Journal of Inequalities in Pure andApplied Mathematics

http://jipam.vu.edu.au/

Volume 7, Issue 4, Article 148, 2006

NEW INEQUALITIES ABOUT CONVEX FUNCTIONS

LAZHAR BOUGOFFA

AL-IMAM MUHAMMAD IBN SAUD ISLAMIC UNIVERSITY

FACULTY OF COMPUTERSCIENCE

DEPARTMENT OFMATHEMATICS

P.O. BOX 84880, RIYADH 11681SAUDI ARABIA .

bougoffa@hotmail.com

Received 11 June, 2006; accepted 15 October, 2006Communicated by B. Yang

ABSTRACT. If f is a convex function andx1, . . . , xn or a1, . . . , an lie in its domain the follow-ing inequalities are proved

n∑i=1

f(xi)− f

(x1 + · · ·+ xn

n

)≥ n− 1

n

[f

(x1 + x2

2

)+ · · ·+ f

(xn−1 + xn

2

)+ f

(xn + x1

2

)]and

(n− 1) [f(b1) + · · ·+ f(bn)] ≤ n [f(a1) + · · ·+ f(an)− f(a)] ,wherea = a1+···+an

n andbi = na−ai

n−1 , i = 1, . . . , n.

Key words and phrases:Jensen’s inequality, Convex functions.

2000Mathematics Subject Classification.26D15.

1. M AIN THEOREMS

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

Theorem 1.1. Let f be a convex function on an intervalI and letw1, . . . , wn be nonnegativereal numbers whose sum is1. Then for allx1, . . . , xn ∈ 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).

ISSN (electronic): 1443-5756

c© 2006 Victoria University. All rights reserved.

166-06

2 LAZHAR BOUGOFFA

The aim of the present note is to establish new inequalities similar to the following knowninequalities:(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∑i=1

f(xi) +n

n− 2f

(x1 + · · ·+ xn

n

)≥ 2

n− 2

∑i<j

f

(xi + xj

2

),

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

(n− 1) [f(b1) + · · ·+ f(bn)] ≤ f(a1) + · · ·+ f(an) + n(n− 2)f(a),

wherea = a1+···+an

nandbi = na−ai

n−1, i = 1, . . . , n, anda1, . . . , an ∈ I.

Our main results are given in the following theorems:

Theorem 1.2. If f is a convex function andx1, x2, . . . , xn lie in its domain, then

(1.3)n∑

i=1

f(xi)− f

(x1 + · · ·+ xn

n

)≥ n− 1

n

[f

(x1 + x2

2

)+ · · ·+ f

(xn−1 + xn

2

)+ f

(xn + x1

2

)].

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

(1.4) f

(x1 + x2

2

)+ · · ·+ f

(xn−1 + xn

2

)+ f

(xn + x1

2

)≤ f(x1)+ f(x2)+ · · ·+ f(xn).

In the summation on the right side of (1.4), the expression∑n

i=1 f(xi) can be written as

n∑i=1

f(xi) =n

n− 1

n∑i=1

f(xi)−1

n− 1

n∑i=1

f(xi),

n∑i=1

f(xi) =n

n− 1

[n∑

i=1

f(xi)−n∑

i=1

1

nf(xi)

].

Replacing∑n

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

f

(x1 + x2

2

)+ · · ·+ f

(xn−1 + xn

2

)+ f

(xn + x1

2

)≤ n

n− 1

[n∑

i=1

f(xi)−n∑

i=1

1

nf(xi)

].

J. Inequal. Pure and Appl. Math., 7(4) Art. 148, 2006 http://jipam.vu.edu.au/

NEW INEQUALITIES ABOUT CONVEX FUNCTIONS 3

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

f

(x1 + x2

2

)+ · · ·+ f

(xn−1 + xn

2

)+ f

(xn + x1

2

)≤ n

n− 1

[n∑

i=1

f(xi)− f

(∑ni=1 xi

n

)],

and this concludes the proof. �

Remark 1.3. Now we consider the simplest case of Theorem 1.2 forn = 3 to obtain thefollowing 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.

Theorem 1.4. If f is a convex function anda1, . . . , an lie in its domain, then

(1.5) (n− 1) [f(b1) + · · ·+ f(bn)] ≤ n [f(a1) + · · ·+ f(an)− f(a)] ,

wherea = a1+···+an

nandbi = na−ai

n−1, i = 1, . . . , n.

Proof. By using the Jensen inequality (1.1),

f(b1) + · · ·+ f(bn) ≤ f(a1) + · · ·+ f(an),

and so,

f(b1) + · · ·+ f(bn) ≤ n

n− 1[f(a1) + · · ·+ f(an)]− 1

n− 1[f(a1) + · · ·+ f(an)] ,

or

f(b1) + · · ·+ f(bn) ≤ n

n− 1[f(a1) + · · ·+ f(an)]− n

n− 1

[1

nf(a1) + · · ·+ 1

nf(an)

],

and so

(1.6) f(b1) + · · ·+ f(bn) ≤ n

n− 1

[f(a1) + · · ·+ f(an)−

(1

nf(a1) + · · ·+ 1

nf(an)

)].

Hence, applying Jensen’s inequality (1.1) to the right hand side of (1.6) we get

f(b1) + · · ·+ f(bn) ≤ n

n− 1

[f(a1) + · · ·+ f(an)− f

(a1 + · · ·+ an

n

)],

and this concludes the proof. �

REFERENCES

[1] D.S. MITRINOVIC, J.E. PECARIC AND A.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 Iasi. I-a, Mat.(N.S), 1965.

J. Inequal. Pure and Appl. Math., 7(4) Art. 148, 2006 http://jipam.vu.edu.au/