Journal of Inequalities in Pure and Applied Mathematics http://jipam.vu.edu.au/ Volume 4, Issue 4, Article 73, 2003 A VARIANT OF JENSEN’S INEQUALITY A.McD. MERCER D EPARTMENT OF M ATHEMATICS AND S TATISTICS , U NIVERSITY OF G UELPH , G UELPH , O NTARIO N1G 2W1, C ANADA . amercer@reach.net Received 14 August, 2003; accepted 27 August, 2003 Communicated by P.S. Bullen A BSTRACT. If f is a convex function the following variant of the classical Jensen’s Inequality is proved X X f x1 + xn − wk kk ≤ f (x1 ) + f (xn ) − wk f (xk ). Key words and phrases: Jensen’s inequality, Convex functions. 2000 Mathematics Subject Classification. 26D15. 1. M AIN T HEOREM Let 0 < x1 ≤ x2 ≤ · · · ≤ xn and let wk (1 ≤ k ≤ n) be positive weights associated with these xk and whose sum is unity. Then Jensen’s inequality [2] reads : Theorem 1.1. If f is a convex function on an interval containing the xk then X X (1.1) f wk xk ≤ wk f (xk ). P P Note: Here and, in all that follows, means n1 . Our purpose in this note is to prove the following variant of (1.1). Theorem 1.2. If f is a convex function on an interval containing the xk then X X f x1 + xn − wk xk ≤ f (x1 ) + f (xn ) − wk f (xk ). Towards proving this theorem we shall need the following lemma: Lemma 1.3. For f convex we have: (1.2) f (x1 + xn − xk ) ≤ f (x1 ) + f (xn ) − f (xk ), ISSN (electronic): 1443-5756 c 2003 Victoria University. All rights reserved. 116-03 (1 ≤ k ≤ n). 2 A.M C D. M ERCER 2. T HE P ROOFS Proof of Lemma 1.3. Write yk = x1 + xn − xk . Then x1 + xn = xk + yk so that the pairs x1 , xn and xk , yk possess the same mid-point. Since that is the case there exists λ such that xk = λx1 + (1 − λ)xn , yk = (1 − λ)x1 + λxn , where 0 ≤ λ ≤ 1 and 1 ≤ k ≤ n. Hence, applying (1.1) twice we get f (yk ) ≤ (1 − λ)f (x1 ) + λf (xn ) = f (x1 ) + f (xn ) − [λf (x1 ) + (1 − λ)f (xn )] ≤ f (x1 ) + f (xn ) − f (λx1 + (1 − λ)xn ) = f (x1 ) + f (xn ) − f (xk ) and since yk = x1 + xn − xk this concludes the proof of the lemma. Proof of Theorem 1.2. We have X X f (x1 + xn − wk xk ) = f wk (x1 + xn − xk ) X ≤ wk f (x1 + xn − xk ) by (1.1) X ≤ wk [f (x1 ) + f (xn ) − f (xk )] by (1.2) X = f (x1 ) + f (xn ) − wf (xk ) and this concludes the proof. 3. T WO E XAMPLES e = x1 + xn − A and G e = x1 xn , where A and G denote the usual arithmetic and Let us write A G geometric means of the xk . (a) Then taking f (x) as the convex function − log x, Theorem 1.2 gives: e≥G e A (b) Taking f (x) as the function log 1−x which is convex if 0 < x ≤ 12 , Theorem 1.2 gives x e A(x) e − x) A(1 ≥ e G(x) e − x) G(1 provided that xk ∈ (0, 21 ] for all k. The example (a) is a special case of a family of inequalities found by a different method in [1]. The example (b) is, of course, an analogue of Ky-Fan’s Inequality [2]. R EFERENCES [1] A.McD. MERCER, A monotonicity property of power means, J. Ineq. Pure and Appl. Math., 3(3) (2002), Article 40. [ONLINE: http://jipam.vu.edu.au/v3n3/014_02.html]. [2] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht, 1993. J. Inequal. Pure and Appl. Math., 4(4) Art. 73, 2003 http://jipam.vu.edu.au/