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
DEPARTMENT OFMATHEMATICS ANDSTATISTICS, UNIVERSITY OFGUELPH,
GUELPH, ONTARION1G 2W1, CANADA.
Received 14 August, 2003; accepted 27 August, 2003 Communicated by P.S. Bullen
ABSTRACT. Iff is a convex function the following variant of the classical Jensen’s Inequality is proved
f
x1+xn−X wkkk
≤f(x1) +f(xn)−X
wkf(xk).
Key words and phrases: Jensen’s inequality, Convex functions.
2000 Mathematics Subject Classification. 26D15.
1. MAIN THEOREM
Let0 < x1 ≤ x2 ≤ · · · ≤ xn and letwk (1 ≤ k ≤ n) be positive weights associated with thesexkand whose sum is unity. Then Jensen’s inequality [2] reads :
Theorem 1.1. Iff is a convex function on an interval containing thexkthen
(1.1) fX
wkxk
≤X
wkf(xk).
Note: Here and, in all that follows,P
meansPn 1 .
Our purpose in this note is to prove the following variant of (1.1).
Theorem 1.2. Iff is a convex function on an interval containing thexkthen
f
x1+xn−X
wkxk
≤f(x1) +f(xn)−X
wkf(xk).
Towards proving this theorem we shall need the following lemma:
Lemma 1.3. Forf convex we have:
(1.2) f(x1+xn−xk)≤f(x1) +f(xn)−f(xk), (1≤k ≤n).
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
116-03
2 A.MCD. MERCER
2. THEPROOFS
Proof of Lemma 1.3. Writeyk =x1+xn−xk. Thenx1+xn=xk+ykso that the pairs x1, xn
and xk, ykpossess the same mid-point. Since that is the case there existsλsuch that xk =λx1+ (1−λ)xn,
yk = (1−λ)x1+λxn, where0≤λ≤1and1≤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 sinceyk =x1+xn−xkthis concludes the proof of the lemma.
Proof of Theorem 1.2. We have f(x1+xn−X
wkxk) =fX
wk(x1+xn−xk)
≤X
wkf(x1+xn−xk) by (1.1)
≤X
wk[f(x1) +f(xn)−f(xk)] by (1.2)
=f(x1) +f(xn)−X
wf(xk)
and this concludes the proof.
3. TWOEXAMPLES
Let us writeAe=x1+xn−AandGe = x1Gxn,whereAandGdenote the usual arithmetic and geometric means of thexk.
(a) Then takingf(x)as the convex function−logx, Theorem 1.2 gives:
Ae≥Ge
(b) Takingf(x)as the functionlog1−xx which is convex if0< x≤ 12,Theorem 1.2 gives
A(x)e
A(1e −x) ≥ G(x)e G(1e −x) provided thatxk∈(0,12]for allk.
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].
REFERENCES
[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. MITRINOVI ´C, J.E. PE ˇCARI ´C ANDA.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/
