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.
volume 7, issue 4, article 148,
2006.
Received 11 June, 2006;
accepted 15 October, 2006.
Communicated by: B. Yang
EMail: bougoffa@hotmail.com
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Abstract
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
New Inequalities About Convex
Functions
Lazhar Bougoffa
and
(n − 1) [f(b1 ) + · · · + f(bn )] ≤ n [f(a1 ) + · · · + f(an ) − f(a)] ,
where a =
a1 +···+an
n
and bi =
na−ai
n−1 ,
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i = 1, . . . , n.
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2000 Mathematics Subject Classification: 26D15.
Key words: Jensen’s inequality, Convex functions.
Contents
1
Main Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
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J. Ineq. Pure and Appl. Math. 7(4) Art. 148, 2006
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1.
Main Theorems
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).
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
n
x1 + · · · + xn
2 X
xi + xj
f (xi ) +
f
≥
f
,
n
−
2
n
n
−
2
2
i=1
i<j
New Inequalities About Convex
Functions
Lazhar Bougoffa
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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
and bi = na−a
, i = 1, . . . , n, and a1 , . . . , an ∈ I.
where a = a1 +···+a
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
n
X
x1 + · · · + xn
(1.3)
f (xi ) − f
n
i=1
x1 + x2
xn−1 + xn
xn + x1
n−1
f
+ ··· + f
+f
.
≥
n
2
2
2
1
,
2
Proof. Using (1.2) with t = we obtain
xn−1 + xn
xn + x1
x1 + x2
+ ··· + f
+f
(1.4) f
2
2
2
≤ f (x1 ) + f (x2 ) + · · · + f (xn ).
P
In the summation on the right side of (1.4), the expression ni=1 f (xi ) can be
written as
n
n
n
X
1 X
n X
f (xi ) =
f (xi ) −
f (xi ),
n
−
1
n
−
1
i=1
i=1
i=1
" n
#
n
n
X
X
X1
n
f (xi ) −
f (xi ) .
f (xi ) =
n − 1 i=1
n
i=1
i=1
New Inequalities About Convex
Functions
Lazhar Bougoffa
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P
Replacing ni=1 f (xi ) with the equivalent expression in (1.4),
xn−1 + xn
xn + x1
x1 + x2
+ ··· + f
+f
f
2
2
2
" n
#
n
X
X
n
1
≤
f (xi ) −
f (xi ) .
n − 1 i=1
n
i=1
Hence, applying Jensen’s inequality (1.1) to the right hand side of the above
resulting inequality we get
x1 + x2
xn−1 + xn
xn + x1
f
+ ··· + f
+f
2
2
2
" n
Pn
#
X
n
i=1 xi
f (xi ) − f
,
≤
n − 1 i=1
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]:
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.
New Inequalities About Convex
Functions
Lazhar Bougoffa
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Theorem 1.3. If f is a convex function and a1 , . . . , an lie in its domain, then
(1.5)
(n − 1) [f (b1 ) + · · · + f (bn )] ≤ n [f (a1 ) + · · · + f (an ) − f (a)] ,
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,
New Inequalities About Convex
Functions
Lazhar Bougoffa
f (b1 ) + · · · + f (bn )
n
1
[f (a1 ) + · · · + f (an )] −
[f (a1 ) + · · · + f (an )] ,
≤
n−1
n−1
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or
f (b1 ) + · · · + f (bn )
n
n
1
1
≤
[f (a1 ) + · · · + f (an )] −
f (a1 ) + · · · + f (an ) ,
n−1
n−1 n
n
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and so
(1.6) f (b1 ) + · · · + f (bn )
1
1
n
f (a1 ) + · · · + f (an ) −
f (a1 ) + · · · + f (an ) .
≤
n−1
n
n
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J. Ineq. Pure and Appl. Math. 7(4) Art. 148, 2006
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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.
New Inequalities About Convex
Functions
Lazhar Bougoffa
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References
[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.
New Inequalities About Convex
Functions
Lazhar Bougoffa
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