International Journal of Analysis and Applications
ISSN 2291-8639
Volume 13, Number 1 (2017), 15-21
http://www.etamaths.com

SOME NEW ESTIMATES OF HERMITE-HADAMARD INEQUALITIES FOR

HARMONICALLY CONVEX FUNCTIONS WITH APPLICATIONS

WEN WANG1,2,∗ AND JIBING QI1

Abstract. In this paper, we first establish an integral identity. Further, using this identity, some

new estimates for Hermite-Hadamard inequalities for harmonically convex functions are established.

Finally, some applications to special mean are showed.

1. Introduction

In this article, let R = (−∞,∞), R++ = (0,∞).
Theory of convex functions and theory of inequalities are closely related to each other. Therefore,

some literature on inequalities can be found for convex functions.
One of the most extensively research on inequalities is Hermite-Hadamard type inequalities.
Definition 1.1 ( [1, 2]) A function f : I ⊂ R → R is convex function on I, if

f(tx + (1 − t)y) ≤ tf(x) + (1 − t)f(y), ∀x,y ∈ I,t ∈ [0, 1]. (1.1)
f is concave function if −f is convex function.

Let f : I ⊂ R → R be a convex function. The following inequality is the well-known Hermite-
Hadamard’s inequality

f

(
a + b

2

)
≤

1

b−a

∫ b
a

f(x)dx ≤
f(a) + f(b)

2
, a,b ∈ I with a < b. (1.2)

Estimates for Hermite-Hadamard inequality for convex functions are studied in a rich literature
[8–16].

Theorem 1.2 ( [1]) Let f : I0 ⊂ R → R be a differentiable mapping on I0 and a,b ∈ I0 with a < b.
If |f′(x)|q is a convex function for q > 1, then∣∣∣∣∣f(a) + f(b)2 − 1b−a

∫ b
a

f(x)dx

∣∣∣∣∣ ≤ (b−a)(|f
′(a)| + |f′(b)|)

8
. (1.3)

Recently, İscan [3] introduced the concept of harmonically convex functions and established Hermite-
Hadamard type inequality for harmonically convex functions.

Definition 1.3 ( [3, 4]) A function f : I ⊂ R\{0} → R is said to be harmonically convex function
on I, if

f

(
1

tx−1 + (1 − t)y−1

)
≤ tf(x) + (1 − t)f(y), ∀x,y ∈ I,t ∈ [0, 1]. (1.4)

f is said to be harmonically concave function if −f is harmonically convex function.
Theorem 1.4 ( [3, 4]) Let f : I ⊂ R\{0} → R be a harmonically convex function a,b ∈ I with

a < b. If f ∈ L[a,b] , then

f

(
2ab

a + b

)
≤

ab

b−a

∫ b
a

f(x)

x2
dx ≤

f(a) + f(b)

2
. (1.5)

Received 1st July, 2016; accepted 18th September, 2016; published 3rd January, 2017.
2010 Mathematics Subject Classification. 26D15, 26A51.

Key words and phrases. Hermite-Hadamard’s inequality; harmonically convex function; estimate; mean; inequality.

c©2017 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

15



16 WEN WANG, JIBING QI

Theorem 1.5 ( [3, 4]) Let f : I ⊂ R\{0}→ R be a harmonically convex function a,b ∈ I with a < b
and f′ ∈ L[a,b]. If |f′(x)|q is harmonically convex on [a,b] for q > 1, 1

p
+ 1

q
= 1, then∣∣∣∣∣f(a) + f(b)2 − abb−a

∫ b
a

f(x)

x2
dx

∣∣∣∣∣
≤
ab(b−a)

2

(
1

p + 1

)1
p

(µ1|f′(a)|q + µ2|f′(b)|q)
1
q , (1.6)

where

µ1 =

[
a2−2q + b1−2q[(b−a)(1 − 2q) −a]

]
2(b−a)2(1 −q)(1 − 2q)

,

µ2 =

[
a2−2q + b1−2q[(b−a)(1 − 2q) − b]

]
2(b−a)2(1 −q)(1 − 2q)

.

For many recent results related to Hermite-Hadamard type inequalities for harmonically functions,
see [3–7].

The aim of this paper is first to establish a integral identity. Then, using this identity, some new
estimates for Hermite-Hadamard inequalities for harmonically convex functions are established by İ.
İscan in [3] are derived.

2. Some lemmas

In order to prove our main results we need some lemmas.
Lemma 2.1 Let f : I ⊂ R\{0}→ R be a differentiable function on I0 and a,b ∈ I0 with a < b. If

f′ ∈ L[0, 1], then for λ ∈ [0, 1], one has

(1 −λ)f
(

ab

(1 −λ)a + λb

)
+ λf

(
ab

(1 −λ)a + λb

)
−

ab

b−a

∫ b
a

f(x)

x2
dx

=
b−a
ab

[∫ 1−λ
0

t

[tb−1 + (1 − t)a−1]
f′
(

1

tb−1 + (1 − t)a−1

)
dt

−
∫ 1
1−λ

t− 1
[tb−1 + (1 − t)a−1]

f′
(

1

tb−1 + (1 − t)a−1

)
dt

]
. (2.1)

Proof. Let x = 1
tb−1+(1−t)a−1 , t ∈ [0, 1] and x ∈ [a,b], then(

1

a
−

1

b

)∫ 1−λ
0

t

(tb−1 + (1 − t)a−1)2
f′
(

1

tb−1 + (1 − t)a−1

)
dt

= tf

(
1

tb−1 + (1 − t)a−1

)∣∣∣∣1−λ
0

−
∫ u
a

f(x)

x2
ab

b−a
dx

= (1 −λ)f
(

ab

(1 −λ)a + λb

)
−

ab

b−a

∫ u
a

f(x)

x2
dx. (2.2)

and (
1

b
−

1

a

)∫ 1
1−λ

t

(tb−1 + (1 − t)a−1)2
f′
(

1

tb−1 + (1 − t)a−1

)
dt

= tf

(
1

tb−1 + (1 − t)a−1

)∣∣∣∣1
1−λ
−
∫ b
u

f(x)

x2
ab

b−a
dx

= λf

(
ab

(1 −λ)a + λb

)
−

ab

b−a

∫ b
u

f(x)

x2
dx, (2.3)

where u = 1
(1−λ)b−1+λa−1 . So (7) follows from (8) and (9).� �



NEW ESTIMATES OF HERMITE-HADAMARD INEQUALITIES 17

Remark 2.2 From (7) we derive the following two identities.

f

(
2ab

a + b

)
−

2ab

b−a

∫ b
a

f(x)

x2
dx

=
2(b−a)
ab

[∫ 1
2

0

t

[tb−1 + (1 − t)a−1]
f′
(

1

tb−1 + (1 − t)a−1

)
dt

−
∫ 1

1
2

t− 1
[tb−1 + (1 − t)a−1]

f′
(

1

tb−1 + (1 − t)a−1

)
dt

]
; (2.4)

f(a) + f(b)

2
−

ab

b−a

∫ b
a

f(x)

x2
dx

=
b−a
2ab

[∫ 1
0

t

[tb−1 + (1 − t)a−1]
f′
(

1

tb−1 + (1 − t)a−1

)
dt

−
∫ 1
0

t− 1
[tb−1 + (1 − t)a−1]

f′
(

1

tb−1 + (1 − t)a−1

)
dt

]
. (2.5)

Proof. Take λ = 1
2

in (7), we can derive (10).
We respectively take λ = 0 and λ = 1 in (7) and add two inequalities, then (11) is obtained. �

Lemma 2.3 By integral calculation, then

C1(a,b,λ) =

∫ 1−λ
0

t(1 − t)[tb + (1 − t)a]2dt

= (1 −λ)3
[

1

5
(b−a)2(1 −λ)2 +

1

2
a(b−a)(1 −λ) +

1

3
λ2
]

; (2.6)

C2(a,b,λ) =

∫ 1−λ
0

t2[tb + (1 − t)a]2dt

= (1 −λ)2
[(

1

20
+

1

5
λ

)
(b−a)2(1 −λ)2 +

(
1

6
+

1

2
λ

)
a(b−a)(1 −λ) +

(
1

6
+

1

3
λ

)
λ2
]

;

(2.7)

C3(a,b,λ) =

∫ 1
1−λ

t(1 − t)[tb + (1 − t)a]2dt =
∫ λ
0

(1 −u)u[(1 −u)b + ua]2du

= −
1

5
(b−a)2λ5 +

1

4
(a2 + 3b2 − 4ab)λ4 +

1

3
(2ab− 3b2)λ3 +

1

2
b2λ2; (2.8)

C4(a,b,λ) =

∫ 1
1−λ

(1 − t)2[tb + (1 − t)a]2dt =
∫ λ
0

u2[(1 −u)b + ua]2du

=
1

5
(b−a)2λ5 +

1

2
b(a− b)λ4 +

1

3
b2λ3, (2.9)

where u = 1 − t.

3. Main results

Our main results are stated as follows.



18 WEN WANG, JIBING QI

Theorem 3.1 Let f : I ⊂ R\{0} → R be a differentiable function and f′ ∈ L[a,b]. If |f′(x)|q is
harmonically convex on [a,b] with 0 ≤ a < b for q ≥ 1, then∣∣∣∣∣(1 −λ)f

(
ab

(1 −λ)a + λb

)
+ λf

(
ab

(1 −λ)a + λb

)
−

ab

b−a

∫ b
a

f(x)

x2
dx

∣∣∣∣∣
≤
b−a
ab

{
(C1(a,b,λ))

1−1
q
[
|f′(b)|qC1(a,b,λ) + |f′(a)|qC2(a,b,λ)

]1
q

+ (C3(a,b,λ))
1−1

q
[
|f′(b)|qC3(a,b,λ) + |f′(a)|qC4(a,b,λ)

]1
q

}
. (3.1)

Proof. From Lemma 2.1 and using Hölder inequality, further, since |f′(x)|q is harmonically convex on
[a,b], we have

I =

∣∣∣∣∣(1 −λ)f
(

ab

(1 −λ)a + λb

)
+ λf

(
ab

(1 −λ)a + λb

)
−

ab

b−a

∫ b
a

f(x)

x2
dx

∣∣∣∣∣
≤
b−a
ab

[∫ 1−λ
0

t

[tb−1 + (1 − t)a−1]

∣∣∣∣f′
(

1

tb−1 + (1 − t)a−1

)
dt

∣∣∣∣
+

∫ 1
1−λ

1 − t
[tb−1 + (1 − t)a−1]

∣∣∣∣f′
(

1

tb−1 + (1 − t)a−1

)
dt

∣∣∣∣
]

≤
b−a
ab


(∫ 1−λ

0

t

[tb−1 + (1 − t)a−1]2
dt

)1−1
q
(∫ 1−λ

0

t

[tb−1 + (1 − t)a−1]2

∣∣∣∣f′
(

1

tb−1 + (1 − t)a−1

)∣∣∣∣q dt
)1

q

+

(∫ 1
1−λ

1 − t
[tb−1 + (1 − t)a−1]2

dt

)1−1
q
∫ 1
1−λ

1 − t
[tb−1 + (1 − t)a−1]2

∣∣∣∣f′
(

1

tb−1 + (1 − t)a−1

)∣∣∣∣q dt
]

≤
b−a
ab



(∫ 1−λ

0

t

[tb−1 + (1 − t)a−1]2
dt

)1−1
q
[∫ 1−λ

0

t
[
t|f′(b)|q + (1 − t)|f′(a)|q

]
[tb−1 + (1 − t)a−1]2

dt

]1
q

+

(∫ 1
1−λ

1 − t
[tb−1 + (1 − t)a−1]2

dt

)1−1
q

[∫ 1
1−λ

(1 − t)
[
t|f′(b)|q + (1 − t)|f′(a)|q

]
[tb−1 + (1 − t)a−1]2

dt

]1
q


 .

Noticing that [tb−1 + (1 − t)a−1]−1 ≤ ta + (1 − t)b, and using (12-15), then, from above inequality we
have

I ≤
b−a
ab



(∫ 1−λ

0

t[tb + (1 − t)a]2dt

)1−1
q
[∫ 1−λ

0

t[tb + (1 − t)a]2
[
t|f′(b)|q + (1 − t)|f′(a)|qdt

]]1q

+

(∫ 1
1−λ

(1 − t)[tb + (1 − t)a]2dt
)1−1

q
[∫ 1

1−λ
(1 − t)[tb + (1 − t)a]2

[
t|f′(b)|q + (1 − t)|f′(a)|q

]
dt

]1
q

}

=
b−a
ab


(C1(a,b,λ))1−1q

[∫ 1−λ
0

[
|f′(b)|qt2[tb + (1 − t)a]2 + |f′(a)|qt(1 − t)[tb + (1 − t)a]2

]
dt

]1
q

+ (C3(a,b,λ))
1−1

q

[∫ 1
1−λ

[
|f′(b)|qt(1 − t)[tb + (1 − t)a]2 + |f′(a)|q(1 − t)2[tb + (1 − t)a]2

]
dt

]1
q

}

=
b−a
ab

{
(C1(a,b,λ))

1−1
q
[
|f′(b)|qC1(a,b,λ) + |f′(a)|qC2(a,b,λ)

]1
q

+ (C3(a,b,λ))
1−1

q
[
|f′(b)|qC3(a,b,λ) + |f′(a)|qC4(a,b,λ)

]1
q

}
.

So the proof is complete. �



NEW ESTIMATES OF HERMITE-HADAMARD INEQUALITIES 19

Corollary 3.2 Assume that all the assumptions of Theorem 3.1 are satisfied. If we take q = 1 and
M = max{|f′(a)|, |f′(b)|}, then∣∣∣∣∣(1 −λ)f

(
ab

(1 −λ)a + λb

)
+ λf

(
ab

(1 −λ)a + λb

)
−

ab

b−a

∫ b
a

f(x)

x2
dx

∣∣∣∣∣
≤
M(b−a)

ab

[
C1(a,b,λ) + C2(a,b,λ) + C3(a,b,λ) + C4(a,b,λ)

]
. (3.2)

Corollary 3.3 Assume that all the assumptions of Theorem 3.1 are satisfied. If we take λ = q = 1
and M = max{|f′(a)|, |f′(b)|}, then∣∣∣∣∣f(a) − abb−a

∫ b
a

f(x)

x2
dx

∣∣∣∣∣ ≤ M(b−a)ab
[

1

4
(b−a)2 +

1

6
b(b− 4a)

]
. (3.3)

Corollary 3.4 Assume that all the assumptions of Theorem 3.1 are satisfied. If we take λ = 0,
q = 1 and M = max{|f′(a)|, |f′(b)|}, then∣∣∣∣∣f(b) − abb−a

∫ b
a

f(x)

x2
dx

∣∣∣∣∣ ≤ M(b−a)ab
[

1

4
(b−a)2 +

2

3
a(b−a)

]
. (3.4)

Corollary 3.5 Assume that all the assumptions of Theorem 3.1 are satisfied. If we take λ = 1
2

,
q = 1 and M = max{|f′(a)|, |f′(b)|}, then∣∣∣∣∣f

(
2ab

a + b

)
−

ab

b−a

∫ b
a

f(x)

x2
dx

∣∣∣∣∣ ≤ M(b−a)ab
×
{
C1(a,b,

1

2
)
[
C1(a,b,

1

2
) + C2(a,b,

1

2
)
]

+ C3(a,b,
1

2
)
[
C3(a,b,

1

2
) + C4(a,b,

1

2
)
]}

. (3.5)

Remark 3.6 By (18) and (19) we obtain the following inequality∣∣∣∣∣f(a) + f(b)2 − abb−a
∫ b
a

f(x)

x2
dx

∣∣∣∣∣ ≤ M(b−a)ab
[

1

12
(4b2 − 6ab−a2)

]
. (3.6)

4. Some Applications for Special Means

Let a,b are two nonnegative number with a < b. Let us recall the following special means of a and
b.

(1) The arithmetic mean A = A(a,b) := a+b
2

; (2) The geometric mean G = G(a,b) :=
√
ab;

(3) The harmonic mean H = H(a,b) := 2ab
a+b

; (4) The logarithmic mean L = L(a,b) := b−a
ln b−ln a;

(5) The -logarithmic mean

Lp = Lp(a,b) :=

(
bp+1 −ap+1

(p + 1)(b−a)

)1
p

, a 6= b,p ∈ R,p 6= 0,−1.

These means are often applied to numerical approximation and in other areas. However, the fol-
lowing simple relationships are known in the literature:

H ≤ G ≤ L ≤ A. (4.1)

It is also known that Lp is monotonically increasing respecting to p ∈ R, denoting L0 = I and
L−1 = L.



20 WEN WANG, JIBING QI

Proposition 1. Let 0 < a < b. Then we obtain the following inequalities:∣∣∣∣H − G2L
∣∣∣∣ ≤ b−aab

×
{
C1(a,b,

1

2
)
[
C1(a,b,

1

2
) + C2(a,b,

1

2
)
]

+ C3(a,b,
1

2
)
[
C3(a,b,

1

2
) + |C4(a,b,

1

2
)
]}

;∣∣∣∣A− G2L
∣∣∣∣ ≤ b−aab

[
1

12
(4b2 − 6ab−a2)

]
.

Proof. The assertion follows from inequalities (20) and (21), respectively, for f(x) = x, x ∈ R++. �

Proposition 2. Let 0 < a < b. Then we derive the following inequalities:∣∣H2 −G2∣∣ ≤ 2(b−a)
a

×
{
C1(a,b,

1

2
)
[
C1(a,b,

1

2
) + C2(a,b,

1

2
)
]

+ C3(a,b,
1

2
)
[
C3(a,b,

1

2
) + |C4(a,b,

1

2
)
]}

;∣∣∣∣A(a2,b2) − G2L
∣∣∣∣ ≤ 2(b−a)a

[
1

12
(4b2 − 6ab−a2)

]
.

Proof. The assertion follows from inequalities (20) and (21), respectively, for f(x) = x2, x ∈ R++. �

Proposition 3. Let 0 < a < b. Then we have the following inequalities:∣∣Hp+2 −LppG2∣∣ ≤ 2bp+2(b−a)ab
×
{
C1(a,b,

1

2
)
[
C1(a,b,

1

2
) + C2(a,b,

1

2
)
]

+ C3(a,b,
1

2
)
[
C3(a,b,

1

2
) + |C4(a,b,

1

2
)
]}

;

∣∣A(ap+2,bp+2) −LppG2∣∣ ≤ 2bn+2(b−a)ab
[

1

12
(4b2 − 6ab−a2)

]
.

Proof. The assertion follows from inequalities (20) and (21), respectively, for f(x) = xp+2, x ∈ R++
and p ∈ (−1,∞)\{0}. �

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1School of Mathematics and Statistics, Hefei Normal University, Hefei 230601,P.R.China

2School of Mathematical Science, University of Science and Technology of China, Hefei 230026, China

∗Corresponding author: wenwang1985@163.com, wwen2014@mail.ustc.edu.cn


	1. Introduction
	2. Some lemmas
	3. Main results
	4.  Some Applications for Special Means
	References