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volume 7, issue 4, article 148, 2006.

Received 11 June, 2006;

accepted 15 October, 2006.

Communicated by:B. Yang

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Journal of Inequalities in Pure and Applied Mathematics

NEW INEQUALITIES ABOUT CONVEX FUNCTIONS

LAZHAR BOUGOFFA

Al-imam Muhammad Ibn Saud Islamic University Faculty of Computer Science

Department of Mathematics P.O. Box 84880, Riyadh 11681 Saudi Arabia.

EMail:bougoffa@hotmail.com

c

2000Victoria University ISSN (electronic): 1443-5756 166-06

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New Inequalities About Convex Functions

Lazhar Bougoffa

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J. Ineq. Pure and Appl. Math. 7(4) Art. 148, 2006

http://jipam.vu.edu.au

Abstract

Iffis a convex function andx₁, . . . , x_nora₁, . . . , a_nlie in its domain the following inequalities are proved

n

X

i=1

f(x_i)−f

x₁+· · ·+x_n n

≥n−1 n

f

x₁+x₂ 2

+· · ·+f

xn−1+x_n 2

+f

x_n+x₁ 2

and

(n−1) [f(b₁) +· · ·+f(b_n)]≤n[f(a₁) +· · ·+f(a_n)−f(a)], wherea=^a¹^+···+a_n ⁿandb_i=^na−a_n−1ⁱ, i= 1, . . . , n.

2000 Mathematics Subject Classification:26D15.

Key words: Jensen’s inequality, Convex functions.

1. Main Theorems

The well-known Jensen’s inequality is given as follows [1]:

Theorem 1.1. Letf be a convex function on an interval I and letw₁, . . . , w_n be nonnegative real numbers whose sum is1. Then for allx₁, . . . , x_n∈I, (1.1) w1f(x1) +· · ·+wnf(xn)≥f(w1x1+· · ·+wnxn).

Recall that a functionf is said to be convex if for anyt∈[0,1]and anyx, y in the domain off,

(1.2) tf(x) + (1−t)f(y)≥f(tx+ (1−t)y).

The aim of the present note is to establish new inequalities similar to the following known inequalities:

(Via Titu Andreescu (see [2, p. 6])) f(x₁) +f(x₂) +f(x₃) +f

x₁+x₂+x₃ 3

≥ 4 3

f

x₁ +x₂ 2

+f

x₂+x₃ 2

+f

x₃+x₁ 2

, wheref is a convex function andx1, x2, x3 lie in its domain,

(Popoviciu inequality [3])

n

X

i=1

f(x_i) + n n−2f

x₁ +· · ·+x_n n

≥ 2

n−2 X

i<j

f

x_i+x_j 2

,

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wheref is a convex function onI andx₁, . . . , x_n∈I,and (Generalized Popoviciu inequality)

(n−1) [f(b1) +· · ·+f(bn)]≤f(a1) +· · ·+f(an) +n(n−2)f(a), wherea= ^a¹^+···+a_n ⁿ andbi = ^na−a_n−1ⁱ, i = 1, . . . , n,anda1, . . . , an ∈I.

Our main results are given in the following theorems:

Theorem 1.2. Iffis a convex function andx₁, x₂, . . . , x_nlie in its domain, then

(1.3)

n

X

i=1

f(x_i)−f

x₁+· · ·+x_n n

≥ n−1 n

f

x₁+x₂ 2

+· · ·+f

x_n−1+x_n 2

+f

x_n+x₁ 2

.

Proof. Using (1.2) witht= ¹₂, we obtain (1.4) f

x₁+x₂ 2

+· · ·+f

xn−1+x_n 2

+f

x_n+x₁ 2

≤f(x₁) +f(x₂) +· · ·+f(x_n).

In the summation on the right side of (1.4), the expressionPn

i=1f(x_i)can be written as

n

X

i=1

f(x_i) = n n−1

n

X

i=1

f(x_i)− 1 n−1

n

X

i=1

f(x_i),

n

X

i=1

f(x_i) = n n−1

" _n X

i=1

f(x_i)−

n

X

i=1

1 nf(x_i)

# .

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ReplacingPn

i=1f(x_i)with the equivalent expression in (1.4), f

x₁+x₂ 2

+· · ·+f

xn−1+x_n 2

+f

x_n+x₁ 2

≤ n

n−1

" _n X

i=1

f(xi)−

n

X

i=1

1 nf(xi)

# . Hence, applying Jensen’s inequality (1.1) to the right hand side of the above resulting inequality we get

f

x₁+x₂ 2

+· · ·+f

xn−1+x_n 2

+f

x_n+x₁ 2

≤ n

n−1

" _n X

i=1

f(x_i)−f Pn

i=1xi

n #

,

and this concludes the proof.

Remark 1. Now we consider the simplest case of Theorem 1.2 for n = 3 to obtain the following variant of via Titu Andreescu [2]:

f(x₁) +f(x₂) +f(x₃)−f

x₁+x₂+x₃ 3

≥ 2 3

f

x₁ +x₂ 2

+f

x₂+x₃ 2

+f

x₃+x₁ 2

. The variant of the generalized Popovicui inequality is given in the following theorem.

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Theorem 1.3. Iff is a convex function anda₁, . . . , a_nlie in its domain, then (1.5) (n−1) [f(b₁) +· · ·+f(b_n)]≤n[f(a₁) +· · ·+f(a_n)−f(a)], wherea= ^a¹^+···+a_n ⁿ andb_i = ^na−a_n−1ⁱ, i= 1, . . . , n.

Proof. By using the Jensen inequality (1.1),

f(b₁) +· · ·+f(b_n)≤f(a₁) +· · ·+f(a_n),

and so,

f(b₁) +· · ·+f(b_n)

≤ n

n−1[f(a₁) +· · ·+f(a_n)]− 1

n−1[f(a₁) +· · ·+f(a_n)], or

f(b₁) +· · ·+f(b_n)

≤ n

n−1[f(a₁) +· · ·+f(a_n)]− n n−1

1

nf(a₁) +· · ·+ 1 nf(a_n)

,

and so

(1.6) f(b₁) +· · ·+f(b_n)

≤ n

n−1

f(a₁) +· · ·+f(a_n)− 1

nf(a₁) +· · ·+ 1 nf(a_n)

.

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Hence, applying Jensen’s inequality (1.1) to the right hand side of (1.6) we get f(b₁) +· · ·+f(b_n)≤ n

n−1

f(a₁) +· · ·+f(a_n)−f

a₁+· · ·+a_n n

, and this concludes the proof.

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References

[1] D.S. MITRINOVI ´C, J.E. PE ˘CARI ´C ANDA.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht, 1993.

[2] KIRAN KEDLAYA, A<B (A is less than B), based on notes for the Math Olympiad Program (MOP) Version 1.0, last revised August 2, 1999.

[3] T. POPOVICIU, Sur certaines inégalitées qui caractérisent les fonctions convexes, An. Sti. Univ. Al. I. Cuza Ia¸si. I-a, Mat. (N.S), 1965.