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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2010, Article ID 128258, 12 pages doi:10.1155/2010/128258 Research Article The Best Lower Bound Depended on Two Fixed Variables for Jensen’s Inequality with Ordered Variables Vasile Cirtoaje Department of Automatic Control and Computers, University of Ploiesti, 100680 Ploiesti, Romania Correspondence should be addressed to Vasile Cirtoaje, vcirtoaje@upg-ploiesti.ro Received 16 June 2010; Accepted 4 November 2010 Academic Editor: R. N. Mohapatra Copyright q 2010 Vasile Cirtoaje. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We give the best lower bound for the weighted Jensen’s discrete inequality with ordered variables applied to a convex function f, in the case when the lower bound depends on f, weights, and two given variables. Furthermore, under the same conditions, we give some sharp lower bounds for the weighted AM-GM inequality and AM-HM inequality. 1. Introduction Let x  {x 1 ,x 2 , ,x n } be a sequence of real numbers belonging to an interval I,andlet p  {p 1 ,p 2 , ,p n } be a sequence of given positive weights associated to x and satisfying p 1  p 2  ··· p n  1. If f is a convex function on I, then the well-known discrete Jensen’s inequality 1 states that Δ  f, p, x  ≥ 0, 1.1 where Δ  f, p, x   p 1 f  x 1   p 2 f  x 2   ··· p n f  x n  − f  p 1 x 1  p 2 x 2  ··· p n x n  1.2 is the so-called Jensen’s difference. The next refinement of Jensen’s inequality was proven in 2, as a consequence of its Theorem 2.1,partii Δ  f, p, x  ≥ max 1≤i<k≤n  p i f  x i   p k f  x k  −  p i  p k  f  p i x i  p k x k p i  p k  ≥ 0. 1.3 2 Journal of Inequalities and Applications By 1.3, for fixed x i and x k ,weget Δ  f, p, x  ≥ p i f  x i   p k f  x k  −  p i  p k  f  p i x i  p k x k p i  p k  : S p,f  x i ,x k  . 1.4 In this paper, we will establish that the best lower bound L p,f x i ,x k  of Jensen’s difference Δf, p, x for x 1 ≤···≤x i ≤···≤x k ≤···≤x n 1.5 has the expression L p,f  x i ,x k   Q i f  x i   R k f  x k  −  Q i  R k  f  Q i x i  R k x k Q i  R k  , 1.6 where Q i  p 1  p 2  ··· p i ,R k  p k  p k1  ··· p n . 1.7 Logically, we need to have L p,f  x i ,x k  ≥ S p,f  x i ,x k  . 1.8 Indeed, this inequality is equivalent to Jensen’s inequality  Q i − p i  f  x i    R k − p k  f  x k    p i  p k  f  p i x i  p k x k p i  p k  ≥  Q i  R k  f  Q i x i  R k x k Q i  R k  . 1.9 2. Main Results Theorem 2.1. Let f be a convex function on I, and let x 1 ,x 2 , ,x n ∈ I n ≥ 3 such that x 1 ≤ x 2 ≤···≤x n . 2.1 For fixed x i and x k (1 ≤ i<k≤ n), Jensen’s difference Δf, p, x is minimal when x 1  x 2  ··· x i−1  x i ,x k1  x k2  ··· x n  x k , x i1  x i2  ··· x k−1  Q i x i  R k x k Q i  R k , 2.2 Journal of Inequalities and Applications 3 that is, Δ  f, p, x  ≥ Q i f  x i   R k f  x k  −  Q i  R k  f  Q i x i  R k x k Q i  R k  : L p,f  x i ,x k  . 2.3 For proving Theorem 2.1, we will need the following three lemmas. Lemma 2.2. Let p, q be nonnegative real numbers, and let f be a convex function on I.Ifa, b, c, d ∈ I such that c, d ∈ a, b and pa  qb  pc  qd, 2.4 then pf  a   qf  b  ≥ pf  c   qf  d  . 2.5 Lemma 2.3. Let f be a convex function on I, and let x 1 ,x 2 , ,x n ∈ I (n ≥ 3) s uch that x 1 ≤ x 2 ≤···≤x n . 2.6 For fixed x i ,x i1 , ,x n ,wherei ∈{2, 3, ,n−1}, Jensen’s difference Δf, p, x is minimal when x 1  x 2  ··· x i−1  x i . 2.7 Lemma 2.4. Let f be a convex function on I, and let x 1 ,x 2 , ,x n ∈ I (n ≥ 3) s uch that x 1 ≤ x 2 ≤···≤x n . 2.8 For fixed x 1 ,x 2 , ,x k ,wherek ∈{2, 3, ,n−1}, Jensen’s difference Δf, p, x is minimal when x k1  x k2  ··· x n  x k . 2.9 Applying Theorem 2.1 for fxe x and using the substitutions a 1  e x 1 , a 2  e x 2 , ,a n  e x n ,weobtain Corollary 2.5. Let 0 <a 1 ≤···≤a i ≤···≤a k ≤···≤a n , 2.10 and let p 1 ,p 2 , ,p n be positive real numbers such that p 1  p 2  ··· p n  1. Then, p 1 a 1  p 2 a 2  ··· p n a n − a p 1 1 a p 2 2 ···a p n n ≥ Q i a i  R k a k −  Q i  R k  a Q i /Q i R k  i a R k /Q i R k  k , 2.11 4 Journal of Inequalities and Applications with equality for a 1  a 2  ··· a i ,a k  a k1  ··· a n , a i1  a i2  ··· a k−1  a Q i /Q i R k  i a R k /Q i R k  k . 2.12 Using Corollary 2.5, we can prove the propositions below. Proposition 2.6. Let 0 <a 1 ≤···≤a i ≤···≤a k ≤···≤a n , 2.13 and let p 1 ,p 2 , ,p n be positive real numbers such that p 1  p 2  ··· p n  1.If P  ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 2Q i R k Q i  R k ,Q i ≤ R k , R k ,Q i ≥ R k , 2.14 then p 1 a 1  p 2 a 2  ··· p n a n − a p 1 1 a p 2 2 ···a p n n ≥ P  √ a k − √ a i  2 , 2.15 with equality for a 1  a 2  ··· a n .WhenQ i  R k , equality holds again for a 1  a 2  ··· a i , a i1  ··· a k−1  √ a i a k , a k  a k1  ··· a n . Proposition 2.7. Let 0 <a 1 ≤···≤a i ≤···≤a k ≤···≤a n , 2.16 and let p 1 ,p 2 , ,p n be positive real numbers such that p 1  p 2  ··· p n  1. Then, p 1 a 1  p 2 a 2  ··· p n a n − a p 1 1 a p 2 2 ···a p n n ≥ 3Q i R k  a k − a i  2  4Q i  2R k  a k   2Q i  4R k  a i , 2.17 with equality if and only if a 1  a 2  ··· a n . Remark 2.8. For p 1  p 2  ··· p n  1/n, from Proposition 2.6 we get the inequality a 1  a 2  ··· a n − n n √ a 1 a 2 ···a n ≥ P  √ a k − √ a i  2 , 2.18 where P  ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 2i  n − k  1  n  i − k  1 ,i k ≤ n  1, n − k  1,i k ≥ n  1. 2.19 Journal of Inequalities and Applications 5 Equality in 2.18 holds for a 1  a 2  ···  a n .Ifi  k  n  1, then equality holds again for a 1  a 2  ··· a i , a i1  ··· a k−1  √ a i a k , a k  a k1  ··· a n . Remark 2.9. For p 1  p 2  ··· p n  1/n, from Proposition 2.7, we get the inequality a 1  a 2  ··· a n − n n √ a 1 a 2 ···a n ≥ 3i  n − k  1  a k − a i  2 2  n  2i − k  1  a k  2  2n  i − 2k  2  a i , 2.20 with equality if and only if a 1  a 2  ··· a n . Applying Theorem 2.1 for fx−ln x,weobtain Corollary 2.10. Let 0 <a 1 ≤···≤a i ≤···≤a k ≤···≤a n , 2.21 and let p 1 ,p 2 , ,p n be positive real numbers such that p 1  p 2  ··· p n  1. Then, p 1 a 1  p 2 a 2  ··· p n a n a p 1 1 a p 2 2 ···a p n n ≥  Q i a i  R k a k  /  Q i  R k  Q i R k a Q i i a R k k , 2.22 with equality for a 1  a 2  ··· a i ,a k  a k1  ··· a n , a i1  a i2  ··· a k−1  Q i a i  R k a k Q i  R k . 2.23 Remark 2.11. For p 1  p 2  ··· p n  1/n, from Corollary 2.10, we get the inequality a 1  a 2  ··· a n n n √ a 1 a 2 ···a n ≥ n      ia i   n − k  1  a k  /  n  i − k  1  ni−k1 a i i a n−k1 k , 2.24 with equality for a 1  a 2  ··· a i ,a k  a k1  ··· a n , a i1  a i2  ··· a k−1  ia i   n − k  1  a k n  i − k  1 . 2.25 If i ≤ n/2andk  n − i  1, then 2.24 becomes a 1  a 2  ··· a n n n √ a 1 a 2 ···a n ≥   a i /a n−i1   a n−i1 /a i 2  2i/n , 2.26 6 Journal of Inequalities and Applications with equality for a 1  a 2  ··· a i ,a n−i1  a n−i2  ··· a n , a i1  a i2  ··· a n−i  a i  a n−i1 2 . 2.27 In the case i  1, from 2.26,weget a 1  a 2  ··· a n n n √ a 1 a 2 ···a n ≥   a 1 /a n   a n /a 1 2  2/n , 2.28 with equality for a 2  a 3  ··· a n−1  a 1  a n 2 . 2.29 Applying Theorem 2.1 for fx1/x, we obtain the following. Corollary 2.12. Let 0 <a 1 ≤···≤a i ≤···≤a k ≤···≤a n , 2.30 and let p 1 ,p 2 , ,p n be positive real numbers such that p 1  p 2  ··· p n  1. Then, p 1 a 1  p 2 a 2  ··· p n a n − 1 p 1 a 1  p 2 a 2  ··· p n a n ≥ Q i R k  a k − a i  2 a i a k  Q i a i  R k a k  , 2.31 with equality for a 1  a 2  ··· a i ,a k  a k1  ··· a n , a i1  a i2  ··· a k−1  Q i a i  R k a k Q i  R k . 2.32 Using Corollary 2.12, we can prove the following proposition. Proposition 2.13. Let 0 <a 1 ≤···≤a i ≤···≤a k ≤···≤a n , 2.33 and let p 1 ,p 2 , ,p n be positive real numbers such that p 1  p 2  ··· p n  1.If P  ⎧ ⎪ ⎨ ⎪ ⎩ Q i ,Q i ≤ 3R k , 4Q i R k Q i  R k ,Q i ≥ 3R k , 2.34 Journal of Inequalities and Applications 7 then p 1 a 1  p 2 a 2  ··· p n a n − 1 p 1 a 1  p 2 a 2  ··· p n a n ≥ P  1 √ a i − 1 √ a k  2 , 2.35 with equality for a 1  a 2  ··· a n . 3. Proof of Lemmas Proof of Lemma 2.2. Since c, d ∈ a, b, there exist λ 1 ,λ 2 ∈ 0, 1 such that c  λ 1 a   1 − λ 1  b, d  λ 2 a   1 − λ 2  b. 3.1 In addition, from pa  qb  pc  qd,weget qλ 2   1 − λ 1  p. 3.2 Applying Jensen’s inequality twice, we obtain f  c   f  λ 1 a   1 − λ 1  b  ≤ λ 1 f  a    1 − λ 1  f  b  , f  d   f  λ 2 a   1 − λ 2  b  ≤ λ 2 f  a    1 − λ 2  f  b  , 3.3 and hence pf  c   qf  d  ≤ p  λ 1 f  a    1 − λ 1  f  b    q  λ 2 f  a    1 − λ 2  f  b    pf  a   qf  b  . 3.4 Proof of Lemma 2.3. We need to show that p 1 f  x 1   p 2 f  x 2   ··· p i f  x i  − f  p 1 x 1  p 2 x 2  ··· p n x n  ≥ Q i f  x i  − f  Q i x i  p i1 x i1  ··· p n x n  . 3.5 Using Jensen’s inequality p 1 f  x 1   p 2 f  x 2   ··· p i f  x i  ≥ Q i f  p 1 x 1  p 2 x 2  ··· p i x i Q i  , 3.6 8 Journal of Inequalities and Applications it suffices to prove that Q i f  p 1 x 1  p 2 x 2  ··· p i x i Q i   f  Q i x i  p i1 x i1  ··· p n x n  ≥ Q i f  x i   f  p 1 x 1  p 2 x 2  ··· p n x n  , 3.7 which can be written as Q i f  X i   f  Y i  ≥ Q i f  x i   f  X  , 3.8 where X i  p 1 x 1  p 2 x 2  ··· p i x i Q i , Y i  Q i x i  p i1 x i1  ··· p n x n , X  p 1 x 1  p 2 x 2  ··· p n x n . 3.9 Since x i ,X ∈ X i ,Y i  and Q i X i  Y i  Q i x i  X, 3.10 by Lemma 2.2, the conclusion follows. Proof of Lemma 2.4. We need to prove that p k f  x k   p k1 f  x k1   ··· p n f  x n  − f  p 1 x 1  p 2 x 2  ··· p n x n  ≥ R k f  x k  − f  p 1 x 1  ··· p k−1 x k−1  R k x k  . 3.11 By Jensen’s inequality, we have p k f  x k   p k1 f  x k1   ··· p n f  x n  ≥ R k f  p k x k  p k1 x k1  ··· p n x n R k  . 3.12 Therefore, it suffices to prove that R k f  p k x k  p k1 x k1  ··· p n x n R k   f  p 1 x 1  ··· p k−1 x k−1  R k x k  ≥ R k f  x k   f  p 1 x 1  p 2 x 2  ··· p n x n  , 3.13 Journal of Inequalities and Applications 9 or, equivalently, R k f  X k   f  Y k  ≥ R k f  x k   f  X  , 3.14 where X k  p k x k  p k1 x k1  ··· p n x n R k , Y k  p 1 x 1  ··· p k−1 x k−1  R k x k , X  p 1 x 1  p 2 x 2  ··· p n x n . 3.15 The inequality 3.14 follows from Lemma 2.2,sincex k ,X ∈ Y k ,X k  and R k X k  Y k  R k x k  X. 3.16 4. Proof of Theorem Proof. By Lemmas 2.3 and 2.4, it follows that for fixed x i , x i1 , ,x k , Jensen’s difference Δf, p, x is minimal when x 1  x 2  ···  x i−1  x i and x k1  x k2  ···  x n  x k ; that is, Δ  f, p, x  ≥ Q i f  x i   p i1 f  x i1   ··· p k−1 f  x k−1   R k f  x k  − f  Q i x i  p i1 x i1  ··· p k−1 x k−1  R k x k  . 4.1 Therefore, towards proving 2.3, we only need to show that p i1 f  x i1   ··· p k−1 f  x k−1    Q i  R k  f  Q i x i  R k x k Q i  R k  ≥ f  Q i x i  p i1 x i1  ··· p k−1 x k−1  R k x k  . 4.2 Since Q i  p i1  ··· p k−1  R k  1, 4.3 this inequality is a consequence of Jensen’s inequality. Thus, the proof is completed. 10 Journal of Inequalities and Applications 5. Proof of Propositions Proof of Proposition 2.6. Using Corollary 2.5, we need to prove that Q i a i  R k a k −  Q i  R k  a Q i /Q i R k  i a R k /Q i R k  k ≥ P  √ a k − √ a i  2 . 5.1 Since this inequality is homogeneous in a i and a k ,andalsoinQ i and R k , without loss of generality, assume that a i  1andQ i  1. Using the notations a k  x 2 and R k  p, where x ≥ 1andp>0, the inequality is equivalent to gx ≥ 0, where g  x   1  px 2 −  1  p  x 2p/1p − P  x − 1  2 , 5.2 with P  ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 2p p  1 ,p≥ 1, p, p ≤ 1. 5.3 We have g   x   2p  x − x p−1/p1  − 2P  x − 1  , g   x   2  p − P  − 2p  p − 1  p  1 x −2/p1 . 5.4 If p ≥ 1, then g   x   2p  p − 1  p  1  1 − x −2/p1  ≥ 0, 5.5 and if p ≤ 1, then g   x   2p  1 − p  p  1 x −2/p1 ≥ 0. 5.6 Since g  x ≥ 0forx ≥ 1, and g  x is increasing, g  x ≥ g10, gx is increasing, and hence gx ≥ g10forx ≥ 1. This concludes the proof. Proof of Proposition 2.7. Using Corollary 2.5, we need to prove that Q i a i  R k a k −  Q i  R k  a Q i /Q i R k  i a R k /Q i R k  k ≥ 3Q i R k  a k − a i  2  4Q i  2R k  a k   2Q i  4R k  a i . 5.7 [...]... Classical and New Inequalities in Analysis, vol 61 of c c c Mathematics and Its Applications, Kluwer Academic Publishers, Dordrecht, The Netherlands, 1993 2 S S Dragomir, J Peˇ ari´ , and L E Persson, “Properties of some functionals related to Jensen’s c c inequality, ” Acta Mathematica Hungarica, vol 70, no 1-2, pp 129–143, 1996 ... 5.12 Also, for Qi ≥ 3Rk , we get Qi Rk √ ai √ ak 2 − P Qi ai Rk ak √ Qi Rk Rk − 3Qi ai Qi Rk ≥ Qi − 3Rk ak 2 Qi Rk Qi Rk Rk − 3Qi ai Qi Rk Qi − 3Rk ai 2 Qi Rk ai The proposition is proved ai ak 0 5.13 12 Journal of Inequalities and Applications References 1 D S Mitrinovi´ , J E Peˇ ari´ , and A M Fink, Classical and New Inequalities in Analysis, vol 61 of c c c Mathematics and Its Applications, Kluwer...Journal of Inequalities and Applications 11 Since this inequality is homogeneous in ai and ak , and also in Qi and Rk , we may set ai 1 and Qi 1 Using the notations ak x and Rk p, where x ≥ 1 and p > 0, the inequality is equivalent to g x ≥ 0, where g x 4 2p x 4p 1 2 px − 1 p xp/ 1 − 3p x − 1 2 p 5.8 We have 1 2 1 2p p x... 0, and g x is strictly Since g x ≥ 0 for x ≥ 1, g x is strictly increasing, g x ≥ g 1 0, g x is strictly increasing, and hence g x ≥ g 1 0 for increasing, g x ≥ g 1 x ≥ 1 Proof of Proposition 2.13 Using Corollary 2.12, we need to prove that Qi Rk ak − ai 2 1 1 ≥P √ −√ ai ak Qi ai Rk ak ai ak 2 5.10 This inequality is true if Qi Rk √ ai √ ak 2 ≥ P Qi ai Rk ak 5.11 For Qi ≤ 3Rk , we have Qi Rk √ ai √ . give the best lower bound for the weighted Jensen’s discrete inequality with ordered variables applied to a convex function f, in the case when the lower bound depends on f, weights, and two given. Corporation Journal of Inequalities and Applications Volume 2010, Article ID 128258, 12 pages doi:10.1155/2010/128258 Research Article The Best Lower Bound Depended on Two Fixed Variables for Jensen’s. weights, and two given variables. Furthermore, under the same conditions, we give some sharp lower bounds for the weighted AM-GM inequality and AM-HM inequality. 1. Introduction Let x  {x 1 ,x 2 ,