Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 4, Article 148, 2006
NEW INEQUALITIES ABOUT CONVEX FUNCTIONS
LAZHAR BOUGOFFA
A L - IMAM M UHAMMAD I BN S AUD I SLAMIC U NIVERSITY
FACULTY OF C OMPUTER S CIENCE
D EPARTMENT OF M ATHEMATICS
P.O. B OX 84880, R IYADH 11681
S AUDI A RABIA .
bougoffa@hotmail.com
Received 11 June, 2006; accepted 15 October, 2006
Communicated by B. Yang
A BSTRACT. If f is a convex function and x1 , . . . , xn or a1 , . . . , an lie in its domain the following inequalities are proved
n
X
x1 + · · · + xn
f (xi ) − f
n
i=1
n−1
x1 + x2
xn−1 + xn
xn + x1
≥
f
+ ··· + f
+f
n
2
2
2
and
(n − 1) [f (b1 ) + · · · + f (bn )] ≤ n [f (a1 ) + · · · + f (an ) − f (a)] ,
a1 +···+an
i
and bi = na−a
where a =
n
n−1 , i = 1, . . . , n.
Key words and phrases: Jensen’s inequality, Convex functions.
2000 Mathematics Subject Classification. 26D15.
1. M AIN T HEOREMS
The well-known Jensen’s inequality is given as follows [1]:
Theorem 1.1. Let f be a convex function on an interval I and let w1 , . . . , wn be nonnegative
real numbers whose sum is 1. Then for all x1 , . . . , xn ∈ I,
(1.1)
w1 f (x1 ) + · · · + wn f (xn ) ≥ f (w1 x1 + · · · + wn xn ).
Recall that a function f is said to be convex if for any t ∈ [0, 1] and any x, y in the domain of
f,
(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
L AZHAR B OUGOFFA
The aim of the present note is to establish new inequalities similar to the following known
inequalities:
(Via Titu Andreescu (see [2, p. 6]))
x1 + x2 + x3
f (x1 ) + f (x2 ) + f (x3 ) + f
3
4
x1 + x2
x2 + x3
x3 + x1
≥
f
+f
+f
,
3
2
2
2
where f is a convex function and x1 , x2 , x3 lie in its domain,
(Popoviciu inequality [3])
n
X
x1 + · · · + xn
2 X
xi + xj
n
f
≥
f
,
f (xi ) +
n−2
n
n − 2 i<j
2
i=1
where f is a convex function on I and x1 , . . . , xn ∈ I, and
(Generalized Popoviciu inequality)
(n − 1) [f (b1 ) + · · · + f (bn )] ≤ f (a1 ) + · · · + f (an ) + n(n − 2)f (a),
n
i
where a = a1 +···+a
and bi = na−a
, i = 1, . . . , n, and a1 , . . . , an ∈ I.
n
n−1
Our main results are given in the following theorems:
Theorem 1.2. If f is a convex function and x1 , x2 , . . . , xn lie in its domain, then
(1.3)
n
X
i=1
x1 + · · · + xn
f (xi ) − f
n
n−1
x1 + x2
xn−1 + xn
xn + x1
≥
f
+ ··· + f
+f
.
n
2
2
2
Proof. Using (1.2) with t = 12 , we obtain
x1 + x2
xn−1 + xn
xn + x1
(1.4) f
+···+f
+f
≤ f (x1 ) + f (x2 ) + · · · + f (xn ).
2
2
2
P
In the summation on the right side of (1.4), the expression ni=1 f (xi ) can be written as
n
X
i=1
n
X
i=1
n
n
n X
1 X
f (xi ) =
f (xi ) −
f (xi ),
n − 1 i=1
n − 1 i=1
" n
#
n
X
X
n
1
f (xi ) =
f (xi ) −
f (xi ) .
n − 1 i=1
n
i=1
Pn
Replacing i=1 f (xi ) with the equivalent expression in (1.4),
x1 + x2
xn−1 + xn
xn + x1
f
+ ··· + f
+f
2
2
2
" n
#
n
X
X
n
1
≤
f (xi ) −
f (xi ) .
n − 1 i=1
n
i=1
J. Inequal. Pure and Appl. Math., 7(4) Art. 148, 2006
http://jipam.vu.edu.au/
N EW I NEQUALITIES A BOUT C ONVEX F UNCTIONS
3
Hence, applying Jensen’s inequality (1.1) to the right hand side of the above resulting inequality
we get
xn−1 + xn
xn + x1
x1 + x2
f
+ ··· + f
+f
2
2
2
" n
Pn
#
X
x
n
i=1 i
≤
f (xi ) − f
,
n − 1 i=1
n
and this concludes the proof.
Remark 1.3. Now we consider the simplest case of Theorem 1.2 for n = 3 to obtain the
following variant of via Titu Andreescu [2]:
x1 + x2 + x3
f (x1 ) + f (x2 ) + f (x3 ) − f
3
2
x1 + x2
x2 + x3
x3 + x1
f
+f
+f
.
≥
3
2
2
2
The variant of the generalized Popovicui inequality is given in the following theorem.
Theorem 1.4. If f is a convex function and a1 , . . . , an lie in its domain, then
(n − 1) [f (b1 ) + · · · + f (bn )] ≤ n [f (a1 ) + · · · + f (an ) − f (a)] ,
(1.5)
where a =
a1 +···+an
n
and bi =
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
1
[f (a1 ) + · · · + f (an )] −
[f (a1 ) + · · · + f (an )] ,
n−1
n−1
or
n
n
1
1
f (b1 ) + · · · + f (bn ) ≤
[f (a1 ) + · · · + f (an )] −
f (a1 ) + · · · + f (an ) ,
n−1
n−1 n
n
and so
n
1
1
(1.6) f (b1 ) + · · · + f (bn ) ≤
f (a1 ) + · · · + f (an ) −
f (a1 ) + · · · + f (an ) .
n−1
n
n
Hence, applying Jensen’s inequality (1.1) to the right hand side of (1.6) we get
n
a1 + · · · + an
f (b1 ) + · · · + f (bn ) ≤
f (a1 ) + · · · + f (an ) − f
,
n−1
n
and this concludes the proof.
R EFERENCES
[1] D.S. MITRINOVIĆ, J.E. PEC̆ARIĆ 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 Iaşi. I-a, Mat. (N.S), 1965.
J. Inequal. Pure and Appl. Math., 7(4) Art. 148, 2006
http://jipam.vu.edu.au/