A refinement of Jensen’s discrete inequality and applications for the celebrated Arithmetic Mean – Geometric Mean – Harmonc Mean inequality and Cauchy-Schwartz-Bunikowski inequality are pointed out

Dragomir, Sever S

Scarmozzino, F. P

English

Victoria University Institutional Repository (VUIR)

A Refinement of Jensen's Discrete Inequality for Differentiable Convex Functions

Victoria University Eprints Repository

A REFINEMENT OF JENSEN’S DISCRETE INEQUALITY FORDIFFERENTIABLE CONVEX FUNCTIONSS.S. DRAGOMIR AND F.P. SCARMOZZINOAbstract. A refinement of Jensen’s discrete inequality and applications forthe celebrated Arithmetic Mean – Geometric Mean – Harmonc Mean inequalityand Cauchy-Schwartz-Bunikowski inequality are pointed out.1. IntroductionThe following inequality is well known in literature as Jensen’s inequality:(1.1) f(1Pnn∑i=1pixi)≤ 1Pnn∑i=1pif (xi) ,provided f : [a, b] → R is a convex function on [a, b] , xi ∈ [a, b] , and pi ≥ 0 withPn :=∑ni=1 pi > 0.Its central role in Analytic Inequality Theory is determined by the fact thatmany other fundamental results such as: the Arithmetic Mean – Geometric Mean– Harmonic Mean inequality, or the Ho¨lder and Minkowski inequalities, or even theKy Fan inequality may be obtained from Jensen’s inequality by appropriate choicesof the function f.There is an extensive literature devoted to Jensen’s inequality concerning differ-ent generalizations, refinements, counterparts and converse results, see, for example[1] – [21].The main aim of this paper is to point out a new refinement of this classicalresult. Two applications in connection with the celebrated A − G − H−meansinequality and the Cauchy-Buniakowski-Schwartz inequality are mentioned as well.2. A Refinement of Jensen’s InequalityThe following refinement of Jensen’s inequality holds.Date: May 24, 2002.1991 Mathematics Subject Classification. Primary 26D15; Secondary 26D10.Key words and phrases. Jensen’s Discrete Inequality, A-G-H means inequality, Cauchy-Schwartz-Bunikowski inequality.12 S.S. DRAGOMIR AND F.P. SCARMOZZINOTheorem 1. Let f : [a, b] → R be a differentiable convex function on (a, b) andxi ∈ (a, b) , pi ≥ 0 with Pn :=∑ni=1 pi > 0. Then one has the inequality(2.1)1Pnn∑i=1pif (xi)− f(1Pnn∑i=1pixi)≥∣∣∣∣∣∣ 1Pnn∑i=1pi∣∣∣∣∣∣f (xi)− f 1Pnn∑j=1pjxj∣∣∣∣∣∣−∣∣∣∣∣∣f ′ 1Pnn∑j=1pjxj∣∣∣∣∣∣ · 1Pnn∑i=1pi∣∣∣∣∣∣xi − 1Pnn∑j=1pjxj∣∣∣∣∣∣∣∣∣∣∣∣ ≥ 0.Proof. Since f is differentiable convex on (a, b) , then for each x, y ∈ (a, b), one hasthe inequality(2.2) f (x)− f (y) ≥ (x− y) f ′ (y) .Using the properties of the modulus, we havef (x)− f (y)− (x− y) f ′ (y) = |f (x)− f (y)− (x− y) f ′ (y)|(2.3)≥ ||f (x)− f (y)| − |x− y| |f ′ (y)||for each x, y ∈ (a, b) .If we choose y = 1Pn∑nj=1 pjxj and x = xi, i ∈ {1, . . . , n} , then we have(2.4) f (xi)− f 1Pnn∑j=1pjxj−xi − 1Pnn∑j=1pjxj f ′ 1Pnn∑j=1pjxj≥∣∣∣∣∣∣∣∣∣∣∣∣f (xi)− f 1Pnn∑j=1pjxj∣∣∣∣∣∣−∣∣∣∣∣∣xi − 1Pnn∑j=1pjxj∣∣∣∣∣∣∣∣∣∣∣∣f ′ 1Pnn∑j=1pjxj∣∣∣∣∣∣∣∣∣∣∣∣for any i ∈ {1, . . . , n} .If we multiply (2.4) by pi ≥ 0, sum over i from 1 to n, and divide by Pn > 0, wededuce1Pnn∑i=1pif (xi)− f 1Pnn∑j=1pjxj− 1Pnn∑i=1pixi − 1Pnn∑j=1pjxj f ′ 1Pnn∑j=1pjxjJENSEN’S DISCRETE INEQUALITY 3≥ 1Pnn∑i=1pi∣∣∣∣∣∣∣∣∣∣∣∣f (xi)− f 1Pnn∑j=1pjxj∣∣∣∣∣∣−∣∣∣∣∣∣xi − 1Pnn∑j=1pjxj∣∣∣∣∣∣∣∣∣∣∣∣f ′ 1Pnn∑j=1pjxj∣∣∣∣∣∣∣∣∣∣∣∣≥∣∣∣∣∣∣ 1Pnn∑i=1pi∣∣∣∣∣∣f (xi)− f 1Pnn∑j=1pjxj∣∣∣∣∣∣− 1Pnn∑i=1pi∣∣∣∣∣∣xi − 1Pnn∑j=1pjxj∣∣∣∣∣∣ ·∣∣∣∣∣∣f ′ 1Pnn∑j=1pjxj∣∣∣∣∣∣∣∣∣∣∣∣ .Since1Pnn∑i=1pixi − 1Pnn∑j=1pjxj = 0,the inequality (2.1) is proved.In particular, we have the following result for unweighted means.Corollary 1. With the above assumptions for f and xi, one has the inequality(2.5)f (x1) + · · ·+ f (xn)n− f(x1 + · · ·+ xnn)≥∣∣∣∣∣ 1nn∑i=1∣∣∣∣xi − f (x1 + · · ·+ xnn)∣∣∣∣−∣∣∣∣f ′(x1 + · · ·+ xnn)∣∣∣∣ · 1nn∑i=1∣∣∣∣∣∣xi − 1nn∑j=1xj∣∣∣∣∣∣∣∣∣∣∣∣ ≥ 0.Remark 1. Similar integral inequalities may be stated as well. We omit the details.3. A Refinement of A.−G.−H. InequalityFor a positive n-tuple x¯ = (x1, . . . , xn) and p¯ = (p1, . . . , pn) with pi ≥ 0 and∑ni=1 pi =: Pn > 0, defineAn (p¯, x¯) :=1Pnn∑i=1pixi (the weighted arithmetic mean),Gn (p¯, x¯) :=(n∏i=1xpii) 1Pn(the weighted geometric mean),Hn (p¯, x¯) :=Pn∑ni=1pixi=[An(p¯,1x¯)]−1(the weighted harmonic mean).The following inequality(3.1) An (p¯, x¯) ≥ Gn (p¯, x¯) ≥ Hn (p¯, x¯)is well known in the literature as the Arithmetic Mean – Geometric Mean – Har-monic Mean (A−G−H)-means inequality.Using Theorem 1, we may improve this result as follows.4 S.S. DRAGOMIR AND F.P. SCARMOZZINOProposition 1. Suppose that x¯, p¯ are as above. Then we have the inequality(3.2)An (p¯, x¯)Gn (p¯, x¯)≥ exp[∣∣∣∣An(p¯, ∣∣∣∣ln( x¯An (p¯, x¯))∣∣∣∣)−An(p¯, ∣∣∣∣x−An (p¯, x¯)An (p¯, x¯)∣∣∣∣)∣∣∣∣] ≥ 1,where for a function h, we denote h (x¯) := (h (x1) , . . . , h (xn)) .Proof. Applying the inequality (2.1) for f (x) = − lnx, we getln[An (p¯, x¯)Gn (p¯, x¯)]≥∣∣∣∣∣ 1Pnn∑i=1pi∣∣∣∣ln( xiAn (p, x))∣∣∣∣−A−1n (p, x) · 1Pnn∑i=1pi |xi −An (p, x)|∣∣∣∣∣ ≥ 0,from where we get the desired inequality (3.2).The following proposition also holds.Proposition 2. Suppose that x¯, p¯ are as above. Then we have the inequality:Gn (p¯, x¯)Hn (p¯, x¯)(3.3)≥ exp[∣∣∣∣An(p¯, ∣∣∣∣ln(Hn (p¯, x¯)x¯)∣∣∣∣)−An(p¯, ∣∣∣∣Hn (p¯, x¯)− x¯x¯∣∣∣∣)∣∣∣∣] ≥ 1,Proof. Follows by Proposition 1 on choosing 1x¯ instead of x¯.4. A Refinement of Cauchy-Buniakowski-Schwartz’s InequalityThe following inequality is well known in the literature as the Cauchy-Buniakowski-Schwartz inequality:(4.1)n∑i=1a2in∑i=1b2i ≥(n∑i=1aibi)2,for any ai, bi ∈ R (i ∈ {1, . . . , n}) .The following refinement of (4.1) holds.Proposition 3. If ai, bi ∈ R, i ∈ {1, . . . , n} , then one has the inequality;(4.2)n∑i=1a2in∑i=1b2i −(n∑i=1aibi)2≥ 1∑ni=1 b2i∣∣∣∣∣n∑i=1∣∣∣∣∣∣∣∣∣∣ a2i b2i(∑nj=1 ajbj)2 (∑nj=1 b2j)2 ∣∣∣∣∣∣∣∣∣∣−2∣∣∣∣∣n∑k=1akbk∣∣∣∣∣ ·n∑i=1|bi|∣∣∣∣∣∣n∑j=1bj∣∣∣∣ ai biaj bj∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ ≥ 0.JENSEN’S DISCRETE INEQUALITY 5Proof. If we apply Theorem 1 for f (x) = x2, we get(4.3)1Pnn∑i=1pix2i −(1Pnn∑i=1pixi)2≥∣∣∣∣∣∣∣1Pnn∑i=1pi∣∣∣∣∣∣∣x2i − 1Pnn∑j=1pjxj2∣∣∣∣∣∣∣− 2∣∣∣∣∣ 1Pnn∑k=1pkxk∣∣∣∣∣ · 1Pnn∑i=1pi∣∣∣∣∣∣xi − 1Pnn∑j=1pjxj∣∣∣∣∣∣∣∣∣∣∣∣ ≥ 0.If in (4.3), we choose pi = b2i , xi =aibi, i ∈ {1, . . . , n} , we get(4.4)∑ni=1 a2i∑ni=1 b2i− (∑ni=1 aibi)2(∑ni=1 b2i )2 ≥∣∣∣∣∣∣ 1∑ni=1 b2in∑i=1b2i ·∣∣∣∣∣∣a2ib2i−(∑nj=1 ajbj∑nj=1 b2j)2∣∣∣∣∣∣−2∣∣∣∣∑nk=1 akbk∑ni=1 b2i∣∣∣∣ ·∑ni=1 b2i∣∣∣aibi − ∑nj=1 ajbj/∑nj=1 b2j ∣∣∣∑ni=1 b2i∣∣∣∣∣∣ ,which is clearly equivalent to (4.2).References[1] E. F. BECKENBACH, An inequality of Jensen, Amer. Math. Monthly, 53 (1946). 501–505.[2] Z. CIESIELSKI, A note on some inequalities of Jensen’s type, Ann. Polon. Math., 4 (1958)269-274.[3] I. CSISZA´R, A note on Jensen’s inequality, Studia Sci. Math. Hungar., 1 (1966) 185–188.[4] R.P. BOAS Jr., The Jensen-Steffensen inequality, Univ. Beograd. Publ. Elektrotehn. Fak. Ser.Mat. Fiz., No. 302-319 (1970), 1–8.[5] P. LAH and M. 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A Refinement of Jensen's Discrete Inequality for Differentiable Convex Functions

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