Volume 9 (2008), Issue 4, Article 98, 10 pp.
AN INTEGRAL INEQUALITY FOR 3-CONVEX FUNCTIONS
VLAD CIOBOTARIU-BOER
“AVRAM I ANCU ” S ECONDARY S CHOOL
C LUJ -NAPOCA , ROMANIA
vlad_ciobotariu@yahoo.com
Received 17 July, 2008; accepted 27 September, 2008
Communicated by J. Pečarić
A BSTRACT. In this paper, an integral inequality and an application of it, that imply the Chebyshev functional for two 3-convex (3-concave) functions, are given.
Key words and phrases: Chebyshev functional, Convex functions, Integral inequality.
2000 Mathematics Subject Classification. Primary 26D15, Secondary 26D10.
1. I NTRODUCTION
For two Lebesgue functions f, g : [a, b] → R, consider the Chebyshev functional
Z b
Z b
Z b
1
1
C(f, g) :=
f (x)g(x)dx −
f (x)dx ·
g(x)dx.
b−a a
(b − a)2 a
a
In 1972, A. Lupaş [2] showed that if f, g are convex functions on the interval [a, b], then
Z b
Z b
12
a+b
a+b
(1.1)
C(f, g) ≥
x−
f (x)dx ·
x−
g(x)dx,
(b − a)4 a
2
2
a
with equality when at least one of the functions f, g is a linear function on [a, b]. He proved this
result using the following lemma:
Lemma 1.1. If f, g : [a, b] → R are convex functions on the interval [a, b], then
(1.2)
[F (e2 ) − F (e)2 ]F (f g) − F (e2 )F (f )F (g)
≥ F (ef )F (eg) − F (e)[F (f )F (eg) + F (ef )F (g)],
where F is an isotonic positive linear functional, defined by one of the following relations:
Rb
Z b
n
X
p(x)f (x)dx
1
a
(1.3)
F (f ) :=
f (x)dx, F (f ) := R b
, F (f ) :=
pi f (xi )
b−a a
p(x)dx
i=1
a
Pn
(xi ∈ [a, b]; pi ≥ 0, i = 1, 2, . . . , n,
p
=
1),
p : [a, b] → R is a positive, integrable
i=1 i
function on [a, b] and e(x) = x, x ∈ [a, b]. If f or g is a linear function, then the equality holds
in (1.2).
203-08
2
V LAD C IOBOTARIU -B OER
In this note, we provide a lower bound for the Chebyshev functional in the case of two 3convex (3-concave) functions f and g.
2. R ESULTS
Note that the inequality (1.2) can be written in the form:
1
F
(e)
F
(g)
(2.1)
F (e) F (e2 ) F (eg) ≥ 0.
F (f ) F (ef ) F (f g) The following lemma holds.
Lemma 2.1. If f, g : [a, b] → R are 3-convex (3-concave) functions on the interval [a, b], then
1
F (e) F (e2 ) F (g) F (e) F (e2 ) F (e3 ) F (eg) (2.2)
≥ 0,
2
3
4
2
F (e ) F (e ) F (e ) F (e g) F (f ) F (ef ) F (e2 f ) F (f g) where ei (x) = xi , x ∈ [a, b], i = 1, 4 and F is defined by (1.3).
If f is 3-convex (3-concave) and g is 3-concave (3-convex) then the reverse of the inequality
in (2.2) holds.
If f or g is a polynomial function of degree at most two, then the equality holds in (2.2).
Proof. Let [x, y, z, t; f ] be the divided difference of a certain function f . If f and g are 3-convex
(3-concave) on the interval [a, b], then we have
(2.3)
[x, y, z, t; f ] · [x, y, z, t; g] ≥ 0,
for all distinct points x, y, z, t from [a, b].
When f is 3-convex (3-concave) and g is 3-concave (3-convex) then the reverse of the inequality in (2.3) holds.
In the following we prove (2.2) in the case when both functions f and g are 3-convex (3concave). The inequality (2.3) is equivalent to
1
1
1
1
1
1
1
1
x
x
y
z
t
y
z
t
(2.4)
·
2
≥ 0,
x
y2
z2
t2 x2
y2
z2
t2 f (x) f (y) f (z) f (t) g(x) g(y) g(z) g(t) with true equality holding when at least one of f and g is a polynomial function of degree at
most two.
Note that the function F defined by (1.3) has the property F (1) = 1. In order to put in
evidence the variable u, we write Fu instead of F .
Now, using the fact that F is a linear positive functional, by applying successively on (2.4)
the functionals Fx , Fy , Fz and then Ft , we obtain the inequality (2.2). For instance, if
1 1 1 2
A = A(x, y, z, t, f, g) := y z t · f (x)g(x),
2 2 2 y z t J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 98, 10 pp.
http://jipam.vu.edu.au/
I NTEGRAL I NEQUALITY FOR 3-C ONVEX F UNCTIONS
3
then
1 1 1 2
Fx (A) = y z t · F (f g),
2 2 2 y z t 1
F (e) F (e2 )
Ft Fz Fy Fx (A) = 6 · F (e) F (e2 ) F (e3 )
F (e2 ) F (e3 ) F (e4 )
· F (f g)
and if
1 1 1 1 1 1 B = B(x, y, z, t, f, g) := y z t · x z t · f (x)g(y),
2 2 2 2 2 2 y z t x z t then
1 1 1 F (f )
1
1
Fx (B) = y z t · g(y) · F (ef ) z t ,
2 2 2 y z t F (e2 f ) z 2 t2 1
F (e) F (g) Ft Fz Fy Fx (B) = 2 · F (e) F (e2 ) F (eg) · F (e2 f )
F (e2 ) F (e3 ) F (e2 g) 2 1
F
(g)
F
(e
)
+ 2 · F (e) F (eg) F (e3 ) · F (ef )
F (e2 ) F (e2 g) F (e4 ) F (g)
F (e) F (e2 )
+ 2 · F (eg) F (e2 ) F (e3 )
F (e2 g) F (e3 ) F (e4 )
· F (f ).
Theorem 2.2. If f, g are 3-convex (3-concave) functions on the interval [a, b], then
180
(2.5) C(f, g) ≥
(b − a)6
Z
b
Z
b
q(x)f (x)dx ·
q(x)g(x)dx
a
a
Z b
Z b
12
a+b
a+b
+
x−
f (x)dx ·
x−
g(x)dx,
(b − a)4 a
2
2
a
where
a+b b−a
a+b b−a
q(x) = x −
− √
x−
+ √
.
2
2
2 3
2 3
If f is 3-convex (3-concave) and g is 3-concave (3-convex) then the reverse of the inequality
in (2.5) holds.
The equality in (2.5) holds when at least one of f or g is a polynomial function of degree at
most two on [a, b].
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 98, 10 pp.
http://jipam.vu.edu.au/
4
V LAD C IOBOTARIU -B OER
Proof. We choose
1
F (f ) =
b−a
(2.6)
Z
b
f (x)dx.
a
Then
(2.7)
a2 + ab + b2
a+b
, F (e2 ) =
,
2
3
a4 + a3 b + a2 b2 + ab3 + b4
a3 + a2 b + ab2 + b3
F (e3 ) =
, F (e4 ) =
.
4
5
F (e) =
Note that the inequality (2.2) can be written as
2 1
F
(e)
F
(e
)
1
F (e) 2
2
2
3 (2.8)
F (e) F (e ) F (e ) · F (f g) − 2 · F (e f )F (e g)
F
(e)
F
(e
)
F (e2 ) F (e3 ) F (e4 ) 2 1
F (e2 ) F (e3 ) F
(e
)
· F (ef )F (eg) − − F (e3 ) F (e4 ) · F (f )F (g)
F (e2 ) F (e4 ) 1
F
(e)
2
2
+ 2
3 · [F (e f )F (eg) + F (ef )F (e g)]
F (e ) F (e )
F (e) F (e2 ) · [F (e2 f )F (g) + F (f )F (e2 g)]
−
F (e2 ) F (e3 ) F (e) F (e2 ) · [F (ef )F (g) + F (f )F (eg)] ≥ 0.
+
F (e3 ) F (e4 ) By calculation, we find
1
F (e) F (e2 )
(2.9)
F (e) F (e2 ) F (e3 )
F (e2 ) F (e3 ) F (e4 )
(b − a)6
,
=
2160
(2.10)
1
F (e2 )
F (e2 ) F (e4 )
(b − a)2 (4a2 + 7ab + 4b2 )
=
,
45
(2.11)
F (e2 ) F (e3 )
F (e3 ) F (e4 )
(b − a)2 (a4 + 4a3 b + 10a2 b2 + 4ab3 + b4 )
=
,
240
(2.12)
1
F (e)
F (e2 ) F (e3 )
(b − a)2 (a + b)
=
,
12
(2.13)
F (e) F (e2 )
F (e2 ) F (e3 )
(b − a)2 (a2 + 4ab + b2 )
=
,
72
(2.14)
F (e) F (e2 )
F (e3 ) F (e4 )
(b − a)2 (a3 + 4a2 b + 4ab2 + b3 )
=
.
60
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 98, 10 pp.
http://jipam.vu.edu.au/
I NTEGRAL I NEQUALITY FOR 3-C ONVEX F UNCTIONS
5
The relations (2.7) – (2.14) give us
Z
Z
Z b
(b − a)5 b
a4 + 4a3 b + 10a2 b2 + 4ab3 + b4 b
(2.15)
f (x)g(x)dx −
f (x)dx
g(x)dx
2160 a
240
a
a
Z b
Z b
Z
Z b
1
4a2 + 7ab + 4b2 b
2
2
≥
x f (x)dx
x g(x)dx +
xf (x)dx
xg(x)dx
12 a
45
a
a
a
Z b
Z b
Z b
Z b
a+b
2
2
−
x f (x)dx
xg(x)dx +
xf (x)dx
x g(x)dx
12
a
a
a
a
Z b
Z b
Z b
Z b
a2 + 4ab + b2
2
2
+
x f (x)dx
g(x)dx +
f (x)dx
x g(x)dx
72
a
a
a
a
Z b
Z b
Z b
Z b
a3 + 4a2 b + 4ab2 + b3
−
xf (x)dx
g(x)dx +
f (x)dx
xg(x)dx ,
60
a
a
a
a
or
180
(2.16) C(f, g) ≥
(b − a)6
Z
b
Z
2
x f (x)dx
a
b
x2 g(x)dx
a
Z
Z b
4(4a2 + 7ab + 4b2 ) b
+
xf (x)dx
xg(x)dx
15
a
a
Z b
Z b
2a4 + 10a3 b + 21a2 b2 + 10ab3 + 2b4
+
·
f (x)dx
g(x)dx
540
a
a
Z b
Z b
Z b
Z b
a+b
2
2
−
x f (x)dx
xg(x)dx +
xf (x)dx
x g(x)dx
12
a
a
a
a
Z b
Z b
Z b
Z b
a2 + 4ab + b2
2
2
x f (x)dx
+
g(x)dx +
f (x)dx
x g(x)dx
72
a
a
a
a
Z b
Z b
a3 + 4a2 b + 4ab2 + b3
xf (x)dx
g(x)dx
−
60
a
a
Z b
Z b
+
f (x)dx
xg(x)dx .
a
a
The last inequality can be written as
2
2
Z b
Z b
180
a+b
a+b
(2.17)
C(f, g) ≥
x−
f (x)dx ·
x−
g(x)dx
(b − a)6 a
2
2
a
"Z 2
Z b
b
15
a+b
−
·
x−
f (x)dx ·
g(x)dx+
(b − a)4
2
a
a
#
2
Z b
Z b
a+b
+
f (x)dx ·
x−
g(x)dx
2
a
a
Z b
Z b
5
+
f (x)dx ·
g(x)dx
4(b − a)2 a
a
Z b
Z b
12
a+b
a+b
+
x−
f (x)dx ·
x−
g(x)dx,
(b − a)4 a
2
2
a
which is equivalent to (2.5).
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 98, 10 pp.
http://jipam.vu.edu.au/
6
V LAD C IOBOTARIU -B OER
Corollary 2.3. Let f and g be as in Theorem 2.2 and assume that
f (x) = −f (a + b − x)
(2.18)
or
g(x) = −g(a + b − x)
(2.19)
for all x from [a, b]. Then Lupaş’ inequality holds.
Proof. Note that the function denoted by q in Theorem 2.2 is symmetric about x =
a+b
,
2
namely
q(x) = q(a + b − x),
for all x from [a, b].
Assume that (2.18) is satisfied. Then we have
Z
Z b
1 b
(2.20)
q(x)f (x)dx =
q(x)[f (x) − f (a + b − x)]dx
2 a
a
Z
Z
1 b
1 b
q(x)f (x)dx −
q(a + b − x)f (a + b − x)dx
=
2 a
2 a
Z
Z
1 b
1 b
=
q(x)f (x)dx −
q(t)f (t)dt = 0.
2 a
2 a
From (2.5) and (2.20), we deduce (1.1).
Note that the condition (2.3) is important. The same results are valid if we suppose that this
(or its reverse) is satisfied. Thus, we obtain a more general result:
Theorem 2.4. If the functions f and g are integrable on the interval [a, b] and satisfy (2.3) (or
its reverse), then we have (2.5) (or its reverse).
The equality in (2.5) holds when at least one of f or g is a polynomial function of degree at
most two on [a, b].
Corollary 2.5. If the function f is integrable on [a, b], then we have
Z
(2.21) (b − a)
b
2
Z
f (x)dx −
f (x)dx
a
12
≥
(b − a)2
2
b
a
Z b a
a+b
x−
2
2
f (x)dx
180
+
(b − a)4
Z
2
b
q(x)f (x)dx
,
a
where q(x) is defined in Theorem 2.2.
Proof. Considering g(x) = f (x) in (2.5), we find the inequality (2.21).
Remark 1. The inequality (2.21) is better than the well-known inequality
Z b
2
Z b
2
(2.22)
(b − a)
f (x)dx ≥
f (x)dx ,
a
a
valid for all integrable functions f on [a, b].
Corollary 2.6. If the functions f, g satisfy the following conditions:
(i) f, g are 3-convex (3-concave) functions on [a, b];
(ii) f, g are differentiable functions on [a, b],
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 98, 10 pp.
http://jipam.vu.edu.au/
I NTEGRAL I NEQUALITY FOR 3-C ONVEX F UNCTIONS
7
then we have
Z b
[f (b) − f (a)][g(b) − g(a)]
(2.23)
f 0 (x)g 0 (x)dx ≥
b−a
a
Z b
1
12
f (a) + f (b)
+
f (x)dx −
b−a b−a a
2
Z b
1
g(a) + g(b)
×
g(x)dx −
.
b−a a
2
Proof. In (1.1), we use the fact that if f, g are 3-convex functions on [a, b], then f 0 , g 0 are convex
functions on [a, b]. We have
Z b
Z b
Z b
1
1
0
0
0
f (x)g (x)dx −
f (x)dx ·
g 0 (x)dx
2
b−a a
(b − a) a
a
Z b
Z b
a+b
a+b
12
0
≥
x−
f (x)dx ·
x−
g 0 (x)dx,
(b − a)4 a
2
2
a
which is equivalent to (2.23).
Remark 2. If, in addition, f and g are convex (concave) on [a, b], then the inequality (2.23) is
better than the inequality
Z b
[f (b) − f (a)][g(b) − g(a)]
(2.24)
f 0 (x) · g 0 (x)dx ≥
,
b−a
a
which is valid for all convex (concave) functions f, g on [a, b].
Remark 3. Lemma 2.1 can be generalized for n-convex functions, obtaining a result similar to
(2.2), from where the inequality
Rb
b−a
b2 −a2
bn −an
.
.
.
g(x)dx
2
n
a
R
b2 −a2
b
b3 −a3
bn+1 −an+1
.
.
.
xg(x)dx
3
n+1
a
2
...
≥ 0,
...
... ...
...
(2.25)
n n
R
b n−1
b −a
bn+1 −an+1
b2n−1 −a2n−1
.
.
.
x
g(x)dx
n
n+1
2n−1
a
Rb
R b n−1
Rb
Rb
f
(x)dx
xf
(x)dx
.
.
.
x
f
(x)dx
f
(x)g(x)dx
a
a
a
a
holds for all integer numbers n ≥ 3.
Some similar results related to the Chebyshev functional are given in [1] – [6].
3. A N A PPLICATION
Let f, g be two 3-time differentiable functions defined on a nonempty interval [a, b].
Denote
m1 = inf f (3) (x),
x∈[a,b]
M1 = sup f (3) (x),
x∈[a,b]
(3)
m2 = inf g (x),
x∈[a,b]
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 98, 10 pp.
M2 = sup g (3) (x).
x∈[a,b]
http://jipam.vu.edu.au/
8
V LAD C IOBOTARIU -B OER
Considering the functions F1 , G1 , F2 , G2 : [a, b] → R, defined by
m1 x3
− f (x),
6
M1 x 3
F2 (x) =
− f (x),
6
F1 (x) =
m2 x3
− g(x),
6
M2 x 3
G2 (x) =
− g(x),
6
G1 (x) =
(3)
(3)
(3)
we note that these are 3-differentiable on [a, b] and F1 (x) ≤ 0, G1 (x) ≤ 0, F2 (x) ≥ 0,
(3)
G2 (x) ≥ 0 for all x ∈ [a, b]. Therefore F1 , G1 are 3-concave on [a, b] and F2 , G2 are 3-convex
on [a, b].
Applying Theorem 2.2 we shall prove the following result:
Theorem 3.1. Let f, g be two 3-differentiable functions on the nonempty interval [a, b]. Then,
we have
Z b
a
+
b
1
(3.1) L(f, g) −
x−
r(x)h(x)dx
6(b − a) a
2
(m1 + M1 )(m2 + M2 )
6
+
· (b − a) 403200
(M1 − m1 )(M2 − m2 )
≤
· (b − a)6 ,
403200
where
12
(3.2) L(f, g) = C(f, g) −
(b − a)4
(3.3)
(3.4)
Z b
a
Z b
a+b
f (x)dx ·
x−
g(x)dx
2
a
Z b
Z b
180
−
q(x)f (x)dx ·
q(x)g(x)dx,
(b − a)6 a
a
a+b
x−
2
m 1 + M1
m 2 + M2
· g(x) +
· f (x),
2
2 !
√
√ !
a + b (b − a) 15
a + b (b − a) 15
r(x) = x −
−
x−
+
.
2
10
2
10
h(x) =
Proof. Applying Theorem 2.2, we have
180
(3.5) C(F1 , G1 ) ≥
(b − a)6
Z
180
(3.6) C(F2 , G2 ) ≥
(b − a)6
Z
b
Z
b
q(x)F1 (x)dx ·
q(x)G1 (x)dx
a
a
Z b
Z b
12
a+b
a+b
+
x−
F1 (x)dx ·
x−
G1 (x)dx,
(b − a)4 a
2
2
a
b
Z
b
q(x)F2 (x)dx ·
q(x)G2 (x)dx
a
a
Z b
Z b
12
a+b
a+b
+
x−
F2 (x)dx ·
x−
G2 (x)dx,
(b − a)4 a
2
2
a
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 98, 10 pp.
http://jipam.vu.edu.au/
I NTEGRAL I NEQUALITY FOR 3-C ONVEX F UNCTIONS
180
(3.7) C(F1 , G2 ) ≤
(b − a)6
Z
180
(3.8) C(F2 , G1 ) ≤
(b − a)6
Z
b
Z
9
b
q(x)F1 (x)dx ·
q(x)G2 (x)dx
a
a
Z b
Z b
a+b
a+b
12
+
x−
F1 (x)dx ·
x−
G2 (x)dx,
(b − a)4 a
2
2
a
b
Z
b
q(x)F2 (x)dx ·
q(x)G1 (x)dx
a
a
Z b
Z b
a+b
12
a+b
+
x−
F2 (x)dx ·
x−
G1 (x)dx,
(b − a)4 a
2
2
a
where q(x) is defined in Theorem 2.2.
By calculation, we find
m1 m2
(3.9) C(F1 , G1 ) = C(f, g) +
(b − a)2 (9a4 + 20a3 b + 26a2 b2 + 20ab3 + 9b4 )
4032
Z b
1
−
x3 [m1 g(x) + m2 f (x)]dx
6(b − a) a
Z
a3 + a2 b + ab2 + b3 b
+
[m1 g(x) + m2 f (x)]dx,
24(b − a)
a
12
(3.10)
(b − a)4
Z b
Z b
a+b
a+b
x−
F1 (x)dx ·
x−
G1 (x)dx
2
2
a
a
Z b
Z b
12
a+b
a+b
=
x−
f (x)dx ·
x−
g(x)dx
(b − a)4 a
2
2
a
m1 m2
+
(b − a)2 (3a2 + 4ab + 3b2 )2
4800
Z 3a2 + 4ab + 3b2 b
a+b
−
x−
[m1 g(x) + m2 f (x)]dx,
20(b − a)
2
a
180
(3.11)
(b − a)6
Z
b
Z
q(x)F1 (x)dx ·
a
180
=
(b − a)6
b
q(x)G1 (x)dx
a
Z
b
Z
q(x)f (x)dx ·
a
b
m1 m2
(b − a)4 (a + b)2
2880
Z b
a+b
−
·
q(x)[m1 g(x) + m2 f (x)]dx.
4(b − a) a
q(x)g(x)dx +
a
From (3.5) and (3.9) – (3.11), we obtain
Z b
m1 m2
1
a+b
6
(3.12) L(f, g) +
(b − a) ≥
x−
r(x)[m1 g(x) + m2 f (x)]dx.
100800
6(b − a) a
2
In a similar way we can prove that the inequality (3.6) is equivalent to
Z b
M1 M2
1
a+b
6
(3.13) L(f, g) +
(b − a) ≥
x−
r(x)[M1 g(x) + M2 f (x)]dx,
100800
6(b − a) a
2
the inequality (3.7) is equivalent to
Z b
m 1 M2
1
a+b
6
(3.14) L(f, g) +
(b − a) ≤
x−
r(x)[m1 g(x) + M2 f (x)]dx,
100800
6(b − a) a
2
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 98, 10 pp.
http://jipam.vu.edu.au/
10
V LAD C IOBOTARIU -B OER
and the inequality (3.8) is equivalent to
Z b
M1 m 2
1
a+b
6
(3.15) L(f, g) +
(b − a) ≤
x−
r(x)[M1 g(x) + m2 f (x)]dx.
100800
6(b − a) a
2
From (3.12) and (3.13) we deduce
Z b
a+b
m 1 m 2 + M1 M2
1
(3.16)
L(f, g) −
x−
r(x)h(x)dx ≥ −
(b − a)6 .
6(b − a) a
2
201600
From (3.14) and (3.15) we find
Z b
1
a+b
m 1 M2 + M1 m 2
(3.17)
L(f, g) −
x−
r(x)h(x)dx ≤ −
(b − a)6 .
6(b − a) a
2
201600
The inequalities (3.16) and (3.17) prove (3.1).
Corollary 3.2. If f, g are 3-time differentiable on [a, b] and symmetric about x = a+b
, then we
2
have
(M1 − m1 )(M2 − m2 )
(m
+
M
)(m
+
M
)
1
1
2
2
6
L(f, g) +
(3.18)
(b
−
a)
(b − a)6 .
≤
403200
403200
Proof. Note that the functions h and r defined on [a, b] by (3.3) and (3.4) are symmetric about
x = a+b
. Hence, their product h · r is symmetric about x = a+b
and
2
2
Z b
a+b
(3.19)
x−
r(x)h(x)dx = 0.
2
a
From (3.1) and (3.19), we obtain (3.18).
R EFERENCES
[1] E.V. ATKINSON, An inequality, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz., (357-380)
(1971), 5–6.
[2] A. LUPAŞ, An integral inequality for convex functions, Univ. Beograd. Publ. Elektrotehn. Fak. Ser.
Mat. Fiz., (381-409) (1972), 17–19.
[3] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalitites in Analysis,
Kluwer Academic Publishers, Dordrecht, 1992.
[4] D.S. MITRINOVIĆ AND P.M. VASIĆ, Analytic Inequalities, Springer-Verlag, Berlin and New-York,
1970.
[5] J.E. PEČARIĆ, F. PROSCHAN AND Y.I. TONG, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press, San Diego, 1992.
[6] P.M. VASIĆ AND I.B. LACKOVIĆ, Notes on convex functions VI: On an inequality for convex
functions proved by A. Lupaş, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz., (634-677)
(1979), 36–41.
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 98, 10 pp.
http://jipam.vu.edu.au/