EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 3, 2023, 1448-1463
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Generalization of Jensen Mercer inequality on Delta
integral with Applications

Sadia Chanan1,2, Nazia Irshad3,∗, Afshan Khan4

1 Department of Mathematics, University of Karachi, University Road, Karachi-75270,
Pakistan
2 Sir Syed University of Engineering and Technology, Main University Road, Karachi 75300,
Pakistan
3 Department of Mathematics, Dawood University of Engineering and Technology, M. A
Jinnah Road, Karachi-74800, Pakistan
4 Department of Mathematics, Nazeer Hussain University, Karachi, Pakistan

Abstract. The aim of this article is to give generalization of Jensen Mercer inequality on delta
integral along with applications to Ky Fan inequality and related results.

2020 Mathematics Subject Classifications: 26A51,26D10

Key Words and Phrases: Jensen Mercer inequality, Time scale Calculus, Ky Fan Inequality

1. Introduction

One of the most well-known inequality in mathematics and statistics is Jensen’s in-
equality for convex functions. Because of their importance, Jensen’s inequality has received
numerous variants, generalizations, and refinements (for reference see [1, 3, 7, 9–12, 14–
18, 21, 22]). In 2003, Mercer established a variant of Jensen’s inequality known as the
Jensen-Mercer inequality [19], which as follows:

Proposition 1. Let ζ : [µ, ν] ⊂ I → R be a convex function and xi ∈ [µ, ν],

s. t.
n∑

i=1

ωi = 1, for 1 ≤ i ≤ n, then

ζ

(
µ+ ν −

n∑
i=1

ωixi

)
≤ ζ (µ) + ζ (v)−

n∑
i=1

ωiζ (xi) .

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v16i3.4784

Email addresses: sadiachanankhan@yahoo.com sadia.khan@ssuet.edu.pk (S. Chanan),
nazia.irshad@duet.edu.pk (N. Irshad), afshan khan19@hotmail.com (A. Khan)

https://www.ejpam.com 1448 © 2023 EJPAM All rights reserved.



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1449

In 1988, Stefan Hilger introduced the idea of theory of time scale calculus in order
to unify discrete and continuous analysis and also extend the traditional differential and
difference equations in [13]. There have been over a thousand publications in this field,
with numerous applications [5] in all branches of science. For interest, readers can see
Bohner and Peterson’s monograph [6] for an introduction to single-variable time scale
calculus and its applications.

Definition 1. A time scale is an arbitrary nonempty closed subset of the real numbers.
Examples of time scales are R,Z and qN0 := {qk|k ∈ N0}. The complex number are not
time scale.

Definition 2. If T is a time scale, then we define forward jump operator
σ : T → R by σ(θ̂) := inf{τ ∈ T|τ > θ̂} for all θ̂ ∈ T, the backward jump operator
ρ : T → R by ρ(θ̂) := sup{τ ∈ T|τ < θ̂} for all θ̂ ∈ T, and the graininess function
µ : T → [0,∞) by µ(θ̂) := σ(θ̂) − θ̂ for all θ̂ ∈ T. Furthermore for a function g : T → R,
we define gσ(θ̂) = g(σ(θ̂)) for all θ̂ ∈ T and gρ(θ̂) = g(ρ(θ̂)) for all θ̂ ∈ T. In this definition
we use inf ∅ = supT (i.e., ρ(θ̂) = θ̂ if θ̂ is the maximum of T) and sup ∅ = inf T (i.e.,
ρ(θ̂) = θ̂ if θ̂ is the minimum of T).

These definitions allow us to characterize every point in a time scale as following
classification of points:

(i) θ̂ right-scattered =⇒ θ̂ < σ(θ̂),

(ii) θ̂ right-dense =⇒ θ̂ = σ(θ̂),

(iii) θ̂ left-scattered =⇒ ρ(θ̂) < θ̂,

(iv) θ̂ left-dense =⇒ ρ(θ̂) = θ̂,

(v) θ̂ isolated =⇒ ρ(θ̂) < θ̂ < σ(θ̂),

(vi) θ̂ dense =⇒ ρ(θ̂) = θ̂ = σ(θ̂).

We define,
If T has a left scattered maximum M1, then we define Tk = T/M1; otherwise Tk = T.
If T has a right scattered maximum M2, then we define Tk = T/M2; otherwise Tk = T.
Finally we define T∗ = Tk

⋂
Tk. The mapping µ, ν : T → [0,∞) defined by

µ(t) = σ(t)− t

and
ν(t) = t− ρ(t)

are called the forward and backward graininess functions, respectively. In the following
consideration, IT = I

⋂
T will denote a time scale interval.



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1450

Definition 3. Let f : T → R be a real function on time scale T. Then for t ∈ Tk, we
define a f∆(t) to be a number with the property that given any ε > 0, there is a neighbour
hood UT of t such that

|(f(ρ(t))− f(s))− f∆(t)[ρ(t)− s]| ≤ ε|ρ(t)− s|

for all s ∈ UT we call f∆(t) the delta derivative of f at t.

For f : T → R , then we define fσ : T → R by fσ(t) = f(σ(t)) for t ∈ T. We define
fρ : T → R by fρ(t) = f(ρ(t)) for t ∈ T.
Following properties holds for t ∈ Tk.

(i) If f is ∆ differentiable at t, then f is continuous at t.

(ii) If f is continuous at t and t is right-scattered, then f is delta differentiable at t with
f∆(t) = (f(t)− f(σ(t))/υ(t))

(iii) If f is right-dense, then f is delta differentiable at t if and only if the lims→t(f(t)−
f(s))/(t− s) exit, then f∆(t) = lims→t(f(t)− f(s))/(t− s).

(iv) If f is ∆ differentiable at t, then fρ(t) = f(t) + f∆(t)υ(t)).

For more details on time scale, we refer the reader to [20].
In [23], Jensen inequality on delta integral is given as follow:

Proposition 2. Let a, b ∈ T, a < b and I ⊂ R, g ∈ C([a, b]T, I) and ω ∈ C([a, b]T,R) is a
probability density function and also ζ ∈ C(I,R) is convex, then

ζ

(∫ b

a
ω(†)g(η)∆η

)
≤
∫ b

a
ω(†)ζ (g(η))∆η. (1)

In this article we generalize Jensen Mercer inequality for ∆−integral . Also give gen-
eralization and refinement of Ky Fan inequality and its related results for ∆−integral.

We establish a Jensen Mercer ∆−integral inequality. Before we further proceed we
recall here a lemma from [19] stated as under:

Lemma 1. Let ζ : [µ, ν] ⊂ I → R be a convex function and xi ∈ [µ, ν]. Then

ζ (µ+ ν − xi) ≤ ζ (µ) + ζ (v)− ζ (xi) , 1 ≤ i ≤ n.

Theorem 1. If g ∈ C([a, b]T, [µ, ν]) and ω ∈ C([a, b]T, [µ, ν]) is a probability density
function and also ζ ∈ C([µ, ν],R) is convex, then

ζ

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

)
≤ ζ (µ) + ζ (v) −

∫ b

a
ω(†)ζ (g(η))∆η. (2)



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1451

Proof.

Since ζ is convex function and

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

)
∈ [µ, ν], therefore by in-

equality (1) and Lemma 1, we have

ζ

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

)
= ζ

(∫ b

a
ω(†)(µ+ ν − g(η))∆η

)
≤
∫ b

a
ω(†) [ζ (µ+ ν − g(η))∆η]

= ζ (µ) + ζ (v)−
∫ b

a
ω(†)ζ (g(η))∆η.

Corollary 1. Let T = R and by considering assumptions of Theorem 1. Then

ζ

(
µ+ ν −

∫ b

a
ω(†)g(η)dη

)
≤ ζ (µ) + ζ (v) −

∫ b

a
ω(†)ζ (g(η)) dη. (3)

1.1. Cases

(i) Let g(η) > 0 on [a, b]T and ζ(†) = tβ is convex and concave on (0,+∞) for β < 0 or
β > 1 and for β ∈ (0, 1) respectively. Then

ζ

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

)β

≤ ζ (µ)β + ζ (v)β −
∫ b

a
ω(†)ζ (g(η))β ∆η, β < 0 or β > 1,

(4)

ζ

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

)β

≥ ζ (µ)β + ζ (v)β −
∫ b

a
ω(†)ζ (g(η))β ∆η, β ∈ (0, 1).

(5)

(ii) Let g(η) > 0 on [a, b]T and ζ(†) = ln(†) is concave on (0,+∞). Then

ln

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

)
≤ ln (µ) + ln (v)−

∫ b

a
ω(†) ln (g(η))∆η.



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1452

(iii) Let T = Z and m ∈ N. Fix a = 1 and b = m + 1, let g : {1, . . . ,m + 1} → (0,∞),
ζ = − lnx is convex on (0,+∞) and by using Theorem 1, we get

ln

µ+ ν −
m∑
†=1

g(†)


= ln

(
µ+ ν −

∫ m+1

1
g(η)∆η

)
≥ ln (µ) + ln (v)−

∫ m+1

1
ln (g(η))∆η

= ln (µν)−

 m∑
†=1

ln(g(†))


= ln (µν)−

ln
 m∏

†=1

g(†)

 ,

and hence

ln

µ+ ν −
m∑
†=1

g(†)

 ≥ ln
(µν) m∏

†=1

g(†)

 .

(iv) Let T = 2M⊬ and M ∈ N. Fix a = 1 and b = 2M and consider a function g : {2l :
0 ≤ l ≤ N} → (0,∞) by using Theorem 1, we get

ln

(
µ+ ν −

∫ 2M

1
g(†)∆

)
= ln

(
µ+ ν −

M−1∑
l=0

2lg(2l)

)

= ln

(
µ+ ν −

∫ 2M

1
g(η)∆η

)
≥ ln (µ) + ln (v)−

∫ 2M

1
ln (g(η))∆η.

= ln (µν)−
∫ 2M

1
ln (g(†))∆†

= ln (µν)−
M−1∑
l=0

2l ln
(
g(2l)

)



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1453

= ln (µν)−

[
ln

M−1∏
l=0

(g(2l))2
l

]

and hence

µ+ ν −

(
M−1∑
l=0

2lg(2l)

)
≥ µν

M−1∏
l=0

(
g(2l)

)2l .

2. Ky Fan inequality and related results

In 1961, the Ky Fan inequality was given in the famous monograph ‘Inequalities’ in [4]
as follow:

Ǧn

Ǧ′
n

≤ Ǎn

Ǎ′
n

, xj ∈
(
0,

1

2

]
(6)

equality holds iff x1 = · · · = xn, which magnetize the attention of several mathematician.
For generalization and refinement of the Ky Fan inequality see papers [2, 8]. (and references
therein)

In this section, we are improving Ky Fan inequality and related results for time scale
calculus.

By considering assumptions of Theorem 1, we define the generalized weighted arith-
metic mean of g ∈ C([a, b]T, [µ, ν]) with weight ω :

Ǎ[µ,ν](g, ω) = µ+ ν −
∫ b

a
ω(†)g(η)∆η, (7)

the generalized weighted geometric mean of the g ∈ C([a, b]T, [µ, ν]) of weight ω:

Ǧ[µ,ν](g, ω) = exp

[
ln (µν)−

∫ b

a
ω(†) ln (g(η))∆η

]
, (8)

the generalized weighted harmonic mean of the g ∈ C([a, b]T, [µ, ν]) of weight ω:

Ȟ[µ,ν](g, ω) =

(
1

µ
+

1

ν
−
∫ b

a
ω(†) 1

(g(η))∆η

)−1

. (9)

Examples

(i) Let T = R. Then

Ǎ[µ,ν](g, ω) = µ+ ν −
∫ b

a
ω(†)g(η)dη, (10)



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1454

G[µ,ν](g, ω) = exp

[
ln (µν)−

∫ b

a
ω(†) ln (g(η)) dη

]
, (11)

and

Ȟ[µ,ν](g, ω) =

(
1

µ
+

1

ν
−
∫ b

a
ω(†) 1

(g(η)) dη

)−1

. (12)

(ii) Let T = Z, a = 1 and b = n + 1, we define ω(i) = ωi and g(i) = gi. The condition

for weight for ω means that
n∑

i=1

ωi > 0. Then, we have

Ǎ[µ,ν](g, ω) = Ǎn(g, ω) = µ+ ν −
n∑

i=1

ωigi, (13)

Ǧ[µ,ν](g, ω) = Ḡn(g, ω) =
(µν)
n∏

i=1

gωi
i

, (14)

and

Ȟ[µ,ν](g, ω) = Ȟn(g, ω) =

(
1

µ
+

1

ν
−

n∑
i=1

ωi
1

(gi)

)−1

. (15)

Now we establish generalized Ky Fan inequality for time scale.

Theorem 2. By considering assumptions of Theorem 1, and g(η) ∈ (0, γ2 ], where 0 < r <
γ < 1, then

Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)
≥

Ǎ[µ,ν](γ − g, ω)

Ǧ[µ,ν](γ − g, ω)
.

Proof. By applying ζ(x) = ln γ−x
x , x ∈ (0, γ2 ] to the Theorem 1, we obtain required

results.

In next theorem we provide refinement of the Ky Fan inequality as follow:

Theorem 3. By considering assumptions of Theorem 1 and 0 < n ≤ g(η) ≤ N, γ > 0,
then

Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)
≥

[
Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)

] N2

(γ −N)2 ≥
Ǎ[µ,ν](γ − x, ω)

Ǧ[µ,ν](γ − x, ω)



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1455

≥

[
Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)

] n2

(γ − n)2 ≥ 1. (16)

Proof. From the inequality
Ǎ[µ,ν](g,ω)

Ǧ[µ,ν](g,ω)
≥ 1 and n,N ∈ (0, γ2 ], the first and last inequali-

ties deduced directly.
Let ϕ : (0, γ) → R, ϕ(r) = ln[(γ−r

r )] + α ln(r) with α ∈ R, we have

ϕ′(r) = − 1

r(γ − r)
+

α

r
, r ∈ (0, γ),

ϕ′′(r) =
1

r2

[
γ(γ − 2r)

(γ − r)2
− α

]
, r ∈ (0, γ).

If φ : (0, γ) → R, defined as φ(r) = γ(γ−2r)
(γ−r)2

, then φ′(r) = 2r(r−1)
(1−r)4

, indicating φ is mono-

tonically strictly decreasing on (0, γ). Consequently for r ∈ (n,N), we have

1− 2N

(1−N)2
= φ(N) ≤ φ(r) ≤ φ(n) =

1− 2n

(1− n)2
. (17)

If α ≤ γ(γ−2N)
(γ−N)2

, we conclude from (17) that the function φ is strictly convex on (n,N).

Applying Theorem 1 to the function

ϕ : (n,N) → R,

ϕ(r) = ln

[(
γ − r

r

)]
+ α ln (r) ,

with α ≤ γ(γ−2N)
(γ−N)2

, we conclude that

ln

(
γ − u

u

)
+ α lnu+ ln

(
γ−ν

ν

)
+ α ln v −

∫ b

a
ω(†)

[
ln

(
γ − g(η)

g(η)

)
+ α ln(g(η))

]
∆η

≥ ln

γ − u−ν +

∫ b

a
ω(†)g(η)∆η

µ+ ν −
∫ b

a
ω(†)g(η)∆η

+ α ln

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

)

ln
Ǧ[µ,ν](γ − g, ω)

Ǧ[µ,ν](g, ω)
+ α ln Ǧ[µ,ν](g, ω) ≥ ln

Ǎ[µ,ν](γ − g, ω)

Ǎ[µ,ν](g, ω)
+ α ln Ǎ[µ,ν](g, ω)(

Ǧ[µ,ν](g, ω)

Ǎ[µ,ν](g, ω)

)α

≥
Ǎ[µ,ν](γ − g, ω)

Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)

Ǧ[µ,ν](γ − g, ω)(
Ǧ[µ,ν](g, ω)

Ǎ[µ,ν](g, ω)

)α−1

≥

(
Ǎ[µ,ν](γ − x, ω)

Ǧ[µ,ν](γ − x, ω)

)
(18)



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1456

from (18) we observe that this inequality is best possible if we have α is maximal, i.e,
α = γ−2N

(γ−N)2 , that leads(
Ǧ[µ,ν](g, ω)

Ǎ[µ,ν](g, ω)

) γ−2N

(γ−N)2
−1

≥
Ǎ[µ,ν](γ − x, ω)

Ǧ[µ,ν](γ − x, ω)

which yield to the second inequality in (16).

We established the third inequality by using the function F (r) = β ln r− ln
[
(γ−r)

r

]
and

the same technique. If β ≥ γ−2n
(γ−n)2

is true, then the function is strictly convex on (n,N).

Remark 1. Since the Ky Fan inequality is also equivalent to

Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)
≥

Ǎ[µ,ν](γ − g, ω)

Ǧ[µ,ν](γ − g, ω)
,

then the first part of the inequality may be seen as refinement of the Ky Fan inequality
while the second part

Ǎ[µ,ν](γ − g, ω)

Ǧ[µ,ν](γ − g, ω)
≥

(
Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)

) n2

(γ−n)2

can be considered as a counter part of the Ky Fan inequality.

Remark 2. (i) Let T = Z, a = 1 and b = n+1, we define ω(i) = ωi and g(i) = gi. The

condition for ω means that
n∑

i=1

ωi > 0. Then, we have

Ǎn(g, ω)

Ǧn(g, ω)
≥
[
Ǎn(g, ω)

Ǧn(g, ω)

] N2

(γ −N)2 ≥ Ǎn(γ − g, ω)

Ǧn(γ − g, ω)
≥
[
Ǎn(g, ω)

Ǧn(g, ω)

] n2

(γ − n)2 ≥ 1.

(ii) Let T = R. Then for weight ω : R → R and for continuous function g : R → R with
g([µ, ν]) ⊂ [n,N ] ⊂ (0, γ2 ], we have

Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)
≥

[
Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)

] N2

(γ −N)2 ≥
Ǎ[µ,ν](γ − g, ω)

Ǧ[µ,ν](γ − g, ω)

≥

[
Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)

] n2

(γ − n)2 ≥ 1.



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1457

Now, we will prove a result related to the inequality Ǎ[µ,ν](γ−x, ω) ≥ Ǧ[µ,ν](γ−x, ω).

Theorem 4. By considering the assumptions of Theorem 1 and also ζ ∈ C([µ, ν],R) is
convex and γ > 0, then

Ǎ[µ,ν](γ − g, ω) ≥ Ǧ[µ,ν](γ − g, ω).

Proof. By applying ζ(x) = x − ln(γ − x) for all x ∈ (0, γ2 ] to the Theorem 1, we get
required result.

Now, we present refinement of Ky Fan inequality via convexity.

Theorem 5. By considering the assumptions of Theorem 1, we get

Ǎ[µ,ν](γ − g, ω)

Ǧ[µ,ν](γ − x, ω)
≤ 1

Ǧ[µ,ν](g, ω) + Ǧ[µ,ν](γ − g, ω)
≤

Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)
. (19)

Proof. By using ζ (x) =
1

1 + ex
, for strictly convex on [0,∞) and strictly concave

on (−∞, 0]. We apply convex function to the inequality (2) and we define g(η) =

ln
γ − g(η)

g(η)
≥ 0, µ = ln

(
γ − µ

µ

)
, ν = ln

(
γ − ν

ν

)
, by which we get,

1

1 + e

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

) ≤ 1

1 + eu
+

1

1 + eν
−
∫ b

a
ω(†) 1

1 + eg(η)

1

1 + exp

(
ln
(
γ−µ
µ

)
+ ln

(γ−ν
ν

)
−
∫ b

a
ω(†) ln

(
γ − g(η)

g(η)

)
∆η

)
≤ 1

1 + exp

(
ln

(
γ − µ

u

)) +
1

1 + exp

(
ln

(
γ − ν

ν

))

−
∫ b

a
ω(†)

 1

1 + exp
(
ln
(
γ−g(η)
g(η)

))
∆η

which gives,

1

1 + exp

(
ln(γ − µ)(γ − ν)−

∫ b

a
ω(†) ln(γ − g(η))∆η

)
− ln(µν) +

∫ b

a
ω(†) ln(g(η))∆η

≤
(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

)



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1458

consequently,
1

1 + exp

(
ln

G[µ,ν]((γ − g), ω)

G[µ,ν](g, ω)

) ≤ A[µ,ν](g, ω)

or
1

Ǧ[µ,ν](g, ω) + Ǧ[µ,ν](γ − g, ω)
≤

Ǎ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)
, this gives the right hand side of (19).

Now by applying the Theorem 1 for the convex function −ζ on (−∞, 0] with g(η) =

ln
g(η)

γ − g(η)
≤ 0, we get left side of the inequality (19).

Now, we will establish Ǎ[µ,ν](g, ω) ≥ Ȟ[µ,ν](g, ω) and Ǎ[µ,ν](1− g, ω) ≥ Ȟ[µ,ν](1− g, ω).

Theorem 6. By considering the assumptions of Theorem 1 also by considering g(η) ∈
(0, γ2 ] ⊂ [µ, ν], then

(i) Ǎ[µ,ν](g, ω) ≥ Ȟ[µ,ν](g, ω).

(ii) Ǎ[µ,ν](γ − g, ω) ≥ Ȟ[µ,ν](γ − g, ω).

Proof. (i) By using ζ(x) = 1
x for all x ∈ (0, γ2 ] to the inequality (2) we get,

1

µ+ ν −
∫ b

a
ω(†)g(η)∆η

≤ 1

µ
+

1

ν
−
∫ b

a
ω(†)

(
1

g(η)

)
∆η

A[µ,ν](g, ω) ≥ 1

1
µ + 1

ν −
∫ b

a
ω(†)

(
1

g(η)

)
∆η

A[µ,ν](g, ω) ≥ H[µ,ν](g, ω).

Proof. (ii) By using ϕ(x) = 1
γ−x for all x ∈ (0, γ2 ] to the inequality (2), we get,

1

γ − (µ+ ν −
∫ b

a
ω(†)g(η)∆η)

≤ 1

γ − µ
+

1

γ − ν
−
∫ b

a
ω(†)

(
1

γ − g(η)

)
∆η

1

(γ − µ) + (γ − ν)−
∫ b

a
ω(†)(γ − g(η))∆η

≤ 1

γ − µ
+

1

γ − ν
−
∫ b

a
ω(†)

(
1

γ − g(η)

)
∆η

Ǎ[µ,ν](γ − g, ω) ≥ Ȟ[µ,ν](γ − g, ω).

Now, we present arithmetic and harmonic mean inequality.



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1459

Theorem 7. By considering the assumptions of Theorem 1, we get

1

Ǎ[µ,ν](g, ω)
− 1

Ǎ[µ,ν](γ − g, ω)
≤ 1

Ȟ[µ,ν](g, ω)
− 1

Ȟ[µ,ν](γ − g, ω)
.

Proof. We establish it by applying the Theorem 1 to the function,

ζ (z) =
1

z
− 1

γ − z
,
(
0 < z ≤ γ

2

)
.

By which we get, 1

µ+ ν −
∫ b

a
ω(†)g(η)∆η

−

 1

(γ − µ) + (γ − ν)−
∫ b

a
ω(†) (γ − g(η))∆η


≤
(
1

µ
− 1

1− µ
+

1

ν
− 1

γ − ν
−
∫ b

a
ω(†)

(
1

g(η)
− 1

γ − g(η)

)
∆η

)
left side of inequality gives,

1

Ǎ[µ,ν](g, ω)
− 1

Ǎ[µ,ν](γ − g, ω)

from the right side of the inequality we obtain,[
1

µ
+

1

ν
−
∫ b

a
ω(†) 1

g(η)
∆η

]
−
[

1

γ − µ
+

1

γ − ν
−
∫ b

a
ω(†)

(
1

(γ − g(η))

)
∆η

]
1[

1

µ
+

1

ν
−
∫ b

a
ω(†) 1

g(η)
∆η

]−1 − 1[
1

γ − µ
+

1

γ − ν
−
∫ b

a
ω(†)

(
1

γ − g(η)

)
∆η

]−1

=
1

Ȟ[µ,ν](g, ω)
− 1

Ȟ[µ,ν](γ − g, ω)

that completes the proof.

Now, we establish geometric and harmonic mean inequality for time scale.

Theorem 8. By considering the assumptions of Theorem 1, we get
Ȟ[µ,ν](g, ω) ≤ Ǧ[µ,ν](g, ω) .

Proof. By using ϕ(x) = ex for all x ∈ [−∞,∞) to the inequality (2) we get

exp

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

)
≤ exp (µ) + exp (ν)−

∫ b

a
ω(†) exp (g(η))∆η.

By replacing µ by ln
(

1
µ

)
, ν by ln

(
1
ν

)
and g(η) by ln

(
1

g(η)

)
, then we obtained result.



S. Chanan, N. Irshad, A. Khan / Eur. J. Pure Appl. Math, 16 (3) (2023), 1448-1463 1460

Theorem 9. By considering the assumptions of Theorem 1, we get
Ȟ[µ,ν](γ − g, ω) ≤ Ǧ[µ,ν](γ − g, ω) .

Proof. By applying ϕ(x) = ex for all x ∈ (0, γ2 ] to the inequality (2),

exp

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

)
≤ exp (µ) + exp (ν)−

∫ b

a
ω(†) exp (g(η))∆η.

By using µ = ln
(

1
γ−µ

)
, ν = ln

(
1

γ−ν

)
and g(η) = ln

(
1

γ−g(η)

)
, then we get required result.

Theorem 10. By considering the assumptions of Theorem 1, we get

Ȟ[µ,ν](γ − g, ω)

Ȟ[µ,ν](g, ω)
≤

Ǧ[µ,ν](γ − g, ω)

Ǧ[µ,ν](g, ω)
. (20)

Proof. Without loss of generality we suppose that g′js are not equal and by using the

strictly convex function ϕ (z) = ln

(
γ − z

z

)
, for all z ∈ (0, γ2 ].

We set,

y =
H[µ,ν](g, ω)

H[µ,ν](g, ω) +H[µ,ν](γ − g, ω)
, y ∈

(
0,

γ

2

]
.

Then we get,

ln


1−

H[µ,ν](g, ω)

H[µ,ν](g, ω) +H[µ,ν](γ − g, ω)

H[µ,ν](g, ω)

H[µ,ν](g, ω) +H[µ,ν](γ − g, ω)


= ln

(
H[µ,ν](γ − g, ω)

H[µ,ν](g, ω)

)

= ln


1
µ + 1

ν −
∫ b

a
ω(†) 1

g(η)
∆η

1
γ−µ + 1

γ−ν −
∫ b

a
ω(†) 1

γ − g(η)
∆η


= ln

(
1

µ
+

1

ν
−
∫ b

a
ω(†) 1

g(η)
∆η

)
− ln

(
1

γ − µ
+

1

γ − µν
−
∫ b

a
ω(†) 1

γ − g(η)
∆η

)
≤
(
ln

1

µν
−
∫ b

a
ω(†) ln( 1

g(η)
)∆η

)
−
(
ln

1

(γ − µ)(γ − ν)
−
∫ b

a
ω(†) ln

(
1

γ − g(η)

)
∆η

)



REFERENCES 1461

right side of the inequality gives,

ln

(
Ǧ[µ,ν](γ − g, ω)

Ǧ[µ,ν](g, ω)

)
.

We get required inequality (20) by taking exponential on both sides.

Theorem 11. By considering the assumptions of Theorem 1, we get

Ȟ[µ,ν](γ − g, ω)

Ǧ[µ,ν](γ − g, ω)
≤ 1

Ǧ[µ,ν](g, ω) + Ǧ[µ,ν](γ − g, ω)
≤

Ȟ[µ,ν](g, ω)

Ǧ[µ,ν](g, ω)
. (21)

Proof. For the left hand inequality we apply function ζ(x) =
1

ex
+1 on (−∞,∞] to the

inequality (2) that is,

1

exp

(
µ+ ν −

∫ b

a
ω(†)g(η)∆η

) + 1

≤
(

1

exp(µ)
+ 1

)
+

(
1

exp(ν)
+ 1

)
−
∫ b

a
ω(†) 1

exp(g(η)) + 1
∆η

by replacing g(η) = ln(γ−g(η)
g(η) ) ≥ 0, µ by ln(γ−µ

µ ) and ν by ln(γ−ν
ν ), we get

1

exp

[
ln

(
Ǧ[µ,ν](γ−g,ω)

Ǧ[µ,ν](γ−g,ω)

)] + 1 ≤
(

1

γ − µ
+

1

γ − ν
−
∫ b

a
ω(†) 1

(γ − g(η))
∆η

)

Ǧ[µ,ν](g, ω) + Ǧ[µ,ν](γ − g, ω)

Ǧ[µ,ν](γ − g, ω)
≤
(

1

γ − µ
+

1

γ − ν
−
∫ b

a
ω(†) 1

(γ − g(η))
∆η

)
Ȟ[µ,ν](γ − g, ω)

Ǧ[µ,ν](γ − g, ω)
≤ 1

Ǧ[µ,ν](γ − g, ω) + Ǧ[µ,ν](g, ω)
.

To prove right-hand of the inequality (21) we apply inequality (2) to the convex function

−ζ on (−∞,∞] with g(η) = ln
(

g(η)
γ−g(η)

)
≤ 0, µ = ln

(
µ

γ−µ

)
, ν = ln

(
ν

γ−ν

)
.

References

[1] M. M. Ali and A. R. Khan. Generalized integral mercer’s inequality and integral
means. J. Inequal. Special Funct., 10(1):60–76, 2019.



REFERENCES 1462

[2] H. Alzer. The inequality of ky fan’s and related results. Acta App. Math., 38:305–354,
1995.

[3] M. K. Bakula and J. Pečarić. On the Jensen’s inequality for convex functions on
the co-ordinates in a rectangle from the plane. Taiwanese J. Math., 10(5):1271–1292,
2006.

[4] E. F. Beckenback and R. Bellman. Inequalities. Springer Verlag, Berlin, Berlin, 1961.

[5] M. Bohner and S. G. Georgiev. Multivariable Dynamic Calculus on Time Scales.
Springer International Publishing Switzerland, Switzerland, 2016.

[6] M. Bohner and A. Peterson. Dynamic equations on time scales. MA: Birkhäuser
Boston Inc., Boston, 2011.

[7] I. Brentic, K. A. Khan, and J. Pečarić. Refinements of Jensen’s Inequality with
applications to cyclic mixed symmetric means and Cauchy means. J. Math. Ineqaul.,
9(4):1309–1321, 2015.

[8] S. Chanan, S. Ahmed A.R. Khan, and N. Raisat. Generalizations of ky fan inequality
and related results. J. Inequal. and Special Functions, 10:123–142, 2019.

[9] S. Chanan, A. R. Khan, and I. Khan. Gabler inequality for functions with nonde-
creasing increments of convex type. Adv. Inequal. Appl., 10(3), 2020.

[10] W. S. Cheung, A. Matković, and J. Pečarić. A variant of jessen’s inequality and
generalized means. J. Ineq. Pure Appl. Math., 7(1):Article 10, 2006.

[11] S. S. Dragomir. Some refinements of jensen’s inequality. J. Math. Anal. Appl.,
168(2):518–522, 1992.

[12] S. S. Dragomir. A new refinement of Jensen’s inequality in linear spaces with appli-
cations. Mathematical and Computer Modelling, 52:153–163, 2010.

[13] S. Hilger. Ein Maßkettenkalkül mit Anwendung auf Zentrumsmannigfaltigkeiten. PhD
thesis, Universität Würzburg, 1988.

[14] S. Hussain and J. Pečarić. An improvement of Jensen’s inequality with some appli-
cations. Asian-European J. Math., 2(1):85–94, 2009.

[15] A. R. Khan and I. Khan. An Extension of Jensen-Mercer Inequality for Functions
with Nondecreasing Increments. J. Inequal. Special Funct, 10(3):1–15, 2019.

[16] A. R. Khan, I. Khan, and S. S. A. Ramji. Generalization and Refinements of Jensen-
Mercer Inequality with Applications. J. Math. Inequal., 15(4):1341–1360, 2021.

[17] A. R. Khan, J. Pečarić, and M. Praljak. A note on generalized mercer’s inequality.
Bull. Malays. Math. Sci. Soc., 40:881–889, 2017.



REFERENCES 1463

[18] M. A. Khan, A. R. Khan, and J. Pečarić. On the refinements of Jensen-Mercer’s
inequality. Rev. Anal. Numer. Theor. Approx., 41(1):62–81, 2012.

[19] A. Mcd. Mercer. A variant of Jensen’s inequality. J. Ineq. Pure and Appl. Math., 4,
2013.

[20] Q. Sheng, M. Fadag, J. Henderson, and J. M. Davis. An exploration of combined
dynamic derivativeson time scales and their applications,. Nonlinear Analysis: Real
World Applications, 7(3):395–413, 2006.

[21] X. L. Tang and J. J. Wen. Some developments of refined jensen’s inequality,. J.
Southwest Univ. Nationalities, 29.

[22] L. C. Wang, X. F. Ma, and L. H. Liu. A note on some new refinements of jensen’s
inequality for convex functions,. J. Ineq. Pure Appl. Math., 10(2):Article 48, 2006.

[23] F.-H. Wong, C.-C. Yeh, and W.-C. Lian. Advances in dynamical systems and appli-
cations. Nonlinear Functional Anal. Appl., 1(1):113–120, 2006.