(x, y) =

f 0 (x) − f 0 ( y) . x−y

We have (n − 1)(1 + y) − 2 1− x − y − xy = 2x y(1 − x)(1 − y) 2x(1 − x)(1 − y) 2(1 + y) − 2 y ≥ = > 0. 2x(1 − x)(1 − y) x(1 − x)(1 − y)

H(x, y) =

Remark 1. We may write the inequality in P 1.30 in the form  Y  ‹ n  n Y p p 1 n 1 − 1 · (1 + a ) ≥ n − . p p i ai n i=1 i=1

Half Convex Function Method

49

On the other hand, by the AM-GM inequality and the Cauchy-Schwarz inequality, we have v !n ‚ Œn  ‹ u X n n n Y p 1 n 1 Xp t1 ai = 1+ p ai ≤ 1 + (1 + ai ) ≤ 1 + . n i=1 n i=1 n i=1 Thus, the following statement follows. • If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers such that a1 + a2 + · · · + an = 1, then      p 1 1 1 n p − 1 · · · p − 1 ≥ ( n − 1) , p −1 a1 a2 an with equality for a1 = a2 = · · · = an = 1/n. Remark 2. By squaring, the inequality in P 1.30 becomes n Y (1 − ai )2 (n − 1)2n . ≥ ai nn i=1

1+ x is convex on (0, 1), by Jensen’s 1− x inequality,   a1 + a2 + · · · + an n ‹ ‹   n 1+ Y 1 + ai n+1 n   n ≥ . = a1 + a2 + · · · + an  1 − ai n−1 i=1 1− n Multiplying these inequality yields the following result (Kee-Wai Lau, 2000).

On the other hand, since the function f (x) = ln

• If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers such that a1 + a2 + · · · + an = 1, then ‹ ‹  ‹  ‹  1 1 1 n 1 − a1 − a2 · · · − an ≥ n − , a1 a2 an n with equality for a1 = a2 = · · · = an = 1/n.

P 1.31. Let a1 , a2 , . . . , an be positive real numbers such that a1 + a2 + · · · + an = n. If  2 p 2 n−1 0< k ≤ 1+ , n then



1 ka1 + a1

‹

‹  ‹ 1 1 ka2 + · · · kan + ≥ (k + 1)n . a2 an (Vasile Cîrtoaje, 2006)

50

Vasile Cîrtoaje

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where

s=

 ‹ 1 f (u) = ln ku + , u

We have f 0 (u) =

ku2 − 1 , u(ku2 + 1)

f 00 (u) =

a1 + a2 + · · · + an = 1, n

u ∈ (0, n). 1 + 4ku2 − k2 u4 . u2 (ku2 + 1)2

For u ∈ (0, 1], we have f 00 (u) > 0, since 1 + 4ku2 − k2 u4 > ku2 (4 − ku2 ) ≥ ku2 (4 − k) ≥ 0. Therefore, f is convex on (0, 1]. By HCF Theorem and Remark 2, it suffices to prove that H(x, y) ≥ 0 for x, y > 0 such that x + (n − 1) y = n, where H(x, y) = Since H(x, y) =

f 0 (x) − f 0 ( y) . x−y

1 + k(x + y)2 − k2 x 6 y 2 k[(x + y)2 − k x 6 y 2 ] > , x y(k x 2 + 1)(k y 2 + 1) x y(k x 2 + 1)(k y 2 + 1)

it suffices to show that x+y≥

p

k x y.

Indeed, by the Cauchy-Schwarz inequality, we have p (x + y)[(n − 1) y + x] ≥ ( n − 1 + 1)2 x y, hence

p 1 p ( n − 1 + 1)2 x y ≥ k x y. n The equality holds for a1 = a2 = · · · = an = 1. x+y≥

P 1.32. If a, b, c, d are nonzero real numbers such that a, b, c, d ≥ then

−1 , 2

a + b + c + d = 4,

‹ 1 1 1 1 1 1 1 1 3 2 + 2 + 2 + 2 + + + + ≥ 16. a b c d a b c d 

Half Convex Function Method

51

Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) ≥ 4 f (s), s = where

a+b+c+d = 1, 4

• ˜ 3 1 −1 11 f (u) = 2 + , u ∈ I = , \ {0}. u u 2 2

Clearly, f is convex for u ∈ I, u ≥ s. Therefore, by HCF Theorem and Remark 4, it suffices to prove that f (x) + 3 f ( y) ≥ 4 f (1) for all x, y ∈ I such that x + 3 y = 4. According to Remark 1, we only need to show that h(x, y) ≥ 0, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

Indeed, we have 3 4 g(u) = − − 2 , u u h(x, y) =

4x y + 3x + 3 y 2(1 + 2x)(6 − x) = ≥ 0. 2 2 x y 3x 2 y 2

The proof is completed. In accord with Remark 3, the equality holds for a = b = c = −1 3 d = 1, and also for a = and b = c = d = (or any cyclic permutation). 2 2

P 1.33. If a1 , a2 , . . . , an are nonnegative real numbers such that a12 + a22 + · · · + an2 = n, then s n 3 3 3 a1 + a2 + · · · + an − n + (a1 + a2 + · · · + an − n) ≥ 0. n−1 (Vasile Cirtoaje, 2007) Solution. Replacing each ai by

p

ai , we have to prove that a1 + a2 + · · · + an = 1, n

f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

p p f (u) = u u + k u,

n , n−1

where

For u ≥ 1, we have f 00 (u) =

k=

s

u ∈ [0, n].

3u − k 3−k p ≥ p > 0. 4u u 4u u

52

Vasile Cîrtoaje

Therefore, f is convex for u ≥ 1. According to HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. Since g(u) = and

f (u) − f (1) u+k =1+ p u−1 u+1

p p p x + y+ xy−k g(x) − g( y) h(x, y) = = p , p p p x−y ( x + y)( x + 1)( y + 1)

we need to show that

p

x+

p

y+

This is true if p

x+y≥

p

x y ≥ k.

s

n . n−1

Indeed, we have

x n +y= . n−1 n−1 The proof is completed. In accord with Remark 3, sthe equality holds for a1 = a2 = · · · = n an = 1, and also for a1 = 0 and a2 = · · · = an = (or any cyclic permutation). n−1 x+y≥

P 1.34. If a, b, c, d, e are nonnegative real numbers such that a2 + b2 + c 2 + d 2 + e2 = 5, then 1 1 1 1 1 + + + + ≤ 1. 7 − 2a 7 − 2b 7 − 2c 7 − 2d 7 − 2e (Vasile Cirtoaje, 2010) p p p p p Solution. Replacing a, b, c, d, e by a, b, c, d, e, we have to prove that f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s), where

1 f (u) = p , 2 u−7

For u ∈ [0, 1], we have

s=

a+b+c+d+e = 1, 5

u ∈ [0, 5].

p 7−6 u f (u) = p > 0. p 2u u(7 − 2 u)3 00

Therefore, f is convex for u ∈ [0, 1]. According to HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + 4 y = 5. Since g(u) =

f (u) − f (1) −2 = p p u−1 5(7 − 2 u)(1 + u)

Half Convex Function Method

53

and p p 2(5 − 2 x − 2 y) g(x) − g( y) = p h(x, y) = p p p p p , x−y ( x + y)(1 + x)(1 + y)(7 − 2 x)(7 − 2 y) we need to show that

p

5 . 2 Indeed, by the Cauchy-Schwarz inequality, we have  ‹ p 1 25 p 2 ( x + y) ≤ 1 + (x + 4 y) = . 4 4 x+

p

y≤

The proof is completed. The equality holds for a = b = c = d = e = 1, and also for 1 a = 2 and b = c = d = e = (or any cyclic permutation). 2 Remark Similarly, we can prove the following generalization. • Let a1 , a2 , . . . , an be nonnegative real numbers such that a12 + a22 + · · · + an2 = n. If n k ≥1+ p , then n−1 1 1 1 n + + ··· + ≤ , k − a1 k − a2 k − an k−1 with equality for a1 = a2 = · · · = an = 1. If k = 1 + p for a1 =

p

n − 1 and a2 = · · · = an = p

P 1.35. If 0 ≤ a1 , a2 , . . . , an < then

1 n−1

n n−1

, then the equality holds also

.

k such that a12 +a22 +· · ·+an2

= n, where 1 < k ≤ 1+

s

n , n−1

1 1 1 n + + ··· + ≥ . k − a1 k − a2 k − an k−1 (Vasile Cirtoaje, 2010)

Solution. Replacing a1 , a2 , . . . , an by

p

a1 ,

p

a2 , . . . ,

f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

1 p , k− u

p

s=

an , we have to prove that a1 + a2 + · · · + an = 1, n

u ∈ [0, k2 ).

54

Vasile Cîrtoaje

From

p 3 u−k f (u) = p , p 4u u(k − u)3 00

it follows that f is convex for u ≥ 1, since p 3 u−k ≥3−k ≥2−

s

n > 0. n−1

According to HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for all 0 ≤ x, y < k2 such that x + (n − 1) y = n. Since g(u) =

f (u) − f (1) 1 = p p u−1 (k − 1)(k − u)(1 + u)

and p p x + y +1−k g(x) − g( y) h(x, y) = = p p p p p p , x−y (k − 1)( x + y)(1 + x)(1 + y)(k − x)(k − y) we need to show that

p

x+

p

y ≥ k − 1.

Indeed, p

x+

p

y≥

p

x+y≥

s

x +y= n−1

s

n ≥ k − 1. n−1

The proof is completed. In accord with Remark 3, sthe equality holds for a1 = a2 = · · · = n an = 1, and also for a1 = 0 and a2 = · · · = an = (or any cyclic permutation). n−1

P 1.36. If a, b, c are nonnegative real numbers, no two of which are zero, then v v v t 48a t 48b t 48c 1+ + 1+ + 1+ ≥ 15. b+c c+a a+b (Vasile Cirtoaje, 2005) Solution. Due to homogeneity, we may assume that a + b + c = 1. Thus, we need to show that a+b+c 1 f (a) + f (b) + f (c) ≥ 3 f (s), s = = , 3 3 where v t 1 + 47u f (u) = , u ∈ [0, 1). 1−u

Half Convex Function Method

55

From f 00 (u) = p

48(47u − 11) (1 − u)5 (1 + 47u)3

,

it follows that f is convex on [1/3, 1). By HCF Theorem, it suffices to show that f (x) + 2 f ( y) ≥ 3 f (1/3) for x, y ≥ 0 such that x + 2 y = 1; that is, v v t 1 + 47x t 49 − 47x +2 ≥ 15. 1− x 1+ x Setting t=

v t 49 − 47x 1+ x

, 1 < t ≤ 7,

the inequality turns into v t 1175 − 23t 2 t2 − 1

≥ 15 − 2t.

By squaring, this inequality becomes 350 − 15t − 61t 2 + 15t 3 − t 4 ≥ 0, (5 − t)2 (2 + t)(7 − t) ≥ 0. The last inequality is clearly true. From x + 2 y = 1 and f (x) + 2 f ( y) = 3 f (1/3), we get x = y = 1/3 and x = 0, y = 1/2. Therefore, in accordance with Remark 3, the equality in the case a + b + c = 1 holds for a = b = c = 1/3, and also for a = 0 and b = c = 1/2 (or any cyclic permutation). For the original inequality, the equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 1.37. If a, b, c are nonnegative real numbers, then v v v t t t 3a2 3b2 3c 2 + + ≤ 1. 7a2 + 5(b + c)2 7b2 + 5(c + a)2 7c 2 + 5(a + b)2 (Vasile Cirtoaje, 2008) Solution. Due to homogeneity, we may assume that a + b + c = 3. Thus, we need to show that a+b+c f (a) + f (b) + f (c) ≥ 3 f (s), s = = 1, 3 where v t 3u2 −u =p , u ∈ [0, 3]. f (u) = − 2 2 7u + 5(3 − u) 4u2 − 10u + 15

56

Vasile Cîrtoaje

From f 00 (u) =

10(u − 1)(15 − 4u) 5(−8u2 + 41u − 30) 5(−8u2 + 38u − 30) ≥ = , 2 5/2 2 5/2 (4u − 10u + 15) (4u − 10u + 15) (4u2 − 10u + 15)5/2

it follows that f is convex on [1, 3]. By HCF Theorem, it suffices to prove the original homogeneous inequality for b = c = 1; that is v v t 3a2 t 3 + 2 ≤ 1. 7a2 + 20 5a2 + 10a + 12 By squaring two times, the inequality becomes Æ a(5a3 + 10a2 + 16a + 50) ≥ 3a (7a2 + 20)(5a2 + 10a + 12), a2 (5a6 + 20a5 − 11a4 + 38a3 − 80a2 − 40a + 68) ≥ 0, a2 (a − 1)2 (5a4 + 30a3 + 44a2 + 96a + 68) ≥ 0. The last inequality is clearly true. The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).

P 1.38. If a, b, c are nonnegative real numbers, then v v v t t t a2 b2 c2 + + ≥ 1. a2 + 2(b + c)2 b2 + 2(c + a)2 c 2 + 2(a + b)2 (Vasile Cirtoaje, 2008) Solution. Due to homogeneity, we may assume that a + b + c = 3. Thus, we need to show that a+b+c f (a) + f (b) + f (c) ≥ 3 f (s), s = = 1, 3 where v t 3u2 u =p , u ∈ [0, 3]. f (u) = 2 2 u + 2(3 − u) u2 − 4u + 6 From f 00 (u) =

2(2u2 − 11u + 12) 2(−11u + 12) ≥ 2 , 2 5/2 (u − 4u + 6) (u − 4u + 6)5/2

it follows that f is convex on [0, 1]. By HCF Theorem, it suffices to prove the original homogeneous inequality for b = c = 1; that is p

2 +p ≥ 1. 2a2 + 4a + 3 a2 + 8 a

Half Convex Function Method

57

By squaring, the inequality becomes Æ a (a2 + 8)(2a2 + 4a + 3) ≥ 3a2 + 8a − 2. For the nontrivial case 3a2 + 8a − 2 > 0, squaring both sides, we get a6 + 2a5 + 5a4 − 8a3 − 14a2 + 16a − 2 ≥ 0, (a − 1)2 [a4 + 4a3 + 9a2 + 4a + (3a2 + 8a − 2)] ≥ 0. The last inequality is clearly true. The equality holds for a = b = c.

P 1.39. Let a, b, c be nonnegative real numbers, no two of which are zero. If k0 =

k ≥ k0 , then 

2a b+c

‹k

+



ln 3 − 1 ≈ 0.585, ln 2

2b c+a

‹k

+



2c a+b

‹k

≥ 3. (Vasile Cirtoaje, 2005)

Solution. For k = 1, the inequality is just the well known Nesbitt’s inequality 2b 2c 2a + + ≥ 3, b+c c+a a+b while for k ≥ 1, the inequality follows from Jensens’s inequality applied to the convex function f (u) = uk : 

2a b+c

‹k

2b + c+a 

‹k

2c + a+b 

‹k

‚ ≥3

2a b+c

+

2b c+a

+

2c a+b

Œk

3

≥ 3.

Consider now that k0 ≤ k < 1. Due to homogeneity, we may assume that a + b + c = 1. Thus, we need to show that f (a) + f (b) + f (c) ≥ 3 f (s), where 2u f (u) = 1−u 

‹k

,

From f 00 (u) =

4k (1 − u)4



2u 1−u

s=

a+b+c 1 = , 3 3

u ∈ [0, 1). ‹k−2

(2u + k − 1),

58

Vasile Cîrtoaje

it follows that f is convex on [1/3, 1) (since u ≥ 1/3 involves 2u + k − 1 ≥ 2/3 + k − 1 = k −1/3 > 0). By HCF Theorem, it suffices to prove the original homogeneous inequality for b = c = 1; that is, to show that h(a) ≥ 3 for all a ≥ 0, where  ‹k 2 k h(a) = a + 2 . a+1 For a > 0, the derivative 0

h (a) = ka

k−1

2 −k a+1 

‹k+1

has the same sign as g(a) = (k − 1) ln a − (k + 1) ln From g 0 (a) =

2 . a+1

2ka + k − 1 , a(a + 1)

it follows that g 0 (a) = 0 for a0 = (1 − k)/(2k) < 1, g 0 (a) < 0 for a ∈ (0, a0 ) and g 0 (a) > 0 for a ∈ (a0 , ∞). Then, g is strictly decreasing on (0, a0 ] and strictly increasing on (a0 , ∞). Since lima→0 g(a) = ∞ and g(1) = 0, there exists a1 ∈ (0, a0 ) such that g(a1 ) = 0, g(a) > 0 for a ∈ (0, a1 ) ∪ (1, ∞) and g(a) < 0 for a ∈ (a1 , 1); hence, h(a) is strictly increasing on [0, a1 ] ∪ [1, ∞) and strictly decreasing on [a1 , 1]. Consequently, h(a) ≥ min{h(0), h(1)}. Since h(0) = 2k+1 ≥ 3 and h(1) = 3, we get h(a) ≥ 3. The proof is completed. The equality holds for a = b = c. If k = k0 , then the equality holds also for a = 0 and b = c (or any cyclic permutation). Remark. For k = 2/3, we can give the following solution (based on the AM-GM inequality): X  2a ‹2/3 X 2a = p 3 b+c 2a · (b + c) · (b + c) X 6a ≥ = 3. 2a + (b + c) + (b + c)

p P 1.40. If a, b, c ∈ [1, 7 + 4 3], then v v v t 2b t 2a t 2c + + ≥ 3. b+c c+a a+b (Vasile Cirtoaje, 2007)

Half Convex Function Method

59

Solution. Denoting s=

a+b+c , 3

we need to show that f (a) + f (b) + f (c) ≥ 3 f (s), where

v t 2u f (u) = , 1 ≤ u < 3s. 3s − u

For u ≥ s, we have 3s − u f (u) = 3s 2u 00



‹3/2

4u − 3s > 0. (3s − u)4

Therefore, f is convex for u ≥ s. By HCF Theorem, it suffices to prove the original inequality for b = c; that is, v s t 2b a +2 ≥ 3. b a+b v tb p Putting t = , the condition a, b ∈ [1, 7 + 4 3] involves a p p 2 − 3 ≤ t ≤ 2 + 3. We need to show that

v t 2t 2 1 2 ≥3− . t2 + 1 t

This is true if

 ‹ 8t 2 1 2 ≥ 3− . t2 + 1 t This is equivalent to the obvious inequality (t − 1)2 (t 2 − 4t + 1) ≥ 0.

The proof is completed. In accord p with Remark 3, the equality holds for a = b = c, and also for a = 1 and b = c = 7 + 4 3 (or any cyclic permutation).

P 1.41. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If 0 < k ≤ k0 ,

k0 =

ln 2 ≈ 1.71, ln 3 − ln 2

then a k (b + c) + b k (c + a) + c k (a + b) ≤ 6.

60

Vasile Cîrtoaje

Solution. For 0 < k ≤ 1, the inequality follows from Jensens’s inequality applied to the convex function f (u) = uk : ˜ • (b + c)a + (c + a)b + (a + b)c k (b + c)a k + (c + a)b k + (a + b)c k ≤ 2(a + b + c) 2(a + b + c)  ‹  ‹ a b + bc + ca k a + b + c 2k =6 ≤6 = 6. 3 3 Consider now that 1 < k ≤ k0 , and write the inequality as f (a) + f (b) + f (c) ≥ 3 f (s), s =

a+b+c = 1, 3

where f (u) = uk (u − 3), u ∈ [0, 3]. For u ≥ 1, we have f 00 (u) = kuk−2 [(k + 1)u − 3k + 3] ≥ kuk−2 [(k + 1) − 3k + 3] = 2k(2 − k)uk−2 > 0; therefore, f is convex for u ≥ 1. By HCF Theorem, it suffices to consider the case where two of a, b, c are equal. Write the desired inequality in the homogeneous form  ‹ a + b + c k+1 k k k a (b + c) + b (c + a) + c (a + b) ≤ 6 . 3 Since this inequality is trivial for b = c = 0, we may consider that b = c = 1. So, we need to show that g(a) ≥ 0 for a ≥ 0, where  ‹ a + 2 k+1 g(a) = 3 − a k − a − 1. 3 We have a+2 g (a) = (k + 1) 3 0



‹k

− ka

k−1

− 1,

1 00 k+1 g (a) = k 3



a+2 3

‹k−1

−

k−1 . a2−k

Since g 00 is strictly increasing, lima→0 g(a) = −∞ and g 00 (1) = 2k(2 − k)/3 > 0, there exists a1 ∈ (0, 1) such that g 00 (a1 ) = 0, g 00 (a) < 0 for a ∈ (0, a1 ), and g 00 (a) > 0 for a > 1. Therefore, g 0 is strictly decreasing on [0, a1 ] and strictly increasing on [a1 , ∞). Since g 0 (0) = (k + 1)(2/3)k − 1 ≥ (k + 1)(2/3)k0 − 1 =

k+1 k−1 −1= > 0, 2 2

g 0 (1) = 0,

there exists a2 ∈ (0, a1 ) such that g 0 (a2 ) = 0, g 0 (a) > 0 for a ∈ [0, a2 ) ∪ (1, ∞), and g 0 (a) < 0 for a ∈ (a2 , 1). Thus, g is strictly increasing on [0, a2 ] ∪ [1, ∞) and strictly decreasing on [a2 , 1]. Consequently, g(a) ≥ min{g(0), g(1)},

Half Convex Function Method

61

and from g(0) = 3(2/3)k+1 − 1 ≥ 3(2/3)k0 +1 − 1 = 1 − 1 = 0,

g(1) = 0,

we get g(a) ≥ 0. This completes the proof. The equality holds for a = b = c = 1. If k = k0 , then the equality holds also for a = 0 and b = c = 3/2 (or any cyclic permutation). Remark 1. Using the Cauchy-Schwarz inequality, we get X

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

Thus, the following statement holds. • Let a, b, c be nonnegative real numbers such that a + b + c = 3. If 0 < k ≤ k0 , then

k0 =

ln 2 ≈ 1.71, ln 3 − ln 2

b c 3 a + + ≥ , bk + c k c k + ak ak + bk 2

with equality for a = b = c = 1. If k = k0 , then the equality holds also for a = 0 and b = c = 3/2 (or any cyclic permutation). Remark 2. Also, the following statement holds. • Let a, b, c be nonnegative real numbers such that a + b + c = 3. If k ≥ k1 ,

k1 =

ln 9 − ln 8 ≈ 0.2905, ln 3 − ln 2

then ak bk ck 3 + + ≥ , b+c c+a a+b 2 with equality for a = b = c = 1. If k = k1 , then the equality holds also for a = 0 and b = c = 3/2 (or any cyclic permutation). For k1 ≤ k ≤ 2, the inequality can be proved using the Cauchy-Schwarz inequality, as follows: X ak (a + b + c)2 9 3 ≥P =P = . 2−k 2−k b+c a (b + c) a (b + c) 2

62

Vasile Cîrtoaje

For k > 2, the inequality can be deduced from the Cauchy-Schwarz inequality and Bernoulli’s inequality, as follows: P k/2
2 P k/2
2 X ak a a ≥ P = , b+c 6 (b + c) ˜ X X• k k/2 a ≥ 1 + (a − 1) = 3. 2

P 1.42. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If n2 , then k≥ 4(n − 1) (a12 + k)(a22 + k) · · · (an2 + k) ≥ (1 + k)n . Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

where f (u) = ln(u2 + k),

u ∈ [0, n].

We have f 0 (u) =

2u , u2 + k

f 00 (u) =

2(k − u2 ) . (u2 + k)2

For u ∈ [0, 1], we have f 00 (u) > 0, since k − u2 ≥ k − 1 ≥

n2 − 1 ≥ 0. 4(n − 1)

Therefore, f is convex on [0, 1]. By HCF Theorem and Remark 2, it suffices to prove that H(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where H(x, y) =

f 0 (x) − f 0 ( y) . x−y

Since

2(k − x y) 2[n2 − 4(n − 1)x y] ≥ , (x 2 + k)( y 2 + k) 4(n − 1)(x 2 + k)( y 2 + k) we only need to show that n2 ≥ 4(n − 1)x y. H(x, y) =

Indeed, we have n2 = [x + (n − 1) y]2 ≥ 4(n − 1)x y. The equality holds for a1 = a2 = · · · = an = 1.

Half Convex Function Method

63

P 1.43. Let a, b, c be nonnegative real numbers such a + b + c = 3. If k ≥ k0 , where p p p 6−2 = (2 + 2)(2 + 3) ≈ 12.74 , k0 = p p 6− 2−1 then p

a+

p

b+

p

c−3≥ k

‚v ta + b 2

+

v tb+c 2

+

s

Œ c+a −3 . 2 (Vasile Cirtoaje, 2008)

Solution. By the Cauchy-Schwarz inequality (1 + 1 + 1)[(a + b) + (b + c) + (c + a)] ≥

€p

a+b+

p

b+c+

p

c+a

Š2

,

we get v ta + b 2

+

v tb+c 2

+

s

c+a ≤ 3. 2

Therefore, it suffices to consider the case k = k0 . Write the inequality as f (a) + f (b) + f (c) ≥ 3 f (s), where

s=

v t3 − u p f (u) = u − k0 , 2

a+b+c = 1, 3

u ∈ [0, 3).

For u ≥ 1, we have 00

4 f (u) = −u

−3/2

k0 + 4



3−u 2

‹−3/2

≥ −1 +

k0 > 0. 4

Therefore, f is convex on [1, 3]. By HCF Theorem and Remark 5, it suffices to consider only the case a ≤ 1 ≤ b = c. Write the original inequality in the homogeneous form v v v ‚v Œ ta + b t b + c sc + a ta + b + c ta + b + c p p p a+ b+ c−3 ≥ k0 + + −3 . 3 2 2 2 3 Due to homogeneity, we may assume that b = c = 1. Moreover, it is convenient to use p the substitution x = a. Thus, we need to show that g(x) ≥ 0 for x ∈ [0, 1], where v v t x2 + 1 t x2 + 2 g(x) = x + 2 − k0 + 3(k0 − 1) − 2k0 . 3 2 We have 0

g (x) = 1 + (k0 − 1)x

v t

v t 2 3 − k x , 0 x2 + 2 x2 + 1

64

Vasile Cîrtoaje k0 g 00 (x) = 2



2 2 x +1

‹3/2 –

x2 + 1 m· 2 x +2

3/2

™ −1 ,

where v u  ‹ t 1 2 3 m= 6 1− ≈ 1.72. k0 Clearly, g 00 (x) has the same sign as h(x), where x2 + 1 − 1. x2 + 2

h(x) = m · Since h(0) =

m − 1 < 0, 2

h(1) =

2m − 1 > 0, 3

there is x 1 ∈ (0, 1) such that h(x) < 0 for x ∈ [0, x 1 ), h(x 1 ) = 0 and h(x) > 0 for x ∈ (x 1 , 1]. Therefore, g 0 is strictly decreasing on [0, x 1 ] and strictly increasing on [x 1 , 1]. Since g 0 (0) = 1 and g 0 (1) = 0, there is x 2 ∈ (0, x 1 ) such that g 0 (x) > 0 for x ∈ [0, x 2 ), g 0 (x 2 ) = 0 and g 0 (x) < 0 for x ∈ (x 2 , 1). Thus, g(x) is strictly p increasing p on [0, x 2 ] and strictly decreasing on [x 2 , 1]. From g(0) = 2 − k0 + (k0 − 1) 6 − k0 2 = 0 and g(1) = 0, it follows that g(x) ≥ 0 for x ∈ [0, 1]. This completes the proof. The equality holds for a = b = c. If k = k0 , then the equality holds also for a = 0 and b = c = 3/2 (or any cyclic permutation).

P 1.44. Let a, b, c be nonnegative real numbers such a + b + c = 3. If k ≤ k1 , where p p p k1 = ( 3 − 1)( 3 + 2) ≈ 2.303 , then p

a+

p

b+

p

c−3≤ k

‚v ta + b 2

+

v tb+c 2

+

s

Œ c+a −3 . 2 (Vasile Cirtoaje, 2008)

Solution. By the Cauchy-Schwarz inequality (1 + 1 + 1)[(a + b) + (b + c) + (c + a)] ≥

€p

a+b+

p

we get v ta + b 2

+

v tb+c 2

+

s

c+a ≤ 3. 2

b+c+

p

c+a

Š2

,

Half Convex Function Method

65

Therefore, it suffices to consider the case k = k1 . Write the inequality as f (a) + f (b) + f (c) ≥ 3 f (s), where

s=

v t3 − u p f (u) = − u + k1 , 2

a+b+c = 1, 3

u ∈ [0, 3).

For 0 ≤ u ≤ 1, we have 4 f 00 (u) = u−3/2 −

k1 4



‹−3/2

3−u 2

≥1−

k1 > 0. 4

Therefore, f is convex on [0, 1]. By HCF Theorem and Remark 5, it suffices to consider only the case a ≥ 1 ≥ b = c. Write the original inequality in the homogeneous form p

a+

p

b+

p

c−3

v ta + b + c 3

≤ k1

‚v ta + b 2

+

v tb+c 2

+

s

v Œ ta + b + c c+a −3 . 2 3

Due to homogeneity, we may assume that b = c = 1. Moreover, it is convenient to use p the substitution x = a. Thus, we need to show that g(x) ≤ 0 for x ≥ 1, where v v t x2 + 1 t x2 + 2 − 2k1 . g(x) = x + 2 − k1 + 3(k1 − 1) 3 2 We have

v t

v t 2 3 g (x) = 1 + (k1 − 1)x − k x , 1 x2 + 2 x2 + 1 ™ 3/2  ‹3/2 – 2 k 2 x + 1 1 g 00 (x) = m· 2 −1 , 2 x2 + 1 x +2 0

where

v u  ‹ t 1 2 3 m= 6 1− ≈ 1.2431 . k1

Clearly, g 00 (x) has the same sign as h(x), where h(x) = m ·

x2 + 1 − 1. x2 + 2

Since h is strictly increasing on [1, ∞) and h(1) =

2m − 1 < 0, 3

lim h1 (x) = m − 1 > 0,

x→∞

66

Vasile Cîrtoaje

there is x 1 ∈ (1, ∞) such that h(x) < 0 for x ∈ [1, x 1 ), h(x 1 ) = 0 and h(x) > 0 for x ∈ (x 1 , ∞). Therefore, g 0 is strictly decreasing on [1, x 1 ] and strictly increasing on [x 1 , ∞). Since g 0 (1) = 0 and lim x→∞ g 0 (x) = 0, it follows that g 0 (x) < 0 for x ∈ (1, ∞). Thus, g(x) is strictly decreasing on [1, ∞), hence g(x) ≤ g(1) = 0. This completes the proof. The equality holds for a = b = c = 1. If k = k0 , then the equality holds also for a = 0 and b = c = 3/2 (or any cyclic permutation).

P 1.45. If a, b, c are positive real numbers such that a bc = 1, then a2 + b2 + c 2 − 3 ≥ 18(a + b + c − a b − bc − ca). Solution. Using the substitution a = ex ,

b = ey,

c = ez ,

we need to show that f (x) + f ( y) + f (z) ≥ 3 f (s),

s=

x + y +z = 0, 3

where f (u) = e2u − 1 − 18(eu − e−u ), u ∈ R. For u ≤ 0, we have

f 00 (u) = 4e2u + 18(e−u − eu ) > 0,

hence f is convex on (−∞, 0]. By HCF Theorem, it suffices to prove the original inequality for b = c := t and a = 1/t 2 , where t > 0. Since a2 + b2 + c 2 − 3 = and a + b + c − a b − bc − ca =

1 (t 2 − 1)2 (2t 2 + 1) 2 + 2t − 3 = t4 t4 −(t 4 − 2t 3 + 2t − 1) −(t − 1)3 (t + 1) = , t2 t2

we get a2 + b2 + c 2 − 3 − 18(a + b + c − a b − bc − ca) =

(t − 1)2 (2t − 1)2 (t + 1)(5t + 1) ≥ 0. t4

The equality holds for a = b = c = 1, and also for a = 4 and b = c = 1/2 (or any cyclic permutation).

Half Convex Function Method

67

P 1.46. If a, b, c are positive real numbers such that a bc = 1, then p p p a2 − a + 1 + b2 − b + 1 + c 2 − c + 1 ≥ a + b + c. Solution. Using the substitution a = ex ,

b = ey,

c = ez ,

we need to show that f (x) + f ( y) + f (z) ≥ 3 f (s), where f (u) =

p

s=

x + y +z = 0, 3

e2u − eu + 1 − eu , u ∈ R.

We claim that f is convex for u ≥ 0. Since e−u f 00 (u) =

4e3u − 6e2u + 9eu − 2 − 1, 4(e2u − eu + 1)3/2

we need to show that (4x 3 − 6x 2 + 9x − 2)2 ≥ 16(x 2 − x + 1)3 , where x = eu ≥ 1. Indeed, (4x 3 − 6x 2 + 9x − 2)2 − 16(x 2 − x + 1)3 = 12x 3 (x − 1) + 9x 2 + 12(x − 1) > 0. By HCF Theorem, it suffices to prove the original inequality for b = c := t and a = 1/t 2 , where t > 0; that is, p p t4 − t2 + 1 1 + 2 t 2 − t + 1 ≥ 2 + 2t, t2 t p

t2 − 1 t4 − t2 + 1 + 1

Since p it suffices to show that

+p

t2 − 1 t4 − t2 + 1

2(1 − t) t2 − t + 1 + t ≥

≥ 0.

t2 − 1 , t2 + 1

t2 − 1 2(1 − t) +p ≥ 0, 2 t +1 t2 − t + 1 + t

which is equivalent to ˜ t +1 2 (t − 1) 2 −p ≥ 0, t +1 t2 − t + 1 + t •

68

Vasile Cîrtoaje ” — p (t − 1) (t + 1) t 2 − t + 1 − t 2 + t − 2 ≥ 0, (t − 1)2 (3t 2 − 2t + 3) ≥ 0. p (t + 1) t 2 − t + 1 + t 2 − t + 2

Clearly, the last inequality is true. The equality holds for a = b = c = 1.

P 1.47. If a, b, c, d ≥

1 p such that a bcd = 1, then 1+ 6 1 1 1 1 4 + + + ≤ . a+2 b+2 c+2 d +2 3 (Vasile Cirtoaje, 2005)

Solution. Using the substitution a = ex ,

b = ey,

c = ez ,

d = ew,

we need to show that f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s), where f (u) = For u ≤ 0, we have f 00 (u) =

s=

x + y +z+w = 0, 4

−1 , u ∈ R. +2

eu

eu (2 − eu ) > 0, (eu + 2)3

hence f is convex on (−∞, 0]. By HCF Theorem, it suffices to prove the original in1 equality for b = c = d := t and a = 1/t 3 , where t ≥ p ; that is, 1+ 6 t3 3 4 + ≤ , 3 2t + 1 t + 2 3 which is equivalent to the obvious inequality (t − 1)2 (5t 2 + 2t − 1) ≥ 0. In accord with Remark 3, the equality holds for a = b = c = d = 1, and also for p 1 a = 19 + 9 6 and b = c = d = p (or any cyclic permutation). 1+ 6

Half Convex Function Method

69

P 1.48. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then  ‹ p 1 1 1 2 2 2 . − − ··· − a1 + a2 + · · · + an − n ≥ 6 3 a1 + a2 + · · · + an − a1 a2 an Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where

s=

x1 + x2 + · · · + x n = 0, n

p f (u) = e2u − 1 − 6 3 (eu − e−u ), u ∈ R.

For u ≤ 0, we have

p f 00 (u) = 4e2u + 6 3(e−u − eu ) > 0,

hence f is convex on (−∞, 0]. By HCF Theorem and Remark 2, it suffices to show that H(x, y) ≥ 0 for x, y ∈ R such that x + (n − 1) y = 0, where H(x, y) = From

f 0 (x) − f 0 ( y) . x−y

p f 0 (u) = 2e2u − 6 3 (eu + e−u ),

we get H(x, y) =


p p 2(e x − e y ) x e + e y − 3 3 + 3 3 e−x e− y . x−y

Since (e x − e y )/(x − y) > 0, we need to prove that p p e x + e y + 3 3 e−x e− y ≥ 3 3. Indeed, by the AM-GM inequality, we have Æ p p p 3 e x + e y + 3 3 e−x e− y ≥ 3 e x · e y · 3 3 e−x e− y = 3 3. The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

P 1.49. If a1 , a2 , . . . , an (n ≥ 4) are positive real numbers such that a1 a2 · · · an = 1, then (n − 1)(a12 + a22 + · · · + an2 ) + n(n + 3) ≥ (2n + 2)(a1 + a2 + · · · + an ).

70

Vasile Cîrtoaje

Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),

s=

x1 + x2 + · · · + x n = 0, n

where f (u) = (n − 1)e2u − (2n + 2)eu , u ∈ R. For u ≥ 0, we have f 00 (u) = 4(n − 1)e2u − (2n + 2)eu = 2eu [2(n − 1)eu − n − 1] ≥ 2eu [2(n − 1) − n − 1] = 2(n − 3)eu > 0. Therefore, f is convex on [0, ∞). By HCF Theorem and Remark 2, it suffices to show that H(x, y) ≥ 0 for x, y ∈ R such that x + (n − 1) y = 0, where H(x, y) =

f 0 (x) − f 0 ( y) . x−y

From f 0 (u) = 2(n − 1)e2u − (2n + 2)eu , we get H(x, y) =

2(e x − e y ) [(n − 1)(e x + e y ) − (n + 1)] . x−y

Since (e x − e y )/(x − y) > 0, we need to prove that (n − 1)(e x + e y ) ≥ n + 1. Using the AM-GM inequality, we have Æ n (n − 1)(e x + e y ) = (n − 1)e x + e y + e y + · · · + e y ≥ n (n − 1)e x · e y · e y · · · e y Æ p n n = n (n − 1)e x+(n−1) y = n n − 1. Thus, it suffices to show that which is equivalent to

p n n n − 1 ≥ n + 1,  ‹ 1 n n−1≥ 1+ . n

This is true for n ≥ 4, since

 ‹ 1 n n−1≥3> 1+ . n The proof is completed. The equality holds for a1 = a2 = · · · = an = 1. Remark. From the proof above, it follows that the following sharper inequality holds in the same conditions (Gabriel Dospinescu and Calin Popa): p n 2n n − 1 2 2 2 a1 + a2 + · · · + an − n ≥ (a1 + a2 + · · · + an − n). n−1

Half Convex Function Method

71

P 1.50. Let a, b, c, d be positive real numbers such that a bcd = 1. If p and q are nonnegative real numbers such that p + q = 3, then then 1 1 1 1 + + + ≥ 1. 1 + pa + qa3 1 + p b + q b3 1 + pc + qc 3 1 + pd + qd 3 (Vasile Cirtoaje, 2007) Solution. Using the substitutions a = ex ,

b = ey,

c = ez ,

d = ew,

we need to show that f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s), where f (u) =

1 1+

peu

+ qe3u

s=

x + y +z+w = 0, 4

, u ∈ R.

We will show that f 00 (u) > 0 for u ≥ 0, hence f is convex on [0, ∞). Since f 00 (u) =

th(t) , (1 + pt + qt 3 )3

where h(t) = 9q2 t 5 + 2pqt 3 − 9qt 2 + p2 t − p,

t = eu , t ≥ 1,

we need to show that h(t) > 0 for t ≥ 1. Indeed, we have h(t) ≥ 9q2 t 3 + 2pqt 3 − 9qt 2 + p2 t − pt = t[(9q2 + 2pq)t 2 − 9qt + p2 − p] and (9q2 + 2pq)t 2 − 9qt + p2 − p ≥ (9q2 + 2pq)(2t − 1) − 9qt + p2 − p = q(18q + 4p − 9)t − 9q2 − 2pq + p2 − p ≥ q(18q + 4p − 9) − 9q2 − 2pq + p2 − p = p2 + 2pq + 9q2 − p − 9q = p2 + 2pq + 9q2 − =

(p + 9q)(p + q) 3

2(p − q)2 + 16q2 > 0. 3

By HCF Theorem, it suffices to prove the original inequality for b = c = d = t and a = 1/t 3 , where t > 0; that is, 3 t9 + ≥ 1, 9 6 t + pt + q 1 + pt + qt 3

72

Vasile Cîrtoaje pt 6 + q 3 ≥ , 1 + pt + qt 3 t 9 + pt 6 + q (3 − pq)t 9 − p2 t 7 + 2pt 6 − q2 t 3 − pqt + 2q ≥ 0, [(p + q)2 − 3pq]t 9 − 3p2 t 7 + 2p(p + q)t 6 − 3q2 t 3 − 3pqt + 2q(p + q) ≥ 0, Ap2 + Bq2 ≥ C pq,

where A = t 9 − 3t 7 + 2t 6 = t 6 (t − 1)2 (t + 2) ≥ 0, B = t 9 − 3t 3 + 2 = (t 3 − 1)2 (t 3 + 2) ≥ 0, C = t 9 − 2t 6 + 3t − 2. Since A ≥ 0 and B ≥ 0, it suffices to consider that C ≥ 0 and to show that 4AB ≥ C 2 . From t 3 − 3t + 2 = (t − 1)2 (t + 2) ≥ 0, we get 3t − 2 ≤ t 3 . Therefore C ≤ t 9 − 2t 6 + t 3 = t 3 (t 3 − 1)2 , hence 4AB − C 2 ≥ 4AB − t 6 (t 3 − 1)4 = t 6 (t − 1)2 (t 3 − 1)2 [4(t + 2)(t 3 + 2) − (t 2 + t + 1)2 ] = t 6 (t − 1)2 (t 3 − 1)2 (3t 4 + 6t 3 − 3t 2 + 6t + 15) ≥ 0. The proof is completed. The inequality holds for a = b = c = d = 1. Remark. Similarly, we can prove the following generalization. • Let a, b, c, d be positive real numbers such that a bcd = 1. If p and q are nonnegative real numbers such that p + q ≥ 3, then then 1 1 1 1 4 + + + ≥ . 3 3 3 3 1 + pa + qa 1 + pb + qb 1 + pc + qc 1 + pd + qd 1+p+q

P 1.51. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 . . . an = 1. If p, q ≥ 0 such that p + q ≥ n − 1, then 1 1+

pa1 + qa12

+

1 1+

pa2 + qa22

+ ··· +

1 n ≥ . 2 1 + pan + qan 1+p+q (Vasile Cirtoaje, 2007)

Half Convex Function Method

73

Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where f (u) =

1 1+

peu

+ qe2u

s=

x1 + x2 + · · · + x n = 0, n

, u ∈ R.

For u ≥ 0, we have eu [4q2 e3u + 3pqe2u + (p2 − 4q)eu − p] (1 + peu + qe2u )3 2u 2 e [4q + 3pq + (p2 − 4q) − p] ≥ (1 + peu + qe2u )3 e2u [(p + 2q)(p + q − 2) + 2q2 + p] = > 0, (1 + peu + qe2u )3

f 00 (u) =

therefore f (u) is convex for u ≥ 0. By HCF Theorem, it suffices to prove the original inequality for a2 = · · · = an := t and a1 = 1/t n−1 , where t > 0. Write this inequality as t 2n−2 n−1 n + ≥ . 2n−2 n−1 2 t + pt + q 1 + pt + qt 1+p+q Applying the Cauchy-Schwarz inequality, it suffices to prove that n (t n−1 + n − 1)2 ≥ , (t 2n−2 + pt n−1 + q) + (n − 1)(1 + pt + qt 2 ) 1 + p + q which is equivalent to pB + qC ≥ A, where A = (n − 1)(t n−1 − 1)2 ≥ 0, A + nE, E = t n−1 + n − 2 − (n − 1)t, n−1 A C = (t n−1 − 1)2 + nF = + nF, F = 2t n−1 + n − 3 − (n − 1)t 2 . n−1 By the AM-GM inequality applied to n − 1 positive numbers, we have E ≥ 0 and F ≥ 0 for n ≥ 3. Since A ≥ 0 and p + q ≥ n − 1, we have B = (t n−1 − 1)2 + nE =

pB + qC − A ≥ pB + qC −

(p + q)A = n(pE + qF ) ≥ 0. n−1

The equality holds for a1 = a2 = · · · = an = 1.

74

Vasile Cîrtoaje

Remark 1. For p = 2k and q = k2 , we get the following result. • Let a1 , a2 , · · · , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If p k ≥ n − 1, then 1 1 n 1 + + ··· + ≥ , (1 + ka1 )2 (1 + ka2 )2 (1 + kan )2 (1 + k)2 with equality for a1 = a2 = · · · = an = 1. In addition, for n = 4 and k = 1, we get the known inequality (Vasile Cirtoaje, 1999): 1 1 1 1 + + + ≥ 1, 2 2 2 (1 + a) (1 + b) (1 + c) (1 + d)2 where a, b, c, d > 0 such that a bcd = 1. Remark 2. For p + q = n − 1, we get the beautiful inequality 1 1+

pa1 + qa12

+

1 1+

pa2 + qa22

+ ··· +

1 ≥ 1, 1 + pan + qan2

n ≥ 3,

which is a generalization of the following inequalities: 1 1 1 + + ··· + ≥ 1, 1 + (n − 1)a1 1 + (n − 1)a2 1 + (n − 1)an 1 1 1 + + · · · + ≥ 1, p p p [1 + ( n − 1)a1 ]2 [1 + ( n − 1)a1 ]2 [1 + ( n − 1)a1 ]2 1 2 + (n − 1)(a1 + a12 )

+

1 2 + (n − 1)(a2 + a22 )

+ ··· +

1 1 ≥ . 2 2 + (n − 1)(an + an ) 2

Remark 3. Similarly, we can prove the following statement: • Let a1 , a2 , . . . , an (n ≥ 4) be positive real numbers such that a1 a2 . . . an = 1. If p, q, r ≥ 0 such that p + q + r ≥ n − 1, then n X i=1

1 1+

pai + qai2

+

r ai3

≥

n . 1+p+q+r

For n = 4 and p + q + r = 3, we get the beautiful inequality 4 X

1

i=1

1 + pai + qai2 + r ai3

≥ 1.

Half Convex Function Method

75

If p = q = r = 1, then we get the known inequality (Vasile Cirtoaje, 1999): 4 X

1

i=1

1 + ai + ai2 + ai3

≥ 1.

Since 2ai2 ≤ ai + ai3 , the best inequality with respect to q if for q = 0; that is, 4 X

1

i=1

1 + pai + r ai3

p + r = 3.

≥ 1,

For p = 1 and p = 2, we get the following strong inequalities: 4 X

1

i=1

1 + ai + 2ai3

4 X

1

i=1

1 + 2ai + ai3

≥ 1,

≥ 1.

Actually, the following generalization holds. • Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1, and let k1 , k2 , . . . , km ≥ 0 such that k1 + k2 + · · · + km ≥ n − 1. If m ≤ n − 1, then n X

1

1 + k1 ai + k2 ai2 i=1

+ · · · + km aim

≥

n 1 + k1 + k2 + · · · + km

.

(*)

For m = n − 1 and k1 = k2 = · · · = km = 1, (*) turns into the known beautiful inequality n X

1

1 + ai + ai2 i=1

+ · · · + ain−1

≥ 1.

Since (m − 1)aik ≤ (m − k)ai + (k − 1)aim ,

k = 2, 3, . . . , m − 1

(by the AM-GM inequality), the best inequality (*) with respect to k2 , · · · , km−1 is for k2 , · · · , km−1 = 0; that is, n X i=1

n 1 , m ≥ 1 + k1 ai + km ai 1 + k1 + km

k1 + km ≥ n − 1, 1 ≤ m ≤ n − 1.

If k1 + km = n − 1 and m = n − 1, then n X

1

i=1

1 + k1 ai + kn−1 ain−1

≥ 1.

76

Vasile Cîrtoaje

For k1 = 1 and k1 = n − 2, we get the following strong inequalities: n X

1

i=1

1 + ai + (n − 2)ain−1

n X

1

i=1

1 + (n − 2)ai + ain−1

≥ 1,

≥ 1.

P 1.52. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1. If k ≥ n2 − 1, then 1 1 1 n . +p + ··· + p ≥p p 1+k 1 + kan 1 + ka1 1 + ka2 Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where

s=

x1 + x2 + · · · + x n = 0, n

1 , u ∈ R. f (u) = p 1 + keu

For u ≥ 0, we have f 00 (u) =

keu (keu − 2) keu (k − 2) ≥ > 0. 4(1 + keu )5/2 4(1 + keu )5/2

Therefore, f is convex on [0, ∞). By HCF Theorem and Remark 5, it suffices to prove the original inequality for a2 = · · · = an := t and a1 = 1/t n−1 , where t ≥ 1. Write this inequality as h(t) ≥ 0, where v t t n−1 n−1 n h(t) = +p −p . n−1 t +k 1 + kt 1+k The derivative h0 (t) =

(n − 1)kt (n−3)/2 (n − 1)k − n−1 3/2 2(t + k) 2(kt + 1)3/2

has the same sign as h1 (t) = t n/3−1 (kt + 1) − t n−1 − k. Denoting m = n/3, m ≥ 2/3, we see that h1 (t) = kt m + t m−1 − t 3m−1 − k = k(t m − 1) − t m−1 (t 2m − 1) = (t m − 1)h2 (t),

Half Convex Function Method

77

where h2 (t) = k − t m−1 − t 2m−1 . For t > 1, we have h02 (t) = t m−2 [−m + 1 − (2m − 1)t m ] < t m−2 [−m + 1 − (2m − 1)] = −(3m − 2)t m−2 ≤ 0, hence h2 (t) is strictly decreasing for t ≥ 1. Since h2 (1) = k − 2 > 0 and lim t→∞ h2 (t) = −∞, there exists t 1 > 1 such that h2 (t 1 ) = 0, h2 (t) > 0 for t ∈ (1, t 1 ), and h2 (t) < 0 for t ∈ (t 1 , ∞). Since h2 , h1 and h0 has the same sign for t > 1, h(t) is strictly increasing for t ∈ [1, t 1 ] and strictly decreasing for t ∈ [t 1 , ∞); this yields h(t) ≥ min{h(1), h(∞)}. n ≥ 0, it follows that h(t) ≥ 0 for all t ≥ 1. The From h(1) = 0 and h(∞) = 1 − p 1+k proof is completed. The equality holds for a1 = a2 = · · · = an = 1. Remark. The following generalization holds (Vasile Cirtoaje, 2005): • Let a1 , a2 , · · · , an be positive real numbers such that a1 a2 · · · an = 1. If k and m are positive numbers such that m ≤ n − 1,

k ≥ n1/m − 1,

then 1 1 1 n + + ··· + ≥ , m m m (1 + ka1 ) (1 + ka2 ) (1 + kan ) (1 + k)m with equality for a1 = a2 = · · · = an = 1. For 0 < m ≤ n − 1 and k = n1/m − 1, we get the beautiful inequality 1 1 1 + + ··· + ≥ 1. m m (1 + ka1 ) (1 + ka2 ) (1 + kan )m

P 1.53. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 . . . an = 1, then 1 1 + a1 + · · · + a1n−1

+

1 1 + a2 + · · · + a2n−1

+ ··· +

1 ≥ 1. 1 + an + · · · + ann−1 (Vasile Cirtoaje, 2007)

78

Vasile Cîrtoaje

Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where f (u) =

s=

x1 + x2 + · · · + x n = 0, n

1 , u ∈ R. 1 + eu + · · · + e(n−1)u

We will show that f is convex for u ≥ 0. Setting t = eu , t ≥ 1, the necessary and sufficient condition f 00 (u) ≥ 0 for u ≥ 0 is equivalent to 2A2 ≥ B(1 + C), where A = t + 2t 2 + · · · + (n − 1)t n−1 , B = t + 4t 2 + · · · + (n − 1)2 t n−1 , C = t + t 2 + · · · + t n−1 . We will prove this inequality by induction on n. For n = 2, the inequality becomes t(t − 1) ≥ 0, which is clearly true for t ≥ 1. Assume now that the inequality is true for n and prove it for n + 1, n ≥ 2. So, we need to show that 2A2 ≥ B(1 + C) involves 2(A + nt n )2 ≥ (B + n2 t n )(1 + C + t n ), which is equivalent to 2A2 − B(1 + C) + t n [n2 (t n − 1) + D] ≥ 0, where 2

D = 2nA − B − n C =

n−1 X

bi t i ,

bi = 3n2 − (2n − i)2 .

i=1

Since 2A − B(1 + C) ≥ 0, it suffices to show that D ≥ 0. Since 2

b1 < b2 < · · · < bn−1 ,

t ≤ t 2 ≤ · · · ≤ t n−1 ,

we may apply Chebyshev’s inequality to get D≥

1 (b1 + b2 + · · · + bn−1 )(t + t 2 + · · · + t n−1 ). n

Thus, it suffices to show that b1 + b2 + · · · + bn−1 ≥ 0. Indeed, b1 + b2 + · · · + bn−1 =

n−1 X n(n − 1)(4n + 1) [3n2 − (2n − i)2 ] = > 0. 6 i=1

Half Convex Function Method

79

By HCF Theorem and Remark 5, it suffices to prove the original inequality for a2 = · · · = an := t and a1 = 1/t n−1 , where t ≥ 1. Setting k = n − 1, k ≥ 1, we need to show that 2

k tk + ≥ 1. 2 1 + t + · · · + tk 1 + tk + · · · + tk For the nontrivial case t > 1, this inequality is equivalent to the following sequence of inequalities: k 1 + t k + · · · + t (k−1)k , ≥ 1 + t + · · · + tk 1 + t k + · · · + t k2 2

k(t − 1) tk − 1 tk − 1 ≥ · , t k+1 − 1 t k − 1 t (k+1)k − 1 2

k(t − 1) tk − 1 ≥ , t k+1 − 1 t (k+1)k − 1 2

t k(k+1) − 1 tk − 1 ≥ , t k+1 − 1 t −1   k 1 + t k+1 + t 2(k+1) + · · · + t (k−1)(k+1) ≥ 1 + t + t 2 + · · · + t (k−1)(k+1) ,  
  k 1 · 1 + t · t k + · · · + t k−1 · t (k−1)k ≥ 1 + t + · · · + t k−1 1 + t k + · · · + t (k−1)k . k

Since 1 < t < · · · < t k−1 and 1 < t k < · · · < t (k−1)k , the last inequality follows from Chebyshev’s inequality. This completes the proof. The equality holds for a1 = a2 = · · · = an = 1.

P 1.54. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1. If p, q ≥ 0 1 such that 0 < p + q ≤ , then n−1 1 1+

pa1 + qa12

+

1 1+

pa2 + qa22

+ ··· +

1 n ≤ . 1 + pan + qan2 1+p+q (Vasile Cirtoaje, 2007)

Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where f (u) =

s=

x1 + x2 + · · · + x n = 0, n

−1 , u ∈ R. 1 + peu + qe2u

80

Vasile Cîrtoaje

For u ≤ 0, we have eu [−4q2 e3u − 3pqe2u + (4q − p2 )eu + p] (1 + peu + qe2u )3 e2u [−4q2 − 3pq + (4q − p2 ) + p] ≥ (1 + peu + qe2u )3 e2u [(p + 4q)(1 − p − q) + 2pq] = ≥ 0, (1 + peu + qe2u )3

f 00 (u) =

therefore f (u) is convex for u ≤ 0. By HCF Theorem, it suffices to prove the original inequality for a2 = · · · = an := t and a1 = 1/t n−1 , where t > 0. Write this inequality as n−1 n t 2n−2 + ≤ , t 2n−2 + pt n−1 + q 1 + pt + qt 2 1+p+q p2 A + q2 B + pqC ≤ pD + qE, where A = t n−1 (t n − nt + n − 1),

B = t 2n − nt 2 + n − 1,

C = t 2n−1 + t 2n − nt n+1 + (n − 1)t n−1 − nt + n − 1, D = t n−1 [(n − 1)t n − nt n−1 + 1],

E = (n − 1)t 2n − nt 2n−2 + 1.

Applying the AM-GM inequality to n positive numbers yields D ≥ 0 and E ≥ 0. Then, since p + q ≤ 1/(n − 1) involves pD + qE ≥ (n − 1)(p + q)(pD + qE), it suffices to show that p2 A + q2 B + pqC ≤ (n − 1)(p + q)(pD + qE). Write this inequality as p2 A1 + q2 B1 + pqC1 ≥ 0, where A1 = (n − 1)D − A = nt n [(n − 2)t n−1 − (n − 1)t n−2 + 1], B1 = (n − 1)E − B = nt 2 [(n − 2)t 2n−2 − (n − 1)t 2n−4 + 1], C1 = (n − 1)(D + E) − C = n y[(n − 2)(t 2n−1 + t 2n−2 ) − 2(n − 1)t 2n−3 + t n + 1]. Applying the AM-GM inequality to n−1 nonnegative numbers yields A1 ≥ 0 and B1 ≥ 0. So, it suffices to show that C1 ≥ 0. Indeed, we have (n − 2)(t 2n−1 + t 2n−2 ) − 2(n − 1)t 2n−3 + t n + 1 = A2 + B2 + C2 , where A2 = (n − 2)t 2n−1 − (n − 1)t 2n−3 + t ≥ 0, B2 = (n − 2)t 2n−2 − (n − 1)t 2n−3 + t n−1 ≥ 0,

Half Convex Function Method

81

C2 = t n − t n−1 − t + 1 = (t − 1)(t n−1 − 1) ≥ 0. The inequalities A2 ≥ 0 and B2 ≥ 0 follows by the AM-GM inequality applied to n − 1 nonnegative numbers. This completes the proof. The equality holds for a1 = a2 = · · · = an = 1. Remark 1. For p + q =

1 , we get the inequality n−1

1 1+

pa1 + qa12

+

1 1+

pa2 + qa22

+ ··· +

1 ≤ n − 1, 1 + pan + qan2

which is a generalization of the following inequalities: 1 1 1 + + ··· + ≤ 1, n − 1 + a1 n − 1 + a2 n − 1 + an 1 2n − 2 + a1 + a12

+

1 2n − 2 + a2 + a22

+ ··· +

1 1 ≤ . 2 2n − 2 + an + an 2

Remark 2. For p = 2k and q = k2 , we get the following result: • Let a1 , a2 , · · · , an be positive real numbers such that a1 a2 · · · an = 1. If 0 0 for t ∈ (0, 1]. Also, we have h1 (t) > 0 and h0 (t) > 0 for t ∈ (0, 1). Therefore, h is strictly increasing on [0, 1], hence h(t) ≤ h(1) = 0. Case 2: h2 (0) < 0. This case is possible for either m = 1 (n = 3) and 1 < k ≤ 5/4, or m > 1 (n ≥ 4). Since h2 (1) = 2 − k > 0, there exists t 1 ∈ (0, 1) such that h2 (t 1 ) = 0, h2 (t) < 0 for t ∈ (0, t 1 ), and h2 (t) > 0 for t ∈ (t 1 , 1). Since h0 has the same sign as h2 on (0, 1), it follows that h is strictly decreasing on [0, t 1 ], and strictly increasing on [t 1 , 1]. n Therefore, h(t) ≤ min{h(0), h(1)}. Since h(0) = n − 1 − p ≤ 0 and h(1) = 0, 1+k

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we have h(t) ≤ 0 for all t ∈ (0, 1]. This completes the proof. The equality holds for a1 = a2 = · · · = an = 1. Remark. The following generalization holds (Vasile Cirtoaje, 2005): • Let a1 , a2 , · · · , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If k and m are positive numbers such that 1 m≥ , n−1 then

 n 1/m k≤ − 1, n−1

1 1 1 n + + ··· + ≤ , m m m (1 + ka1 ) (1 + ka2 ) (1 + kan ) (1 + k)m

with equality for a1 = a2 = · · · = an = 1. For n ≥ 3, m ≥

 n 1/m 1 and k = − 1, we get the beautiful inequality n−1 n−1 1 1 1 + + ··· + ≤ n − 1. m m (1 + ka1 ) (1 + ka2 ) (1 + kan )m

P 1.56. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 . . . an = 1, then  ‹ 1 1 1 n−1 n−1 n−1 . a1 + a2 + · · · + an + n(n − 2) ≥ (n − 1) + + ··· + a1 a2 an Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where

s=

x1 + x2 + · · · + x n = 0, n

f (u) = e(n−1)u − (n − 1)e−u , u ∈ R.

For u ≥ 0, we have f 00 (u) = (n − 1)2 e(n−1)u − (n − 1)e−u = (n − 1)e−u [(n − 1)e nu − 1] ≥ 0; therefore, f (u) is convex on [0, ∞). By HCF Theorem and Remark 2, it suffices to show that H(x, y) ≥ 0 for x, y ∈ R such that x + (n − 1) y = 0, where H(x, y) =

f 0 (x) − f 0 ( y) . x−y

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From

f 0 (u) = (n − 1)[e(n−1)u + e−u ],

we get  (n − 1)(e x − e y )  (n−2)x e + e(n−3)x+ y + · · · + e x+(n−3) y + e(n−2) y − e−x− y x−y  (n − 1)(e x − e y )  (n−2)x = e + e(n−3)x+ y + · · · + e x+(n−3) y ) . x−y

H(x, y) =

Since (e x − e y )/(x − y) > 0, we have H(x, y) > 0. The equality holds for a1 = a2 = · · · = an = 1.

P 1.57. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1. If k ≥ n, then  ‹ 1 1 1 k k k a1 + a2 + · · · + an + kn ≥ (k + 1) + + ··· + . a1 a2 an (Vasile Cirtoaje, 2006) Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),

s=

x1 + x2 + · · · + x n = 0, n

where f (u) = e ku − (k + 1)e−u , u ∈ R. For u ≥ 0, we have   f 00 (u) = k2 e ku − (k + 1)e−u = e−u k2 e(k+1)u − k − 1 ≥ e−u (k2 − k − 1) > 0; therefore, f is convex on [0, ∞). By HCF Theorem and Remark 5, it suffices to to prove the original inequality for a2 = · · · = an := b ≥ 1 and a1 := a ≤ 1, a b n−1 = 1; that is a k + (n − 1)b k −

k + 1 (k + 1)(n − 1) − + kn ≥ 0. a b

By the weighted AM-GM inequality, we have 1

a k + (kn − k − 1) ≥ [1 + (kn − k − 1)](a k ) 1+(kn−k−1) = Then, we still have to show that  ‹  ‹ 1 1 (n − 1) b k − − (k + 1) − 1 ≥ 0, b a

k(n − 1) . b

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85

which is equivalent to h(b) ≥ 0 for b ≥ 1, where h(b) = (n − 1)(b k+1 − 1) − (k + 1)(b n − b). Since h0 (b) = (n − 1)b k − nb n−1 + 1 ≥ (n − 1)b n − nb n−1 + 1 k+1 = nb n−1 (b − 1) − (b n − 1)   = (b − 1) (b n−1 − b n−2 ) + (b n−1 − b n−3 ) + · · · + (b n−1 − 1) ≥ 0, h is increasing on [1, ∞), hence h(b) ≥ h(1) = 0. The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

P 1.58. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 . . . an = 1, then ‹  ‹  ‹  1 a2 1 an 1 a1 + 1− + ··· + 1 − ≤ n − 1. 1− n n n (Vasile Cîrtoaje, 2006) Solution. Let k=

n , n−1

k > 1,

and m = ln k, 0 < m ≤ ln 2 < 1. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where

s=

x1 + x2 + · · · + x n = 0, n

u

f (u) = −k−e , u ∈ R. From

u

f 00 (u) = meu k−e (1 − meu ), it follows that f 00 (u) > 0 for u ≤ 0, since 1 − meu ≥ 1 − m ≥ 1 − ln 2 > 0. Therefore, f is convex on (−∞, 0]. By HCF Theorem and Remark 5, it suffices to prove the original inequality for a2 = · · · = an := t and a1 = t −n+1 , where 0 < t ≤ 1. Write this inequality as h(t) ≤ n − 1,

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where h(t) = k−t

−n+1

We have h0 (t) = (n − 1)mt −n k−t h01 (t) = k−t

−n+1

−t

+ (n − 1)k−t , −n+1

t ∈ (0, 1].

h1 (t), h1 (t) = 1 − t n k−t

−n+1

−t

,

h2 (t), h2 (t) = [m(n − 1 + t n ) − nt n−1 ,

h02 (t) = nt n−2 (mt − n + 1) ≤ nt n−2 (m − n + 1) ≤ nt n−2 (m − 1) < 0, hence h2 is strictly decreasing on [0, 1]. Since h2 (0) = (n − 1)m > 0 and h2 (1) = n(m − 1) < 0, there is t 1 ∈ (0, 1) such that h2 (t 1 ) = 0, h2 (t) > 0 for t ∈ [0, t 1 ) and h2 (t) < 0 for t ∈ (t 1 , 1]. Therefore, h1 is strictly increasing on (0, t 1 ] and is strictly decreasing on [t 1 , 1]. Since lim t→0 h1 (u) = −∞ and h1 (1) = 0, there is t 2 ∈ (0, t 1 ) such that h1 (t 2 ) = 0, h1 (t) < 0 for t ∈ (0, t 2 ) and h1 (t) > 0 for t ∈ (t 2 , 1). Thus, h is strictly decreasing on (0, t 2 ] and is strictly increasing on [t 2 , 1]. Since lim t→0 h(u) = n − 1 and h(1) = n − 1, we have h(t) ≤ n − 1 for all t ∈ (0, 1]. This completes the proof. The equality holds for a1 = a2 = · · · = an = 1.

P 1.59. If a, b, c are positive real numbers such that a bc = 1, then 1 1 1 + + ≤ 1. p p p 1 + 1 + 3a 1 + 1 + 3b 1 + 1 + 3c (Vasile Cîrtoaje, 2008) Solution. Using the substitution a = e−x ,

b = e− y ,

c = e−z ,

we need to show that f (x) + f ( y) + f (z) ≥ 3 f (s), where f (u) =

s=

x + y +z = 0, 3

p −3 = eu − e2u + 3eu , u ∈ R. p 1 + 1 + 3e−u

For u ≥ 0, which involves t = eu ≥ 1, we have   4t 2 + 18t + 9 00 f (u) = t 1 − >0 p 4(t + 3) t(t + 3) since 16t(t + 3)3 − (4t 2 + 18t + 9)2 = 9(4t 2 + 12t − 9) > 0.

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87

Therefore, f is half convex for u ≥ 0. By HCF Theorem, it suffices to prove that f (x) + 2 f ( y) ≥ 3 f (0), where x, y ∈ R such that x + 2 y = 0. Substituting a = e x and b = e y , we need to show that p p a − a2 + 3a + 2(b − b2 + 3b ) ≥ −3, where a, b > 0 such that a b2 = 1. Write this inequality as 2b3 + 3b2 + 1 ≥

p

3b2 + 1 + 2b2

p

b2 + 3b.

Squaring and dividing by b2 , the inequality becomes 9b2 + 4b + 3 ≥ 4

Æ

(b2 + 3b)(3b2 + 1).

Since 2

Æ

(b2 + 3b)(3b2 + 1) ≤ (b2 + 3b) + (3b2 + 1) = 4b2 + 3b + 1,

we get 9b2 + 4b + 3 − 4

Æ

(b2 + 3b)(3b2 + 1) ≥ 9b2 + 4b + 3 − 2(4b2 + 3b + 1) = (b − 1)2 ≥ 0.

The equality holds for a = b = c = 1. Remark. Similarly, we can prove the following generalization. • Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If 00 f (u) = t p 4(t + k) t(t + k) f (u) =

since (4t 2 + 6kt + k2 )2 − 16t(t + k)3 = k2 (k2 − 4kt − 4t 2 ) ≥ k2 (k2 − 4k − 4) > 0. Therefore, f is convex on (−∞, 0]. By HCF Theorem, it suffices to prove that f (x) + (n − 1) f ( y) ≥ n f (0), where x, y ∈ R such that x + (n − 1) y = 0. Substituting a = e x and b = e y , we need to show that p p p a2 + ka − a + (n − 1)( b2 + k b − b) ≥ n( 1 + k − 1), where a, b > 0 such that a b n−1 = 1. Write this inequality as Æ Æ 4n(n − 1)b n−1 + 1 + (n − 1) 4n(n − 1)b2n−1 + b2n ≥ (n − 1)b n + 2n(n − 1)b n−1 + 1. By Minkowski’s inequality, we have Æ Æ 4n(n − 1)b n−1 + 1 + (n − 1) 4n(n − 1)b2n−1 + b2n ≥ Æ ≥ 4n(n − 1)b n−1 [1 + (n − 1)b n/2 ]2 + [1 + (n − 1)b n ]2 . Thus, it suffices to show that 4n(n − 1)b n−1 [1 + (n − 1)b n/2 ]2 + [1 + (n − 1)b n ]2 ≥ [(n − 1)b n + 2n(n − 1)b n−1 + 1]2 , which is equivalent to 4n(n − 1)2 b

3n−2 2

”

n

2 + (n − 2)b 2 − nb

n−2 2

—

≥ 0.

This inequality follows immediately by the AM-GM inequality. Thus, the proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

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89

P 1.61. If a, b, c are positive real numbers such that a bc = 1, then b6 c6 a6 + + ≥ 1. 1 + 2a5 1 + 2b5 1 + 2c 5 (Vasile Cîrtoaje, 2008) Solution. Using the substitution a = ex ,

b = ey,

c = ez ,

we need to show that f (x) + f ( y) + f (z) ≥ 3 f (s),

s=

x + y +z = 0, 3

where

e6u , u ∈ R. 1 + 2e5u For u ≤ 0, which involves w = eu ∈ (0, 1], we have f (u) =

f 00 (u) =

2w6 (2 − w5 )(9 − 2w5 ) > 0. (1 + 2w5 )3

Therefore, f is convex for u ≤ 0. By HCF Theorem, it suffices to prove the original inequality for b = c := t and a = 1/t 2 , where t > 0; that is, 1 t 2 (t 10 + 2)

+

2t 6 ≥ 1. 1 + 2t 5

Since 1 + 2t 5 ≤ 1 + t 4 + t 6 , it suffices to show that 1 2x 3 + ≥ 1, x(x 5 + 2) 1 + x 2 + x 3

x=

p

t.

This inequality can be written as follows: x 3 (x 6 − x 5 − x 3 + 2x − 1) + (x − 1)2 ≥ 0, x 3 (x − 1)2 (x 4 + x 3 + x 2 − 1) + (x − 1)2 ≥ 0, (x − 1)2 [x 7 + x 5 + (x 6 − x 3 + 1)] ≥ 0. Clearly, the last inequality is true. The equality holds for a = b = c = 1.

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P 1.62. If a, b, c are positive real numbers such that a bc = 1, then p p p 25a2 + 144 + 25b2 + 144 + 25c 2 + 144 ≤ 5(a + b + c) + 24. (Vasile Cîrtoaje, 2008) Solution. Using the substitution a = ex ,

b = ey,

c = ez ,

we need to show that f (x) + f ( y) + f (z) ≥ 3 f (s), where f (u) = 5eu −

p

s=

x + y +z = 0, 3

25e2u + 144, u ∈ R.

We will show that f is convex for u ≤ 0. From   5w(25w2 + 288) 00 f (u) = 5w 1 − , (25w2 + 144)3/2

w = eu ∈ (0, 1],

we need to show that (25w2 + 144)3 ≥ 25w2 (25w2 + 288)2 . Setting 25w2 = 144z, z ∈ (0, 25/144], we have (25w2 +144)3 −25w2 (25w2 +288)2 = 1443 (z+1)3 −1443 z(z+2)2 = 1443 (1−z−z 2 ) > 0. By HCF Theorem, it suffices to prove the original inequality for a = t 2 and b = c = 1/t, where t > 0; that is, p p 5t 3 + 24t + 10 ≥ 25t 6 + 144t 2 + 2 25 + 144t 2 . Squaring and dividing by 4t give 60t 3 + 25t 2 − 36t + 120 ≥

Æ

(25t 4 + 144)(144t 2 + 25).

Squaring again and dividing by 120, the inequality becomes 25t 5 − 36t 4 + 105t 3 − 112t 2 − 72t + 90 ≥ 0, (t − 1)2 (25t 3 + 14t 2 + 108t + 90) ≥ 0. Since the last inequality is true, the proof is completed. The equality holds for a = b = c = 1.

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91

P 1.63. If a, b, c are positive real numbers such that a bc = 1, then p p p 16a2 + 9 + 16b2 + 9 + 16c 2 + 9 ≥ 4(a + b + c) + 3. (Vasile Cîrtoaje, 2008) Solution. Using the substitution a = ex ,

b = ey,

c = ez ,

we need to show that f (x) + f ( y) + f (z) ≥ 3 f (s), where f (u) =

p

s=

x + y +z = 0, 3

16e2u + 9 − 4eu , u ∈ R.

We will show that f is convex for u ≥ 0. From   4w(16w2 + 18) 00 f (u) = 4w −1 , (16w2 + 9)3/2

w = eu ≥ 1,

we need to show that 16w2 (16w2 + 18)2 ≥ (16w2 + 9)3 . Setting 16w2 = 9z, z ≥ 16/9, we have Indeed, 16w2 (16w2 + 18)2 − (16w2 + 9)3 = 729z(z + 2)2 − 729(z + 1)3 = 729(z 2 + z − 1) > 0. By HCF Theorem, it suffices to prove the original inequality for a = t 2 and b = c = 1/t, where t > 0; that is, p p 16a6 + 9a2 + 2 16 + 9a2 ≥ 4t 3 + 3t + 8. Squaring and dividing by 4a give Æ (16a4 + 9)(9a2 + 16) ≥ 6a3 + 16a2 − 9a + 12. Squaring again and dividing by 12a, the inequality becomes 9a5 − 16a4 + 9a3 + 12a2 − 32a + 18 ≥ 0, (a − 1)2 (9a3 + 2a2 + 4a + 18) ≥ 0. Since the last inequality is true, the proof is completed. The equality holds for a = b = c = 1.

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P 1.64. If a, b, c, d are real numbers such that a + b + c + d = 4, then a2

b c d a + 2 + 2 + 2 ≤ 1. −a+4 b − b+4 c −c+4 d −d +4 (Sqing, 2015)

Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) ≥ 4 f (s), where f (u) = From f 0 (u) =

−u , u2 − u + 4

u2 − 4 , (u2 − u + 4)2

f 00 (u) =

s=

a+b+c+d = 1, 4

u ∈ R. 2(−u3 + 12u − 4) , (u2 − u + 4)3

it follows that f is increasing on (−∞, −2] ∪ [2, ∞), decreasing on [−2, 2] and convex on [1, 2]. Define the function   f (u), u ≤ 2 f0 (u) = .  f (2), u > 2 Since f0 (u) ≤ f (u) for u ∈ R and f0 (1) = f (1), it suffices to show that f0 (a) + f0 (b) + f0 (c) + f0 (d) ≥ 4 f0 (s),

s=

a+b+c+d = 1. 4

It is easy to check that f0 is convex on [1, ∞). Therefore, by HCF Theorem and Remark 5, we only need to show that f0 (x)+3 f0 ( y) ≥ 4 f0 (1) for all x, y ∈ R such that x ≤ 1 ≤ y and x + 3 y = 4. There are two cases to consider: y ≤ 2 and y > 2. Case 1: y ≤ 2. We need to show that f (x) + 3 f ( y) ≥ 3 f (1) for all x, y ∈ R such that x + 3 y = 4. According to Remark 1, this is true if h(x, y) ≥ 0 for x + 3 y = 4. We have g(u) = h(x, y) =

f (u) − f (1) u−4 = , 2 u−1 4(u − u + 4)

g(x) − g( y) 4(x + y) − x y 3( y − 2)2 + 4 = = > 0. x−y 4(x 2 − x + 4)( y 2 − y + 4) 4(x 2 − x + 4)( y 2 − y + 4)

Case 2: y > 2. From y > 2 and x + 3 y = 4, we get x < −2 and f0 (x) + 3 f0 ( y) − 4 f0 (1) = f (x) + 3 f (2) − 4 f (1) = The equality holds for a = b = c = d = 1.

x2

−x > 0. − x +4

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P 1.65. If a, b, c are nonnegative real numbers such that a + b + c = 12, then (a2 + 10)(b2 + 10)(c 2 + 10) ≥ 13310. (Vasile Cîrtoaje, 2006) Solution. Let f (u) = ln(u2 + 10), From f 00 (u) =

u ∈ [0, ∞).

2(10 − u2 ) , (u2 + 10)2

p p it follows that f is convex on [0, 10] and concave on [ 10, ∞). According to LCRCFTheorem, the sum f (a) + f (b) + f (c) is minimum when two of a, b, c are equal. Therefore, it suffices to prove the original inequality for b = c. So, we need to show that (a2 + 10)(b2 + 10)2 ≥ 13310 for a + 2b = 12. Write this inequality as follows: [(12 − 2b)2 + 10](b2 + 10)2 ≥ 13310, (2b2 − 24b + 77)(b4 + 20b2 + 100) ≥ 6655, 2b6 − 24b5 + 117b4 − 480b3 + 1740b2 − 2400b + 1045 ≥ 0, (b − 1)2 (2b4 − 20b3 + 75b2 − 310b + 1045) ≥ 0, (b − 1)2 [2b2 (b − 5)2 + 5(5b2 − 62b + 209)] ≥ 0. The last inequality is true since 31 5b − 62b + 209 = 5 b − 5 2



‹2

+

84 > 0. 5

The proof is completed. The equality holds for a = 10 and b = c = 1 (or any cyclic permutation). Remark. Similarly, we can prove the following generalization. • If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = 2n(n − 1), then (a12 + k)(a22 + k) · · · (an2 + k) ≥ k(k + 1)n , k = (n − 1)(2n − 1). The equality holds for a1 = k and a2 = · · · = an = 1 (or any cyclic permutation).

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P 1.66. Let a, b, c be nonnegative real numbers. If k0 ≤ k ≤ 3, then

k0 =

ln 2 ≈ 1.71, ln 3 − ln 2

a+b+c a (b + c) + b (c + a) + c (a + b) ≤ 2 2 k

k



k

‹k+1

.

Solution. Due to homogeneity, we may assume that a + b + c = 2. Write the inequality as f (a) + f (b) + f (c) ≤ 2, where f (u) = uk (2 − u), u ∈ [0, ∞). From f 00 (u) = kuk−2 [2k − 2 − (k + 1)u], ˜ • ˜ • 2k − 2 2k − 2 and concave on , 2 . By it follows that therefore, f is convex on 0, k+1 k+1 LCRCF-Theorem, the sum f (a) + f (b) + f (c) is maximum when a = 0 or 0 < a ≤ b = c. Case 1: a = 0. We need to show that bc(b k−1 + c k−1 ) ≤ 2 for b + c = 2. Since 0 < (k − 1)/2 ≤ 1, Bernoulli’s inequality gives b k−1 + c k−1 = (b2 )(k−1)/2 + (c 2 )(k−1)/2 ≤ 1 + =3−k+

k−1 2 k−1 2 (b − 1) + 1 + (b − 1) 2 2

k−1 2 (b + c 2 ). 2

Thus, it suffices to show that (3 − k)bc +

k−1 bc(b2 + c 2 ) ≤ 2. 2

Since bc ≤



b+c 2

‹2

= 1,

we only need to show that bc(b2 + c 2 ) ≤ 2. Indeed, we have 8[2 − bc(b2 + c 2 )] = (b + c)4 − 8bc(b2 + c 2 ) = (b − c)4 ≥ 0.

Half Convex Function Method

95

Case 2: 0 < a ≤ b = c. We only need to prove the original homogeneous inequality for b = c = 1 and 0 < a ≤ 1; that is,  a k+1 − a k − a − 1 ≥ 0. 1+ 2  a k+1 Since 1 + is increasing and a k is decreasing when k increases, it suffices to prove 2 that g(a) ≥ 0 for 0 < a ≤ 1, where  a k0 +1 g(a) = 1 + − a k0 − a − 1. 2 We have

k0 + 1  a k0 − k0 a k0 −1 − 1, 1+ 2 2 k0 + 1  1 00 a k0 k0 − 1 − 2−k . g (a) = 1+ k0 4 2 a 0

g 0 (a) =

Since g 00 is increasing on (0, 1], lim x→0 g 00 (a) = −∞ and  ‹ k0 + 1 3 k0 k0 + 1 3 − k0 1 00 g (1) = − k0 + 1 = − k0 + 1 = > 0, k0 4 2 2 2 there exists a1 ∈ (0, 1) such that g 00 (a1 ) = 0, g 00 (a) < 0 for a ∈ (0, a1 ), and g 00 (a) > 0 for a ∈ (a1 , 1]. Therefore, g 0 is strictly decreasing on [0, a1 ] and strictly increasing on  k0 − 1 k0 + 1  [a1 , 1]. Since g 0 (0) = > 0 and g 0 (1) = (3/2)k0 − 2 = 0, there exists 2 2 a2 ∈ (0, a1 ) such that g 0 (a2 ) = 0, g 0 (a) > 0 for a ∈ [0, a2 ), and g 0 (a) < 0 for a ∈ (a2 , 1). Thus, g is strictly increasing on [0, a2 ], and strictly decreasing on [a2 , 1]. Consequently, g(a) ≥ min{g(0), g(1)}, and from g(0) = 0,

g(1) = (3/2)k0 +1 − 3 = 0,

we get g(a) ≥ 0. This completes the proof. The equality holds for a = 0 and b = c (or any cyclic permutation). If k = k0 , then the equality holds also for a = b = c.

P 1.67. If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an = n, then  ‹ 1 1 1 (n + 1)2 + + ··· + ≥ 4(n + 2)(a12 + a22 + · · · + an2 ) + n(n2 − 3n − 6). a1 a2 an (Vasile Cîrtoaje, 2006)

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Vasile Cîrtoaje

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n(n2 − 3n − 6), where f (u) =

(n + 1)2 − 4(n + 2)u2 , u

u ∈ (0, ∞).

From

2(n + 1)2 − 8(n + 2), u3 it follows that f is strictly convex on (0,c] and strictly concave on [c, ∞), where v t 2 3 (n + 1) c= . 4(n + 2) f 00 (u) =

According to LCRCF-Theorem and Remark 6, it suffices to consider the case a1 = a2 = · · · = an−1 = x, 0 < x ≤ 1,

an = n − (n − 1)x,

when the inequality becomes as follows:  ‹ 1 2 n−1 (n + 1) + ≥ 4(n + 2)[(n − 1)x 2 + an2 ) + n(n2 − 3n − 6), x an n(n − 1)(2x − 1)2 [(n + 2)(n − 1)x 2 − (n + 2)(2n − 1)x + (n + 1)2 ] ≥ 0. The last inequality is true since 2n − 1 (n+2)(n−1)x −(n+2)(2n−1)x +(n+1) = (n+2)(n−1) x − 2n − 2 2

2

The equality holds for a1 = a2 = · · · = an−1 = tation).



‹2

+

3(n − 2) ≥ 0. 4(n − 1)

n+1 1 and an = (or any cyclic permu2 2

Chapter 2

Half Convex Function Method for Ordered Variables 2.1

Theoretical Basis

Half Convex Function Theorem for Ordered Variables (Vasile Cirtoaje, 2007). Let f (u) be a function defined on a real interval I and convex on Iu≥s / Iu≤s , where s ∈ I. The inequality a + a + ··· + a  1 2 n f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f n holds for all a1 , a2 , . . . , an ∈ I such that a1 + a2 + · · · + an = ns and at least n − m of a1 , a2 , . . . , an are smaller/greater than or equal to s if and only if f (x) + m f ( y) ≥ (1 + m) f (s) for all x, y ∈ I such that x + m y = (1 + m)s. Proof (for the right convexity). The necessity is obvious. Without loss of generality, assume that a1 ≤ a2 ≤ · · · ≤ an , a1 ≤ · · · ≤ an−m ≤ s. Since at least n − m of a1 , a2 , . . . , an are smaller than or equal to s , there is an integer k ∈ {n − m, . . . , n − 1} such that a1 ≤ · · · ≤ ak ≤ s ≤ ak+1 ≤ · · · ≤ an . Since f is convex on Iu≥s , we can apply Jensen’s inequality to get f (ak+1 ) + · · · + f (an ) ≥ (n − k) f (z), 97

z=

ak+1 + · · · + an . n−k

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Vasile Cîrtoaje

Thus, it suffices to show that f (a1 ) + · · · + f (ak ) + (n − k) f (z) ≥ n f (s). Let b1 , . . . , bk ∈ I defined by ai + mbi = (1 + m)s,

i = 1, . . . , k.

We claim that z ≥ b1 ≥ · · · ≥ bk ≥ s. Indeed, we have b1 ≥ · · · ≥ bk , bk − s =

s − ak ≥ 0, m

mb1 = (1 + m)s − a1 = (m − n + 1)s + (a2 + · · · + ak ) + (ak+1 + · · · + an ) ≤ (m − n + 1)s + (k − 1)s + ak+1 + · · · + an = (m − n + k)s + (ak+1 + · · · + an ) = (m − n + k)s + (n − k)z = (m − n + k)(s − z) + mz ≤ mz. By hypothesis, we have f (a1 ) + m f (b1 ) ≥ (1 + m) f (s), ··· f (ak ) + m f (bk ) ≥ (1 + m) f (s), hence f (a1 ) + · · · + f (ak ) + m[ f (b1 ) + · · · + f (bk )] ≥ k(1 + m) f (s). Consequently, it suffices to show that k(1 + m) f (s) − m[ f (b1 ) + · · · + f (bk )] + (n − k) f (z) ≥ n f (s), which is equivalent to p f (z) + (k − p) f (s) ≥ f (b1 ) + · · · + f (bk ),

p=

n−k ≤ 1. m

By Jensen’s inequality, we have p f (z) + (1 − p) f (s) ≥ f (w),

w = pz + (1 − p)s ≥ s.

Thus, we only need to show that f (w) + (k − 1) f (s) ≥ f (b1 ) + · · · + f (bk ).

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Since the decreasingly ordered vector A~k = (w, s, . . . , s) majorizes the decreasingly ordered vector B~k = (b1 , b2 , . . . , bk ), this inequality follows from Karamata’s inequality for convex functions. The proof for the left convexity is similar. Remark 1. Let us denote g(u) =

f (u) − f (s) , u−s

h(x, y) =

g(x) − g( y) . x−y

In many applications, it is useful to replace the hypothesis f (x) + m f ( y) ≥ (1 + m) f (s) in Half Convex Function Theorem for Ordered Variables (HCF-OV Theorem) by the equivalent condition h(x, y) ≥ 0 for all x, y ∈ I such that x + m y = (1 + m)s. This equivalence is true since f (x) + m f ( y) − (1 + m) f (s) = [ f (x) − f (s)] + m[ f ( y) − f (s)] = (x − s)g(x) + m( y − s)g( y) m = (x − y)[g(x) − g( y)] 1+m m (x − y)2 h(x, y). = 1+m Remark 2. Assume that f is differentiable on I, and let H(x, y) =

f 0 (x) − f 0 ( y) . x−y

Then, the desired inequality of Jensen’s type in HCF-OV Theorem holds true by replacing the hypothesis f (x) + m f ( y) ≥ (1 + m) f (s) with the more restrictive condition H(x, y) ≥ 0 for all x, y ∈ I such that x + m y = (1 + m)s. To prove this claim, we will show that the new condition implies f (x) + m f ( y) ≥ (1 + m) f (s) for all x, y ∈ I such that x + m y = (1 + m)s. Write this inequality as f1 (x) ≥ (1 + m) f (s), where  ‹ (1 + m)s − x f1 (x) = f (x) + m f ( y) = f (x) + m f . m

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Vasile Cîrtoaje

From f10 (x) = f 0 (x) − f 0



(1 + m)s − x m

‹

= f 0 (x) − f 0 ( y) =

1+m (x − s)H(x, y), m

it follows that f1 is decreasing for x ≤ s and increasing for x ≥ s; therefore, f1 (x) ≥ f1 (s) = (1 + m) f (s). Remark 3. HCF-OV Theorem is also valid in the case when I = [a, b] \ {u0 } or I = (a, b) \ {u0 }, where a, b, u0 are real numbers such that a < u0 < b. Clearly, two cases are possible: (1) u0 < s - when f is right convex, i.e. convex on Iu≥s ; (2) u0 > s - when f is left convex, i.e. convex on Iu≤s . Remark 4. In HCF-OV Theorem for the right convexity (when HCF-OV Theorem is called RHCF-OV Theorem), it suffices to consider x ≤ s ≤ y. This claim is true because in the proof of HCF-OV Theorem, the hypothesis f (x) + m f ( y) ≥ (1 + m) f (s) is used to get the inequalities f (ai ) + m f (bi ) ≥ (1 + m) f (s),

i = 1, 2, · · · , k,

where ai ≤ s ≤ bi . Similarly, in HCF-OV Theorem for the left convexity (when HCF-OV Theorem is called LHCF-OV Theorem), it suffices to consider x ≥ s ≥ y. Remark 5. HCF-OV Theorem is a generalization of HCF-Theorem, because the last theorem can be obtained from the first theorem for m = n − 1. Remark 6. If a1 ≥ · · · ≥ am ≥ s ≥ am+1 ≥ · · · ≥ an , then at least n − m of a1 , a2 , . . . , an are smaller than or equal to s. Also, if a1 ≥ · · · ≥ an−m ≥ s ≥ an−m+1 ≥ · · · ≥ an , then at least n − m of a1 , a2 , . . . , an are greater than or equal to s. Thus, from HCF-OV Theorem and Remark 4, we get the following corollaries.

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RHCF-OV Corollary. Let f (u) be a function defined on a real interval I and convex on Iu≥s , where s ∈ I. The inequality a + a + ··· + a  1 2 n f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f n holds for all a1 , a2 , . . . , an ∈ I satisfying a1 ≥ · · · ≥ am ≥ s ≥ am+1 ≥ · · · ≥ an ,

a1 + a2 + · · · + an = ns,

if and only if f (x) + m f ( y) ≥ (1 + m) f (s) for all x, y ∈ I such that x ≤ s ≤ y and x + m y = (1 + m)s. LHCF-OV Corollary. Let f (u) be a function defined on a real interval I and convex on Iu≤s , where s ∈ I. The inequality a + a + ··· + a  1 2 n f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f n holds for all a1 , a2 , . . . , an ∈ I satisfying a1 ≤ · · · ≤ am ≤ s ≤ am+1 ≤ · · · ≤ an ,

a1 + a2 + · · · + an = ns,

if and only if f (x) + m f ( y) ≥ (1 + m) f (s) for all x, y ∈ I such that x ≥ s ≥ y and x + m y = (1 + m)s. Remark 7. The inequality in RHCF-OV Corollary becomes an equality for a1 = a2 = · · · = an = s and also for a1 = · · · = am = y,

am+1 = · · · = an−1 = s,

an = x (x < y),

where x, y ∈ I satisfy the equations x + m y = (1 + m)s,

f (x) + m f ( y) = (1 + m) f (s).

For x 6= y, these equations are equivalent to x + m y = (1 + m)s, h(x, y) = 0. Remark 8. The inequality in LHCF-OV Corollary becomes an equality for a1 = a2 = · · · = an = s

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and also for a1 = x,

a2 = · · · = an−m = s,

an−m+1 = · · · = an = y (x > y),

where x, y ∈ I satisfy the equations x + m y = (1 + m)s,

f (x) + m f ( y) = (1 + m) f (s).

For x 6= y, these equations are equivalent to x + m y = (1 + m)s, h(x, y) = 0.

HCF Method for Ordered Variables

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103

Applications

2.1. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. (a) If at least n − m of a1 , a2 , . . . , an are smaller than or equal to 1, then m(a13 + a23 + · · · + an3 − n) ≥ (2m + 1)(a12 + a22 + · · · + an2 − n); (b) If at least n − m of a1 , a2 , . . . , an are greater than or equal to 1, then a13 + a23 + · · · + an3 − n ≤ (m + 2)(a12 + a22 + · · · + an2 − n).

2.2. If a, b, c, d are real numbers such that a + b + c + d = 4,

a ≥ b ≥ 1 ≥ c ≥ d, then

(3a2 − 2)(a − 1)2 + (3b2 − 2)(b − 1)2 + (3c 2 − 2)(c − 1)2 + (3d 2 − 2)(d − 1)2 ≥ 0.

2.3. If a1 , a2 , . . . , a2n ≥

−2n − 1 such that n−1

a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,

a1 + a2 + · · · + a2n = 2n,

then 3 ≥ 2n. a13 + a23 + · · · + a2n

2.4. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = n. (a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then a14 + a24 + · · · + an4 − n ≥ 6(a12 + a22 + · · · + an2 − n); (b) If a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an , then a14 + a24 + · · · + an4 − n ≥

14 2 (a + a22 + · · · + an2 − n); 3 1

(c) If a1 ≥ · · · ≥ an−2 ≥ 1 ≥ an−1 ≥ an , then a14 + a24 + · · · + an4 − n ≥

2(n2 − 3n + 3) 2 (a1 + a22 + · · · + an2 − n). n2 − 5n + 7

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2.5. Let a, b, c, d, e be nonnegative real numbers such that a + b + c + d + e = 5. (a) If a ≥ b ≥ 1 ≥ c ≥ d ≥ e, then 21(a2 + b2 + c 2 + d 2 + e2 ) ≥ a4 + b4 + c 4 + d 4 + e4 + 100; (b) If a ≥ b ≥ c ≥ 1 ≥ d ≥ e, then 13(a2 + b2 + c 2 + d 2 + e2 ) ≥ a4 + b4 + c 4 + d 4 + e4 + 60. 2.6. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. (a) If a1 ≥ · · · ≥ an−1 ≥ 1 ≥ an , then 7(a13 + a23 + · · · + an3 ) ≥ 3(a14 + a24 + · · · + an4 ) + 4n; (b) If a1 ≥ · · · ≥ an−2 ≥ 1 ≥ an−1 ≥ an , then 13(a13 + a23 + · · · + an3 ) ≥ 4(a14 + a24 + · · · + an4 ) + 9n. 2.7. If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an = n,

a1 ≥ · · · ≥ an−m ≥ 1 ≥ an−m+1 ≥ · · · ≥ an , then (m + 1)2



‹ 1 1 1 + + ··· + − n ≥ 4m(a12 + a22 + · · · + an2 − n). a1 a2 an

2.8. If a1 , a2 , . . . , an are positive real numbers such that 1 1 1 + + ··· + = n, a1 a2 an

a1 ≤ · · · ≤ an−m ≤ 1 ≤ an−m+1 ≤ · · · ≤ an , then a12

+ a22

+ · · · + an2

p ‹  m (a1 + a2 + · · · + an − n). −n≥2 1+ 1+m

2.9. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. (a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then 1 a12

+2

+

1 a22

+2

+ ··· +

an2

1 n ≥ ; +2 3

(b) If a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an , then 1 2a12

+3

+

1 2a22

+3

+ ··· +

1 n ≥ . +3 5

2an2

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2.10. If a1 , a2 , . . . , a2n are nonnegative real numbers such that a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,

a1 + a2 + · · · + a2n = 2n,

then 1 na12

+ n2

+n+1

+

1 na22

+ n2

+n+1

+ ··· +

1 2 na2n

+ n2

+n+1

≤

2n . (n + 1)2

2.11. If a, b, c, d, e, f are nonnegative real numbers such that a ≥ b ≥ c ≥ 1 ≥ d ≥ e ≥ f, then

a + b + c + d + e + f = 6,

3f + 4 3a + 4 3b + 4 3c + 4 3d + 4 3e + 4 + 2 + 2 + 2 + 2 + ≤ 6. 2 3a + 4 3b + 4 3c + 4 3d + 4 3e + 4 3 f 2 + 4

2.12. If a, b, c, d, e, f are nonnegative real numbers such that a ≥ b ≥ 1 ≥ c ≥ d ≥ e ≥ f,

a + b + c + d + e + f = 6,

then f2−1 b2 − 1 c2 − 1 d2 − 1 e2 − 1 a2 − 1 + + + + + ≥ 0. (2a + 7)2 (2b + 7)2 (2c + 7)2 (2d + 7)2 (2e + 7)2 (2 f + 7)2

2.13. If a, b, c, d, e, f are nonnegative real numbers such that a ≥ b ≥ c ≥ d ≥ 1 ≥ e ≥ f,

a + b + c + d + e + f = 6,

then f2−1 b2 − 1 c2 − 1 d2 − 1 e2 − 1 a2 − 1 + + + + + ≤ 0. (2a + 5)2 (2b + 5)2 (2c + 5)2 (2d + 5)2 (2e + 5)2 (2 f + 5)2

2.14. If a, b, c, d are nonnegative real numbers such that a ≥ b ≥ 1 ≥ c ≥ d, then

a + b + c + d = 4,

1 1 1 1 4 + 3 + 3 + 3 ≤ . + 5 2b + 5 2c + 5 2d + 5 7

2a3

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2.15. If a1 , a2 , . . . , a8 are nonnegative real numbers such that a1 ≥ a2 ≥ a3 ≥ a4 ≥ 1 ≥ a5 ≥ a6 ≥ a7 ≥ a8 ,

a1 + a2 + · · · + a8 = 8,

then (a12 + 1)(a22 + 1) · · · (a82 + 1) ≥ (a1 + 1)(a2 + 1) · · · (a8 + 1).

2.16. If a, b, c, d are positive real numbers such that a ≥ b ≥ 1 ≥ c ≥ d, then

a bcd = 1,

‹ 1 1 1 1 a + b + c + d − 4 ≥ 18 a + b + c + d − − − − . a b c d 2

2

2



2

2.17. If a, b, c, d are positive real numbers such that a ≥ b ≥ 1 ≥ c ≥ d,

a bcd = 1,

then p

a2 − a + 1 +

p

b2 − b + 1 +

p

c2 − c + 1 +

p

d 2 − d + 1 ≥ a + b + c + d.

2.18. If a, b, c, d are positive real numbers such that a ≥ 1 ≥ b ≥ c ≥ d, then a3

a bcd = 1,

1 1 1 1 2 + 3 + 3 + 3 ≥ . + 3a + 2 b + 3b + 2 c + 3c + 2 d + 3d + 2 3

2.19. Let a1 , a2 , . . . , an be positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , If k ≥ 1, then

a1 a2 · · · an = 1.

1 1 1 n + + ··· + ≥ . 1 + ka1 1 + ka2 1 + kan 1+k

HCF Method for Ordered Variables

107

2.20. If a1 , a2 , . . . , a9 are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ a9 , then

a1 a2 · · · a9 = 1,

1 1 1 + + ··· + ≥ 1. (a1 + 2)2 (a2 + 2)2 (a9 + 2)2

2.21. If a1 , a2 , . . . , an (n ≥ 4) are positive real numbers such that a1 ≥ a2 ≥ a3 ≥ 1 ≥ a4 ≥ · · · ≥ an , then

a1 a2 · · · an = 1,

1 1 1 n + + ··· + ≥ . (a1 + 1)2 (a2 + 1)2 (an + 1)2 4

2.22. 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 + 3) (a2 + 3) (an + 3) 16

2.23. Let a1 , a2 , . . . , an be positive real numbers such that a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,

a1 a2 · · · an = 1.

If p, q ≥ 0 such that p + q ≤ 1, then 1 1+

pa1 + qa12

+

1 1+

pa2 + qa22

+ ··· +

1 n ≤ . 2 1 + pan + qan 1+p+q

2.24. If a1 , a2 , . . . , an are positive real numbers such that a1 ≤ a2 ≤ 1 ≤ a3 ≤ · · · ≤ an , then

a1 a2 · · · an = 1,

1 1 1 n + + ··· + ≤ . 2 2 2 (a1 + 5) (a2 + 5) (an + 5) 36

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2.25. If a1 , a2 , . . . , an are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

1 p

1 + 3a1

+p

1 1 + 3a2

a1 a2 · · · an = 1,

+ ··· + p

1 1 + 3an

≥

n . 2

2.26. If a1 , a2 , . . . , an are positive real numbers such that a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

1 p

1 + 2a1

+p

1 1 + 2a2

a1 a2 · · · an = 1,

+ ··· + p

1

n ≤p . 3 1 + 2an

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109

Solutions

P 2.1. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. (a) If at least n − m of a1 , a2 , . . . , an are smaller than or equal to 1, then m(a13 + a23 + · · · + an3 − n) ≥ (2m + 1)(a12 + a22 + · · · + an2 − n); (b) If at least n − m of a1 , a2 , . . . , an are greater than or equal to 1, then a13 + a23 + · · · + an3 − n ≤ (m + 2)(a12 + a22 + · · · + an2 − n). (Vasile Cirtoaje, 2007) Solution. (a) Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

where f (u) = mu3 − (2m + 1)u2 ,

u ∈ [0, n].

From f 00 (u) = 2(3mu − 2m − 1), it follows that f is convex on [1, n]. By HCF-OV Theorem (or RHCF-OV Corollary) and Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + m y = m + 1. We have g(u) =

f (u) − f (1) = m(u2 + u + 1) − (2m + 1)(u + 1) = mu2 − (m + 1)u − m − 1, u−1 h(x, y) =

g(x) − g( y) = m(x + y) − m − 1 = (m − 1)x ≥ 0. x−y

From x + m y = m + 1 and h(x, y) = 0, we get x = 0, y = (m + 1)/m. Therefore, in accordance with Remark 7, the equality holds for a1 = a2 = · · · = an = 1, and also for a1 = · · · = am =

m+1 , m

am+1 = · · · = an−1 = 1,

an = 0

(or any thereof permutation). (b) Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

where f (u) = (m + 2)u2 − u3 ,

u ∈ [0, n].

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From f 00 (u) = 2(m + 2 − 3u), it follows that f is convex on [0, 1]. By HCF-OV Theorem (or LRHCF-OV Corollary) and Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + m y = m + 1. We have g(u) =

f (u) − f (1) = (m + 2)(u + 1) − (u2 + u + 1) = −u2 + (m + 1)u + m + 1, u−1 h(x, y) =

g(x) − g( y) = −(x + y) + m + 1 = (m − 1) y ≥ 0. x−y

From x +m y = m+1 and h(x, y) = 0, we get x = m+1, y = 0. Therefore, in accordance with Remark 8, the equality holds for a1 = a2 = · · · = an = 1, and also for a1 = m + 1,

a2 = · · · = an−m = 1,

an−m+1 = · · · = an = 0

(or any thereof permutation). Remark 1. For m = n − 1, we get the following results: • If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then (n − 1)(a13 + a23 + · · · + an3 − n) ≥ (2n − 1)(a12 + a22 + · · · + an2 − n), with equality for a1 = a2 = · · · = an = 1, and also for a1 = 0 and a2 = a3 = · · · = an = n (or any cyclic permutation). n−1 • If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then a13 + a23 + · · · + an3 − n ≤ (n + 1)(a12 + a22 + · · · + an2 − n), with equality for a1 = a2 = · · · = an = 1, and also for a1 = n and a2 = a3 = · · · = an = 0 (or any cyclic permutation). Remark 2. For m = 1, we get the following results: • If a1 , a2 , . . . , an are nonnegative real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,

a1 + a2 + · · · + an = n,

then a13 + a23 + · · · + an3 + 2n ≥ 3(a12 + a22 + · · · + an2 ), with equality for a1 = a2 = · · · = an = 1, and also for a1 = 2, a2 = · · · = an−1 = 1, an = 0. • If a1 , a2 , . . . , an are nonnegative real numbers such that a1 ≥ · · · ≥ an−1 ≥ 1 ≥ an ,

a1 + a2 + · · · + an = n,

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111

then a13 + a23 + · · · + an3 + 2n ≤ 3(a12 + a22 + · · · + an2 ), with equality for a1 = a2 = · · · = an = 1, and also for a1 = 2, a2 = · · · = an−1 = 1, an = 0. Remark 3. Replacing n with 2n and choosing m = n, we get the following results: • If a1 , a2 , . . . , a2n are nonnegative real numbers such that a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,

a1 + a2 + · · · + a2n = 2n,

then 2 3 − 2n), − 2n) ≥ (2n + 1)(a12 + a22 + · · · + a2n n(a13 + a23 + · · · + a2n

with equality for a1 = a2 = · · · = a2n = 1, and also for a1 = · · · = an =

n+1 , n

an+2 = · · · = a2n−1 = 1,

a2n = 0.

• If a1 , a2 , . . . , a2n are nonnegative real numbers such that a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,

a1 + a2 + · · · + a2n = 2n,

then 2 3 − 2n), − 2n ≤ (n + 2)(a12 + a22 + · · · + a2n a13 + a23 + · · · + a2n

with equality for a1 = a2 = · · · = a2n = 1, and also for a1 = n + 1,

a2 = · · · = an = 1,

an+1 = · · · = a2n = 0.

P 2.2. If a, b, c, d are real numbers such that a ≥ b ≥ 1 ≥ c ≥ d,

a + b + c + d = 4,

then (3a2 − 2)(a − 1)2 + (3b2 − 2)(b − 1)2 + (3c 2 − 2)(c − 1)2 + (3d 2 − 2)(d − 1)2 ≥ 0. (Vasile Cirtoaje, 2007)

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Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),

s=

a+b+c+d = 1, 4

where f (u) = (3u2 − 2)(u − 1)2 ,

u ∈ R.

From f 00 (u) = 2(18u2 − 18u + 1), it follows that f 00 (u) > 0 for u ≥ 1, hence f is convex for u ≥ 1. Therefore, we may apply HCF-OV Theorem or RHCF-OV Corollary for n = 4 and m = 2. By these, it suffices to show that f (x) + 2 f ( y) ≥ 3 f (1) for all real x, y such that x + 2 y = 3. Using Remark 1, we only need to show that h(x, y) ≥ 0, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

We have g(u) = 3(u3 + u2 + u + 1) − 6(u2 + u + 1) + u + 1 = 3u3 − 3u2 − 2u − 2, h(x, y) = 3(x 2 + x y + y 2 ) − 3(x + y) − 2 = (3 y − 4)2 ≥ 0. From x + 2 y = 3 and h(x, y) = 0, we get x = 1/3, y = 4/3. Therefore, in accordance with Remark 7, the equality holds for a = b = c = d = 1, and also for a = b = 4/3, c = 1 and d = 1/3. Remark. Similarly, we can prove the following generalization: • Let a1 , a2 , . . . , a2n be real numbers such that a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n , If k =

n2

a1 + a2 + · · · + a2n = 2n.

n , then −n+1 2 (a12 − k)(a1 − 1)2 + (a22 − k)(a2 − 1)2 + · · · + (a2n − k)(a2n − 1)2 ≥ 0,

with equality for a1 = a2 = · · · = a2n = 1, and also for a1 = · · · = an =

n2 , n2 − n + 1

an+1 = · · · = a2n−1 = 1,

a2n =

n2

1 . −n+1

HCF Method for Ordered Variables

P 2.3. If a1 , a2 , . . . , a2n ≥

113

−2n − 1 such that n−1

a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,

a1 + a2 + · · · + a2n = 2n,

then 3 a13 + a23 + · · · + a2n ≥ 2n.

(Vasile Cirtoaje, 2007) Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (a2n ) ≥ 2n f (s), where f (u) = u3 ,

u≥

s=

a1 + a2 + · · · + a2n = 1, 2n

−2n − 1 . n−1

From f 00 (u) = 6u, it follows that f is convex for u ≥ 1. Therefore, we may apply HCF-OV Theorem or RHCF-OV Corollary for 2n numbers and m = n. By Remark 1, it suffices to −2n − 1 show that h(x, y) ≥ 0 for all x, y ≥ such that x + n y = n + 1. We have n−1 g(u) = h(x, y) =

f (u) − f (1) = u2 + u + 1, u−1

g(x) − g( y) (n − 1)x + 2n + 1 = x + y +1= ≥ 0. x−y n−1

n+2 −2n − 1 and y = . Therefore, n−1 n−1 in accordance with Remark 7, the equality holds for a1 = a2 = · · · = a2n = 1, and also for n+2 −2n − 1 a1 = · · · = an = , an+1 = · · · = a2n−1 = 1, a2n = . n−1 n−1 From x + n y = n + 1 and h(x, y) = 0, we get x =

P 2.4. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = n. (a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then a14 + a24 + · · · + an4 − n ≥ 6(a12 + a22 + · · · + an2 − n); (b) If a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an , then a14 + a24 + · · · + an4 − n ≥

14 2 (a + a22 + · · · + an2 − n); 3 1

114

Vasile Cîrtoaje (c) If a1 ≥ · · · ≥ an−2 ≥ 1 ≥ an−1 ≥ an , then a14 + a24 + · · · + an4 − n ≥

2(n2 − 3n + 3) 2 (a1 + a22 + · · · + an2 − n). n2 − 5n + 7 (Vasile Cirtoaje, 2009)

Solution. Consider the inequality a14 + a24 + · · · + an4 − n ≥ k(a12 + a22 + · · · + an2 − n),

k ≤ 6,

and write it as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

where f (u) = u4 − ku2 ,

u ≥ 0.

From f 00 (u) = 2(6u2 − k), it follows that f is convex for u ≥ 1. Therefore, we may apply HCF-OV Theorem or RHCF-OV Corollary for m = 1, m = 2 and m = n − 2, respectively. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x +m y = m+1. We have f (u) − f (1) g(u) = = u3 + u2 + u + 1 − k(u + 1), u−1 h(x, y) =

g(x) − g( y) = x 2 + x y + y 2 + x + y − k + 1. x−y

(a) We have k = 6, m = 1 and x + y = 2, which involve h(x, y) = 1 − x y =

1 (x − y)2 ≥ 0. 4

From x + y = 2 and h(x, y) = 0, we get x = y = 1. Therefore, in accordance with Remark 7, the equality holds for a1 = a2 = · · · = an = 1. (b) We have k = 14/3, m = 2 and x + 2 y = 3, which involve h(x, y) =

1 (3 y − 5)2 ≥ 0. 3

From x + 2 y = 3 and h(x, y) = 0, we get x = −1/3 and y = 5/3. Therefore, in accordance with Remark 7, the equality holds for a1 = a2 = · · · = an = 1, and also for a1 = a2 =

5 , 3

a3 = · · · = an−1 = 1,

an =

−1 . 3

HCF Method for Ordered Variables

(c) We have k =

115

2(n2 − 3n + 3) , m = n − 2 and x + (n − 2) y = n − 1, which involve n2 − 5n + 7

h(x, y) =

n2

1 [(n2 − 5n + 7) y − n2 + 3n − 1]2 ≥ 0. − 5n + 7

−n2 + 5n − 4 n2 − 3n + 1 and . n2 − 5n + 7 n2 − 5n + 7 Therefore, in accordance with Remark 7, the equality holds for a1 = a2 = · · · = an = 1, and also for From x + (n − 2) y = n − 1 and h(x, y) = 0, we get x =

a1 = · · · = an−2 =

n2 − 3n + 1 , n2 − 5n + 7

an−1 = 1,

an =

−n2 + 5n − 4 . n2 − 5n + 7

P 2.5. Let a, b, c, d, e be nonnegative real numbers such that a + b + c + d + e = 5. (a) If a ≥ b ≥ 1 ≥ c ≥ d ≥ e, then 21(a2 + b2 + c 2 + d 2 + e2 ) ≥ a4 + b4 + c 4 + d 4 + e4 + 100; (b) If a ≥ b ≥ c ≥ 1 ≥ d ≥ e, then 13(a2 + b2 + c 2 + d 2 + e2 ) ≥ a4 + b4 + c 4 + d 4 + e4 + 60. (Vasile Cirtoaje, 2009) Solution. Consider the inequality k(a2 + b2 + c 2 + d 2 + e2 − 5) ≥ a4 + b4 + c 4 + d 4 + e4 − 5,

k ≥ 6,

and write it as f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s),

s=

a+b+c+d+e = 1, 5

where f (u) = ku2 − u4 ,

u ≥ 0.

From f 00 (u) = 2(k − 6u2 ), it follows that f is convex on [0, 1]. Therefore, we may apply HCF-OV Theorem or LHCF-OV Corollary for m = 3 and m = 2, respectively. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + m y = m + 1. We have f (u) − f (1) g(u) = = k(u + 1) − (u3 + u2 + u + 1), u−1

116

Vasile Cîrtoaje h(x, y) =

g(x) − g( y) = k − (x 2 + x y + y 2 + x + y + 1). x−y

(a) We have k = 21, m = 3 and x + 3 y = 4, which involve y ≤ 4/3 and h(x, y) = 20 − (x 2 + x y + y 2 + x + y) = y(22 − 7 y) ≥ 0. From x + 3 y = 4 and h(x, y) = 0, we get x = 4 and y = 0. Therefore, in accordance with Remark 8, the equality holds for a = b = c = d = e = 1 and also for a = 4, b = 1, c = d = e = 0. (b) We have k = 13, m = 2 and x + 2 y = 3, which involve y ≤ 3/2 and h(x, y) = 12 − (x 2 + x y + y 2 + x + y) = y(10 − 3 y) ≥ 0. From x + 2 y = 3 and h(x, y) = 0, we get x = 3 and y = 0. Therefore, in accordance with Remark 8, the equality holds for a = b = c = d = e = 1 and also for a = 3, b = c = 1, d = e = 0.

P 2.6. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. (a) If a1 ≥ · · · ≥ an−1 ≥ 1 ≥ an , then 7(a13 + a23 + · · · + an3 ) ≥ 3(a14 + a24 + · · · + an4 ) + 4n; (b) If a1 ≥ · · · ≥ an−2 ≥ 1 ≥ an−1 ≥ an , then 13(a13 + a23 + · · · + an3 ) ≥ 4(a14 + a24 + · · · + an4 ) + 9n. (Vasile Cirtoaje, 2009) Solution. Consider the inequality k(a13 + a23 + · · · + an3 − n) ≥ a14 + a24 + · · · + an4 − n,

k > 2,

and write it as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

where f (u) = ku3 − u4 ,

u ≥ 0.

From f 00 (u) = 6u(k − 2u2 ), it follows that f is convex on [0, 1]. Therefore, we may apply HCF-OV Theorem or LHCF-OV Corollary for m = 1 and m = 2, respectively. By

HCF Method for Ordered Variables

117

Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + m y = m + 1. We have f (u) − f (1) g(u) = = k(u2 + u + 1) − (u3 + u2 + u + 1), u−1 g(x) − g( y) h(x, y) = = −(x 2 + x y + y 2 ) + (k − 1)(x + y + 1). x−y (a) We have k = 7/3, m = 1 and x + y = 2, which involve h(x, y) = x y ≥ 0. From x + y = 2 and h(x, y) = 0, we get x = 2 and y = 0. Therefore, in accordance with Remark 8, the equality holds for a1 = a2 = · · · = an = 1, and also for a1 = 2,

a2 = · · · = an−1 = 1,

an = 0.

(b) We have k = 13/4, m = 2 and x + 2 y = 3, which involve h(x, y) = 3 y(9 − 4 y) = 3 y(3 + 2x) ≥ 0. From x + 2 y = 3 and h(x, y) = 0, we get x = 3 and y = 0. Therefore, in accordance with Remark 8, the equality holds for a1 = a2 = · · · = an = 1, and also for a1 = 3,

a2 = · · · = an−2 = 1,

an−1 = an = 0.

P 2.7. If a1 , a2 , . . . , an are positive real numbers such that a1 ≥ · · · ≥ an−m ≥ 1 ≥ an−m+1 ≥ · · · ≥ an , then (m + 1)2



a1 + a2 + · · · + an = n,

‹ 1 1 1 + + ··· + − n ≥ 4m(a12 + a22 + · · · + an2 − n). a1 a2 an (Vasile Cirtoaje, 2007)

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

s=

a1 + a2 + · · · + an = 1, n

(m + 1)2 − 4mu2 , u ∈ (0, n). u

118

Vasile Cîrtoaje

For u ∈ (0, 1], we have f 00 (u) =

2(m + 1)2 − 8m ≥ 2(m + 1)2 − 8m = 2(m − 1)2 ≥ 0. u3

Since f is convex on (0, 1], we may apply HCF-OV Theorem or LHCF-OV Corollary. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y > 0 such that x + m y = m + 1. We have f (u) − f (1) −(m + 1)2 g(u) = = − 4m(u + 1), u−1 u h(x, y) =

[m + 1 − 2m y]2 (m + 1)2 − 4m = ≥ 0. xy xy

1+m 1+m and y = . Therefore, in 2 2m accordance with Remark 8, the equality holds for a1 = a2 = · · · = an = 1, and also for From x + m y = m + 1 and h(x, y) = 0, we get x =

a1 =

1+m , 2

a2 = a3 = · · · = an−m = 1,

an−m+1 = · · · = an =

1+m . 2m

Remark 1. For m = 1, we get the following elegant statement: • If a1 , a2 , . . . , an are positive real numbers such that a1 ≥ · · · ≥ an−1 ≥ 1 ≥ an , then

a1 + a2 + · · · + an = n,

1 1 1 + + ··· + ≥ a12 + a22 + · · · + an2 , a1 a2 an

with equality for a1 = a2 = · · · = an = 1 Remark 2. Replacing n with 2n and choosing m = n, we get the statement: • If a1 , a2 , . . . , a2n are positive real numbers such that a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n , then (n + 1)2



a1 + a2 + · · · + a2n = 2n,

‹ 1 1 1 2 + + ··· + − 2n ≥ 4n(a12 + a22 + · · · + a2n − 2n), a1 a2 a2n

with equality for a1 = a2 = · · · = a2n = 1, and also for a1 =

1+n , 2

a2 = a3 = · · · = an = 1,

an+1 = · · · = a2n =

1+n . 2n

HCF Method for Ordered Variables

119

P 2.8. If a1 , a2 , . . . , an are positive real numbers such that 1 1 1 + + ··· + = n, a1 a2 an

a1 ≤ · · · ≤ an−m ≤ 1 ≤ an−m+1 ≤ · · · ≤ an , then a12

+ a22

p  ‹ m −n≥2 1+ (a1 + a2 + · · · + an − n). 1+m

+ · · · + an2

(Vasile Cirtoaje, 2007) Solution. Replacing each ai by 1/ai , we need to prove that a1 ≥ · · · ≥ an−m ≥ 1 ≥ an−m+1 ≥ · · · ≥ an ,

a1 + a2 + · · · + an = n

involves f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where

s=

a1 + a2 + · · · + an = 1, n

p  ‹ 1 1 m , u ∈ (0, n). f (u) = 2 − 2 1 + u 1+m u

For u ∈ (0, 1], we have 00

f (u) =

€ 6−4 1+

p Š m 1+m u

u4

≥

€ 6−4 1+

p Š m m+1

u4

p 2( m − 1)2 = ≥ 0. (1 + m)u4

Thus, f is convex on (0, 1]. By HCF-OV Theorem (or LHCF-OV Corollary) and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y > 0 such that x + m y = 1 + m, where h(x, y) = We have

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

p ‹  2 m 1 −1 g(u) = 2 + 1 + u 1+m u

and h(x, y) = We only need to show that

1 xy



p ‹ 1 1 2 m + −1− . x y 1+m

p 1 1 2 m + ≥1+ . x y 1+m

Indeed, using the Cauchy-Schwarz inequality, we get p p p (1 + m)2 (1 + m)2 2 m 1 1 + ≥ = =1+ . x y x + my 1+m 1+m

120

Vasile Cîrtoaje

1+m 1+m p and y = p . In 1+ m m+ m accordance with Remark 8, we have f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (1) for a1 = a2 = · · · = an = 1, and also for From x + m y = 1 + m and h(x, y) = 0, we get x =

a1 =

1+m p , 1+ m

a2 = a3 = · · · = an−m = 1,

an−m+1 = · · · = an =

1+m p . m+ m

Therefore, the original inequality becomes an equality for a1 = a2 = · · · = an = 1, and also for p p m+ m 1+ m , a2 = a3 = · · · = an−m = 1, an−m+1 = · · · = an = . a1 = 1+m 1+m Remark. Replacing n with 2n and choosing m = n, we get the statement: • If a1 , a2 , . . . , a2n are positive real numbers such that a1 ≤ · · · ≤ an ≤ 1 ≤ an+1 ≤ · · · ≤ a2n , then a12

+ a22

1 1 1 + + ··· + = 2n, a1 a2 a2n

p ‹  n − 2n ≥ 2 1 + (a1 + a2 + · · · + a2n − 2n). 1+n

2 + · · · + a2n

with equality for a1 = a2 = · · · = a2n = 1, and also for a1 =

p 1+ n , 1+n

a2 = a3 = · · · = an = 1,

an+1 = · · · = a2n =

p n+ n . 1+n

P 2.9. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. (a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then 1 a12

+2

+

1 a22

+2

+ ··· +

1 n ≥ ; an2 + 2 3

(b) If a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an , then 1 2a12

+3

+

1 2a22

+3

+ ··· +

1 n ≥ . 2an2 + 3 5 (Vasile Cirtoaje, 2007)

HCF Method for Ordered Variables

121

Solution. Consider the inequality 1 a12

+k

+

1 a22

+k

+ ··· +

an2

1 n ≥ , +k 1+k

k ∈ [0, 3];

and write it as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

1 , u2 + k

s=

a1 + a2 + · · · + an = 1, n

u ∈ [0, n].

For u ∈ [1, n], we have f 00 (u) =

2(3u2 − k) 2(3 − k) ≥ 2 ≥ 0, 2 3 (u + k) (u + k)3

hence f is convex on [1, n]. Therefore, we may apply HCF-OV Theorem or RHCF-OV Corollary for m = 1 and m = 2, respectively. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + m y = m + 1. Since g(u) = h(x, y) =

f (u) − f (1) −u − 1 = , u−1 (1 + k)(u2 + k)

xy+x + y−k g(x) − g( y) = , x−y (1 + k)(x 2 + k)( y 2 + k)

we only need to show that x y + x + y − k ≥ 0. (a) We have k = 2, m = 1 and x + y = 2, which involve x y + x + y − k = x y ≥ 0. From x + y = 2 and h(x, y) = 0, we get x = 0 and y = 2. Therefore, in accordance with Remark 7, the equality holds for a1 = a2 = · · · = an = 1, and also for a1 = 2,

a2 = · · · = an−1 = 1,

an = 0.

(b) We have k = 3/2, m = 2 and x + 2 y = 3, which involve xy+x + y−k=

x(1 + 2 y) x(4 − x) = ≥ 0. 2 2

From x + 2 y = 3 and h(x, y) = 0, we get x = 0 and y = 3/2. Therefore, in accordance with Remark 7, the equality holds for a1 = a2 = · · · = an = 1, and also for a1 = a2 = 3/2,

a3 = · · · = an−1 = 1,

an = 0.

122

Vasile Cîrtoaje

P 2.10. If a1 , a2 , . . . , a2n are nonnegative real numbers such that a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,

a1 + a2 + · · · + a2n = 2n,

then 1 na12

+ n2

+n+1

+

1 na22

+ n2

+n+1

+ ··· +

1 2 na2n

+ n2

+n+1

≤

2n . (n + 1)2

(Vasile Cirtoaje, 2007) Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (a2n ) ≥ 2n f (s), where f (u) =

−1 , nu2 + n2 + n + 1

s=

a1 + a2 + · · · + a2n = 1, 2n

u ∈ [0, 2n].

For u ∈ [0, 1], we have f 00 (u) =

2nu(n2 + n + 1 − 3nu2 ) 2nu(n2 + n + 1 − 3n) ≥ ≥ 0, (nu2 + n2 + n + 1)3 (nu2 + n2 + n + 1)3

hence f is convex on [0, 1]. Therefore, we may apply HCF-OV Theorem or LHCF-OV Corollary for m = n. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + n y = n + 1. Since g(u) = h(x, y) =

f (u) − f (1) n(u + 1) = , 2 u−1 (n + 1) (nu2 + n2 + n + 1)

g(x) − g( y) n(n2 + n + 1 − nx − n y − nx y) = , x−y (n + 1)2 (nx 2 + n2 + n + 1)(n y 2 + n2 + n + 1)

we only need to show that n2 + n + 1 − n(x + y + x y) ≥ 0. Indeed, n2 + n + 1 − n(x + y + x y) = (n y − 1)2 ≥ 0. From x + n y = n + 1 and h(x, y) = 0, we get x = n and y = 1/n. Therefore, in accordance with Remark 7, the equality holds for a1 = a2 = · · · = a2n = 1, and also for a1 = n,

a2 = · · · = an = 1,

an+1 = · · · = an = 1/n.

HCF Method for Ordered Variables

123

P 2.11. If a, b, c, d, e, f are nonnegative real numbers such that a ≥ b ≥ c ≥ 1 ≥ d ≥ e ≥ f, then

a + b + c + d + e + f = 6,

3f + 4 3b + 4 3c + 4 3d + 4 3e + 4 3a + 4 + 2 + 2 + 2 + 2 + ≤ 6. 2 3a + 4 3b + 4 3c + 4 3d + 4 3e + 4 3 f 2 + 4 (Vasile Cirtoaje, 2009)

Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) + f (e) + f ( f ) ≥ 6 f (s), where f (u) =

−3u − 4 , 3u2 + 4

s=

a+b+c+d+e+ f = 1, 6

u ∈ [0, 6].

For u ∈ [0, 1], we have f 00 (u) =

6(16 − 9u3 ) + 216u(1 − u) ≥ 0, (3u2 + 4)3

hence f is convex on [0, 1]. Therefore, we may apply HCF-OV Theorem or LHCF-OV Corollary for m = 3. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + 3 y = 4. We have g(u) = h(x, y) =

f (u) − f (1) 3u = 2 , u−1 3u + 4

g(x) − g( y) 3(4 − 3x y) 3(x − 2)2 = = ≥ 0. x−y (3x 2 + 4)(3 y 2 + 4) (3x 2 + 4)(3 y 2 + 4)

From x + 3 y = 4 and h(x, y) = 0, we get x = 2 and y = 2/3. Therefore, in accordance with Remark 8, the equality holds only for a = b = c = d = e = f = 1, and also for a = 2, b = c = 1, d = e = f = 2/3.

P 2.12. If a, b, c, d, e, f are nonnegative real numbers such that a ≥ b ≥ 1 ≥ c ≥ d ≥ e ≥ f,

a + b + c + d + e + f = 6,

then f2−1 b2 − 1 c2 − 1 d2 − 1 e2 − 1 a2 − 1 + + + + + ≥ 0. (2a + 7)2 (2b + 7)2 (2c + 7)2 (2d + 7)2 (2e + 7)2 (2 f + 7)2 (Vasile Cirtoaje, 2009)

124

Vasile Cîrtoaje

Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) + f (e) + f ( f ) ≥ 6 f (s), where f (u) =

u2 − 1 , (2u + 7)2

For u ∈ [0, 1], we have f 00 (u) =

s=

a+b+c+d+e+ f = 1, 6

u ∈ [0, 6].

2(37 − 28u) ≥ 0, (2u + 7)4

hence f is convex on [0, 1]. Therefore, we may apply HCF-OV Theorem or LHCF-OV Corollary for m = 4. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + 4 y = 5. We have g(u) = h(x, y) =

f (u) − f (1) u+1 = , u−1 (2u + 7)2

g(x) − g( y) 21 − 4x − 4 y − 4x y (x − 4)2 = = ≥ 0. x−y (2x + 7)2 (2 y + 7)2 (2x + 7)2 (2 y + 7)2

From x + 4 y = 5 and h(x, y) = 0, we get x = 4 and y = 1/4. Therefore, in accordance with Remark 8, the equality holds only for a = b = c = d = e = f = 1, and also for a = 4, b = 1, c = d = e = f = 1/4.

P 2.13. If a, b, c, d, e, f are nonnegative real numbers such that a ≥ b ≥ c ≥ d ≥ 1 ≥ e ≥ f,

a + b + c + d + e + f = 6,

then f2−1 a2 − 1 b2 − 1 c2 − 1 d2 − 1 e2 − 1 + + + + + ≤ 0. (2a + 5)2 (2b + 5)2 (2c + 5)2 (2d + 5)2 (2e + 5)2 (2 f + 5)2 (Vasile Cirtoaje, 2009) Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) + f (e) + f ( f ) ≥ 6 f (s), where f (u) =

1 − u2 , (2u + 5)2

s=

u ∈ [0, 6].

a+b+c+d+e+ f = 1, 6

HCF Method for Ordered Variables

125

For u ≥ 1, we have f 00 (u) =

2(20u − 13) ≥ 0, (2u + 5)4

hence f is convex on [1, 6]. Therefore, we may apply HCF-OV Theorem or RHCF-OV Corollary for m = 4. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + 4 y = 5. We have g(u) = h(x, y) =

f (u) − f (1) −u − 1 = , u−1 (2u + 5)2

g(x) − g( y) 4x y + 4x + 4 y − 5 x(8 − x) = = ≥ 0. 2 2 x−y (2x + 5) (2 y + 5) (2x + 5)2 (2 y + 5)2

From x + 4 y = 5 and h(x, y) = 0, we get x = 0 and y = 5/4. Therefore, in accordance with Remark 7, the equality holds only for a = b = c = d = e = f = 1, and also for a = b = c = d = 5/4, e = 1, f = 0.

P 2.14. If a, b, c, d are nonnegative real numbers such that a ≥ b ≥ 1 ≥ c ≥ d, then

a + b + c + d = 4,

1 1 1 4 1 + 3 + 3 + 3 ≤ . + 5 2b + 5 2c + 5 2d + 5 7 (Vasile Cirtoaje, 2009)

2a3

Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) ≥ 4 f (s), where f (u) =

−1 , 2u3 + 5

From f 00 (u) =

s=

a+b+c+d = 1, 4

u ∈ [0, 4].

12u(5 − 4u3 ) , (2u3 + 5)3

it follows that f 00 (u) ≥ 0 for u ∈ [0, 1], hence f is convex on [0, 1]. Therefore, we may apply HCF-OV Theorem or LHCF-OV Corollary for n = 4 and m = 2. Thus, it suffices to show that f (x) + 2 f ( y) ≥ 3 f (1) for all x, y ≥ 0 such that x + 2 y = 3. Using Remark 1, we only need to show that h(x, y) ≥ 0. We have g(u) =

f (u) − f (1) 2(u2 + u + 1) = , u−1 7(2u3 + 5)

126

Vasile Cîrtoaje h(x, y) =

g(x) − g( y) 2E = , 3 x−y 7(2x + 5)(2 y 3 + 5)

where E = −2x 2 y 2 − 2x y(x + y) − 2(x 2 + x y + y 2 ) + 5(x + y) + 5. Since E = (1 − 2 y)2 (2 + 3 y − 2 y 2 ) = (1 − 2 y)2 (2 + x y) ≥ 0, the proof is completed. From x + 2 y = 3 and h(x, y) = 0, we get x = 2, y = 1/2. Therefore, in accordance with Remark 8, the equality holds for a = b = c = d = 1, and also for a = 2, b = 1, c = d = 1/2. Remark. Similarly, we can prove the following generalization: • If a1 , a2 , . . . , a2n are nonnegative real numbers such that a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,

a1 + a2 + · · · + a2n = 2n.

then 1 a13 + n +

1 n

+

1 a23 + n +

1 n

+ ··· +

1 3 a2n +n+

1 n

≥

2n2 , n2 + n + 1

with equality for a1 = a2 = · · · = a2n = 1, and also for a1 = n,

a2 = · · · = an = 1,

an+1 = · · · = a2n = 1/n.

P 2.15. If a1 , a2 , . . . , a8 are nonnegative real numbers such that a1 ≥ a2 ≥ a3 ≥ a4 ≥ 1 ≥ a5 ≥ a6 ≥ a7 ≥ a8 ,

a1 + a2 + · · · + a8 = 8,

then (a12 + 1)(a22 + 1) · · · (a82 + 1) ≥ (a1 + 1)(a2 + 1) · · · (a8 + 1). (Vasile Cirtoaje, 2008) Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (a8 ) ≥ 8 f (s),

s=

a1 + a2 + · · · + a8 = 1, 8

where f (u) = ln(u2 + 1) − ln(u + 1),

u ∈ [0, 8].

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127

For u ∈ [0, 1], we have f 00 (u) =

1 (u2 − u4 ) + 4u(1 − u2 ) + u2 + 3 2(1 − u2 ) + = > 0. (u2 + 1)2 (u + 1)2 (u2 + 1)2 (u + 1)2

Therefore, f is convex on [0, 1]. According to HCF-OV Theorem or LHCF-OV Corollary applied for n = 8 and m = 4, it suffices to show that f (x) + 4 f ( y) ≥ 5 f (1) for all x, y ≥ 0 such that x + 4 y = 5. Using Remark 2, we only need to show that H(x, y) ≥ 0 for x, y ≥ 0 such that x + 4 y = 5, where H(x, y) =

f 0 (x) − f 0 ( y) 2(1 − x y) 1 = 2 + . 2 x−y (x + 1)( y + 1) (x + 1)( y + 1)

Since 2(x 2 + 1) ≥ (x + 1)2 and 2( y 2 + 1) ≥ ( y + 1)2 , it suffices to prove that 8(1 − x y) + (x + 1)( y + 1) ≥ 0. Indeed, 8(1 − x y) + (x + 1)( y + 1) = 28x 2 − 38x + 14 = 28(x − 19/28)2 + 31/28 > 0. The proof is completed. The equality holds for a1 = a2 = · · · = a8 .

P 2.16. If a, b, c, d are positive real numbers such that a ≥ b ≥ 1 ≥ c ≥ d, then

a bcd = 1,

‹ 1 1 1 1 . a + b + c + d − 4 ≥ 18 a + b + c + d − − − − a b c d 2

2

2

2



(Vasile Cirtoaje, 2008) Solution. Using the substitution a = ex ,

b = ey,

c = ez ,

d = ew,

we need to show that f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s), where x ≥ y ≥ 0 ≥ z ≥ w,

s=

x + y +z+w = 0, 4

f (u) = e2u − 1 − 18(eu − e−u ), u ∈ R. For u ≤ 0, we have

f 00 (u) = 4e2u + 18(e−u − eu ) > 0,

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Vasile Cîrtoaje

hence f is convex on (−∞, 0]. By HCF-OV Theorem or LHCF-OV Corollary applied for n = 4 and m = 2, it suffices to show that f (x) + 2 f ( y) ≥ 3 f (0) for all real x, y such that x + 2 y = 0; that is, to show that  ‹ 1 2 2 2 a + 2b − 3 − 18 a + 2b − − ≥0 a b for all a, b > 0 such that a b2 = 1. This inequality is equivalent to (b2 − 1)2 (2b2 + 1) 18(b − 1)3 (b + 1) + ≥ 0, b4 b2 (b − 1)2 (2b − 1)2 (b + 1)(5b + 1) ≥ 0. b4 The proof is completed. The equality holds for a = b = c = d = 1, and also for a = 4, b = 1, c = d = 1/2.

P 2.17. If a, b, c, d are positive real numbers such that a ≥ b ≥ 1 ≥ c ≥ d,

a bcd = 1,

then p

a2 − a + 1 +

p

b2 − b + 1 +

p

c2 − c + 1 +

p

d 2 − d + 1 ≥ a + b + c + d. (Vasile Cirtoaje, 2008)

Solution. Using the substitution a = ex ,

b = ey,

c = ez ,

d = ew,

we need to show that f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s), where x ≥ y ≥ 0 ≥ z ≥ w, f (u) =

p

s=

x + y +z+w = 0, 4

e2u − eu + 1 − eu , u ∈ R.

We claim that f is convex for u ≥ 0. Since e−u f 00 (u) =

4e3u − 6e2u + 9eu − 2 − 1, 4(e2u − eu + 1)3/2

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129

we need to show that (4t 3 − 6t 2 + 9t − 2)2 ≥ 16(t 2 − t + 1)3 , where t = eu ≥ 1. Indeed, (4t 3 − 6t 2 + 9t − 2)2 − 16(t 2 − t + 1)3 = 12t 3 (t − 1) + 9t 2 + 12(t − 1) > 0. By HCF-OV Theorem or RHCF-OV Corollary applied for n = 4 and m = 2, it suffices to show that f (x) + 2 f ( y) ≥ 3 f (0) for all real x, y such that x + 2 y = 0; that is, to show that p p a2 − a + 1 + 2 b2 − b + 1 ≥ a + 2b for all a, b > 0 such that a b2 = 1. This inequality is equivalent to p p b4 − b2 + 1 1 + 2 b2 − b + 1 ≥ 2 + 2b, 2 b b p

b2 − 1 b4 − b2 + 1 + 1

Since p it suffices to show that

+p

b2 − 1 b4 − b2 + 1

2(1 − b) b2 − b + 1 + b ≥

≥ 0.

b2 − 1 , b2 + 1

b2 − 1 2(1 − b) +p ≥ 0, 2 b +1 b2 − b + 1 + b

which is equivalent to ˜ b+1 2 ≥ 0, − p b2 + 1 b2 − b + 1 + b ” — p (b − 1) (b + 1) b2 − b + 1 − b2 + b − 2 ≥ 0, (b − 1)

•

(b − 1)2 (3b2 − 2b + 3) ≥ 0. p (b + 1) b2 − b + 1 + b2 − b + 2 Clearly, the last inequality is true. The equality holds for a = b = c = d = 1.

P 2.18. If a, b, c, d are positive real numbers such that a ≥ 1 ≥ b ≥ c ≥ d, then a3

a bcd = 1,

1 1 1 1 2 + 3 + 3 + 3 ≥ . + 3a + 2 b + 3b + 2 c + 3c + 2 d + 3d + 2 3 (Vasile Cirtoaje, 2007)

130

Vasile Cîrtoaje

Solution. Using the substitution a = ex ,

b = ey,

c = ez ,

d = ew,

we need to show that f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s), where x ≥ y ≥ 0 ≥ z ≥ w, f (u) =

s=

x + y +z+w = 0, 4

1 , u ∈ R. e3u + 3eu + 2

We claim that f is convex for u ≥ 0. Indeed, denoting t = eu , t ≥ 1, we have 3t(3t 5 + 2t 3 − 6t 2 + 3t − 2) (t 3 + 3t + 2)3 3t(t − 1)(3t 4 + 3t 3 + 5t 2 − t + 2) = ≥ 0. (t 3 + 3t + 2)3

f 00 (u) =

By HCF-OV Theorem or RHCF-OV Corollary applied for n = 4 and m = 1, it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0; that is, to show that 1 1 1 + ≥ a3 + 3a + 2 b3 + 3b + 2 3 for all a, b > 0 such that a b = 1. This inequality is equivalent to (a − 1)4 (a2 + a + 1) ≥ 0, which is clearly true. The equality holds for a = b = c = d = 1.

P 2.19. Let a1 , a2 , . . . , an be positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , If k ≥ 1, then

a1 a2 · · · an = 1.

1 1 1 n + + ··· + ≥ . 1 + ka1 1 + ka2 1 + kan 1+k (Vasile Cirtoaje, 2007)

HCF Method for Ordered Variables

131

Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where x1 ≥ 0 ≥ x2 ≥ · · · ≥ x n,

s=

x1 + x2 + · · · + x n = 0, n

f (u) =

1 , u ∈ R. 1 + keu

f 00 (u) =

keu (keu − 1) ≥ 0, (1 + keu )3

For u ≥ 0, we have

therefore f (u) is convex for u ≥ 0. By HCF-OV Theorem or RHCF-OV Corollary applied for m = 1, it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0; that is, to show that 1 1 2 + ≥ 1 + ka 1 + k b 1+k for all a, b > 0 such that a b = 1. This is true since 1 1 2 k(k − 1)(a − 1)2 + − = ≥ 0. 1 + ka 1 + k b 1 + k (1 + ka)(a + k) The equality holds for a1 = a2 = · · · = an = 1.

P 2.20. If a1 , a2 , . . . , a9 are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ a9 , then

a1 a2 · · · a9 = 1,

1 1 1 + + ··· + ≥ 1. (a1 + 2)2 (a2 + 2)2 (a9 + 2)2 (Vasile Cirtoaje, 2007)

Solution. Using the substitution ai = e x i ,

i = 1, 2, . . . , 9,

we can write the inequality as f (x 1 ) + f (x 2 ) + · · · + f (x 9 ) ≥ 9 f (s),

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Vasile Cîrtoaje

where x1 ≥ 0 ≥ x2 ≥ · · · ≥ x9, f (u) =

(eu

For u ∈ [0, ∞), we have f 00 (u) =

s=

x1 + x2 + · · · + x9 9

1 , + 2)2

= 0,

u ∈ R.

4eu (eu − 1) ≥ 0, (eu + 2)4

hence f is convex on [0, ∞). According to HCF-OV Theorem or RHCF-OV Corollary (case m = 1), it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0; that is, to show that 1 1 2 + ≥ 2 2 (a + 2) (b + 2) 9 for all a, b > 0 such that a b = 1. Write this inequality as b2 1 2 + ≥ , 2 2 (2b + 1) (b + 2) 9 which is equivalent to the obvious inequality (b − 1)4 ≥ 0. The equality holds for a1 = a2 = · · · = a9 = 1. Remark. Similarly, we can prove the following generalization. • Let a1 , a2 , · · · , an be positive real numbers such that a1 a2 · · · an = 1.

a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , If p, q ≥ 0 such that p+q ≥1+ then

1 1+

pa1 + qa12

+

1 1+

pa2 + qa22

with equality for a1 = a2 = · · · = an = 1.

2pq , p + 4q

+ ··· +

1 n ≥ . 2 1 + pan + qan 1+p+q

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133

P 2.21. If a1 , a2 , . . . , an (n ≥ 4) are positive real numbers such that a1 a2 · · · an = 1,

a1 ≥ a2 ≥ a3 ≥ 1 ≥ a4 ≥ · · · ≥ an , then

1 1 1 n + + ··· + ≥ . 2 2 2 (a1 + 1) (a2 + 1) (an + 1) 4 (Vasile Cirtoaje, 2007)

Solution. Using the substitution ai = e x i ,

i = 1, 2, . . . , n,

we can write the inequality as f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where x1 ≥ x2 ≥ x3 ≥ 0 ≥ x4 ≥ · · · ≥ x n, f (u) =

(eu

For u ∈ [0, ∞), we have f 00 (u) =

1 , + 1)2

s=

x1 + x2 + · · · + x n = 0, n

u ∈ R.

2eu (2eu − 1) > 0, (eu + 1)4

hence f is convex on [0, ∞). According to HCF-OV Theorem or RHCF-OV Corollary (case m = 3), it suffices to show that f (x) + 3 f ( y) ≥ 4 f (0) for all real x, y such that x + 3 y = 0; that is, to show that 1 3 + ≥1 2 (a + 1) (b + 1)2 for all a, b > 0 such that a b3 = 1. The inequality is equivalent to 3 b6 + ≥ 1. 3 2 (b + 1) (b + 1)2 Using the Cauchy-Schwarz inequality, it suffices to show that (b3 + 3)2 ≥ 1, (b3 + 1)2 + 3(b + 1)2 which is equivalent to the obvious inequality (b − 1)2 (4b + 5) ≥ 0. The equality holds for a1 = a2 = · · · = an = 1.

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Vasile Cîrtoaje

P 2.22. If a1 , a2 , . . . , an are positive real numbers such that a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

a1 a2 · · · an = 1,

1 1 n 1 + + ··· + ≤ . 2 2 2 (a1 + 3) (a2 + 3) (an + 3) 16 (Vasile Cirtoaje, 2007)

Solution. Using the substitution ai = e x i ,

i = 1, 2, . . . , n,

we can write the inequality as f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where s=

x1 ≤ 0 ≤ x2 ≤ · · · ≤ x n, f (u) =

−1 , + 3)2

(eu

x1 + x2 + · · · + x n = 0, n u ∈ R.

For u ∈ (−∞, 0], we have f 00 (u) =

2eu (3 − 2eu ) > 0, (eu + 3)4

hence f is convex on (−∞, 0]. According to HCF-OV Theorem or LHCF-OV Corollary (case m = 1), it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0; that is, to show that 1 1 1 + ≤ 2 2 (a + 3) (b + 3) 8 for all a, b > 0 such that a b = 1. Write this inequality as b2 1 1 + ≤ , 2 2 (3b + 1) (b + 3) 8 which is equivalent to the obvious inequality (b2 − 1)2 + 12b(b − 1)2 ≥ 0. The equality holds for a1 = a2 = · · · = an = 1. Remark. Similarly, we can prove the following generalization:

HCF Method for Ordered Variables

135

• Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1.

a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , p If k ≥ 1 + 2, then

1 1 1 n + + ··· + ≤ , (a1 + k)2 (a2 + k)2 (an + k)2 (1 + k)2 with equality for a1 = a2 = · · · = an = 1.

P 2.23. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1.

a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , If p, q ≥ 0 such that p + q ≤ 1, then 1 1+

pa1 + qa12

+

1 1+

pa2 + qa22

+ ··· +

n 1 ≤ . 1 + pan + qan2 1+p+q (Vasile Cirtoaje, 2007)

Solution. Using the substitution ai = e x i ,

i = 1, 2, . . . , n,

we can write the inequality as f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where x1 ≤ 0 ≤ x2 ≤ · · · ≤ x n, f (u) =

s=

x1 + x2 + · · · + x n = 0, n

−1 , 1 + peu + qe2u

u ∈ R.

For u ≤ 0, we have eu [−4q2 e3u − 3pqe2u + (4q − p2 )eu + p] (1 + peu + qe2u )3 e2u [−4q2 − 3pq + (4q − p2 ) + p] ≥ (1 + peu + qe2u )3 e2u [(p + 4q)(1 − p − q) + 2pq] = ≥ 0, (1 + peu + qe2u )3

f 00 (u) =

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Vasile Cîrtoaje

therefore f (u) is convex for u ≤ 0. According to HCF-OV Theorem or LHCF-OV Corollary (case m = 1), it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0; that is, to show that 1 1 2 + ≤ 2 2 1 + pa + qa 1 + pb + qb 1+p+q for all a, b > 0 such that a b = 1. Write this inequality as (a − 1)2 [q(1 − p − q)a2 + (p + 2q − p2 − pq − 2q2 )a + q(1 − p − q)] ≥ 0, which is true because p + 2q − p2 − pq − 2q2 ≥ (p + 2q)(p + q) − p2 − pq − 2q2 = 2pq ≥ 0. The equality holds for a1 = a2 = · · · = an = 1.

P 2.24. If a1 , a2 , . . . , an are positive real numbers such that a1 ≤ a2 ≤ 1 ≤ a3 ≤ · · · ≤ an , then

a1 a2 · · · an = 1,

1 1 1 n + + ··· + ≤ . 2 2 2 (a1 + 5) (a2 + 5) (an + 5) 36 (Vasile Cirtoaje, 2007)

Solution. Using the substitution ai = e x i ,

i = 1, 2, . . . , n,

we can write the inequality as f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where x1 ≤ x2 ≤ 0 ≤ x3 ≤ · · · ≤ x n, f (u) =

s=

−1 , (eu + 5)2

x1 + x2 + · · · + x n = 0, n u ∈ R.

For u ∈ (−∞, 0], we have f 00 (u) =

2eu (5 − 2eu ) > 0, (eu + 5)4

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137

hence f is convex on (−∞, 0]. According to HCF-OV Theorem or LHCF-OV Corollary (case m = 2), it suffices to show that f (x) + 2 f ( y) ≥ 3 f (0) for all real x, y such that x + 2 y = 0; that is, to show that 1 2 1 + ≤ 2 2 (a + 5) (b + 5) 12 for all a, b > 0 such that a b2 = 1. Since 1 b4 b4 b2 = ≤ = , (a + 5)2 (5b2 + 1)2 (4b2 + 2b)2 4(2b + 1)2 it suffices to show that

2 1 b2 + ≤ , 4(2b + 1)2 (b + 5)2 12

which is equivalent to the obvious inequality (b − 1)2 (b2 + 16b + 1) ≥ 0. The equality holds for a1 = a2 = · · · = an = 1. Remark. Similarly, we can prove the following generalization: • Let a1 , a2 , . . . , an be positive real numbers such that a1 ≤ a2 ≤ 1 ≤ a3 ≤ · · · ≤ an , If k ≥ 2 +

a1 a2 · · · an = 1.

p 6, then 1 1 1 n + + ··· + ≤ , 2 2 2 (a1 + k) (a2 + k) (an + k) (1 + k)2

with equality for a1 = a2 = · · · = an = 1.

P 2.25. If a1 , a2 , . . . , an are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

1 p

1 + 3a1

+p

1 1 + 3a2

a1 a2 · · · an = 1,

+ ··· + p

1 1 + 3an

≥

n . 2 (Vasile Cirtoaje, 2007)

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Vasile Cîrtoaje

Solution. Using the substitution ai = e x i ,

i = 1, 2, . . . , n,

we can write the inequality as f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where x1 ≥ 0 ≥ x2 ≥ · · · ≥ x n,

s=

x1 + x2 + · · · + x n = 0, n

1 f (u) = p , 1 + 3eu

u ∈ R.

For u ≥ 0, we have

eu (3eu − 2) > 0, 2(1 + 3eu )5/2 hence f is convex on [0, ∞). According to HCF-OV Theorem or RHCF-OV Corollary (case m = 1), it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0; that is, to show that f 00 (u) =

1 1 +p ≥1 p 1 + 3a 1 + 3b for all a, b > 0 such that a b = 1. Write this inequality as s 1 a + ≥ 1. p a+3 1 + 3a 1 Substituting p = t, 0 < t < 1, the inequality becomes 1 + 3a v t 1 − t2 ≥ 1 − t. 8t 2 + 1 By squaring, we get t(1 − t)(2t − 1)2 ≥ 0, which is true. The equality holds for a1 = a2 = · · · = an = 1.

P 2.26. If a1 , a2 , . . . , an are positive real numbers such that a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

1 p

1 + 2a1

+p

1 1 + 2a2

a1 a2 · · · an = 1,

+ ··· + p

1

n ≤p . 3 1 + 2an (Vasile Cirtoaje, 2007)

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139

Solution. Using the substitution ai = e x i ,

i = 1, 2, . . . , n,

we can write the inequality as f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where x1 ≤ 0 ≤ x2 ≤ · · · ≤ x n,

s=

x1 + x2 + · · · + x n = 0, n

−1 f (u) = p , 1 + 2eu For u ≤ 0, we have f 00 (u) =

u ∈ R.

eu (1 − eu ) > 0, (1 + 2eu )5/2

hence f is convex on (−∞, 0]. According to HCF-OV Theorem or LHCF-OV Corollary (case m = 1), it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0; that is, to show that v v t 3 t 3 + ≤2 1 + 2a 1 + 2b for all a, b > 0 such that a b = 1. By the Cauchy-Schwarz inequality, we get v v v ‹ ‹ t 3 t 3 t 3 3 + ≤ +1 1+ = 2. 1 + 2a 1 + 2b 1 + 2a 1 + 2b The equality holds for a1 = a2 = · · · = an = 1.

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Vasile Cîrtoaje

Chapter 3

Partially Convex Function Method 3.1

Theoretical Basis

Right Partially Convex Function Theorem (Vasile Cirtoaje, 2012). Let f be a function defined on a real interval I and convex on [s, s0 ], where s, s0 ∈ I, s < s0 . In addition, f (u) is decreasing on Iu≤s0 and min f (u) = f (s0 ). u∈I

The inequality a + a + ··· + a  1 2 n f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f n holds for all a1 , a2 , · · · , an ∈ I satisfying a1 + a2 + · · · + an = ns if and only if f (x) + (n − 1) f ( y) ≥ n f (s) for all x, y ∈ I such that x ≤ s ≤ y and x + (n − 1) y = ns. Proof. The necessity is obvious. By Lemma below, to prove the sufficiency, it suffices to consider that a1 , a2 , . . . , an ∈ J, where J = Iu≤s0 . Because f (u) is convex on Ju≥s , the desired inequality follows from HCF Theorem applied to the interval J. Lemma. Let f be a function defined on a real interval I and convex on [s, s0 ], where s, s0 ∈ I, s < s0 . In addition, f (u) is decreasing on Iu≤s0 and min f (u) = f (s0 ). u∈I

If the inequality f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s) 141

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Vasile Cîrtoaje

holds for all a1 , a2 , · · · , an ∈ Iu≤s0 such that a1 + a2 + · · · + an = ns, then it holds for all a1 , a2 , . . . , an ∈ I such that a1 + a2 + · · · + an = ns. Proof. For i = 1, 2, . . . , n, define the numbers bi ∈ Iu≤s0 as follows ( bi =

a i , a i ≤ s0 s0 ,

a i > s0 .

We have bi ≤ ai and f (bi ) ≤ f (ai ) for i = 1, 2, · · · , n. Therefore, b1 + b2 + · · · + bn ≤ a1 + a2 + · · · + an = ns and f (b1 ) + f (b2 ) + · · · + f (bn ) ≤ f (a1 ) + f (a2 ) + · · · + f (an ). Thus, it suffices to show that f (b1 ) + f (b2 ) + · · · + f (bn ) ≥ n f (s) for all b1 , b2 , . . . , bn ∈ Iu≤s0 such that b1 + b2 +· · ·+ bn ≤ ns. By hypothesis, this inequality is true for b1 , b2 , . . . , bn ∈ Iu≤s0 and b1 + b2 + · · · + bn = ns. Since f (u) is decreasing on Iu≤s0 , the more we have f (b1 ) + f (b2 ) + · · · + f (bn ) ≥ n f (s) for b1 , b2 , . . . , bn ∈ Iu≤s0 and b1 + b2 + · · · + bn ≤ ns. Similarly, we can prove Left Partially Convex Function Theorem (Vasile Cirtoaje, 2012). Left Partially Convex Function Theorem. Let f be a function defined on a real interval I and convex on [s0 , s], where s0 , s ∈ I, s0 < s. In addition, f (u) is increasing on Iu≥s0 and satisfies min f (u) = f (s0 ). u∈I

The inequality f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f

a + a + ··· + a  1 2 n n

holds for all a1 , a2 , · · · , an ∈ I satisfying a1 + a2 + · · · + an = ns if and only if f (x) + (n − 1) f ( y) ≥ n f (s) for all x, y ∈ I such that x ≥ s ≥ y and x + (n − 1) y = ns. Remark 1. Let us denote g(u) =

f (u) − f (s) , u−s

h(x, y) =

g(x) − g( y) . x−y

Partially Convex Function Method

143

As shown in Remark 1 from 1.1, we may replace the hypothesis condition in Right Partially Convex Function Theorem (RPCF Theorem) and Left Partially Convex Function Theorem (LPCF Theorem), namely f (x) + (n − 1) f ( y) ≥ n f (s), by the condition h(x, y) ≥ 0 for all x, y ∈ I such that x + (n − 1) y = ns. Remark 2. Assume that f is differentiable on I, and let H(x, y) =

f 0 (x) − f 0 ( y) . x−y

As shown in Remark 2 from 1.1, the inequalities in RPCF Theorem and LPCF Theorem hold true by replacing the hypothesis f (x) + (n − 1) f ( y) ≥ n f (s) with the more restrictive condition H(x, y) ≥ 0 for all x, y ∈ I such that x + (n − 1) y = ns. Remark 3. The inequality in RPCF Theorem or LPCF Theorem becomes an equality for a1 = a2 = · · · = an = s and also for a1 = x,

a2 = · · · = an = y

(or any cyclic permutation), where x, y ∈ I satisfy the equations x + (n − 1) y = ns,

f (x) + (n − 1) f ( y) = n f (s).

For x 6= y, these equations are equivalent to x + (n − 1) y = ns, h(x, y) = 0. Remark 4. RPCF Theorem is also valid in the case when I = [a, b] \ {u0 } or I = (a, b) \ {u0 }, where a, b, u0 are real numbers such that a < s < s0 < u0 < b. Similarly, LPCF Theorem is also valid in the case when I = [a, b]\{u0 } or I = (a, b)\{u0 }, where a, b, u0 are real numbers such that a < u0 < s0 < s < b.

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Vasile Cîrtoaje

Remark 5. The desired inequality in RPCF Theorem holds true by replacing the condition f is decreasing on Iu≤s0 with the sufficient condition ns − (n − 1)s0 ≤ inf I. To prove this claim, we define the function ( f (u), u ≤ s0 , u ∈ I f0 (u) = f (s0 ), u ≥ s0 , u ∈ I, which is convex for u ∈ I, u ≥ s. Taking into account that f0 (s) = f (s) and f0 (u) ≤ f (u) for all u ∈ I, it suffices to prove that f0 (x 1 ) + f0 (x 2 ) + · · · + f0 (x n ) ≥ n f0 (s) for all x 1 , x 2 , · · · , x n ∈ I satisfying x 1 + x 2 + · · · + x n = ns. According to HCF Theorem and Remark 5 from Section 1, we only need to show that f0 (x) + (n − 1) f0 ( y) ≥ n f0 (s)

(*)

for all x, y ∈ I such that x ≤ s ≤ y and x +(n−1) y = ns. The case y > s0 is not possible because x = ns − (n − 1) y < ns − (n − 1)s0 ≤ inf I involves x 6∈ I. For the possible case y ≤ s0 , (*) becomes f (x) + (n − 1) f ( y) ≥ n f (s), which holds (by hypothesis) for all x, y ∈ I such that x + (n − 1) y = ns. Similarly, we can show that the desired inequality in LPCF Theorem holds true by replacing the condition f is increasing on Iu≥s0 with the sufficient condition ns − (n − 1)s0 ≥ sup I.

Partially Convex Function Method

3.2

145

Applications

3.1. If a, b, c are real numbers such that a + b + c = 3, then 16a − 5 16b − 5 16c − 5 + + ≤ 1. 32a2 + 1 32b2 + 1 32c 2 + 1

3.2. If a, b, c, d are real numbers such that a + b + c + d = 4, then 18b − 5 18c − 5 18d − 5 18a − 5 + + + ≤ 4. 12a2 + 1 12b2 + 1 12c 2 + 1 12d 2 + 1

3.3. If a, b, c, d, e, f are real numbers such that a + b + c + d + e + f = 6, then 5f − 1 5a − 1 5b − 1 5c − 1 5d − 1 5e − 1 + + + + + ≤ 4. 5a2 + 1 5b2 + 1 5c 2 + 1 5d 2 + 1 5e2 + 1 5 f 2 + 1

3.4. If a1 , a2 , · · · , an (n ≥ 3) are real numbers such that a1 + a2 + · · · + an = n, then n(n + 1) − 2a1 n2 + (n − 2)a12

+

n(n + 1) − 2a2 n2 + (n − 2)a22

+ ··· +

n(n + 1) − 2an ≤ n. n2 + (n − 2)an2

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

3.6. Let a1 , a2 , · · · , an 6= −k be real numbers such that a1 + a2 + · · · + an = n, where n k≥ p . 2 n−1 Then,

an (an − 1) a1 (a1 − 1) a2 (a2 − 1) + + ··· + ≥ 0. 2 2 (a1 + k) (a2 + k) (an + k)2

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Vasile Cîrtoaje

3.7. Let a1 , a2 , . . . , an 6= −k be real numbers such that a1 + a2 + · · · + an = n. If k ≥1+ p then

a12 − 1 (a1 + k)2

+

a22 − 1 (a2 + k)2

n n−1

+ ··· +

,

an2 − 1 (an + k)2

≥ 0.

3.8. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n. If k > 1, then a1 a1k

+ a2 + · · · + an

+

a2 a1 + a2k

+ · · · + an

+ ··· +

an ≤ 1. a1 + a2 + · · · + ank

3.9. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n. If k > 1, then 1 a1k

+ a2 + · · · + an

+

1 a1 + a2k

+ · · · + an

+ ··· +

1 ≤ 1. a1 + a2 + · · · + ank

3.10. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then an − 1 a1 − 1 a2 − 1 + + ··· + ≤ 0. 2 2 a1 + a2 + · · · + an2 a1 + a2 + · · · + an a1 + a2 + · · · + an 3.11. Let a1 , a2 , · · · , an be positive real numbers such that a1 + a2 + · · · + an = n. If n , then 0 0. (32u2 + 1)3 Therefore, f is convex on [1/2, 1], hence on [s0 , 1]. According to LPCF Theorem, we only need to show that f (x) + 2 f ( y) ≥ 3 f (1) for all real x, y such that x + 2 y = 3. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

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Vasile Cîrtoaje

Indeed, we have g(u) = h(x, y) =

32(2u − 1) , 3(32u2 + 1)

64(1 + 16x + 16 y − 32x y) 64(4x − 5)2 = ≥ 0. 3(32x 2 + 1)(32 y 2 + 1) 3(32x 2 + 1)(32 y 2 + 1)

Thus, the proof is completed. From x + 2 y = 3 and h(x, y) = 0, we get x = 5/4 and, y = 7/8. Therefore, in accordance with Remark 3, the equality holds for a = b = c = 1, and also for a = 5/4 and b = c = 7/8 (or any cyclic permutation).

P 3.2. If a, b, c, d are real numbers such that a + b + c + d = 4, then 18b − 5 18c − 5 18d − 5 18a − 5 + + + ≤ 4. 12a2 + 1 12b2 + 1 12c 2 + 1 12d 2 + 1 (Vasile Cîrtoaje, 2012) Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) ≥ 4 f (s), where

5 − 18u , 12u2 + 1

f (u) =

s=

a+b+c+d = 1, 4

u ∈ R.

From

6(36u2 − 20u − 3) , (12u2 + 1)2    p p  5 − 52 5 − 52 ∪[s0 , ∞) and decreasing on , s0 , it follows that f is increasing on −∞, 18 18 p 5 + 52 ≈ 0.678. Since where s0 = 18 f 0 (u) =

lim f (u) = 0

u→−∞

and f (s0 ) < f (5/18) = 0, it follows that min f (u) = f (s0 ). u∈R

For u ∈ [5/18, 1], we have 1 00 −216u3 + 180u2 + 54u − 5 f (u) = 24 (12u2 + 1)3 216u2 (1 − u) + 36u(1 − u) + (18u − 5) = > 0. (32u2 + 1)3

Partially Convex Function Method

153

Therefore, f is convex on [1/2, 1], hence on [s0 , 1]. According to LPCF Theorem and Remark 1, we only need to show that h(x, y) ≥ 0 for x, y ∈ R such that x + 3 y = 4. We have f (u) − f (1) 6(2u − 1) g(u) = = , u−1 12u2 + 1 h(x, y) =

g(x) − g( y) 12(1 + 6x + 6 y − 12x y) 12(2x − 3)2 = = ≥ 0. x−y (12x 2 + 1)(12 y 2 + 1) (12x 2 + 1)(12 y 2 + 1)

Thus, the proof is completed. From x + 3 y = 4 and h(x, y) = 0, we get x = 3/2 and, y = 5/6. Therefore, in accordance with Remark 3, the equality holds for a = b = c = d = 1, and also for a = 3/2 and b = c = d = 5/6 (or any cyclic permutation).

P 3.3. If a, b, c, d, e, f are real numbers such that a + b + c + d + e + f = 6, then 5f − 1 5b − 1 5c − 1 5d − 1 5e − 1 5a − 1 + 2 + 2 + 2 + 2 + ≤ 4. 2 5a + 1 5b + 1 5c + 1 5d + 1 5e + 1 5 f 2 + 1 (Vasile Cîrtoaje, 2012) Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) + f (e) + f ( f ) ≥ 4 f (s), where f (u) =

1 − 5u , 5u2 + 1

s=

a+b+c+d+e+ f = 1, 6

u ∈ R.

From

5(5u2 − 2u − 1) , (5u2 + 1)2    p  p 1− 6 1− 6 , s0 , it follows that f is increasing on −∞, ∪[s0 , ∞) and decreasing on 5 5 p 1+ 6 where s0 = ≈ 0.69. Since 5 f 0 (u) =

lim f (u) = 0

u→−∞

and f (s0 ) < f (1/5) = 0, it follows that min f (u) = f (s0 ). u∈R

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Vasile Cîrtoaje

For u ∈ [1/5, 1], we have −25u3 + 15u2 + 15u − 1 1 00 f (u) = 10 (5u2 + 1)3 2 25u (1 − u) + 10u(1 − u) + (5u − 1) = > 0. (3u2 + 4)3 Therefore, f is convex on [1/5, 1], hence on [s0 , 1]. According to LPCF Theorem and Remark 1, we only need to show that h(x, y) ≥ 0 for x, y ∈ R such that x + 5 y = 6. We have f (u) − f (1) 5(2u − 1) g(u) = = , u−1 3(5u2 + 1) h(x, y) =

g(x) − g( y) 5(2 + 5x + 5 y − 10x y) 10(x − 2)2 = = ≥ 0. x−y 3(5x 2 + 1)(5 y 2 + 1) 3(5x 2 + 1)(5 y 2 + 1)

Thus, the proof is completed. In accord with Remark 3, the equality holds for a = b = c = d = e = f = 1, and also for a = 2 and b = c = d = e = f = 4/5 (or any cyclic permutation).

P 3.4. If a1 , a2 , · · · , an (n ≥ 3) are real numbers such that a1 + a2 + · · · + an = n, then n(n + 1) − 2a1 n2

+ (n − 2)a12

+

n(n + 1) − 2a2 n2

+ (n − 2)a22

+ ··· +

n(n + 1) − 2an ≤ n. n2 + (n − 2)an2 (Vasile Cîrtoaje, 2008)

Solution. The desired inequality is true for a1 > n(n + 1) − 2a1 n2 + (n − 2)a12 and

n(n + 1) − 2ai n2

+ (n − 2)ai2


(n − 1)(ai + 1)2 ≥ 0.

Partially Convex Function Method

155

Consider further that a1 , a2 , · · · , an ≤

n(n + 1) and rewrite the desired inequality as 2

f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where

2u − n(n + 1) , f (u) = (n − 2)u2 + n2

s=

a1 + a2 + · · · + an = 1, n

 ˜ n(n + 1) u ∈ I = −∞, . 2

We have f 0 (u) n2 + n(n + 1)u − u2 = 2(n − 2) [(n − 2)u2 + n2 ]2 and

f1 (u) f 00 (u) = , 2(n − 2) [(n − 2)u2 + n2 ]3

where f1 (u) = 2(n − 2)u3 − 3n(n + 1)(n − 2)u2 − 2n2 (2n − 3)u + n3 (n + 1). From of f 0 , it follows that f is decreasing on (−∞, s0 ] and increasing • the expression ˜ n(n + 1) , where on s0 , 2 s0 =

p n (n + 1 − n2 + 2n + 5 ) ∈ (−1, 0); 2

therefore, min f (u) = f (s0 ). u∈I

On the other hand, for −1 ≤ u ≤ 1, we have f1 (u) > −2(n − 2) − 3n(n + 1)(n − 2) − 2n2 (2n − 3) + n3 (n + 1) = n2 (n − 3)2 + 4(n + 1) > 0, and hence f 00 (u) > 0. Since [s0 , 1] ⊂ [−1, 1], f is convex on [s0 , 1]. By LPCF Theorem and Remark 1, we only need to show that h(x, y) ≥ 0 for x, y ∈ R and x + (n − 1) y = n, where g(x) − g( y) f (u) − f (1) h(x, y) = , g(u) = . x−y u−1 Indeed, we have g(u) =

(n − 2)u + n (n − 2)u2 + n2

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Vasile Cîrtoaje

and n2 − n(x + y) − (n − 2)x y h(x, y) = n−2 [(n − 2)x 2 + n2 ][(n − 2) y 2 + n2 ] (n − 1)(n − 2) y 2 = ≥ 0. [(n − 2)x 2 + n2 ][(n − 2) y 2 + n2 ] The proof is completed. In accord with Remark 3, the equality holds for a1 = a2 = · · · = an = 1, and also for a1 = n and a2 = · · · = an = 0 (or any cyclic permutation).

P 3.5. If a, b, c, d are real numbers such that a + b + c + d = 4, then a(a − 1) b(b − 1) c(c − 1) d(d − 1) + + 2 + ≥ 0. 3a2 + 4 3b2 + 4 3c + 4 3d 2 + 4 (Vasile Cîrtoaje, 2012) Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) ≥ 4 f (s), where f (u) =

u2 − u , 3u2 + 4

s=

a+b+c+d = 1, 4

u ∈ R.

From

3u2 + 8u − 4 , (3u2 + 4)2    p p  −4 − 2 7 −4 − 2 7 it follows that f is increasing on −∞, ∪[s0 , ∞) and decreasing on , s0 , 3 3 p −4 + 2 7 ≈ 0.43. Since where s0 = 3 f 0 (u) =

lim f (u) = 1/3

u→−∞

and f (s0 ) < f (0) = 0, it follows that min f (u) = f (s0 ). u∈R

For u ∈ [0, 1], we have 1 00 −9u3 − 36u2 + 36u + 14 f (u) = 2 (3u2 + 4)3 9u2 (1 − u) + 45u(1 − u) + (16 − 9u) = > 0. (3u2 + 4)3

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157

Therefore, f is convex on [0, 1], hence on [s0 , 1]. According to LPCF Theorem and Remark 1, we only need to show that h(x, y) ≥ 0 for x, y ∈ R such that x + 3 y = 4. We have f (u) − f (1) u) g(u) = = 2 , u−1 3u + 4 h(x, y) =

g(x) − g( y) 4 − 3x y (x − 2)2 = = ≥ 0. x−y (3x 2 + 4)(3 y 2 + 4) (12x 2 + 1)(12 y 2 + 1)

The proof is completed. From x + 3 y = 4 and h(x, y) = 0, we get x = 2 and, y = 2/3. In accord with Remark 3, the equality holds for a = b = c = d = 1, and also for a = 2 and b = c = d = 2/3 (or any cyclic permutation). Remark. Similarly, we can prove the following generalization. • If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n, then a1 (a1 − 1) 4(n − 1)a12

+ n2

+

a2 (a2 − 1) 4(n − 1)a22

+ n2

+ ··· +

an (an − 1) ≥ 0, 4(n − 1)an2 + n2

n with equality for a1 = a2 = · · · = an = 1, and also for a1 = and a2 = a3 = · · · = an = 2 n (or any cyclic permutation). 2(n − 1)

P 3.6. Let a1 , a2 , · · · , an 6= −k be real numbers such that a1 + a2 + · · · + an = n, where n k≥ p . 2 n−1 Then,

an (an − 1) a1 (a1 − 1) a2 (a2 − 1) + + ··· + ≥ 0. 2 2 (a1 + k) (a2 + k) (an + k)2 (Vasile Cîrtoaje, 2008)

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

s=

a1 + a2 + · · · + an = 1, n

u(u − 1) , u ∈ I = R \ {−k}. (u + k)2

From f 0 (u) =

(2k + 1)u − k , (u + k)3

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Vasile Cîrtoaje

it follows that f is increasing on (−∞, −k) ∪ [s0 , ∞) and decreasing on (−k, s0 ], where s0 =

k < 1. 2k + 1

Since lim f (u) = 1

u→−∞

and f (s0 ) < f (1) = 0, we have min f (u) = f (s0 ). u∈I

From

1 00 k(k + 2) − (2k + 1)u f (u) = , 2 (u + k)4 ˜ • k(k + 2) , hence on [s0 , 1]. According to LPCF Theoit follows that f is convex on 0, 2k + 1 rem, Remark 4 and Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ∈ I which satisfy x + (n − 1) y = n, where h(x, y) =

g(x) − g( y) , x−y

Indeed, we have g(u) = and h(x, y) =

g(u) =

f (u) − f (1) . u−1

u (u + k)2

k2 − x y ≥ 0, (x + k)2 ( y + k)2

because

[2(n − 1) y − n]2 n2 −xy = ≥ 0. 4(n − 1) 4(n − 1) n The equality holds for a1 = a2 = · · · = an = 1. If k = p , then the equality holds 2 n−1 n n (or any cyclic permutation). also for a1 = and a2 = · · · = an = 2 2(n − 1) k2 − x y ≥

P 3.7. Let a1 , a2 , . . . , an 6= −k be real numbers such that a1 + a2 + · · · + an = n. If k ≥1+ p then

a12 − 1 (a1 + k)2

+

a22 − 1 (a2 + k)2

n n−1

+ ··· +

, an2 − 1 (an + k)2

≥ 0. (Vasile Cîrtoaje, 2008)

Partially Convex Function Method

159

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

s=

a1 + a2 + · · · + an = 1, n

u2 − 1 , u ∈ I = R \ {−k}. (u + k)2

From f 0 (u) =

2(ku + 1) , (u + k)3

it follows that f is increasing on (−∞, −k) ∪ [s0 , ∞) and decreasing on (−k, s0 ], where s0 =

−1 ∈ (−1, 0). k

Since lim f (u) = 1

u→−∞

and f (s0 ) < f (−1) = 0, we have min f (u) = f (s0 ). u∈I

From f 00 (u) =

2(k2 − 3 − 2ku) , (u + k)4

it follows that f is convex on [−1, 1], hence on [s0 , 1], since k2 − 3 − 2ku ≥ k2 − 3 − 2k = (k + 1)(k − 3) ≥ 0. According to LPCF Theorem, Remark 4 and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ∈ I which satisfy x + (n − 1) y = n. We have g(u) = h(x, y) =

f (u) − f (1) u+1 = , u−1 (u + k)2

g(x) − g( y) (k − 1)2 − 1 − x − y − x y = ≥ 0, x−y (x + k)2 ( y + k)2

since [(n − 1) y − 1]2 n2 −1− x − y − xy = ≥ 0. n−1 n−1 n The equality holds for a1 = a2 = · · · = an = 1. If k = 1 + p , then the equality n−1 1 holds also for a1 = n − 1 and a2 = · · · = an = (or any cyclic permutation). n−1 (k − 1)2 − 1 − x − y − x y ≥

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Vasile Cîrtoaje

P 3.8. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n. If k > 1, then a1 a1k

+ a2 + · · · + an

+

a2 a1 + a2k

+ · · · + an

+ ··· +

an ≤ 1. a1 + a2 + · · · + ank (Vasile Cîrtoaje, 2006)

Solution. According to the proof of P 2.106 from Volume 2, it suffices to consider the case where a1 + a2 + · · · + an = n and k > n + 1. Write the desired inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

uk

−u , −u+n

We have

s=

a1 + a2 + · · · + an = 1, n

u ∈ I = [0, n].

f 0 (u) =

(k − 1)uk − n (uk − u + n)2

f 00 (u) =

f1 (u) , (uk − u + n)3

and

where f1 (u) = k(k − 1)uk−1 (uk − u + n) − 2(kuk−1 − 1)[(k − 1)uk − n]. From the expression of f 0 , it follows that f is decreasing on [0, s0 ] and increasing on [s0 , n], where  n 1/k s0 = < 1. k−1 On the other hand, for u ∈ [s0 , 1], we have (k − 1)uk − n ≥ (k − 1)s0k − n = 0, and hence f1 (u) ≥ k(k − 1)uk−1 (uk − u + n) − 2kuk−1 [(k − 1)uk − n] = kuk−1 [−(k − 1)(uk + u) + n(k + 1)] ≥ kuk−1 [−2(k − 1) + 2(k + 1)] = 4kuk−1 > 0. Thus, f 00 (u) > 0, and hence f is convex on [s0 , 1]. By LPCF Theorem, we need to show that f (x) + (n − 1) f ( y) ≥ n f (1) for all positive x, y which satisfy x ≥ 1 ≥ y ≥ 0 and x + (n − 1) y = n. Consider the nontrivial case where x > 1 > y ≥ 0 and write the inequality f (x) + (n − 1) f ( y) ≥ n f (1) as follows: xk

(n − 1) y x + k ≤ 1, −x −n y − y+n

Partially Convex Function Method

161

xk − x + n ≥ xk − x ≥

x( y k − y + n) , yk − ny + n

(n − 1) y( y − y k ) . yk − ny + n

Since y < 1, it suffices to show that xk − x ≥

(n − 1)( y − y k ) , yk − y + 1

which is equivalent to h(x) ≥

y − yk , (1 − y)( y k − y + 1)

where

xk − x . x −1 By the weighted AM-GM inequality, we have h(x) =

h0 (x) =

(k − 1)x k + 1 − k x k−1 > 0, (x − 1)2

and hence h is strictly increasing. Since x − 1 = (n − 1)(1 − y) ≥ 1 − y, we get h(x) ≥ h(2 − y) =

(2 − y)k − 2 + y . 1− y

Thus, it suffices to show that (2 − y)k − 2 + y ≥

y − yk . yk − y + 1

Putting 1 − y = t, 0 < t ≤ 1, we write this inequality as (2 − y)k − 1 + y ≥ (1 + t)k − t ≥

1 , yk − y + 1

1 , (1 − t)k + t

(1 − t 2 )k + t(1 + t)k ≥ 1 + t 2 + t(1 − t)k . By Bernoulli’s inequality, (1 − t 2 )k + t(1 + t)k > 1 − kt 2 + t(1 + kt) = 1 + t. So, we only need to show that t(1 − t) ≥ t(1 − t)k , which is clearly true. The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

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P 3.9. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n. If k > 1, then 1 a1k

+ a2 + · · · + an

+

1 a1 + a2k

+ · · · + an

+ ··· +

1 ≤ 1. a1 + a2 + · · · + ank (Vasile Cîrtoaje, 2006)

Solution. Clearly, it suffices to consider the case a1 + a2 +· · ·+ an = n, when the desired inequality can be written as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) =

uk

−1 , −u+n

From f 0 (u) =

s=

a1 + a2 + · · · + an = 1, n

u ∈ I = [0, n].

kuk−1 − 1 , (uk − u + n)2

it follows that f is decreasing on [0, s0 ] and increasing on [s0 , n], where 1

s0 = k 1−k < 1. We will show that f is convex on [s0 , 1]. For u ∈ [s0 , 1], we have f 00 (u) =

g(u) −k(k + 1)u2k−2 + k(k + 3)uk−1 + nk(k − 1)uk−2 − 2 > k , k 3 (u − u + n) (u − u + n)3

where g(u) = −k(k + 1)u2k−2 + k(k + 3)uk−1 − 2. Denoting t = kuk−1 , we have 1 ≤ t ≤ k and k g(u) = −(k + 1)t 2 + k(k + 3)t − 2k = (k + 1)(t − 1)(k − t) + (k − 1)(t + k) > 0. By LPCF Theorem, it suffices to show that xk

1 n−1 + k ≤1 −x +n y − y+n

for x ≥ 1 ≥ y ≥ 0 and x + (n − 1) y = n. Since this inequality is trivial for x = y = 1, assume next that x > 1 > y ≥ 0, and write the desired inequality as follows: xk − x + n ≥

yk − y + n , yk − y + 1

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163

xk − x ≥

(n − 1)( y − y k ) , yk − y + 1

y − yk xk − x ≥ . x −1 (1 − y)( y k − y + 1) Let h(x) =

xk − x , x > 1. By the weighted AM-GM inequality, we have x −1 h0 (x) =

(k − 1)x k + 1 − k x k−1 > 0. (x − 1)2

Therefore, h is increasing. Since x − 1 = (n − 1)(1 − y) ≥ 1 − y, hence x ≥ 2 − y > 1, we get (2 − y)k + y − 2 . h(x) ≥ h(2 − y) = 1− y Thus, it suffices to show that (2 − y)k + y − 2 ≥

y − yk , yk − y + 1

which is equivalent to (2 − y)k + y − 1 ≥

yk

1 . − y +1

Substituting t = 1 − y, 0 < t ≤ 1, the inequality becomes (1 + t)k − t ≥

1 , (1 − t)k + t

(1 − t 2 )k + t(1 + t)k ≥ 1 + t 2 + t(1 − t)k . By Bernoulli’s inequality, (1 − t 2 )k + t(1 + t)k ≥ 1 − kt 2 + t(1 + kt) = 1 + t. Thus, it suffices to show that t(1 − t) ≥ t(1 − t)k , which is clearly true. The proof is completed. The equality holds for a1 = a2 = · · · = an = 1. Remark. Using this result, we can get the following statement. • Let x 1 , x 2 , . . . , x n be nonnegative real numbers such that x 1 + x 2 + · · · + x n ≥ n. If k > 1, then x 1k − x 1 x 1k + x 2 + · · · + x n

+

x 2k − x 2 x 1 + x 2k + · · · + x n

+ ··· +

x nk − x n x 1 + x 2 + · · · + x nk

≥ 0.

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This inequality is equivalent to 1 x 1k

+ x2 + · · · + x n

+

1 x1 +

x 2k

+ · · · + xn

Using the substitutions s=

+ ··· +

1 n ≤ . k x1 + x2 + · · · + x n x1 + x2 + · · · + x n

x1 + x2 + · · · + x n , n

s ≥ 1,

and

xi , i = 1, 2, . . . , n, s which yields a1 + a2 + · · · + an = n, the desired inequality becomes ai =

1

X s k−1 a1k

+ a2 + · · · + a n

≤ 1.

Since s k−1 ≥ 1, it suffices to show that 1

X a1k

+ a2 + · · · + an

≤ 1,

which is just the inequality from P 3.9. Since x 1 x 2 · · · x n ≥ 1 involves x 1 + x 2 + · · · + x n ≥ n, the inequality is also true under the more restrictive condition x 1 x 2 · · · x n ≥ 1. For n = 3 and k = 5/2, we get the inequality from IMO-2005: • If x, y, z are nonnegative real numbers such that x yz ≥ 1, then y5 − y2 x5 − x2 z5 − z2 + + ≥ 0. x 5 + y 2 + z2 x 2 + y 5 + z2 x 2 + y 2 + z5

P 3.10. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then an − 1 a1 − 1 a2 − 1 + + ··· + ≤ 0. 2 2 a1 + a2 + · · · + an2 a1 + a2 + · · · + an a1 + a2 + · · · + an (Vasile Cîrtoaje, 2006) Solution. For n = 2, the inequality is equivalent to (a1 − a2 )2 ≥ 0. Consider further that n ≥ 3 and write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

Partially Convex Function Method where f (u) =

u2

165

1−u , −u+n

u ∈ I = [0, n].

From f 0 (u) =

u2 − 2u − n + 1 , (u2 − u + n)2

it follows that f is decreasing on [0, s0 ] and increasing on [s0 , n], where s0 = 1 +

p

n, s0 ∈ (1, n).

We will show that f is convex on [1, s0 ]. For u ∈ [1, s0 ], we have f 00 (u) =

(u2

2g(u) , − u + n)3

g(u) = −u3 + 3u2 + 3(n − 1)u − 2n + 1.

Since g 0 (u) = −3(u2 − 2u − n + 1) ≥ 0, g is increasing on [1, s0 ], hence g(u) ≥ g(1) = n > 0 and f 00 (u) > 0. By RPCF Theorem, it suffices to show that (n − 1)(1 − y) 1− x + ≥0 2 x −x +n y2 − y + n for 0 ≤ x ≤ 1 ≤ y and x + (n − 1) y = n. Since (n − 1)(1 − y) = x − 1, we have  ‹ (n − 1)(1 − y) 1− x 1 1 + = (x − 1) − 2 + x2 − x + n y2 − y + n x − x + n y2 − y + n (x − 1)(x − y)(x + y − 1) = (x 2 − x + n)( y 2 − y + n) n(x − 1)2 (x + y − 1) = ≥ 0. (n − 1)(x 2 − x + n)( y 2 − y + n) The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.

P 3.11. Let a1 , a2 , · · · , an be positive real numbers such that a1 + a2 + · · · + an = n. If n 0 0. Thus, it suffices to prove the desired inequality for k=

n , n−1

1 < k ≤ 2.

Rewrite the desired inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

where f (u) = −uk/u , We have

u ∈ I = (0, n). k

f 0 (u) = ku u −2 (ln u − 1), k

f 00 (u) = ku u −4 [u + (1 − ln u)(2u − k + k ln u)]. For n = 2, when k = 2 and I = (0, 2), f is convex for u ∈ I, u ≥ 1, because 1 − ln u > 0,

2u − k + k ln u = 2u − 2 + 2 ln u ≥ 2u − 2 ≥ 0.

Therefore, we may apply HCF Theorem. Consider now that n ≥ 3. From the expression of f 0 , it follows that f is decreasing on (0, s0 ] and increasing on [s0 , n), where s0 = e. In addition, we claim that f is convex on [1, s0 ]. Indeed, since 1 − ln u ≥ 0,

2u − k + k ln u ≥ 2 − k > 0,

we have f 00 > 0 for u ∈ [1, s0 ]. Therefore, by HCF Theorem (for n = 2) and RPCF Theorem (for n ≥ 3), we only need to show that x k/x + (n − 1) y k/ y ≤ n for 0 < x ≤ 1 ≤ y and x + (n − 1) y = n. We have k/x ≥ k > 1. Also, from k n n = > = 1, y (n − 1) y x + (n − 1) y we get 0
0. u4

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Vasile Cîrtoaje

According to LPCF Theorem and Remark 4, we only need to show that that f (x) + 4 f ( y) ≥ 5 f (1) for all nonzero real x, y such that x + 4 y = 5. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where h(x, y) =

g(x) − g( y) , x−y

From g(u) = 5 h(x, y) =



g(u) =

f (u) − f (1) . u−1

‹ 9 5 − 2 , u u

5(5x + 5 y − 9x y) ≥0 x2 y2

and 5x + 5 y − 9x y = (6 y − 5)2 ≥ 0, it follows that h(x, y) ≥ 0. The proof is completed. In accordance with Remark 3, the equality holds for a = b = c = d = e = 1, and also for a = 5/3 and b = c = d = e = 5/6 (or any cyclic permutation). Remark. Similarly, we can prove the following generalization. • Let a1 , a2 , . . . , an be nonzero real numbers such that a1 + a2 + · · · + an = n. If n , then k= p n+ n−1 ‹  ‹  ‹  k 2 k 2 k 2 + 1− + ··· + 1 − ≥ n(1 − k)2 , 1− a1 a2 an with equality for a1 = a2 = · · · = an = 1, and also for a1 = · · · = an =

n (or any cyclic permutation). p n−1+ n−1

n and a2 = a3 = p 1+ n−1

P 3.13. If a, b, c are nonnegative real numbers such that a + b + c = 3, then (1 − a + a4 )(1 − b + b4 )(1 − c + c 4 ) ≥ 1. Solution. Write the inequality as f (a) + f (b) + f (c) ≥ 3 f (s),

s=

a+b+c = 1, 3

where f (u) = ln(1 − u + u4 ),

u ∈ [0, 3].

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169

From f 0 (u) =

4u3 − 1 , 1 − u + u4

p 3 it follows that f is decreasing on [0, s0 1] and increasing on [s0 , 4], where s0 = 1/ 4. Also, f is convex for u ∈ [s0 , 1] because f 00 (u) =

−4u6 − 4u3 + 12u2 − 1 −4u2 − 4u2 + 12u2 − 1 4u2 − 1 ≥ = > 0. (1 − u + u4 )2 (1 − u + u4 )2 (1 − u + u4 )2

According to LPCF Theorem, we only need to show that f (x) + 2 f ( y) ≥ 3 f (1) for all x, y ≥ 0 such that x + 2 y = 3. Using Remark 2, it suffices to prove that H(x, y) ≥ 0, where f 0 (x) − f 0 ( y) . H(x, y) = x−y We have (x + y)(x − y)2 − 1 + 4(x 2 + y 2 + x y) − 2x y(x + y) − 4x 3 y 3 (1 − x + x 4 )(1 − y + y 4 ) −1 + 4(x 2 + y 2 + x y) − 2x y(x + y) − 4x 3 y 3 ≥ (1 − x + x 4 )(1 − y + y 4 ) −1 + 4(x + y)2 − 2x y(x + y) − 4x y − 4x 3 y 3 = . (1 − x + x 4 )(1 − y + y 4 )

H(x, y) =

Thus, we only need to show that −1 + 4(x + y)2 − 2x y(x + y) − 4x y − 4x 3 y 3 ≥ 0. p From 3 = x + 2 y ≥ 2 2x y and (1 − x)(1 − y) ≤ 0, we get xy ≤

9 , 8

x + y ≥ 1 + x y.

Therefore, − 1 + 4(x + y)2 − 2x y(x + y) − 4x y − 4x 3 y 3 ≥ ≥ − 1 + 2(x + y)[2(x + y) − x y] − 4x y − 4x 3 y 3 ≥ − 1 + 2(1 + x y)[2(1 + x y) − x y] − 4x y − 4x 3 y 3 =3 + 2x y + 2x 2 y 2 − 4x 3 y 3 ≥ 3 + 2x y + 2x 2 y 2 − 5x 2 y 2 =3 + 2x y − 3x 2 y 2 ≥ 3 + 2x y − 4x y = 3 − 2x y > 0. The proof is completed. In accordance with Remark 3, the equality holds for a = b = c = 1.

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P 3.14. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then (1 − a + a3 )(1 − b + b3 )(1 − c + c 3 )(1 − d + d 3 ) ≥ 1. (Vasile Cîrtoaje, 2012) Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),

s=

a+b+c+d = 1, 4

where f (u) = ln(1 − u + u3 ), From f 0 (u) =

u ∈ [0, 4].

3u2 − 1 , 1 − u + u3

p it follows that f is decreasing on [0, s0 ] and increasing on [s0 , 4], where s0 = 1/ 34. Also, f is convex for u ∈ [s0 , 1] because f 00 (u) =

−3u4 + 6u − 1 −3u + 6u − 1 3u − 1 ≥ = > 0. 3 2 3 2 (1 − u + u ) (1 − u + u ) (1 − u + u3 )2

According to LPCF Theorem, we only need to show that f (x) + 3 f ( y) ≥ 4 f (1) for all x, y ≥ 0 such that x + 3 y = 4. Using Remark 2, it suffices to prove that H(x, y) ≥ 0, where f 0 (x) − f 0 ( y) . H(x, y) = x−y We have H(x, y) =

(x − y)2 + 3(x + y) − 1 − 3x 2 y 2 3(x + y) − 1 − 3x 2 y 2 ≥ . (1 − x + x 3 )(1 − y + y 3 ) (1 − x + x 3 )(1 − y + y 3 )

Thus, we only need to show that 3(x + y) − 1 − 3x 2 y 2 ≥ 0. From 4 = x + 3 y ≥ 2

p

3x y and (1 − x)(1 − y) ≤ 0, we get xy ≤

4 , 3

x + y ≥ 1 + x y.

Therefore, 3(x + y) − 1 − 3x 2 y 2 ≥ 3(1 + x y) − 1 − 3x 2 y 2 ≥ 3(1 + x y) − 1 − 4x y = 2 − x y > 0. The proof is completed. The equality holds for a = b = c = d = 1.

Partially Convex Function Method

171

P 3.15. If a1 , a2 , . . . , a10 are real numbers such that a1 + a2 + · · · + a10 = 10, then 2 (1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a10 + a10 ) ≥ 1.

(Vasile Cîrtoaje, 2006) Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (a10 ) ≥ 10 f (s),

s=

a1 + a2 + · · · + a10 =1 10

where f (u) = ln(1 − u + u2 ), From f 0 (u) =

u ∈ R.

2u − 1 , 1 − u + u2

it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = 1/2. Also, from 1 + 2u(1 − u) f 00 (u) = , (1 − u + u2 )2 it follows that f 00 (u) > 0 for u ∈ [s0 , 1], hence f is convex on [s0 , 1]. According to LPCF Theorem, we only need to show that f (x) + 9 f ( y) ≥ 10 f (1) for all real x, y such that x + 9 y = 10. Using Remark 2, it suffices to prove that H(x, y) ≥ 0, where H(x, y) = Since H(x, y) =

f 0 (x) − f 0 ( y) . x−y

1 + x + y − 2x y (1 − x + x 2 )(1 − y + y 2 )

and 1 + x + y − 2x y = 18 y 2 − 8 y + 1 = 2 y 2 + (4 y − 1)2 > 0, the conclusion follows. The equality holds for a1 = a2 = · · · = a10 = 1. Remark. By replacing a1 , a2 , . . . , a10 respectively with 1 − a1 , 1 − a2 , . . . , 1 − a10 , we get the following statement. • If a1 , a2 , . . . , a10 are real numbers such that a1 + a2 + · · · + a10 = 0, then 2 (1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a10 + a10 ) ≥ 1,

with equality for a1 = a2 = · · · = an = 0.

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Vasile Cîrtoaje

P 3.16. If a, b, c, d, e are nonzero real numbers such that a + b + c + d + e = 5, then ‹  ‹  1 1 1 1 1 1 1 1 1 1 + + + + . 5 2 + 2 + 2 + 2 + 2 + 45 ≥ 14 a b c d e a b c d e (Vasile Cîrtoaje, 2013) Solution. Write the desired inequality as f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s), s = where f (u) =

a+b+c+d+e = 1, 5

5 14 − + 9, u ∈ I = R \ {0}. 2 u u

From

2(7u − 5) , u3 it follows that f is increasing on (−∞, 0) ∪ [s0 , ∞) and decreasing on (0, s0 ], where f 0 (u) =

s0 =

5 . 7

Since lim f (u) = 9

u→−∞

and f (s0 ) < f (1) = 0, we have min f (u) = f (s0 ). u∈I

From

2(15 − 14u) , u4 it follows that f is convex on [s0 , 1]. By LPCF Theorem, Remark 4 and Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ∈ I which satisfy x + 4 y = 5, where f 00 (u) =

h(x, y) =

g(x) − g( y) , x−y

Indeed, we have g(u) = h(x, y) =

g(u) =

f (u) − f (1) . u−1

9 5 − 2, u u

(6 y − 5)2 5x + 5 y − 9x y = ≥ 0. x2 y2 x2 y2

In accordance with Remark 3, the equality holds for a = b = c = d = e = 1, and also 5 5 for a = and b = c = d = e = (or any cyclic permutation). 3 6

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173

P 3.17. If a, b, c are positive real numbers such that a bc = 1, then 7 − 6a 7 − 6b 7 − 6c + + ≥ 1. 2 + a2 2 + b2 2 + c 2 (Vasile Cîrtoaje, 2008) Solution. Using the substitution a = e x , b = e y , c = ez , we need to show that f (x) + f ( y) + f (z) ≥ 3 f (s), where f (u) = From f 0 (u) =

s=

x + y +z = 0, 3

7 − 6eu , u ∈ R. 2 + e2u

2(3eu + 2)(eu − 3) , (2 + e2u )2

it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = ln 3. Also, we have 2t · h(t) f 00 (u) = , t = eu , (2 + t 2 )3 where h(t) = −3t 4 + 14t 3 + 36t 2 − 28t − 12. We will show that h(t) > 0 for t ∈ [1, 3], hence f is convex on [0, s0 ]. We have h(t) = 3(t 2 − 1)(9 − t 2 ) + 14t 3 + 6t 2 − 28t + 15 = 3(t 2 − 1)(9 − t 2 ) + 14t 2 (t − 1) + 14(t − 1)2 + 6t 2 + 1 > 0. By RPCF Theorem, we only need to prove that f (x) + 2 f ( y) ≥ 3 f (0) for all real x, y such that x + 2 y = 0. That is, to show that the original inequality holds for b = c := t and a = 1/t 2 , where t > 0. Write this inequality as t 2 (7t 2 − 6) 2(7 − 6t) + ≥ 1, 2t 4 + 1 2 + t2 (t − 1)2 (t − 2)2 (5t 2 + 6t + 3) ≥ 0. Since the last inequality is true, the proof is completed. In accordance with Remark 3, the equality holds for a = b = c = 1, and also for a = 1/4 and b = c = 2 (or any cyclic permutation).

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Vasile Cîrtoaje

P 3.18. If a, b, c are positive real numbers such that a bc = 1, then 1 1 1 1 + + ≤ . a + 5bc b + 5ca c + 5a b 2 (Vasile Cîrtoaje, 2008) Solution. Write the inequality as a2

a b c 1 + 2 + 2 ≤ . +5 b +5 c +5 2

Using the substitution a = e x , b = e y , c = ez , we need to show that f (x) + g( y) + g(z) ≥ 3 f (s), where f (u) =

s=

x + y +z = 0, 3

−eu , u ∈ R. e2u + 5

From f 0 (u) =

eu (e2u − 5) , (e2u + 5)2

it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = Also, from f 00 (u) =

eu (−e4u + 30e2u − 25) , (e2u + 5)3

it follows that f is convex on [0, s0 ], because 0 ≤ u ≤ s0 involves 1 ≤ eu ≤

1 ln 5. 2

p 5, hence

−e4u + 30e2u − 25 ≥ e2u (5 − e2u ) + 25(e2u − 1) > 0. By RPCF Theorem, we only need to prove the original inequality for b = c := t and a = 1/t 2 , where t > 0. Write this inequality as t2 2t 1 + ≤ , 5t 4 + 1 t 2 + 5 2 (t − 1)2 (5t 4 − 10t 3 − 2t 2 + 6t + 5) ≥ 0, (t − 1)2 [5(t − 1)4 + 2t(5t 2 − 16t + 13)] ≥ 0. Since the last inequality is true, the proof is completed. The equality holds for a = b = c = 1.

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175

P 3.19. If a, b, c are positive real numbers such that a bc = 1, then 1 1 3 1 + + ≤ . 4 − 3a + 4a2 4 − 3b + 4b2 4 − 3c + 4c 2 5 (Vasile Cirtoaje, 2008) Solution. Let a = ex ,

b = ey,

c = ez .

We need to show that f (x) + f ( y) + f (z) ≥ 3 f (s), where f (u) =

s=

x + y +z = 0, 3

−1 , u ∈ R. 4 − 3eu + 4e2u

From f 0 (u) =

eu (8eu − 3) , (4 − 3eu + 4e2u )2

it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = ln

3 < 0. 8

We claim that f is convex on [s0 , 0]. Since f 00 (u) =

eu (−64e3u + 36e2u + 55eu − 12) , (4 − 3eu + 4e2u )3

we need to show that −64x 3 + 36x 2 + 55x − 12 ≥ 0, where x = eu ∈ [3/8, 1]. Indeed, −64x 3 + 36x 2 + 55x − 12 > −72x 3 + 36x 2 + 48x − 12 = 12(1 − x)(6x 2 + 3x − 1) ≥ 0. By LPCF Theorem, we only need to prove the original inequality for b = c := t and a = 1/t 2 , where t > 0. Write this inequality as follows: t4 2 3 + ≤ , 4 2 2 4t − 3t + 4 4 − 3t + 4t 5 28t 6 − 21t 5 − 48t 4 + 27t 3 + 42t 2 − 36t + 8 ≥ 0, (t − 1)2 (28t 4 + 35t 3 − 6t 2 − 20t + 8) ≥ 0. It suffices to show that 7(4t 4 + 5t 3 − t 2 − 3t + 1) ≥ 0.

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Vasile Cîrtoaje

Indeed, 4t 4 + 5t 3 − t 2 − 3t + 1 = t 2 (2t − 1)2 + 9t 3 − 2t 2 − 3t + 1 and 9t 3 − 2t 2 − 3t + 1 = t(3t − 1)2 + (2t − 1)2 > 0. The equality holds for a = b = c = 1. Remark. Since 1 1 1 ≥ = , 4 − 3a + 4a2 4 − 3a + 4a2 + (1 − a)2 5(1 − a + a2 ) we get the following known inequality 1 1 1 + + ≤ 3. 2 2 1−a+a 1− b+ b 1 − c + c2

P 3.20. If a, b, c are positive real numbers such that a bc = 1, then 1 1 1 3 + + ≤ . (3a + 1)(3a2 − 5a + 3) (3b + 1)(3b2 − 5b + 3) (3c + 1)(3c 2 − 5c + 3) 4 Solution. Let a = ex ,

b = ey,

c = ez .

We need to show that f (x) + f ( y) + f (z) ≥ 3 f (s), where f (u) =

(3eu

From f 0 (u) =

s=

x + y +z = 0, 3

−1 , u ∈ R. + 1)(3e2u − 5eu + 3)

(3eu − 2)(9eu − 2) , (3eu + 1)2 (3e2u − 5eu + 3)2

it follows that f is increasing on (−∞, s1 ] ∪ [s0 , ∞) and decreasing on [s1 , s0 ], where s1 = ln 2 − ln 9, s0 = ln 2 − ln 3,

s1 < s0 < 0.

Since lim f (u) = f (s0 ) = −1/3,

u→−∞

Partially Convex Function Method

177

we get min f (u) = f (s0 ). u∈R

We claim that f is convex on [s0 , 0]. We have f 00 (u) =

(3t

t · h(t) , − 5t + 3)3

+ 1)3 (3t 2

where t = eu , h(t) = −729t 5 + 1188t 4 − 648t 3 + 387t 2 − 160t + 12. Since the polynomial h(t) has the real roots t 1 ≈ 0.0933, t 2 ≈ 0.5072, t 3 ≈ 1.11008, it follows that h(t) > 0 for t ∈ [2/3, 1], hence f is convex for u ∈ [s0 , 0]. By LPCF Theorem, we only need to prove the original inequality for b = c := t and a = 1/t 2 , where t ∈ (0, 1]. Write this inequality as follows: 2 3 t6 + ≤ . 2 4 2 2 (t + 3)(3t − 5t + 3) (3t + 1)(3t − 5t + 3) 4 Since t 2 + 3 ≥ 2(t + 1) and 3t 4 − 5t 2 + 3 ≥ t(3t 2 − 5t + 3), it suffices to prove that t5 2 3 + ≤ . 2(t + 1)(3t 2 − 5t + 3) (3t + 1)(3t 2 − 5t + 3) 4 This is equivalent to the obvious inequality (1 − t)2 (1 + 15t + 5t 2 − 14t 3 − 6t 4 ) ≥ 0. The equality holds for a = b = c = 1.

P 3.21. Let a1 , a2 , · · · , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If p, q ≥ 0 such that p + 4q ≥ n − 1, then 1 − a1 1+

pa1 + qa12

+

1 − a2 1+

pa2 + qa22

+ ··· +

1 − an ≥ 0. 1 + pan + qan2 (Vasile Cîrtoaje, 2008)

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Vasile Cîrtoaje

Solution. For q = 0, we get a known inequality (see Remark 2 from the proof of P 1.51). Consider further that q > 0. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where

s=

x1 + x2 + · · · + x n = 0, n

f (u) =

1 − eu , u ∈ R. 1 + peu + qe2u

f 0 (t) =

eu (qe2u − 2qeu − p − 1) , (1 + peu + qe2u )2

From

it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where v t p+1 . s0 = ln r0 > 0, r0 = 1 + 1 + q Also, we have f 00 (u) =

t · h(t) , (1 + pt + qt 2 )3

where h(t) = −q2 t 4 + q(p + 4q)t 3 + 3q(p + 2)t 2 + (p − 4q + p2 )t − p − 1,

t = eu .

We will show that h(t) ≥ 0 for t ∈ [1, r0 ], hence f is convex on [0, s0 ]. We have h0 (t) = −4q2 t 3 + 3q(p + 4q)t 2 + 6q(p + 2)t + p − 4q + p2 , h00 (t) = 6q[−2qt 2 + (p + 4q)t + p + 2]. Since h00 (t) = 6q[2(−qt 2 + 2qt + p + 1) + p(t − 1)] ≥ 12q(−qt 2 + 2qt + p + 1) ≥ 0, h0 (t) is increasing, h0 (t) ≥ h0 (1) = p2 + 9pq + 8q2 + p + 8q > 0, h is increasing, hence h(t) ≥ h(1) = p2 + 4pq + 3q2 + 2q − 1 = (p + 2q)2 − (q − 1)2 = (p + q + 1)(p + 3q − 1). Therefore, f is convex on [0, s0 ] if p + 3q − 1 ≥ 0. Indeed, p + 3q − 1 ≥ p + 3q −

p + 2q p + 4q = > 0. n−1 2

Partially Convex Function Method

179

By RPCF Theorem, we only need to prove the original inequality for a2 = · · · = an := t and a1 = 1/t n−1 , where t ≥ 1. Write this inequality as (n − 1)(1 − t) t n−1 (t n−1 − 1) + ≥ 0, t 2n−2 + pt n−1 + q 1 + pt + qt 2 or pA + qB ≥ C, where A = t n−1 (t n − nt + n − 1), B = t 2n − t n+1 − (n − 1)(t − 1), C = t n−1 [(n − 1)t n − nt n−1 + 1]. Since p+4q ≥ n−1 and C ≥ 0 (by the AM-GM inequality applied to n positive numbers), it suffices to show that (p + 4q)C pA + qB ≥ , n−1 which is equivalent to p[(n − 1)A − C] + q[(n − 1)B − 4C] ≥ 0. Clearly, this is true if (n − 1)A − C ≥ 0 and (n − 1)B − 4C ≥ 0 for t ≥ 1. By the AM-GM inequality, we have (n − 1)A − C = nt n−1 [t n−1 − (n − 1)t + n − 2] ≥ 0. For n = 3 and n = 4.... Consider further that n ≥ 5. Since t − 1 ≤ t n−1 (t − 1), we have B ≥ t 2n − t n+1 − (n − 1)t n−1 (t − 1) = t n−1 [t n+1 − t 2 − (n − 1)t + n − 1]   t2 + 1 n−1 n+1 2 ≥t t − t − (n − 1) +n−1 2 1 = t n−1 [2t n+1 − (n + 1)t 2 + n − 1] 2 hence 2(n − 1)B − 8C ≥ (n − 1)t n−1 [2t n+1 − (n + 1)t 2 + n − 1] − 8C = (n − 1)t n−1 h(t),

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Vasile Cîrtoaje

where h(t) = 2t n+1 − 8t n +

8nt n−1 − 8 − (n + 1)t 2 + n − 1. n−1

We have h0 (t) = 2th1 (t),

h1 (t) = (n + 1)t n−1 − 4nt n−2 + 4nt n−3 − n − 1,

h01 (t) = t n−4 g(t),

g(t) = (n2 − 1)t 2 + 4n(n − 3) − 4n(n − 2)t,

where Æ

(n2 − 1)t 2 · 4n(n − 3) − 4n(n − 2)t p — p ”Æ =4 n (n2 − 1)(n − 3) − (n − 2) n t.

g(t) ≥ 2

Since (n2 − 1)(n − 3) − n(n − 2)2 = n(n − 5) + 3 > 0, we have g(t) > 0, h01 (t) > 0, h1 (t) is increasing for t ≥ 1, h1 (t) ≥ h1 (1) = 0, h0 (t) ≥ 0, h(t) is increasing for t ≥ 1, h(t) ≥ h(1) = 0, hence (n − 1)B − 4C ≥ 0. Thus, the proof is completed. In accordance with Remark 3, the equality holds for a1 = a2 = · · · = an = 1. Remark 1. For p = 0 and q = 1, we get the following inequality (Vasile Cîrtoaje, 2006): 1−a 1− b 1−c 1−d 1−e + + + + ≥ 0, 2 2 2 2 1+a 1+ b 1+c 1+d 1 + e2 where a, b, c, d, e are positive real numbers such that a bcd e = 1. Replacing a, b, c, d, e by 1/a, 1/b, 1/c, 1/d, 1/e, we get the inequality 1+a 1+ b 1+c 1+d 1+e + + + + ≤ 5, 1 + a2 1 + b2 1 + c 2 1 + d 2 1 + e2 where a, b, c, d, e are positive real numbers such that a bcd e = 1. Notice that the inequality 1 − a1 1 + a12

+

1 − a2 1 + a22

+

1 − a3 1 + a32

+

1 − a4 1 + a42

+

1 − a5 1 + a52

+

1 − a6 1 + a62

≥0

is not true for all positive numbers a1 , a2 , a3 , a4 , a5 , a6 satisfying a1 a2 a3 a4 a5 a6 = 1. Indeed, for a2 = a3 = a4 = a5 = a6 = 2, the inequality becomes 1 − a1 1 + a12 which is false for a1 > 0.

− 1 ≥ 0,

Partially Convex Function Method

181

P 3.22. If a, b, c are positive real numbers such that a bc = 1, then 1−a 1− b 1−c + + ≥ 0. 2 2 17 + 4a + 6a 17 + 4b + 6b 17 + 4c + 6c 2 (Vasile Cîrtoaje, 2008) Solution. Using the substitution a = e x , b = e y , c = ez , we need to show that f (x) + g( y) + g(z) ≥ 3 f (s), where f (u) =

s=

x + y +z = 0, 3

1 − eu , u ∈ R, 1 + peu + qe2u

with p=

4 , 17

q=

6 . 17

As we have shown in the proof of the preceding P 3.21, f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = ln r0 > 0,

r0 = 1 +

v t

v t9 p+1 1+ =1+ . q 2

In addition, f is convex on [0, s0 ] if p + 3q − 1 ≥ 0. Indeed, p + 3q − 1 =

5 > 0. 17

By RPCF Theorem, we only need to prove the original inequality for b = c := t and a = 1/t 2 , where t > 0. Write this inequality as follows: t 2 (t 2 − 1) 2(1 − t) + ≥ 0, 4 2 t + pt + q 1 + pt + qt 2 pA + qB ≥ C, where A = t 2 (t − 1)2 (t + 2), B = (t − 1)2 (t 4 + 2t 3 + 2t 2 + 2t + 2), C = t 2 (t − 1)2 (2t + 1).

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Vasile Cîrtoaje

Indeed, we have pA + qB − C =

3(t − 1)2 (t − 2)2 (2t 2 + 2t + 1) ≥ 0. 17

In accordance with Remark 3, the equality holds for a = b = c = 1, and also for a = 1/4 and b = c = 2 (or any cyclic permutation).

P 3.23. If a1 , a2 , · · · , a8 are positive real numbers such that a1 a2 · · · a8 = 1, then 1 − a8 1 − a1 1 − a2 + + · · · + ≥ 0. (1 + a1 )2 (1 + a2 )2 (1 + a8 )2 (Vasile Cîrtoaje, 2006) Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , 8, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x 8 ) ≥ 8 f (s), where f (u) =

s=

x1 + x2 + · · · + x8 = 0, 8

1 − eu , u ∈ R. (1 + eu )2

From f 0 (t) =

eu (eu − 3) , (1 + eu )3

it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = ln 3 > 1. Also, we have f 00 (u) =

eu (8eu − e2u − 3) . (1 + eu )4

We will show that 8eu − e2u − 3 > 0 for u ∈ [0, s0 ], hence f is convex on [0, s0 ]. Indeed, 8eu − e2u − 3 ≥ 8eu − 3eu − 3 = 5eu − 3 > 5 − 3 > 0. By RPCF Theorem, we only need to prove the original inequality for a2 = · · · = a8 := t and a = 1/t 7 , where t ≥ 1. For the nontrivial case t > 1, write this inequality as follows: t 7 (t 7 − 1) 7(t − 1) ≥ . (t 7 + 1)2 (t + 1)2

Partially Convex Function Method

183

t 7 (t 7 − 1)(t + 1)2 ≥ 7, (t − 1)(t 7 + 1)2 t 7 (t 6 + t 5 + t 4 + t 3 + t 2 + t + 1) ≥ 7. (t 6 − t 5 + t 4 − t 3 + t 2 − t + 1)2 Since t 6 − t 5 + t 4 − t 3 + t 2 − t + 1 = t 4 (t 2 − t + 1) − (t − 1)(t 2 + 1) > t 4 (t 2 − t + 1), it suffices to show that t6 + t5 + t4 + t3 + t2 + t + 1 ≥ 7, t(t 2 − t + 1)2 which is equivalent to the obvious inequality (t − 1)6 ≥ 0. Thus, the proof is completed. The equality holds for a1 = a2 = · · · = a8 = 1. Remark. The inequality 1 − a9 1 − a1 1 − a2 + + · · · + ≥0 (1 + a1 )2 (1 + a2 )2 (1 + a9 )2 is not true for all positive numbers a1 , a2 , · · · , a9 satisfying a1 a2 · · · a9 = 1. Indeed, for a2 = a3 = · · · = a9 = 3, the inequality becomes 1 − a1 − 1 ≥ 0, (1 + a1 )2 which is false for a1 > 0.

˜ −13 13 P 3.24. Let a, b, c be positive real numbers such that a bc = 1. If k ∈ p , p , then 3 3 3 3 •

a+k b+k c+k 3(1 + k) + 2 + 2 ≤ . 2 a +1 b +1 c +1 2 (Vasile Cîrtoaje, 2012) Solution. The desired inequality is equivalent to X (a − 1)2 a2 + 1

≥k

X

‹ 2 − 3 . a2 + 1

184

Vasile Cîrtoaje

13 Thus, it suffices to prove this inequality for only |k| = p . On the other hand, replacing 3 3 a, b, c by 1/a, 1/b, 1/c, the inequality becomes X (a − 1)2 a2 + 1

 X ≥ k 3−

‹ 2 . a2 + 1

Thus, we only need to prove the desired inequality for 13 k= p . 3 3 Using the substitution a = e x , b = e y , c = ez , we need to show that f (x) + g( y) + g(z) ≥ 3 f (s), where f (u) =

s=

x + y +z = 0, 3

−eu − k , u ∈ R. e2u + 1

From f 0 (t) =

e2u + 2keu − 1 , (e2u + 1)2

it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = ln r0 < 0,

1 r0 = p . 3 3

Also, we have f 00 (u) =

t · h(t) , (1 + t 2 )3

where h(t) = −t 4 − 4kt 3 + 6t 2 + 4kt − 1,

t = eu .

We will show that h(t) > 0 for t ∈ [r0 , 1], hence f is convex on [s0 , 0]. Indeed, since 52t 52 4kt = p ≥ > 1, 3 3 27 we have h(t) = −t 4 + 6t 2 − 1 + 4kt(1 − t 2 ) ≥ −t 4 + 6t 2 − 1 + (1 − t 2 ) = t 2 (5 − t 2 ) > 0.

Partially Convex Function Method

185

By LPCF Theorem, we only need to prove the original inequality for b = c := t and a = 1/t 2 , where t > 0. Write this inequality as t 2 (kt 2 + 1) 2(t + k) 3(1 + k) + 2 ≤ , t4 + 1 t +1 2 3t 6 − 4t 5 + t 4 + t 2 − 4t + 3 − k(1 − t 2 )3 ≥ 0, (t − 1)2 [(3 + k)t 4 + 2(1 + k)t 3 + 2t 2 + 2(1 − k)t + 3 − k] ≥ 0 p p p p (t − 1)2 (t − 2 + 3)2 [(27 + 13 3)t 2 + 24(2 + 3)t + 33 + 17 3] ≥ 0. The last inequality is clearly true. Thus, the proof is completed. The equality holds p 13 for a = b = c = 1. If k = p , then the equality holds also for a = 7 + 4 3 and 3 3 p −13 b = c = 2 − 3 (or any cyclic permutation). If k = p , then the equality holds also 3 3 p p for a = 7 − 4 3 and b = c = 2 + 3 (or any cyclic permutation).

p P 3.25. If a, b, c are positive real numbers and 0 < k ≤ 2 + 2 2, then b3 c3 a+b+c a3 + + ≥ . ka2 + bc k b2 + ca kc 2 + a b k+1 (Vasile Cîrtoaje, 2011) p p Solution. For k < 2 + 2 2, the proof is similar to the one of the main case k = 2 + 2 2. For this reason, we consider further only the case where p k = 2 + 2 2. Due to homogeneity, we may assume that a bc = 1. On this hypothesis, X

 X  a4 a3 1 X a 1 X a4 − a − a = − = . ka2 + bc k + 1 ka3 + 1 k + 1 k+1 ka3 + 1

Thus, we can write the inequality as a4 − a b4 − b c4 − c + + ≥ 0. ka3 + 1 k b3 + 1 kc 3 + 1 Using the substitution a = e x , b = e y , c = ez ,

186

Vasile Cîrtoaje

we need to show that f (x) + g( y) + g(z) ≥ 3 f (s), where f (u) = From f 0 (t) =

s=

x + y +z = 0, 3

e4u − eu , u ∈ R. ke3u + 1

ke6u + 2(k + 2)e3u − 1 , (ke3u + 1)2

it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = ln r0 < 0,

r0 =

v p u t 3 −k − 2 + (k + 1)(k + 4) k

≈ 0.4149.

Also, we have f 00 (u) =

t · h(t) , (kt 3 + 1)3

where h(t) = k2 t 9 − k(4k + 1)t 6 + (13k + 16)t 3 − 1,

t = eu .

We will show that h(t) > 0 for t ∈ [r0 , 1], hence f is convex on [s0 , 0]. Indeed, we have h(t) ≥ 0 for t ∈ [t 1 , t 2 ], where t 1 ≈ 0.2345 and t 2 ≈ 1.02. Since [r0 , 1] ⊂ [t 1 , t 2 ], the conclusion follows. By LPCF Theorem, we only need to prove the original inequality for b = c. Due to homogeneity, we may consider that b = c = 1. Thus, we need to show that a3 2 a+2 + ≥ , 2 ka + 1 a + k k+1 which is equivalent to the obvious inequality p (a − 1)2 (a − 2)2 ≥ 0. Thus, the proof is completed. In accordance with Remark 3, the equality holds for p a a = b = c. If k = 2 + 2 2, then the equality holds also for p = b = c (or any cyclic 2 permutation).

P 3.26. If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that a1 , a2 , . . . , an ≥

−1 , n−2

a1 + a2 + · · · + an = n,

Partially Convex Function Method then

a12 a12 − a1 + 1

+

187

a22 a22 − a2 + 1

+ ··· +

an2 an2 − an + 1

≤ n. (Vasile Cirtoaje, 2012)

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

where   1−u −1 n2 − n − 1 f (u) = 2 , u∈I= , . u −u+1 n−2 n−2 From f 0 (u) =

u(u − 2) , (u2 − u + 1)2

f 00 (u) =

2(3u2 − u3 − 1) 2u2 (2 − u) + 2(u2 − 1) = , (u2 − u + 1)3 (u2 − u + 1)3

it follow that f is convex on [1, s0 ], decreasing for 1 ≤ u ≤ s0 and increasing for u ≥ s0 , where s0 = 2 ∈ I. Since f is not decreasing on Iu≤s0 , we can not apply RPCF Theorem in its original form. However, according to Remark 5, we may replace this condition with the condition ns − (n − 1)s0 ≤ inf I. Indeed, we have ns − (n − 1)s0 = n − 2(n − 1) = −n + 2 ≤

−1 = inf I. n−2

So, it suffices to show that f (x)+(n−1) f ( y) ≥ n f (1) for all x, y ∈ I such that x ≤ 1 ≤ y and x + (n − 1) y = n. According to Remark 1, we only need to show that h(x, y) ≥ 0, where f (u) − f (1) g(x) − g( y) g(u) = , h(x, y) = . u−1 x−y We have g(u) = h(x, y) =

(x 2

u2

−1 , −u+1

x + y −1 (n − 2)x + 1 = ≥ 0. 2 2 − x + 1)( y − y + 1) (n − 1)(x − x + 1)( y 2 − y + 1)

The proof is completed. The equality holds for a1 = a2 = · · · = an = 1, and also for −1 n−1 a1 = and a2 = a3 = · · · = an = (or any cyclic permutation). n−2 n−2

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Vasile Cîrtoaje

P 3.27. If a1 , a2 , . . . , an (n ≥ 3) are nonzero real numbers such that a1 , a2 , . . . , an ≥ then

1 a12

+

1 a22

−n , n−2

+ ··· +

a1 + a2 + · · · + an = n,

1 1 1 1 ≥ + + ··· + . an2 a1 a2 an (Vasile Cirtoaje, 2012)

Solution. According to P 2.25-(a) in Volume 1, the inequality is true for n = 3. Assume further that n ≥ 4 and write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where

s=

a1 + a2 + · · · + an = 1, n

• ˜ 1 1 −n n(2n − 3) f (u) = 2 − , u ∈ I = , \ {0}. u u n−2 n−2

From f 0 (u) =

u−2 , u3

f 00 (u) =

2(3 − u) , u4

it follow that f is convex on [1, 3], decreasing for 1 ≤ u ≤ s0 and increasing for u ≥ s0 , where s0 = 2 ∈ I. We see that f is not decreasing on Iu≤s0 . Therefore, according to Remark 4 and Remark 5, we may replace this condition in RPCF Theorem with the condition ns − (n − 1)s0 ≤ inf I. Indeed, for n ≥ 4, we have ns − (n − 1)s0 = n − 2(n − 1) = −n + 2 ≤

−n = inf I. n−2

So, it suffices to show that f (x)+(n−1) f ( y) ≥ n f (1) for all x, y ∈ I such that x ≤ 1 ≤ y and x + (n − 1) y = n. According to Remark 1, we only need to show that h(x, y) ≥ 0, where f (u) − f (1) g(x) − g( y) g(u) = , h(x, y) = . u−1 x−y We have g(u) =

x+y −1 (n − 2)x + n , h(x, y) = 2 2 = ≥ 0. 2 u x y (n − 1)x 2 y 2

The proof is completed. In accordance with Remark 3, the equality holds for a1 = a2 = −n n · · · = an = 1, and also for a1 = and a2 = a3 = · · · = an = (or any cyclic n−2 n−2 permutation). Remark. Similarly, we can prove the following generalization.

Partially Convex Function Method • Let a1 , a2 , · · · , an ≥

189

−n such that a1 + a2 + · · · + an = n. If n ≥ 3 and k ≥ 0, then n−2 1 − a1 k + a12

+

1 − a2 k + a22

+ ··· +

1 − an ≥ 0, k + an2

−n with equality for a1 = a2 = · · · = an = 1, and also for a1 = and a2 = a3 = · · · = n−2 n (or any cyclic permutation). an = n−2

P 3.28. If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that a1 , a2 , . . . , an ≥ then

−(3n − 2) , n−2

a1 + a2 + · · · + an = n,

1 − an 1 − a1 1 − a2 + + ··· + ≥ 0. 2 2 (1 + a1 ) (1 + a2 ) (1 + an )2 (Vasile Cîrtoaje, 2014)

Solution. According to P 2.25-(b) in Volume 1, the inequality is true for n = 3. Assume further that n ≥ 4 and write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where

s=

a1 + a2 + · · · + an = 1, n

  −(3n − 2) 4n2 − 7n + 2 1−u f (u) = , u∈I= , \ {−1}. (1 + u)2 n−2 n−2

From f 0 (u) =

u−3 , (u + 1)3

f 00 (u) =

2(5 − u) , (u + 1)4

it follow that f is convex on [1, 5], decreasing for 1 ≤ u ≤ s0 and increasing for u ≥ s0 , where s0 = 3 ∈ I. We see that f is not decreasing on Iu≤s0 . Therefore, according to Remark 4 and Remark 5, we may apply RPCF Theorem by replacing this condition with the sufficient condition ns − (n − 1)s0 ≤ inf I. Indeed, for n ≥ 4, we have ns − (n − 1)s0 = n − 3(n − 1) = −2n + 3 ≤

−(3n − 2) = inf I. n−2

Thus, it suffices to show that f (x) + (n − 1) f ( y) ≥ n f (1) for all x, y ∈ I such that x ≤ 1 ≤ y and x + (n − 1) y = n. According to Remark 1, we only need to show that h(x, y) ≥ 0, where g(u) =

f (u) − f (1) , u−1

h(x, y) =

g(x) − g( y) . x−y

190

Vasile Cîrtoaje

We have g(u) =

x + y +2 (n − 2)x + 3n − 2 −1 , h(x, y) = = ≥ 0. 2 2 2 (u + 1) (1 + x) (1 + y) (n − 1)(1 + x)2 (1 + y)2

The proof is completed. In accordance with Remark 3, the equality holds for a1 = a2 = −(3n − 2) n+2 · · · = an = 1, and also for a1 = and a2 = a3 = · · · = an = (or any cyclic n−2 n−2 permutation).

P 3.29. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If 2 n ≥ 3 and k ≥ 2 − , then n 1 − an 1 − a1 1 − a2 + + ··· + ≥ 0. (1 − ka1 )2 (1 − ka2 )2 (1 − kan )2 (Vasile Cîrtoaje, 2012) Solution. According to P 3.96 in Volume 1, the inequality is true for n = 3. Assume further that n ≥ 4 and write the inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s), where f (u) = From f 0 (u) =

s=

a1 + a2 + · · · + an = 1, n

1−u , u ∈ I = [0, 3] \ {1/k}. (1 − ku)2

−ku + 2k − 1 , (1 − ku)3

f 00 (u) =

2k(−ku + 3k − 2) , (1 − ku)4

it follow that f is convex on [1, s0 ], increasing on [0, 1/k) ∪ [s0 , 3] and decreasing on (1/k, s0 ], where 3n − 4 s0 = 2 − 1/k ≥ > 1. 2(n − 1) Since f is not decreasing on Iu≤s0 , we can not apply RPCF Theorem in its original form. However, according to Remark 5, we may replace this condition with the condition s1 ≤ inf I, where s1 = ns − (n − 1)s0 . Indeed, we have s1 ≤ n − (n − 1) ·

3n − 4 4−n = ≤ 0 = inf I. 2(n − 1) 2

Partially Convex Function Method

191

So, it suffices to show that f (x)+(n−1) f ( y) ≥ n f (1) for all x, y ∈ I such that x ≤ 1 ≤ y and x + (n − 1) y = n. According to Remark 1, we only need to show that h(x, y) ≥ 0, where f (u) − f (1) g(x) − g( y) g(u) = , h(x, y) = . u−1 x−y Since g(u) = h(x, y) =

−1 , (1 − ku)2

k[k(x + y) − 2] , (1 − k x)2 (1 − k y)2

we need to show that k(x + y) − 2 ≥ 0. Indeed, k(x + y) =

kn k[(n − 2)x + n] ≥ ≥ 2. n−1 n−1

2 The proof is completed. The equality holds for a1 = a2 = · · · = an = 1. If k = 2 − , n n then the equality also holds for a1 = 0 and a2 = a3 = · · · = an = (or any cyclic n−1 permutation).

192

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Chapter 4

Partially Convex Function Method for Ordered Variables 4.1

Theoretical Basis

Right Partially Convex Function Theorem for Ordered Variables (Vasile Cirtoaje, 2012). Let f be a function defined on a real interval I and convex on [s, s0 ], where s, s0 ∈ I, s < s0 . In addition, f (u) is decreasing on Iu≤s0 and min f (u) = f (s0 ). u∈I

The inequality f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f

a + a + ··· + a  1 2 n n

holds for all a1 , a2 , . . . , an ∈ I such that a1 + a2 + · · · + an = ns and at least n − m of a1 , a2 , . . . , an are smaller than or equal to s if and only if f (x) + m f ( y) ≥ (1 + m) f (s) for all x, y ∈ I such that x ≤ s ≤ y and x + m y = (1 + m)s. Proof. The necessity is obvious. By Lemma from section 3, to prove the sufficiency, it suffices to consider that a1 , a2 , . . . , an ∈ J, where J = Iu≤s0 . Because f (u) is convex on Ju≥s , the desired inequality follows from HCF-OV Theorem applied to the interval J. Similarly, we can prove Left Partially Convex Function Theorem for Ordered Variables (Vasile Cirtoaje, 2012). 193

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Left Partially Convex Function Theorem for Ordered Variables. Let f be a function defined on a real interval I and convex on [s0 , s], where s0 , s ∈ I, s0 < s. In addition, f (u) is increasing on Iu≥s0 and satisfies min = f (s0 ). u∈I

The inequality f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f

a + a + ··· + a  1 2 n n

holds for all a1 , a2 , · · · , an ∈ I such that a1 + a2 + · · · + an = ns and at least n − m of a1 , a2 , . . . , an are greater than or equal to s if and only if f (x) + m f ( y) ≥ (1 + m) f (s) for all x, y ∈ I such that x ≥ s ≥ y and x + m y = (1 + m)s. Remark 1. Let us denote g(u) =

f (u) − f (s) , u−s

h(x, y) =

g(x) − g( y) . x−y

We may replace the hypothesis condition in RPCF-OV Theorem and LPCF-OV Theorem, namely f (x) + m f ( y) ≥ (1 + m) f (s), by the condition h(x, y) ≥ 0 for all x, y ∈ I such that x + m y = (1 + m)s. Remark 2. Assume that f is differentiable on I, and let H(x, y) =

f 0 (x) − f 0 ( y) . x−y

Then, the desired inequality of Jensen’s type in RPCF-OV Theorem and LPCF-OV Theorem holds true by replacing the hypothesis f (x) + m f ( y) ≥ (1 + m) f (s) with the more restrictive condition H(x, y) ≥ 0 for all x, y ∈ I such that x + m y = (1 + m)s. Remark 3. RPCF-OV Theorem is a generalization of LPCF Theorem, because the last theorem can be obtained from the first theorem for m = n − 1. Similarly, LPCF-OV Theorem is a generalization of LPCF Theorem.

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195

Remark 4. If a1 ≥ · · · ≥ am ≥ s ≥ am+1 ≥ · · · ≥ an , then at least n − m of a1 , a2 , . . . , an are smaller than or equal to s. Also, if a1 ≥ · · · ≥ an−m ≥ s ≥ an−m+1 ≥ · · · ≥ an , then at least n − m of a1 , a2 , . . . , an are greater than or equal to s. Thus, from RPCF-OV Theorem and LPCF-OV Theorem, we get the following corollaries. RPCF-OV Corollary. Let f be a function defined on a real interval I and convex on [s, s0 ], where s, s0 ∈ I, s < s0 . In addition, f (u) is decreasing on Iu≤s0 and min f (u) = f (s0 ). u∈I

The inequality f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f

a + a + ··· + a  1 2 n n

holds for all a1 , a2 , . . . , an ∈ I satisfying a1 ≥ · · · ≥ am ≥ s ≥ am+1 ≥ · · · ≥ an ,

a1 + a2 + · · · + an = ns,

if and only if f (x) + m f ( y) ≥ (1 + m) f (s) for all x, y ∈ I such that x ≤ s ≤ y and x + m y = (1 + m)s. LPCF-OV Corollary. Let f be a function defined on a real interval I and convex on [s0 , s], where s0 , s ∈ I, s0 < s. In addition, f (u) is increasing on Iu≥s0 and satisfies min f (u) = f (s0 ). u∈I

The inequality f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f

a + a + ··· + a  1 2 n n

holds for all a1 , a2 , . . . , an ∈ I satisfying a1 ≤ · · · ≤ am ≤ s ≤ am+1 ≤ · · · ≤ an ,

a1 + a2 + · · · + an = ns,

if and only if f (x) + m f ( y) ≥ (1 + m) f (s) for all x, y ∈ I such that x ≥ s ≥ y and x + m y = (1 + m)s.

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Remark 5. The inequality in RPCF-OV Corollary becomes an equality for a1 = a2 = · · · = an = s and also for a1 = · · · = am = y,

am+1 = · · · = an−1 = s,

an = x (x < y),

where x, y ∈ I satisfy the equations x + m y = (1 + m)s,

f (x) + m f ( y) = (1 + m) f (s).

For x 6= y, these equations are equivalent to x + m y = (1 + m)s, h(x, y) = 0. Remark 6. The inequality in LPCF-OV Corollary becomes an equality for a1 = a2 = · · · = an = s and also for a1 = x,

a2 = · · · = an−m = s,

an−m+1 = · · · = an = y (x > y),

where x, y ∈ I satisfy the equations x + m y = (1 + m)s,

f (x) + m f ( y) = (1 + m) f (s).

For x 6= y, these equations are equivalent to x + m y = (1 + m)s, h(x, y) = 0. Remark 7. RPCF-OV Theorem and RPCF-OV Corollar are also valid in the case when I = [a, b] \ {u0 } or I = (a, b) \ {u0 }, where a, b, u0 are real numbers such that a < s < s0 < u0 < b. Similarly, LPCF Theorem is also valid in the case when I = [a, b]\{u0 } or I = (a, b)\{u0 }, where a, b, u0 are real numbers such that a < u0 < s0 < s < b.

PCF Method for Ordered Variables

4.2

197

Applications

4.1. If a, b, c, d are real numbers such that a ≤ 1 ≤ b ≤ c ≤ d, then

a + b + c + d = 4,

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

3a2

4.2. If a, b, c, d are real numbers such that a ≥ b ≥ 1 ≥ c ≥ d, then

a + b + c + d = 4,

16b − 5 16c − 5 16d − 5 4 16a − 5 + + + ≤ . 32a2 + 1 32b2 + 1 32c 2 + 1 32d 2 + 1 3

4.3. If a, b, c, d, e are real numbers such that a ≥ b ≥ 1 ≥ c ≥ d ≥ e, then

a + b + c + d + e = 5,

18a − 5 18b − 5 18c − 5 18d − 5 18e − 5 + + + + ≤ 5. 2 2 2 2 12a + 1 12b + 1 12c + 1 12d + 1 12e2 + 1

4.4. If a, b, c, d, e are real numbers such that a ≥ b ≥ 1 ≥ c ≥ d ≥ e, then

a + b + c + d + e = 5,

a(a − 1) b(b − 1) c(c − 1) d(d − 1) e(e − 1) + + 2 + + 2 ≥ 0. 3a2 + 4 3b2 + 4 3c + 4 3d 2 + 4 3e + 4

4.5. Let a1 , a2 , . . . , a2n 6= −k be real numbers such that a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n , If k ≥

a1 + a2 + · · · + a2n = 2n.

n+1 p , then 2 n a2n (a2n − 1) a1 (a1 − 1) a2 (a2 − 1) + + ··· + ≥ 0. 2 2 (a1 + k) (a2 + k) (a2n + k)2

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4.6. Let a1 , a2 , . . . , a2n 6= −k be real numbers such that a1 + a2 + · · · + a2n = 2n.

a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n , n+1 If k ≥ 1 + p , then n a12 − 1 (a1 + k)2

+

a22 − 1 (a2 + k)2

+ ··· +

2 a2n −1

(a2n + k)2

≥ 0.

4.7. If a1 , a2 , · · · , an are positive real numbers such that a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

2/a1

a1

2/a2

+ a2

a1 + a2 + · · · + an = n,

+ · · · + an2/an ≤ n.

4.8. If a1 , a2 , · · · , a11 are real numbers such that a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ a11 ,

a1 + a2 + · · · + a11 = 11,

then 2 ) ≥ 1. (1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a11 + a11

4.9. If a1 , a2 , · · · , a8 are nonzero real numbers such that a1 ≥ a2 ≥ a3 ≥ a4 ≥ 1 ≥ a5 ≥ a6 ≥ a7 ≥ a8 , then

 5

1 a12

+

1 a22

+ ··· +

1 a82



a1 + a2 + · · · + a8 = 8,

‹ 1 1 1 + 72 ≥ 14 + + ··· + . a1 a2 a8 

4.10. If a, b, c, d are positive real numbers such that a ≥ b ≥ 1 ≥ c ≥ d, then

a bcd = 1,

7 − 6a 7 − 6b 7 − 6c 7 − 6d 4 + + + ≥ . 2 2 2 2 2+a 2+ b 2+c 2+d 3

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199

4.11. If a, b, c are positive real numbers such that a ≥ 1 ≥ b ≥ c, then

a bc = 1,

7 − 4a 7 − 4b 7 − 4c + + ≥ 3. 2 + a2 2 + b2 2 + c 2

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

a1 a2 · · · an = 1.

If p, q ≥ 0 such that p + 3q ≥ 1, then 1 − a1 1+

pa1 + qa12

+

1 − a2 1+

pa2 + qa22

+ ··· +

1 − an ≥ 0. 1 + pan + qan2

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PCF Method for Ordered Variables

4.3

201

Solutions

P 4.1. If a, b, c, d are real numbers such that a ≤ 1 ≤ b ≤ c ≤ d, then

a + b + c + d = 4,

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

3a2

Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) ≥ 4 f (s), where f (u) =

−u , 3u2 + 1

From

s=

a+b+c+d = 1, 4

u ∈ R.

3u2 − 1 , (3u2 + 1)2

f 0 (u) =

it follows p that f is increasing on (−∞, −s0 ]∪[s0 , ∞) and decreasing on [−s0 , s0 ], where s0 = 1/ 3. Since lim f (u) = 0 u→−∞

and f (s0 ) < 0, it follows that

min f (u) = f (s0 ). u∈R

From f 00 (u) =

18u(1 − u2 ) , (3u2 + 1)3

it follows that f is convex on [0, 1], hence on [s0 , 1]. Therefore, we apply LPCF-OV Theorem or LPCF-OV Corollary for n = 4 and m = 1. We only need to show that f (x) + f ( y) ≥ 2 f (1) for all real x, y such that x + y = 2. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where h(x, y) =

g(x) − g( y) , x−y

Indeed, we have g(u) = h(x, y) =

g(u) =

f (u) − f (1) . u−1

3u − 1 , 4(3u2 + 1)

3(1 + x + y − 3x y) 9(1 − x y) = ≥ 0, 2 2 2 4(3x + 1)(3 y + 1) 4(3x + 1)(3 y 2 + 1)

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since 4(1 − x y) = (x + y)2 − 4x y = (x − y)2 ≥ 0. Thus, the proof is completed. The equality holds for a = b = c = d = 1. Remark. Similarly, we can prove the following generalization. • If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n,

a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then

a1 3a12

+1

+

a2 3a22

+1

+ ··· +

an n ≤ , 3an2 + 1 4

with equality for a1 = a2 = · · · = an = 1.

P 4.2. If a, b, c, d are real numbers such that a ≥ b ≥ 1 ≥ c ≥ d, then

a + b + c + d = 4,

16a − 5 16b − 5 16c − 5 16d − 5 4 + + + ≤ . 2 2 2 2 32a + 1 32b + 1 32c + 1 32d + 1 3 (Vasile Cîrtoaje, 2012)

Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),

s=

a+b+c+d = 1, 4

where

5 − 16u , u ∈ R. 32u2 + 1 As shown in the proof of P 3.1, f is convex on [s0 , 1], increasing for u ≥ s0 and f (u) =

min f (u) = f (s0 ), u∈R

where

p 5 + 33 s0 = ≈ 0.6715. 16 Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for n = 4 and m = 2. We only need to show that f (x) + 2 f ( y) ≥ 3 f (1) for all real x, y such that x + 2 y = 3. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

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203

Indeed, we have g(u) = h(x, y) =

32(2u − 1) , 3(32u2 + 1)

64(1 + 16x + 16 y − 32x y) 64(4x − 5)2 = ≥ 0. 3(32x 2 + 1)(32 y 2 + 1) 3(32x 2 + 1)(32 y 2 + 1)

Thus, the proof is completed. From x + 2 y = 3 and h(x, y) = 0, we get x = 5/4 and, y = 7/8. Therefore, in accordance with Remark 6, the equality holds for a = b = c = d = 1, and also for a = 5/4, b = 1 and c = d = 7/8. Remark. Similarly, we can prove the following generalization. • If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that a1 ≥ · · · ≥ an−2 ≥ 1 ≥ an−1 ≥ an , then

16a1 − 5 32a12

+1

+

16a2 − 5 32a22

+1

+ ··· +

a1 + a2 + · · · + an = n, 16an − 5 n ≤ , 2 32an + 1 3

with equality for a1 = a2 = · · · = an = 1, and also for a1 = 5/4, a2 = · · · = an−2 = 1 and an−1 = an = 7/8.

P 4.3. If a, b, c, d, e are real numbers such that a ≥ b ≥ 1 ≥ c ≥ d ≥ e, then

a + b + c + d + e = 5,

18a − 5 18b − 5 18c − 5 18d − 5 18e − 5 + + + + ≤ 5. 2 2 2 2 12a + 1 12b + 1 12c + 1 12d + 1 12e2 + 1 (Vasile Cîrtoaje, 2012)

Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s), where f (u) =

5 − 18u , 12u2 + 1

s=

a+b+c+d+e = 1, 5

u ∈ R.

As shown in the proof of P 3.2, f is convex on [s0 , 1], increasing for u ≥ s0 and min f (u) = f (s0 ), u∈R

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where

p 5 + 52 s0 = ≈ 0.678. 18

Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for n = 5 and m = 3. We only need to show that f (x) + 3 f ( y) ≥ 4 f (1) for all real x, y such that x + 3 y = 4. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

Indeed, we have g(u) = h(x, y) =

6(2u − 1) , 12u2 + 1

12(1 + 6x + 6 y − 12x y) 12(2x − 3)2 = ≥ 0. (12x 2 + 1)(12 y 2 + 1) (12x 2 + 1)(12 y 2 + 1)

Thus, the proof is completed. From x + 3 y = 4 and h(x, y) = 0, we get x = 3/2 and, y = 5/6. Therefore, in accordance with Remark 6, the equality holds for a = b = c = d = e = 1, and also for a = 3/2, b = 1 and c = d = e = 5/6. Remark. Similarly, we can prove the following generalization. • If a1 , a2 , . . . , an (n ≥ 4) are real numbers such that a1 ≥ · · · ≥ an−3 ≥ 1 ≥ an−2 ≥ an−1 ≥ an , then

18a1 − 5 12a12

+1

+

18a2 − 5 12a22

+1

+ ··· +

a1 + a2 + · · · + an = n, 18an − 5 ≤ n, 12an2 + 1

with equality for a1 = a2 = · · · = an = 1, and also for a1 = 3/2, a2 = · · · = an−3 = 1 and an−2 = an−1 = an = 5/6.

P 4.4. If a, b, c, d, e are real numbers such that a ≥ b ≥ 1 ≥ c ≥ d ≥ e,

a + b + c + d + e = 5,

then a(a − 1) b(b − 1) c(c − 1) d(d − 1) e(e − 1) + + 2 + + 2 ≥ 0. 3a2 + 4 3b2 + 4 3c + 4 3d 2 + 4 3e + 4 (Vasile Cîrtoaje, 2012)

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205

Solution. Write the inequality as f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s),

s=

a+b+c+d+e = 1, 5

where

u2 − u , u ∈ R. 3u2 + 4 As shown in the proof of P 3.5, f is convex on [s0 , 1], increasing for u ≥ s0 and f (u) =

min f (u) = f (s0 ), u∈R

where

p −4 + 2 7 s0 = ≈ 0.43. 3 Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for n = 5 and m = 3. We only need to show that f (x) + 3 f ( y) ≥ 4 f (1) for all real x, y such that x + 3 y = 4. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where h(x, y) =

g(x) − g( y) , x−y

g(u) =

f (u) − f (1) . u−1

Indeed, we have g(u) = h(x, y) =

u) , +4

3u2

4 − 3x y (x − 2)2 = ≥ 0. (3x 2 + 4)(3 y 2 + 4) (12x 2 + 1)(12 y 2 + 1)

Thus, the proof is completed. From x + 3 y = 4 and h(x, y) = 0, we get x = 2 and, y = 2/3. Therefore, in accordance with Remark 6, the equality holds for a = b = c = d = e = 1, and also for a = 2, b = 1, c = d = e = 2/3.

Remark. Similarly, we can prove the following generalizations. • If a1 , a2 , . . . , an (n ≥ 4) are real numbers such that a1 ≥ · · · ≥ an−3 ≥ 1 ≥ an−2 ≥ an−1 ≥ an , then

a1 (a1 − 1) 3a12

+4

+

a2 (a2 − 1) 3a22

+4

+ ··· +

a1 + a2 + · · · + an = n, an (an − 1) ≥ 0, 3an2 + 4

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with equality for a1 = a2 = · · · = an = 1, and also for a1 = 2, a2 = · · · = an−3 = 1 and an−2 = an−1 = an = 2/3. • If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an ,

a1 + a2 + · · · + an = n,

then a1 (a1 − 1) 4(n − 2)a12

+ (n − 1)2

+

a2 (a2 − 1) 4(n − 2)a22

+ (n − 1)2

+ ··· +

an (an − 1) ≥ 0, 4(n − 2)an2 + (n − 1)2

with equality for a1 = a2 = · · · = an = 1, and also for a1 = an =

n−1 . 2(n − 2)

n−1 , a2 = 1 and a3 = · · · = 2

P 4.5. Let a1 , a2 , . . . , a2n 6= −k be real numbers such that a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n , If k ≥

a1 + a2 + · · · + a2n = 2n.

n+1 p , then 2 n a2n (a2n − 1) a1 (a1 − 1) a2 (a2 − 1) + + ··· + ≥ 0. 2 2 (a1 + k) (a2 + k) (a2n + k)2 (Vasile Cîrtoaje, 2012)

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (a2n ) ≥ 2n f (s), where f (u) =

s=

a1 + a2 + · · · + a2n = 1, 2n

u(u − 1) , u ∈ I = R \ {−k}. (u + k)2

As shown in the proof of P 3.6, f is convex on [s0 , 1], increasing for u ≥ s0 and min f (u) = f (s0 ), u∈I

where s0 =

k < 1. 2k + 1

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207

Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for 2n real numbers and m = n. We only need to show that f (x) + n f ( y) ≥ (n + 1) f (1) for all real x, y such that x + n y = n + 1. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where h(x, y) =

g(x) − g( y) , x−y

Indeed, we have g(u) = h(x, y) =

g(u) =

f (u) − f (1) . u−1

u , (u + k)2

k2 − x y ≥ 0, (x + k)2 ( y + k)2

because

[2n y − n − 1]2 (n + 1)2 −xy = ≥ 0. 4n 4n n+1 The equality holds for a1 = a2 = · · · = an = 1. If k = p , then the equality holds also 2 n n+1 n+1 for a1 = , a2 = · · · = an = 1 and an+1 = · · · = a2n = . 2 2n k2 − x y ≥

P 4.6. Let a1 , a2 , . . . , a2n 6= −k be real numbers such that a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,

a1 + a2 + · · · + a2n = 2n.

n+1 If k ≥ 1 + p , then n a12 − 1 (a1 + k)2

a22 − 1

+

(a2 + k)2

+ ··· +

2 a2n −1

(a2n + k)2

≥ 0. (Vasile Cîrtoaje, 2012)

Solution. Write the inequality as f (a1 ) + f (a2 ) + · · · + f (a2n ) ≥ 2n f (s), where f (u) =

s=

a1 + a2 + · · · + a2n = 1, 2n

u2 − 1 , u ∈ I = R \ {−k}. (u + k)2

As shown in the proof of P 3.7, f is convex on [s0 , 1], increasing for u ≥ s0 and min f (u) = f (s0 ), u∈I

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where

−1 ∈ (−1, 0). k Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for 2n real numbers and m = n. We only need to show that f (x) + n f ( y) ≥ (n + 1) f (1) for all real x, y such that x + n y = n + 1. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where s0 =

h(x, y) =

g(x) − g( y) , x−y

Indeed, we have g(u) = h(x, y) =

g(u) =

f (u) − f (1) . u−1

u+1 , (u + k)2

(k − 1)2 − 1 − x − y − x y ≥ 0, (x + k)2 ( y + k)2

because (k − 1)2 − 1 − x − y − x y ≥

(n y − 1)2 (n + 1)2 −1− x − y − xy = ≥ 0. n n

n+1 The equality holds for a1 = a2 = · · · = an = 1. If k = 1 + p , then the equality holds n 1 also for a1 = n, a2 = · · · = an = 1 and an+1 = · · · = a2n = . n

P 4.7. If a1 , a2 , · · · , an are positive real numbers such that a1 + a2 + · · · + an = n,

a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then

2/a1

a1

2/a2

+ a2

+ · · · + an2/an ≤ n. (Vasile Cîrtoaje, 2012)

Solution. Rewrite the desired inequality as f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),

s=

a1 + a2 + · · · + an = 1, n

where f (u) = −u2/u , We have

u ∈ I = (0, n). 2

f 0 (u) = 2u u −2 (ln u − 1),

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209

2

f 00 (u) = 2u u −4 [u + 2(1 − ln u)(u − 1 + ln u)]. From the expression of f 0 , it follows that f is decreasing on (0, s0 ] and increasing on [s0 , n), where s0 = e. In addition, we claim that f is convex on [1, s0 ]. Indeed, since 1 − ln u ≥ 0 and u − 1 + ln u ≥ u − 1 ≥ 0, we have f 00 > 0 for u ∈ [1, s0 ]. Therefore, we may apply RPCF-OV Theorem or RPCF-OV Corollary for m = 1. We only need to show that f (x) + f ( y) ≥ 2 f (1) for all x, y > 0 such that x + y = 2. The inequality f (x) + f ( y) ≥ 2 f (1) is equivalent to x 2/x + y 2/ y ≤ 2, which is just the inequality in P 3.27 from Volume 2. The equality holds for a1 = a2 = · · · = an = 1.

P 4.8. If a1 , a2 , · · · , a11 are real numbers such that a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ a11 ,

a1 + a2 + · · · + a11 = 11,

then 2 ) ≥ 1. (1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a11 + a11

(Vasile Cîrtoaje, 2012) Solution. Rewrite the desired inequality as f (a1 ) + f (a2 ) + · · · + f (a11 ) ≥ 11 f (s),

s=

a1 + a2 + · · · + a11 = 1, 11

where f (u) = ln(1 − u + u2 ), From f 0 (u) =

u ∈ R.

2u − 1 , 1 − u + u2

it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = 1/2. Also, from 1 + 2u(1 − u) f 00 (u) = , (1 − u + u2 )2 it follows that f 00 (u) > 0 for u ∈ [s0 , 1], hence f is convex on [s0 , 1]. Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for n = 11 and m = 9. We only need to

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show that f (x) + 9 f ( y) ≥ 9 f (1) for all real x, y such that x + 9 y = 10. Using Remark 2, it suffices to prove that H(x, y) ≥ 0, where H(x, y) = Since H(x, y) =

f 0 (x) − f 0 ( y) . x−y

1 + x + y − 2x y (1 − x + x 2 )(1 − y + y 2 )

and 1 + x + y − 2x y = 18 y 2 − 8 y + 1 = 2 y 2 + (4 y − 1)2 > 0, the conclusion follows. The equality holds for a1 = a2 = · · · = a11 = 1. Remark. By replacing a1 , a2 , . . . , a11 respectively with 1 − a1 , 1 − a2 , . . . , 1 − a11 , we get the following statement. • If a1 , a2 , . . . , a11 are real numbers such that a1 ≤ a2 ≤ 0 ≤ a3 ≤ · · · ≤ a11 ,

a1 + a2 + · · · + a11 = 0,

then 2 ) ≥ 1, (1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a11 + a11

with equality for a1 = a2 = · · · = an = 0.

P 4.9. If a1 , a2 , · · · , a8 are nonzero real numbers such that a1 ≥ a2 ≥ a3 ≥ a4 ≥ 1 ≥ a5 ≥ a6 ≥ a7 ≥ a8 , then

 5

1 a12

+

1 a22

+ ··· +

1



a82

+ 72 ≥ 14



a1 + a2 + · · · + a8 = 8,

‹ 1 1 1 + + ··· + . a1 a2 a8 (Vasile Cîrtoaje, 2012)

Solution. Write the desired inequality as f (a1 ) + f (a2 ) + · · · + f (a8 ) ≥ 8 f (s), s = where

a1 + a2 + · · · + a8 = 1, 8

5 14 − + 9, u ∈ I = R \ {0}. 2 u u As shown in the proof of P 3.16, f is convex on [s0 , 1], increasing for u ≥ s0 and f (u) =

min f (u) = f (s0 ), u∈I

PCF Method for Ordered Variables

211

where

5 . 7 Taking into account Remark 7, we may apply LPCF-OV Theorem or LPCF-OV Corollary for n = 8 and m = 4. We only need to show that f (x) + 4 f ( y) ≥ 5 f (1) for all x, y ∈ I such that x + 4 y = 5. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where s0 =

h(x, y) =

g(x) − g( y) , x−y

g(u) =

Indeed, we have g(u) = g(u) = h(x, y) =

f (u) − f (1) . u−1

5 9 − 2, u u

5x + 5 y − 9x y (6 y − 5)2 = ≥ 0. x2 y2 x2 y2

In accordance with Remark 6, the equality holds for a1 = a2 = · · · = a8 = 1, and also 5 5 for a1 = , a2 = a3 = a4 = 1 and a5 = a6 = a7 = a8 = (or any cyclic permutation). 3 6

P 4.10. If a, b, c, d are positive real numbers such that a ≥ b ≥ 1 ≥ c ≥ d, then

a bcd = 1,

7 − 6a 7 − 6b 7 − 6c 7 − 6d 4 + + + ≥ . 2 + a2 2 + b2 2 + c 2 2 + d 2 3 (Vasile Cîrtoaje, 2012)

Solution. Using the substitution a = e x , b = e y , c = ez , d = e w we need to show that f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s), where x ≥ y ≥ 0 ≥ z ≥ w,

s=

x + y +z+w = 0, 4

7 − 6eu , u ∈ R. 2 + e2u As shown in the proof of P 3.17, f is convex on [0, s0 ], is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = ln 3. Therefore, we may apply RPCF-OV Theorem or f (u) =

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RPCF-OV Corollary for n = 4 and m = 2. We only need to show that f (x) + 2 f ( y) ≥ 3 f (0) for all real x, y such that x + 2 y = 0. That is, to prove that 7 − 6a 2(7 − 6b) + ≥1 2 + a2 2 + b2 for all a, b > 0 such that a b2 = 1. This is equivalent to (b − 1)2 (b − 2)2 (5b2 + 6bt + 3) ≥ 0, which is clearly true. In accordance with Remark 5, the equality holds for a = b = c = d = 1, and also for a = b = 2, c = 1 and d = 1/4.

P 4.11. If a, b, c are positive real numbers such that a ≥ 1 ≥ b ≥ c, then

a bc = 1,

7 − 4a 7 − 4b 7 − 4c + + ≥ 3. 2 + a2 2 + b2 2 + c 2 (Vasile Cîrtoaje, 2012)

Solution. Using the substitution a = e x , b = e y , c = ez , we need to show that f (x) + f ( y) + f (z) ≥ 3 f (s), where x ≥ 1 ≥ y ≥ z, f (u) = From f 0 (u) =

s=

x + y +z = 0, 3

7 − 4eu , u ∈ R. 2 + e2u

2eu (2eu + 1)(eu − 4) , (2 + e2u )2

it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = ln 4. Also, we have 4t · h(t) f 00 (u) = , t = eu , (2 + t 2 )3 where h(t) = −t 4 + 7t 3 + 12t 2 − 14t − 4.

PCF Method for Ordered Variables

213

We will show that h(t) ≥ 0 for t ∈ [1, 4], hence f is convex on [0, s0 ]. Indeed, h(t) = (t − 1)[t 2 (−t + 6) + 18t + 4] ≥ 0. Therefore, we may apply RPCF-OV Theorem or RPCF-OV Corollary for n = 3 and m = 1. We only need to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0. That is, to prove that 7 − 4a 7 − 4b + ≥2 2 + a2 2 + b2 for all a, b > 0 such that a b = 1. This is equivalent to (a − 1)4 ≥ 0. The equality holds for a = b = c = 1.

P 4.12. Let a1 , a2 , · · · , an be positive real numbers such that a1 a2 · · · an = 1.

a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , If p, q ≥ 0 such that p + 3q ≥ 1, then 1 − a1 1+

pa1 + qa12

+

1 − a2 1+

pa2 + qa22

+ ··· +

1 − an ≥ 0. 1 + pan + qan2 (Vasile Cîrtoaje, 2012)

Solution. For q = 0, we need to show that p ≥ 1 involves 1 − an 1 − a1 1 − a2 + + ··· + ≥ 0. 1 + pa1 1 + pa2 1 + pan This is just the inequality from P 2.19. Consider next that q > 0. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s), where x1 ≥ 1 ≥ x2 ≥ · · · ≥ x n, f (u) =

s=

x1 + x2 + · · · + x n = 0, n

1 − eu , u ∈ R. 1 + peu + qe2u

As shown in the proof of P 3.21, f is convex on [0, s0 ] if p + 3q − 1 ≥ 0, where v t p+1 s0 = ln r0 > 0, r0 = 1 + 1 + . q

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Vasile Cîrtoaje

Clearly, the condition p+3q−1 ≥ 0 is satisfied by hypothesis. In addition, f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞). Therefore, we may apply RPCF-OV Theorem or RPCF-OV Corollary for m = 1. We only need to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0. That is, to prove that 1−a 1− b + ≥0 2 1 + pa + qa 1 + p b + q b2 for all a, b > 0 such that a b = 1. This is equivalent to (a − 1)2 [(p − 1)a + q(a2 + a + 1)] ≥ 0, which is true because (p − 1)a + q(a2 + a + 1) ≥ (p − 1)a + q(3a) = (p + 3q − 1)a ≥ 0. The equality holds for a1 = a2 = · · · = an = 1.

Chapter 5

Bibliography

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