# Petrović-Type Inequalities for Harmonich-convex Functions

Authors: Imran Abbas Baloch; Yu-Ming Chu
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

In the article, we establish several Petrović-type inequalities for the harmonich-convex (concave) function if h is a submultiplicative (super-multiplicative) function, provide some new majorizaton type inequalities for harmonic convex function, and prove the superadditivity, subadditivity, linearity, and monotonicity properties for the functionals derived from the Petrović type inequalities.


---

## Body

## 1. Introduction

LetI
⊆
R be a nonempty interval. Then a real-valued function f
:
I
→
R is said to be convex (concave) on I if the inequality(1)
f
t
x
+
1
−
t
y
≤
≥
t
f
x
+
1
−
t
f
y
,holds for allx
,
y
∈
I and t
∈
0
,
1.It is well known that the convex (concave) functions have wide applications in pure and applied mathematics [1–21], many remarkable properties and inequalities can be found in the literature [22–48] via the theory of convexity. Recently, a great deal of generalizations, extensions, and variants have been made for convexity, for example, GA-convexity [49], GG-convexity [50], s-convexity [51, 52], preinvex convexity [53], strong convexity [54–57], Schur convexity [58–60], and others [61–67].The following well-known Petrović inequality was established by Petrović [68] in 1932.Theorem 1.
Let
a
>
0
and
f
:
0
,
a
→
R
be a convex function. Then the inequality
(2)
∑
k
=
1
n
p
k
f
x
k
≤
f
∑
k
=
1
n
p
k
x
k
+
∑
k
=
1
n
p
k
−
1
f
0
,
holds for all nonnegative
n-tuples x
1
,
x
2
,
…
,
x
n
and
p
1
,
p
2
,
…
,
p
n
such that
x
i
≤
∑
k
=
1
n
p
k
x
k
≤
a
i
=
1
,
2
,
…
,
n.Now, we recall the definitions of various convexities and the majorization relation between twon-tuples.Definition 2.
LetI
⊆
R be an interval. Then a real-valued function f
:
I
→
R is said to be submultiplicative (supermultiplicative) on I if the inequality(3)
f
x
,
y
≤
≥
f
x
f
y
,
holds for allx
,
y
∈
I.Definition 3.
Letn
≥
2, a
=
a
1
,
a
2
,
…
,
a
n and b
=
b
1
,
b
2
,
…
,
b
n be two n-tuples. Then b is said to be majorized by a or a majorizes b (in symbol a
≻
b) if(4)
∑
i
=
1
k
a
i
≥
∑
i
=
1
k
b
i
,
fori
=
1
,
2
,
…
,
n
−
1 and(5)
∑
i
=
1
n
a
i
=
∑
i
=
1
n
b
i
,
wherea
i denotes the ith largest component in a.Definition 4.
LetI
⊆
R be an interval. Then a real-valued function f
:
I
→
R is said to be harmonic convex (concave) on I if the inequality(6)
f
x
y
t
x
+
1
−
t
y
≤
≥
t
f
y
+
1
−
t
f
x
,
holds for allx
,
y
∈
I with x
,
y
≠
0
,
0 and t
∈
0
,
1.Definition 5.
Letα
∈
0
,
1, m
∈
0
,
1, and I
⊆
R be an interval. Then a real-valued function f
:
I
→
R is said to be harmonic α
,
m-convex (concave) on I if the inequality(7)
f
m
x
y
1
−
t
x
+
m
t
y
≤
≥
t
α
f
x
+
m
1
−
t
α
f
y
,
holds for allx
,
y
∈
I with x
,
y
≠
0
,
0 and t
∈
0
,
1.Definition 6.
Lets
,
m
∈
0
,
1 and I
⊆
R be an interval. Then a real-valued function f
:
I
→
R is said to be harmonic s
,
m-convex (concave) in the second sense if the inequality(8)
f
m
x
y
1
−
t
x
+
m
t
y
≤
≥
t
s
f
x
+
m
1
−
t
s
f
y
,
holds for allx
,
y
∈
I with x
,
y
≠
0
,
0 and t
∈
0
,
1.Definition 7.
Lets
,
m
∈
0
,
1, p
∈
R
∖
0, and I
⊆
R be an interval. Then a real-valued function f
:
I
→
R is said to be harmonic p
,
s
,
m-convex (concave) on I if the inequality(9)
f
m
x
y
1
−
t
x
p
+
t
m
y
p
1
/
p
≤
≥
t
s
f
x
+
m
1
−
t
s
f
y
,
holds for allt
∈
0
,
1 and x
,
y
∈
I with x
,
y
≠
0
,
0 and m
y
∈
I.Definition 8.
LetI
∈
R be an interval and h
:
0
,
1
→
R be a nonnegative real-valued function. Then a real-valued function f
:
I
→
R is said to be harmonic h-convex (concave) if the inequality(10)
f
x
y
t
x
+
1
−
t
y
≤
≥
h
t
f
y
+
h
1
−
t
f
x
,holds for allt
∈
0
,
1 and x
,
y
∈
I with x
,
y
≠
0
,
0.Abbas Baloch [69] proved that.Lemma 9.
Let
I
⊆
0
,
∞
and
f
:
I
→
R
be a real-valued function. Then
f
is harmonic convex on
I
if and only if the function
x
f
x
is convex on
I
.The following three examples also can be found in the literature [67].Example 1.
Letf
x
=
x. Then f is harmonic convex on 0
,
∞ and harmonic concave on −
∞
,
0.Example 2.
Letf
:
0
,
∞
→
R be defined by(11)
f
x
=
x
,
0
<
x
<
2
,
4
−
4
x
,
x
≥
2
.
Thenf is harmonic convex on 0
,
∞ due to x
f
x is convex on 0
,
∞.Example 3.
The function(12)
g
x
=
x
−
1
2
+
1
x
,
is harmonic convex on0
,
∞ due to x
g
x is convex on 0
,
∞.Lemma 10.
(see [70]). Let
I
⊆
R
be an interval and
f
:
I
→
R
be a real-valued function. Then
f
is convex on
I
if and only if the bivariate function
(13)
h
f
x
,
y
=
f
y
−
f
x
y
−
x
,
is increasing with respect to its variablesx and y.


## 2. Main Results

Lemma 11.
Let
c
,
a
∈
R
with
c
<
a, x
1
,
x
2
,
…
,
x
n
and
p
1
,
p
2
,
…
,
p
n
be two n-tuples such that
(14)
c
<
x
i
≤
∑
k
=
1
n
p
k
x
k
≤
a
,
for
i
=
1
,
2
,
…
,
n, and
h
:
c
,
a
→
0
,
∞
and
f
:
c
,
a
→
R
be two real-valued functions. Then one has
(15)
∑
k
=
1
n
p
k
f
x
k
≤
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
f
∑
k
=
1
n
p
k
x
k
,
if the function
f
x
/
h
x
−
c
is increasing on
c
,
a.Proof.
It follows from (14) and the monotonicity of the function f
x
/
h
x
−
c on the interval c
,
a that(16)
f
∑
k
=
1
n
p
k
x
k
h
∑
k
=
1
n
p
k
x
k
−
c
≥
f
x
j
h
x
j
−
c
,
h
∑
k
=
1
n
p
k
x
k
−
c
f
x
j
≤
h
x
j
−
c
f
∑
k
=
1
n
p
k
x
k
,
for allj
=
1
,
2
,
…
,
n.
From above inequality we clearly see that(17)
h
∑
k
=
1
n
p
k
x
k
−
c
∑
j
=
1
n
p
j
f
x
j
≤
f
∑
k
=
1
n
p
k
x
k
∑
j
=
1
n
p
j
h
x
j
−
c
,
∑
k
=
1
n
p
k
f
x
k
≤
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
f
∑
k
=
1
n
p
k
x
k
.Theorem 12.
Let
c
,
a
∈
R
with
c
<
a
,
x
1
,
x
2
,
…
,
x
n
, and
p
1
,
p
2
,
…
,
p
n
be two n-tuples such that
(18)
c
<
x
i
≤
∑
k
=
1
n
p
k
x
k
≤
a
,
for
i
=
1
,
2
,
…
,
n
, and
h
:
c
,
a
→
0
,
∞
be a supermultiplicative function such that
(19)
h
α
+
h
1
−
α
≤
1
,
for all
α
∈
0
,
1
. Then one has
(20)
∑
k
=
1
n
p
k
x
k
f
x
k
≤
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
∑
k
=
1
n
p
k
x
k
f
∑
k
=
1
n
p
k
x
k
+
c
∑
k
=
1
n
p
k
−
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
f
c
,
if
f
:
c
,
a
→
R
is a harmonic
h
-convex function.Proof.
It follows from Lemma1 and the harmonic h-convexity of the function f on c
,
a that the function g
z
=
z
f
z is h-convex on c
,
a. Let x
,
y
∈
c
,
a such that x
=
α
y
+
1
−
α
c and(21)
S
h
x
=
g
x
−
g
c
h
x
−
c
.
Then we clearly see thaty
>
x and(22)
S
h
x
=
g
α
y
+
1
−
α
c
−
g
c
h
α
y
−
c
≤
h
α
g
y
+
h
1
−
α
−
1
g
c
h
α
y
−
c
.
Sinceh is a supermultiplicative function, we have(23)
S
h
x
≤
h
α
g
y
+
h
1
−
α
−
1
g
c
h
α
h
y
−
c
.
Making use of (19) we get(24)
S
h
x
≤
g
y
−
g
c
h
y
−
c
=
S
h
y
,
which implies that the functionS
h
x
=
g
x
−
g
c
/
h
x
−
c is increasing on c
,
a. Then from Lemma 11 one has(25)
∑
k
=
1
n
p
k
g
x
k
−
g
c
≤
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
g
∑
k
=
1
n
p
k
x
k
−
g
c
,
which is equivalent to(26)
∑
k
=
1
n
p
k
g
x
k
≤
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
g
∑
k
=
1
n
p
k
x
k
+
∑
k
=
1
n
p
k
−
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
g
c
.
Therefore, inequality (20) can be obtained immediately by replacing g
x
=
x
f
x in the last inequality above.
Leth
x
=
x. Then h is a supermultiplicative function and inequality (19) becomes an identity, and Theorem 12 leads to Theorem 13 immediately.Theorem 13.
Let
c
,
a
∈
R
with
c
<
a
,
x
1
,
x
2
,
…
,
x
n
, and
p
1
,
p
2
,
…
,
p
n
be two n-tuples such that
(27)
c
<
x
i
≤
∑
k
=
1
n
p
k
x
k
≤
a
,
for
i
=
1
,
2
,
…
,
n
. Then the inequality
(28)
∑
k
=
1
n
p
k
x
k
f
x
k
≤
∑
k
=
1
n
p
k
x
k
−
c
∑
k
=
1
n
p
k
x
k
−
c
∑
k
=
1
n
p
k
x
k
f
∑
k
=
1
n
p
k
x
k
+
c
∑
k
=
1
n
p
k
−
∑
k
=
1
n
p
k
x
k
−
c
∑
k
=
1
n
p
k
x
k
−
c
f
c
,
holds if
f
:
c
,
a
→
R
is a harmonic convex function.Remark 14.
Letc
,
a
∈
R with c
<
a, x
1
,
x
2
,
…
,
x
n, and p
1
,
p
2
,
…
,
p
n be two n-tuples such that(29)
c
<
x
i
≤
∑
k
=
1
n
p
k
x
k
≤
a
,
fori
=
1
,
2
,
…
,
n. Then from the proof of Theorem 12 we clearly see that the following statements are true:(1)
Ifh
:
c
,
a
→
0
,
∞ and f
:
c
,
a
→
R are two functions such that f
x
/
h
x
−
c is decreasing on c
,
a, then the reverse inequality of (15) holds.
(2)
Ifh
:
c
,
a
→
0
,
∞ is a submultiplicative function such that h
α
+
h
1
−
α
≥
1 for all α
∈
0
,
1 and f
:
c
,
a
→
R is a harmonic h-concave function, then the reverse inequality of (20) holds.
(3)
Iff
:
c
,
a
→
R is a harmonic concave function, then the reverse inequality of (28) holds.Theorem 15.
Let
I
∈
R
∖
0
,
a
1
,
a
2
,
…
,
a
n
,
b
1
,
b
2
,
…
,
b
n
∈
I
,
a
=
a
1
,
a
2
,
…
,
a
n
,
b
=
b
1
,
b
2
,
…
,
b
n
such that
a
≻
b
, and
f
:
I
→
R
be a harmonic convex function. Then the inequality
(30)
∑
i
=
1
n
a
i
f
a
i
≥
∑
i
=
1
n
b
i
f
b
i
,
holds.Proof.
Letg
z
=
z
f
z and(31)
λ
i
=
g
b
i
−
g
a
i
b
i
−
a
i
.
Then from Lemmas1 and 2 together with the harmonic convexity of f that g is convex on I and the sequence λ
i
i
=
1
n is decreasing. Let(32)
A
k
=
∑
i
=
1
k
a
i
,
B
k
=
∑
i
=
1
k
b
i
,
A
0
=
B
0
.
ThenA
n
=
B
n due to a
≻
b and(33)
∑
i
=
1
n
a
i
f
a
i
−
∑
i
=
1
n
b
i
f
b
i
=
∑
i
=
1
n
g
a
i
−
∑
i
=
1
n
g
b
i
=
∑
i
=
1
n
λ
i
a
i
−
b
i
=
∑
i
=
1
n
λ
i
A
i
−
A
i
−
1
−
B
i
+
B
i
−
1
=
∑
i
=
1
n
λ
i
A
i
−
B
i
−
∑
i
=
1
n
λ
i
A
i
−
1
−
B
i
−
1
=
∑
i
=
1
n
−
1
λ
i
A
i
−
B
i
−
∑
i
=
1
n
−
1
λ
i
+
1
A
i
−
B
i
=
∑
i
=
1
n
−
1
λ
i
−
λ
i
+
1
A
i
−
B
i
.
Therefore, inequality (4.0) follows from the monotonicity of the sequenceλ
i
i
=
1
n
−
1 and A
i
≥
B
i for all i
=
1
,
2
,
…
,
n
−
1 together with the last identity above.
The weighted version of Theorem15 can be stated as follows.Theorem 16.
Let
I
∈
R
∖
0
,
a
1
,
a
2
,
…
,
a
n
,
b
1
,
b
2
,
…
,
b
n
∈
I
,
a
=
a
1
,
a
2
,
…
,
a
n
,
b
=
b
1
,
b
2
,
…
,
b
n
such that
a
≻
b
,
p
=
p
1
,
p
2
,
…
,
p
n
∈
0
,
∞
n
and
f
:
I
→
R
be a harmonic convex function. Then the inequality
(34)
∑
i
=
1
n
p
i
a
i
f
a
i
≥
∑
i
=
1
n
p
i
b
i
f
b
i
,
holds.


## 3. Related Results

In this section, we present several interesting properties for the functionals derived from the Petrović-type inequalities given in Section2.Letc
,
a
∈
R with c
<
a, x
=
x
1
,
x
2
,
…
,
x
n
∈
c
,
a
n, and p
=
p
1
,
p
2
,
…
,
p
n
∈
0
,
∞
n be two n-tuples such that(35)
x
i
≤
∑
k
=
1
n
p
k
x
k
≤
a
,fori
=
1
,
2
,
…
,
n, and h
:
c
,
a
→
0
,
∞ and f
:
c
,
a
→
R be two real-valued functions. Then the Petrović-type functional F
1
x
,
p
;
f is defined by(36)
F
1
x
,
p
;
f
=
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
f
∑
k
=
1
n
p
k
x
k
−
∑
k
=
1
n
p
k
f
x
k
.Remark 17.
From Lemma11 we clearly see that(37)
F
1
x
,
p
;
f
≥
0
,
iff
x
/
h
x
−
c is increasing on c
,
a and(38)
F
1
x
,
p
;
f
≤
0
,
iff
x
/
h
x
−
c is decreasing on c
,
a.
LetI
∈
R
∖
0, a
=
a
1
,
a
2
,
…
,
a
n
,
b
=
b
1
,
b
2
,
…
,
b
n
∈
I
n, p
=
p
1
,
p
2
,
…
,
p
n
∈
0
,
∞
n, and f
:
I
→
R be a real-valued function. Then the Petrović-type functional F
2
a
,
b
,
p
;
f is defined by(39)
F
2
a
,
b
,
p
;
f
=
∑
i
=
1
n
p
i
a
i
f
a
i
−
∑
i
=
1
n
p
i
b
i
f
b
i
.Remark 18.
Theorem16 leads to the conclusion that(40)
F
2
a
,
b
,
p
;
f
≥
0
,
ifa
≻
b and f is harmonic convex on I and(41)
F
2
a
,
b
,
p
;
f
≤
0
,
ifa
≻
b and f is harmonic concave on I.Theorem 19.
Let
c
,
a
∈
R
with
c
<
a, x
=
x
1
,
x
2
,
…
,
x
n
∈
c
,
a
n
and
p
=
p
1
,
p
2
,
…
,
p
n
,
q
=
q
1
,
q
2
,
…
,
q
n
∈
0
,
∞
n
be two n-tuples such that
(42)
x
i
≤
∑
k
=
1
n
p
k
x
k
≤
a
,
x
i
≤
∑
k
=
1
n
q
k
x
k
≤
a
,
for
i
=
1
,
2
,
…
,
n
, and
h
:
c
,
a
→
1
,
∞
and
f
:
c
,
a
→
R
be two real-valued functions, and
F
1
be defined by (36). Then the following statements are true:
(1)
If
∑
k
=
1
n
p
k
+
q
k
x
k
∈
c
,
a
and
f
x
/
h
x
−
c
is increasing on
c
,
a
, then
(43)
F
1
x
,
p
+
q
;
f
≥
F
1
x
,
p
;
f
+
F
1
x
,
q
;
f
, that is
F
1
p
=
F
1
x
,
p
;
f
is superadditive on
0
,
∞
n.
(2)
If
p
≥
q
(namely,
p
j
≥
q
j
for all
j
=
1
,
2
,
…
,
n
) and
∑
k
=
1
n
p
k
−
q
k
x
k
≥
x
j
for all
j
=
1
,
2
,
…
,
n
, and
f
x
/
h
x
−
c
is increasing on
c
,
a
, then
(44)
F
1
x
,
p
;
f
≥
F
1
x
,
q
;
f
≥
0
,
that is
F
1
p
=
F
1
x
,
p
;
f
is increasing with respect to positive n-tuples.
(3)
If
∑
k
=
1
n
p
k
+
q
k
x
k
∈
c
,
a
and
f
x
/
h
x
−
c
is decreasing on
c
,
a
, then the reverse inequality of (43) holds, that is
F
1
p
=
F
1
x
,
p
;
f
is subadditive on
0
,
∞
n
. If
p
≥
q and ∑
k
=
1
n
p
k
−
q
k
x
k
≥
x
j
for all
j
=
1
,
2
,
…
,
n
, and
f
x
/
h
x
−
c
is decreasing on
c
,
a
, then the reverse inequality of (44) holds, that is
F
1
p
=
F
1
x
,
p
;
f
is decreasing with respect to positiven n-tuples.Proof.
(1) It follows from (36) that(45)
F
1
x
,
p
+
q
;
f
=
∑
k
=
1
n
p
k
+
q
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
+
q
k
x
k
−
c
⋅
f
∑
k
=
1
n
p
k
+
q
k
x
k
−
∑
k
=
1
n
p
k
+
q
k
f
x
k
=
∑
k
=
1
n
p
k
h
x
k
−
c
+
∑
k
=
1
n
q
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
+
∑
k
=
1
n
q
k
x
k
−
c
⋅
∑
k
=
1
n
p
k
x
k
+
∑
k
=
1
n
q
k
x
k
−
∑
k
=
1
n
p
k
f
x
k
−
∑
k
=
1
n
q
k
f
x
k
.
Making use of the monotonicity of the functionf
x
/
h
x
−
c and Lemma 11 one has(46)
∑
k
=
1
n
p
k
f
x
k
+
∑
k
=
1
n
q
k
f
x
k
≤
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
f
∑
k
=
1
n
p
k
x
k
+
∑
k
=
1
n
q
k
h
x
k
−
c
h
∑
k
=
1
n
q
k
x
k
−
c
f
∑
k
=
1
n
q
k
x
k
≤
h
∑
k
=
1
n
p
k
x
k
−
c
+
h
∑
k
=
1
n
q
k
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
+
∑
k
=
1
n
q
k
x
k
−
c
f
∑
k
=
1
n
p
k
x
k
+
∑
k
=
1
n
q
k
x
k
.
Combining (45) and (46) gives(47)
F
1
x
,
p
+
q
;
f
≥
∑
k
=
1
n
p
k
h
x
k
−
c
+
∑
k
=
1
n
q
k
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
+
h
∑
k
=
1
n
q
k
x
k
−
c
f
∑
k
=
1
n
p
k
x
k
+
f
∑
k
=
1
n
q
k
x
k
−
∑
k
=
1
n
p
k
x
k
−
∑
k
=
1
n
q
k
x
k
=
∑
k
=
1
n
p
k
h
x
k
−
c
+
∑
k
=
1
n
q
k
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
+
h
∑
k
=
1
n
q
k
x
k
−
c
f
∑
k
=
1
n
p
k
x
k
+
∑
k
=
1
n
p
k
h
x
k
−
c
+
∑
k
=
1
n
q
k
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
+
h
∑
k
=
1
n
q
k
x
k
−
c
f
∑
k
=
1
n
q
k
x
k
−
∑
k
=
1
n
p
k
x
k
−
∑
k
=
1
n
q
k
x
k
≥
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
+
h
∑
k
=
1
n
q
k
x
k
−
c
f
∑
k
=
1
n
p
k
x
k
+
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
+
h
∑
k
=
1
n
q
k
x
k
−
c
f
∑
k
=
1
n
q
k
x
k
−
∑
k
=
1
n
p
k
f
x
k
−
∑
k
=
1
n
q
k
f
x
k
≥
∑
k
=
1
n
p
k
h
x
k
−
c
h
∑
k
=
1
n
p
k
x
k
−
c
f
∑
k
=
1
n
p
k
x
k
+
∑
k
=
1
n
q
k
h
x
k
−
c
h
∑
k
=
1
n
q
k
x
k
−
c
f
∑
k
=
1
n
q
k
x
k
−
∑
k
=
1
n
p
k
f
x
k
−
∑
k
=
1
n
q
k
f
x
k
.
Therefore,(48)
F
1
x
,
p
+
q
;
f
≥
F
1
x
,
p
;
f
+
F
1
x
,
q
;
f
,
follows from (47).
(2) It follows from part (1) and Remark17 that(49)
F
1
x
,
p
;
f
=
F
1
x
,
p
−
q
+
q
;
f
≥
F
1
x
,
q
;
f
+
F
1
x
,
p
−
q
;
f
≥
F
1
x
,
q
;
f
≥
0
.
(3) Part (3) can be proved by similar methods as in proving parts (1) and (2).Theorem 20.
LetI
∈
R
∖
0, a
=
a
1
,
a
2
,
…
,
a
n
,
b
=
b
1
,
b
2
,
…
,
b
n
∈
I
n with a
≻
b, p
=
p
1
,
p
2
,
…
,
p
n
,
q
=
q
1
,
q
2
,
…
,
q
n
∈
0
,
∞
n, f
:
I
→
R be a harmonic convex function, and F
2
a
,
b
,
p
;
fbe defined by (39). Then the following statements are true:(1) F
2
a
,
b
,
p
;
f is linear with respect to p, that is
(50)
F
2
a
,
b
,
p
+
q
;
f
=
F
2
a
,
b
,
p
;
f
+
F
2
a
,
b
,
q
;
f
.
(2) If p
≥
q
, then
(51)
F
2
a
,
b
,
p
;
f
≥
F
2
a
,
b
,
q
;
f
≥
0
,
that is F
2
a
,
b
,
p
;
f is increasing with respect to p
.
(3) If f is harmonic concave on I
, then inequalities (50) and (51) also hold true.Proof.
(1) It follows from (39) that(52)
F
2
a
,
b
,
p
+
q
;
f
=
∑
i
=
1
n
p
i
+
q
i
a
i
f
a
i
−
∑
i
=
1
n
p
i
+
q
i
b
i
f
b
i
=
∑
i
=
1
n
p
i
a
i
f
a
i
−
∑
i
=
1
n
p
i
b
i
f
b
i
+
∑
i
=
1
n
q
i
a
i
f
a
i
−
∑
i
=
1
n
q
i
b
i
f
b
i
=
F
2
a
,
b
,
p
;
f
+
F
2
a
,
b
,
q
;
f
.
(2) From part (1) and Remark18 we get(53)
F
2
a
,
b
,
p
;
f
=
F
2
a
,
b
,
p
−
q
+
q
;
f
=
F
2
a
,
b
,
p
−
q
;
f
+
F
2
a
,
b
,
q
;
f
≥
F
2
a
,
b
,
q
;
f
≥
0
.
(3) Part (3) can be proved by similar methods in proving parts (1) and (2).



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