International Journal of Analysis and Applications
ISSN 2291-8639
Volume 10, Number 1 (2016), 40-47
http://www.etamaths.com

FEJÉR–HADAMARD INEQULALITY FOR CONVEX FUNCTIONS ON THE

COORDINATES IN A RECTANGLE FROM THE PLANE

G. FARID∗, M. MARWAN, AND ATIQ UR REHMAN

Abstract. We give Fejér–Hadamard inequality for convex functions on coordinates in the rectangle

from the plane. We define some mappings associated to it and discuss their properties.

1. Introduction

A real valued function f : I → R, where I is an interval in R, is called convex if

f(αx + (1 −α)y) ≤ αf(x) + (1 −α)f(y),

where α ∈ [0, 1], for all x,y ∈ I.
Convex functions play a vital role in the theory of inequalities. A lot of inequalities are established

using convex functions, e.g. see for convex functions in [1, 2, 6, 7]. The most classical and fundamental
inequality is Hermite-Hadamard inequality, this is stated as follows:

(1) f

(
a + b

2

)
≤

1

b−a

∫ b
a

f(x)dx ≤
f(a) + f(b)

2

holds for every convex function f : I → R and a, b ∈ I with a < b.
This inequality is present in many textbooks and monographs devoted to convex functions and it is
also extensively studied by many researchers. With the help of (1) researchers have produced many
integral and differential inequalities (see [8, 9]), and operators. Very interesting historical remarks
concerning the inequality (1) can be found in [13] (see also [14, pp. 62] ).
In 1906, Fejér (see [16, page 138] and [11]) established the following weighted generalization of the
Hermite–Hadamard inequality for symmetric functions.

(2) f

(
a + b

2

)∫ b
a

g(x)dx ≤
∫ b
a

f(x)g(x)dx ≤
f(a) + f(b)

2

∫ b
a

g(x)dx

holds for every convex function f : I → Ra,b ∈ I, and g : [a,b] → R+ symmetric about (a + b)/2.
In [5] Dragomir gave the Hermite-Hadamard inequality on a rectangle in plane, by defining convex

functions on coordinates.

Definition 1.1. Let us consider the two-dimensional interval ∆ := [a,b] × [c,d] in R2 with a < b
and c < d. A function f : ∆ → R will be called convex on the coordinates if the partial mappings
fy : [a,b] → R, fy(u) := f(u,y), and fx : [a,b] → R, fx(v) := f(x,v), are convex where defined for all
y ∈ [c,d] and x ∈ [a,b].

One can note that every convex mapping f : ∆ → R is convex on the coordinates but the converse
is not true. For example, f(x,y) = xy is convex on coordinates in R2 but it is not convex.

2010 Mathematics Subject Classification. 26A51, 26D15, 65D30.
Key words and phrases. convex functions; Hadamard inequality; convex functions on coordinates.

c©2016 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

40



FEJÉR–HADAMARD INEQULALITY 41

Theorem 1.2. Let f : ∆ → R be convex on co-ordinate in ∆.
Then we have:

f

(
a + b

2
,
c + d

2

)
≤

1

2

[
1

b−a

∫ b
a

f(x,
c + d

2
)dx +

1

d− c

∫ d
c

f(
a + b

2
,y)dy

]

≤
1

(b−a)(d− c)

∫ b
a

∫ d
c

f(x,y)dydx

≤
1

4

[
1

(b−a)

∫ b
a

f(x,c)dx +
1

(b−a)

∫ b
a

f(x,d)dx

+
1

d− c

∫ d
c

f(a,y)dy +
1

d− c

∫ d
c

f(b,y)dy

]

≤
1

4
[f(a,c) + f(a,d) + f(b,c) + f(b,d)] .

There in [5] some mappings connected to above inequality are also considered and their properties
are discussed. In this paper we are interested to give the Fejér–Hadamard inequality for a rectangle
in plane via convex functions on coordinates. We also study some properties of mappings associated
with the Fejér–Hadamard inequality for convex functions on coordinates.

2. Main results

Theorem 2.1. Let ∆ := [a,b] × [c,d] ⊂ R2 and f : ∆ → R be a convex function on coordinates in ∆.
Also let g1 : [a,b] → R+ and g2 : [c,d] → R+ be two integrable and symmetric functions about (a + b)/2
and (c + d)/2 respectively. Then one has the following inequalities

(3)

f

(
a + b

2
,
c + d

2

)
≤

1

2

[
1

G1

∫ b
a

f

(
x,
c + d

2

)
g1(x)dx +

1

G2

∫ d
c

f

(
a + b

2
,y

)
g2(y)dy

]

≤
1

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx

≤
1

4

[
1

G1

∫ b
a

g1(x)f(x,c)dx +
1

G1

∫ b
a

g1(x)f(x,d)dx

+
1

G2

∫ d
c

g2(y)f(a,y)dy +
1

G2

∫ d
c

g2(y)f(b,y)dy

]

≤
1

4
[f(a,c) + f(a,d) + f(b,c) + f(b,d)] ,

where

G1 =

∫ b
a

g1(x)dx and G2 =

∫ d
c

g2(y)dy.

These inequalities are sharp.

Proof. Since f : ∆ → R is convex on coordinates, it follows that functions fx and fy are convex on
[c,d] and [a,b] respectively. Thus from (2), we have

(4) f

(
x,
c + d

2

)
≤

1

G2

∫ d
c

f(x,y)g2(y)dy ≤
f(x,c) + f(x,d)

2

and

(5) f

(
a + b

2
,y

)
≤

1

G1

∫ b
a

f(x,y)g1(x)dx ≤
f(a,y) + f(b,y)

2
.



42 FARID, MARWAN, AND REHMAN

Multiplying (4) by g1(x)

g1(x)f

(
x,
c + d

2

)
≤

1

G2

∫ d
c

f(x,y)g1(x)g2(y)dy ≤ g1(x)
f(x,c) + f(x,d)

2
.

Now integrating on [a,b], we get

(6)

∫ b
a

g1(x)f

(
x,
c + d

2

)
dx ≤

1

G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx

≤
1

2

[∫ b
a

g1(x)f(x,c)dx +

∫ b
a

g1(x)f(x,d)dx

]
.

Now multiplying (5) by g2(y) and integrating on [c,d], we get

(7)

∫ d
c

g2(y)f

(
a + b

2
,y

)
dy ≤

1

G1

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx

≤
1

2

[∫ d
c

g2(y)f(a,y)dy +

∫ d
c

g2(y)f(b,y)dy

]
.

Since G1,G2 > 0, dividing inequalities (6), (7) by G1, G2 respectively and adding we get second and
third inequalities in (3).
From first part of (4) and (5), we have

f

(
a + b

2
,
c + d

2

)
≤

1

G2

∫ d
c

f

(
a + b

2
,y

)
g2(y)dy

and

f

(
a + b

2
,
c + d

2

)
≤

1

G1

∫ b
a

f

(
x,
c + d

2

)
g1(x)dx.

Adding the above two inequalities we get the first inequality in (3).
Now from second part of (4) and (5), we can get

1

G1

∫ b
a

f(x,c)g1(x)dx ≤
f(a,c) + f(b,c)

2
,

1

G1

∫ b
a

f(x,d)g1(x)dx ≤
f(a,d) + f(b,d)

2
,

1

G2

∫ d
c

f(a,y)g2(y)dy ≤
f(a,c) + f(a,d)

2
,

1

G2

∫ d
c

f(b,y)g2(y)dy ≤
f(b,c) + f(b,d)

2
.

By adding the above four inequalities, we get last inequality in (3).
If in (3) we choose f(x) = xy, then (3) becomes an equality, which shows that inequalities in (3) are
sharp. �

Remark 2.2. If we put g1 ≡ 1 and g2 ≡ 1 in above theorem, then we get Theorem 1.2, which is the
main theorem of [5].

For a mapping f : ∆ → R, we define the mapping Ĥ : [0, 1]2 → R, as follows:

(8) Ĥ(t,s) =
1

G1G2

∫ b
a

∫ d
c

f

(
tx + (1 − t)

a + b

2
,sy + (1 −s)

c + d

2

)
g1(x)g2(y)dydx.

The properties of this mapping are studied in the following theorem. We need a following Lemma to
give desire results, which is due to Levin and Stečkin [16, pp. 200].



FEJÉR–HADAMARD INEQULALITY 43

Lemma 2.3. Let f be a convex function on [a,b], g be a function symmetric about (a + b)/2 and
nonincreasing on [a, (a + b)/2]. Then∫ b

a

f(x)g(x)dx ≥
1

b−a

∫ b
a

f(x)dx

∫ b
a

g(x)dx.

Theorem 2.4. Suppose that f : ∆ → R is convex on the coordinates in ∆. Then the mapping Ĥ,
defined in (8), is convex on the coordinates on [0, 1]2. Further if g1 is nonincreasing on [a, (a + b)/2]
and g2 is nonincreasing on [c, (c + d)/2], then

inf
(t,s)∈[0,1]2

Ĥ(t,s) = f

(
a + b

2
,
c + d

2

)
= Ĥ(0, 0)

and

sup
(t,s)∈[0,1]2

Ĥ(t,s) =
1

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx = Ĥ(1, 1).

Proof. For convexity, fix s ∈ [0, 1]. Then for all α,β ≥ 0 with α + β = 1, and t1, t2 ∈ [0, 1] we have

Ĥ(αt1 + βt2,s) =
1

G1G2

×
∫ b
a

∫ d
c

f

(
(αt1 + βt2)x + (1 − (αt1 + βt2))

a + b

2
,sy + (1 −s)

c + d

2

)
g1(x)g2(y)dydx

which gives us

Ĥ(αt1 + βt2,s) =
1

G1G2

∫ b
a

∫ d
c

f

(
α

(
t1x + (1 − t1)

a + b

2

)
+ β

(
t1x + (1 − t1)

a + b

2

)
,sy + (1 −s)

c + d

2

)
g1(x)g2(y)dydx

≤
α

G1G2

∫ b
a

∫ d
c

f

(
t1x + (1 − t1)

a + b

2
,sy + (1 −s)

c + d

2

)
g1(x)g2(y)dydx

+
β

G1G2

∫ b
a

∫ d
c

f

(
t2x + (1 − t2)

a + b

2
,sy + (1 −s)

c + d

2

)
g1(x)g2(y)dydx

= αĤ(t1,s) + βĤ(t2,s).

If t ∈ [0, 1] is fixed, then for all s1,s2 ∈ [0, 1] and α,β ≥ 0 with α + β = 1, we also have:

Ĥ(t,αs1 + βs2) ≤ αĤ(t,s1) + βĤ(t,s2)

and the statement is proved.
Now to prove the remaining part of the theorem, we take

Ĥ(t,s) =
1

G1G2

∫ b
a

∫ d
c

f

(
tx + (1 − t)

a + b

2
,sy + (1 −s)

c + d

2

)
g2(y)g1(x)dydx.

Since f is convex on the coordinates and 1
G2

∫ d
c
g2(y)dy = 1, we apply Jensen’s inequality for integrals

on second coordinate to get

Ĥ(t,s) ≥
1

G1

∫ b
a

f

(
tx + (1 − t)

a + b

2
,

1

G2

∫ d
c

(
sy + (1 −s)

c + d

2

)
g2(y)dy

)
g1(x)dx.

Now it follows from Lemma 2.3, that

Ĥ(t,s) ≥
1

G1

∫ b
a

f

(
tx + (1 − t)

a + b

2
,
c + d

2

)
g1(x)dx.(9)



44 FARID, MARWAN, AND REHMAN

Since 1
G1

∫ b
a
g1(x) dx = 1, Jensen’s inequality for integrals leads to

Ĥ(t,s) ≥ f

(
1

G1

∫ b
a

(
tx + (1 − t)

a + b

2

)
g1(x)dx,

c + d

2

)
.

Now by Lemma 2.3, we have

Ĥ(t,s) ≥ f
(
a + b

2
,
c + d

2

)
.

This gives us the lower bound of Ĥ. To get upper bound, we use convexity on second coordinates of
f to get

Ĥ(t,s) ≤
1

G1G2

∫ b
a

[
s

∫ d
c

f

(
tx + (1 − t)

a + b

2
,y

)
g2(y)dy

+ (1 −s)f
(
tx + (1 − t)

a + b

2
,
c + d

2

)
g2(y)dy

]
g1(x)dx.

This gives

Ĥ(t,s) ≤
s

G1G2

∫ b
a

∫ d
c

f

(
tx + (1 − t)

a + b

2
,y

)
g1(x)g2(y)dydx

+
(1 −s)
G1G2

∫ b
a

∫ d
c

f

(
tx + (1 − t)

a + b

2
,
c + d

2

)
g1(x)g2(y)dydx

≤
s

G1G2

∫ b
a

∫ d
c

[
tf(x,y) + (1 − t)f

(
a + b

2
,y

)]
g1(x)g2(y)dydx

+
1 −s
G1G2

∫ b
a

∫ d
c

[
tf

(
x,
c + d

2

)
+ (1 − t)f

(
a + b

2
,y

)]
g1(x)g2(y)dydx.

On simplification, we have

Ĥ(t,s) ≤
st

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx +
s(1 − t)
G2

∫ d
c

f

(
a + b

2
,y

)
g2(y)dy

+
(1 −s)t
G1

∫ b
a

f

(
x,
c + d

2

)
g1(x)dx + (1 −s)(1 − t)f

(
a + b

2
,
c + d

2

)
.

From inequalities (6) and (7), we have

1

G2

∫ d
c

f

(
a + b

2
,y

)
g2(y)dy ≤

1

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx

and

1

G1

∫ b
a

f

(
x,
c + d

2

)
g1(x) ≤

1

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx.

Using above inequalities, we deduce that

Ĥ(t,s) ≤ [st + s(1 − t) + (1 −s)t + (1 −s)(1 − t)]
1

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx

=
1

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx, (t,s) ∈ [0, 1]2.

Now we have to show the monotonicity of the mapping Ĥ(t,s). For this firstly, we will show that

Ĥ(t,s) ≥ Ĥ(t, 0) for all (t,s) ∈ [0, 1]2. By (9), we have:

Ĥ(t,s) ≥
1

G1

∫ b
a

f

(
tx + (1 − t)

a + b

2
,
c + d

2

)
g1(x)dx = Ĥ(t, 0)



FEJÉR–HADAMARD INEQULALITY 45

for all (t,s) ∈ [0, 1]2.
Now let 0 ≤ s1 ≤ s2 ≤ 1. By convexity of mapping Ĥ(t, .) for all t ∈ [0, 1], we have

Ĥ(t,s2) − Ĥ(t,s1)
s2 −s1

≥
Ĥ(t,s1) − Ĥ(t, 0)

s1
≥ 0.

This completes the proof. �

Remark 2.5. If we put g1 ≡ 1 and g2 ≡ 1, in Theorem 2.4 then we get Theorem 2 of [5].

If the function f is convex on ∆, instead of coordinated convex, then we have the following theorem.

Theorem 2.6. Suppose that f : ∆ → R is convex on ∆.
(i) The mapping Ĥ is convex on ∆.

(ii) Let ĥ : [0, 1] → R be the mapping defined as ĥ(t) = Ĥ(t,t). Then ĥ is convex, monotonic nonde-
creasing on [0, 1] and one has the bounds:

inf
t∈[0,1]

ĥ(t) = f

(
a + b

2
,
c + d

2

)
= Ĥ(0, 0)

and

sup
t∈[0,1]

ĥ(t) =
1

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx = Ĥ(1, 1).

Proof. (i) For convexity, let (t1,s1), (t2,s2) ∈ [0, 1]2 and α,β ≥ 0 with α + β = 1. Then

Ĥ(αt1 + βt2,αs1 + βs2) =
1

G1G2

×
∫ b
a

∫ d
c

f

[
α

(
t1x + (1 − t1)

a + b

2
,s1y + (1 −s1)

c + d

2

)
+β

(
t2x + (1 − t2)

a + b

2
,s2y + (1 −s2)

c + d

2

)]
g1(x)g2(y)dydx

≤
α

G1G2

∫ b
a

∫ d
c

f

(
t1 + (1 − t1)

a + b

2
,s1y + (1 −s1)

c + d

2

)
g1(x)g2(y)dydx

+
β

G1G2

∫ b
a

∫ d
c

f

(
t2x + (1 − t2)

a + b

2
,s2y + (1 −s2)

c + d

2

)
g1(x)g2(y)dydx

= αĤ(t1,s1) + βĤ(t2,s2).

Which shows that H is convex on [0, 1]2.

(ii) Now we prove the convexity of ĥ on [0, 1]. For this, let t1, t2 ∈ [0, 1] and α,β ≥ 0 with α + β = 1.
Then

ĥ(αt1 + βt2) = Ĥ(αt1 + βt2,αt1 + βt2)

= Ĥ(α(t1, t1) + β(t2, t2))

≤ αĤ(t1, t1) + βĤ(t2, t2)

= αĥ(t1) + βĥ(t2),

which shows the convexity of ĥ on [0, 1]. Now to prove the remaining part of the theorem, we take

ĥ(t) =
1

G1G2

∫ b
a

∫ d
c

f

(
tx + (1 − t)

a + b

2
, ty + (1 − t)

c + d

2

)
g2(y)g1(x)dydx.

Since f is convex on the coordinates and 1
G2

∫ d
c
g2(y)dy = 1, we apply Jensen’s inequality for integrals

on second coordinate to get

ĥ(t) ≥
1

G1

∫ b
a

f

(
tx + (1 − t)

a + b

2
,

1

G2

∫ d
c

[
ty + (1 − t)

c + d

2

]
g2(y)dy

)
g1(x)dx.



46 FARID, MARWAN, AND REHMAN

Now it follows from Lemma 2.3, that

ĥ(t) ≥
1

G1

∫ b
a

f

(
tx + (1 − t)

a + b

2
,
c + d

2

)
g1(x)dx.(10)

Since 1
G1

∫ b
a
g1(x) dx = 1, Jensen’s inequality for integrals leads to

ĥ(t) ≥ f

(
1

G1

∫ b
a

[
tx + (1 − t)

a + b

2

]
g1(x)dx,

c + d

2

)
.

Now by Lemma 2.3, we have

ĥ(t) ≥ f
(
a + b

2
,
c + d

2

)
.

This gives us the lower bound of ĥ(.). To get upper bound, we use convexity on second coordinates of
f to get

ĥ(t) ≤
1

G1G2

∫ b
a

[
t

∫ d
c

f

(
tx + (1 − t)

a + b

2
,y

)
g2(y)dy

+ (1 − t)f
(
tx + (1 − t)

a + b

2
,
c + d

2

)
g2(y)dy

]
g1(x)dx.

This gives

ĥ(t) ≤
t

G1G2

∫ b
a

∫ d
c

f

(
tx + (1 − t)

a + b

2
,y

)
g1(x)g2(y)dydx

+
(1 − t)
G1G2

∫ b
a

∫ d
c

f

(
tx + (1 − t)

a + b

2
,
c + d

2

)
g1(x)g2(y)dydx

≤
t

G1G2

∫ b
a

∫ d
c

[
tf(x,y) + (1 − t)f

(
a + b

2
,y

)]
g1(x)g2(y)dydx

+
1 − t
G1G2

∫ b
a

∫ d
c

[
tf

(
x,
c + d

2

)
+ (1 − t)f

(
a + b

2
,y

)]
g1(x)g2(y)dydx.

On simplification, we have

ĥ(t) ≤
t2

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx +
t(1 − t)
G2

∫ d
c

f

(
a + b

2
,y

)
g2(y)dy

+
(1 − t)t
G1

∫ b
a

f

(
x,
c + d

2

)
g1(x)dx + (1 − t)2f

(
a + b

2
,
c + d

2

)
.

From inequalities (6) and (7), we have

1

G2

∫ d
c

f

(
a + b

2
,y

)
g2(y)dy ≤

1

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx

and

1

G1

∫ b
a

f

(
x,
c + d

2

)
g1(x) ≤

1

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx.

Using above inequalities, we deduce that

ĥ(t) ≤
[
t2 + t(1 − t) + (1 − t)t + (1 − t)2

] 1
G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx

=
1

G1G2

∫ b
a

∫ d
c

f(x,y)g1(x)g2(y)dydx, t ∈ [0, 1].



FEJÉR–HADAMARD INEQULALITY 47

Now we have to show the monotonicity of the mapping ĥ for this firstly, we show that Ĥ(t,t) ≥ Ĥ(t, 0)
for all t ∈ [0, 1]. By (10), we have:

ĥ(t) ≥
1

G1

∫ b
a

f

(
tx + (1 − t)

a + b

2
,
c + d

2

)
g1(x)dx = Ĥ(t, 0)

for all t ∈ [0, 1].
Now let 0 ≤ t1 ≤ t2 ≤ 1. By convexity of mapping Ĥ(t, .) for all t ∈ [0, 1], we have

ĥ(t2) − ĥ(t1)
t2 − t1

≥
ĥ(t1) − ĥ(0)

t1
≥ 0.

This completes the proof. �

Remark 2.7. If we put g1 ≡ 1 and g2 ≡ 1 in Theorem 2.6, then we get Theorem 3 of [5].

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COMSATS Institute of Information Technology, Attock Campus, Pakistan

∗Corresponding author: faridphdsms@hotmail.com