# Refinements of Aczél-Type Inequality and Their Applications

Authors: Jingfeng Tian; Wen-Li Wang
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We present some new sharpened versions of Aczél-type inequality. Moreover, as applications, some refinements of integral type of Aczél-type inequality are given.


---

## Body

## 1. Introduction

Letn be a positive integer, and let a
i, b
i
(
i
=
1,2
,
…
,
n
) be real numbers such that a
1
2
-
∑
i
=
2
n
‍
a
i
2
>
0 or b
1
2
-
∑
i
=
2
n
‍
b
i
2
>
0. Then, the famous Aczél inequality [1] can be stated as follows:
(1)
(
a
1
2
-
∑
i
=
2
n
‍
a
i
2
)
(
b
1
2
-
∑
i
=
2
n
‍
b
i
2
)
≤
(
a
1
b
1
-
∑
i
=
2
n
‍
a
i
b
i
)
2
.Aczél’s inequality plays a very important role in the theory of functional equations in non-Euclidean geometry. Due to the importance of Aczél’s inequality (1), it has received considerable attention by many authors and has motivated a large number of research papers giving it various generalizations, improvements, and applications (see [2–21] and the references therein).In 1959, Popoviciu [10] first obtained an exponential extension of the Aczél inequality as follows.Theorem B.
Letp
≥
q
>
1, (
1
/
p
)
+
(
1
/
q
)
=
1, and let a
i, b
i
(
i
=
1,2
,
…
,
n
) be positive numbers such that a
1
p
-
∑
i
=
2
n
‍
a
i
p
>
0 and b
1
q
-
∑
i
=
2
n
‍
b
i
q
>
0. Then
(2)
(
a
1
p
-
∑
i
=
2
n
‍
a
i
p
)
1
/
p
(
b
1
q
-
∑
i
=
2
n
‍
b
i
q
)
1
/
q
≤
a
1
b
1
-
∑
i
=
2
n
‍
a
i
b
i
.Later, in 1982, Vasić and Pečarić [16] established the following reversed version of inequality (2).Theorem C.
Letq
<
0, p
>
0, (
1
/
p
)
+
(
1
/
q
)
=
1, and let a
i, b
i
(
i
=
1,2
,
…
,
n
) be positive numbers such that a
1
p
-
∑
i
=
2
n
‍
a
i
p
>
0 and b
1
q
-
∑
i
=
2
n
‍
b
i
q
>
0. Then
(3)
(
a
1
p
-
∑
i
=
2
n
‍
a
i
p
)
1
/
p
(
b
1
q
-
∑
i
=
2
n
‍
b
i
q
)
1
/
q
≥
a
1
b
1
-
∑
i
=
2
n
‍
a
i
b
i
.In another paper, Vasić and Pečarić [15] generalized inequality (2) in the following form.Theorem D.
Leta
r
j
>
0, β
j
>
0, a
1
j
β
j
-
∑
r
=
2
n
‍
a
r
j
β
j
>
0, r
=
1,2
,
…
,
n, j
=
1,2
,
…
,
m, and let ∑
j
=
1
m
‍
(
1
/
β
j
)
≥
1. Then
(4)
∏
j
=
1
m
(
a
1
j
β
j
-
∑
r
=
2
n
a
r
j
β
j
)
1
/
β
j
≤
∏
j
=
1
m
a
1
j
-
∑
r
=
2
n
∏
j
=
1
m
a
r
j
.In 2012, Tian [13] presented the reversed version of inequality (4) as follows.Theorem E.
Leta
r
j
>
0, β
1
≠
0, β
j
<
0
(
j
=
2,3
,
…
,
m
), ∑
j
=
1
m
‍
(
1
/
β
j
)
≤
1, a
1
j
β
j
-
∑
r
=
2
n
‍
a
r
j
β
j
>
0, r
=
1,2
,
…
,
n, j
=
1,2
,
…
,
m. Then
(5)
∏
j
=
1
m
(
a
1
j
β
j
-
∑
r
=
2
n
a
r
j
β
j
)
1
/
β
j
≥
∏
j
=
1
m
a
1
j
-
∑
r
=
2
n
∏
j
=
1
m
a
r
j
.Moreover, in [13] Tian established an integral type of inequality (5).Theorem F.
Letβ
1
>
0, β
j
<
0
(
j
=
2,3
,
…
,
m
), ∑
j
=
1
m
‍
(
1
/
β
j
)
=
1, let t
j
>
0
(
j
=
1,2
,
…
,
m
), and let f
j
(
x
)
(
j
=
1,2
,
…
,
m
) be positive Riemann integrable functions on [
a
,
b
] such that t
j
β
j
-
∫
a
b
‍
f
j
β
j
(
x
)
d
x
>
0. Then
(6)
∏
j
=
1
m
‍
(
t
j
β
j
-
∫
a
b
‍
f
j
β
j
(
x
)
d
x
)
1
/
β
j
≥
∏
j
=
1
m
‍
t
j
-
∫
a
b
‍
∏
j
=
1
m
‍
f
j
(
x
)
d
x
.Remark 1.
In fact, the integral form of inequality (4) is also valid; that is, one has the following.Theorem G.
Letβ
j
>
0
(
j
=
1,2
,
…
,
m
), ∑
j
=
1
m
‍
(
1
/
β
j
)
=
1, let t
j
>
0
(
j
=
1,2
,
…
,
m
), and let f
j
(
x
)
(
j
=
1,2
,
…
,
m
) be positive Riemann integrable functions on [
a
,
b
] such that t
j
β
j
-
∫
a
b
‍
f
j
β
j
(
x
)
d
x
>
0. Then
(7)
∏
j
=
1
m
‍
(
t
j
β
j
-
∫
a
b
‍
f
j
β
j
(
x
)
d
x
)
1
/
β
j
≤
∏
j
=
1
m
‍
t
j
-
∫
a
b
‍
∏
j
=
1
m
‍
f
j
(
x
)
d
x
.The main purpose of this work is to give new refinements of inequalities (4) and (5). As applications, new refinements of inequalities (6) and (7) are also given.


## 2. Refinements of Aczél-Type Inequality

In order to present our main results, we need some lemmas as follows.Lemma 2 (see [6]).
Leta
i
,
x
i
(
i
=
1,2
,
…
,
n
) be real numbers such that a
i
≥
0 and x
i
>
-
1. If ∑
i
=
1
n
‍
a
i
≤
1, then
(8)
∏
i
=
1
n
‍
(
1
+
x
i
)
a
i
≤
1
+
∑
i
=
1
n
‍
a
i
x
i
.
If either a
i
≥
1
(
i
=
1,2
,
…
,
n
) or a
i
≤
0
(
i
=
1,2
,
…
,
n
) and if all x
i are positive or negative with x
i
>
-
1, then the reverse inequality of (8) holds.Lemma 3 (see [15]).
Leta
i
j
>
0
(
i
=
1,2
,
…
,
n
,
j
=
1,2
,
…
,
m
).(a)
Ifλ
j
≥
0 and if ∑
j
=
1
m
‍
λ
j
≥
1, then
(9)
∑
i
=
1
n
‍
∏
j
=
1
m
‍
a
i
j
λ
j
≤
∏
j
=
1
m
‍
(
∑
i
=
1
n
‍
a
i
j
)
λ
j
.
(b)
Ifλ
j
≤
0
(
j
=
1,2
,
…
,
m
), then
(10)
∑
i
=
1
n
‍
∏
j
=
1
m
‍
a
i
j
λ
j
≥
∏
j
=
1
m
‍
(
∑
i
=
1
n
‍
a
i
j
)
λ
j
.
(c)
Ifλ
1
>
0, λ
j
≤
0
(
j
=
2,3
,
…
,
m
), and ∑
j
=
1
m
‍
λ
j
≤
1, then
(11)
∑
i
=
1
n
‍
∏
j
=
1
m
‍
a
i
j
λ
j
≥
∏
j
=
1
m
‍
(
∑
i
=
1
n
‍
a
i
j
)
λ
j
.Lemma 4 (see [18]).
Let0
≤
x
<
1, α
>
0. Then
(12)
(
1
-
x
)
1
/
α
≤
1
-
x
max
⁡
⁡
{
α
,
1
}
.Lemma 5.
Let0
<
β
1
≤
β
2
≤
⋯
≤
β
m, ∑
j
=
1
m
‍
(
1
/
β
j
)
≥
1, m
≥
2, let 0
<
x
j
<
1
(
j
=
1,2
,
…
,
m
), and let ξ
(
m
)
=
{
m
/
2
i
f
m
e
v
e
n
(
m
-
1
)
/
2
i
f
m
o
d
d.
Then(13)
∏
j
=
1
m
‍
(
1
-
x
j
β
j
)
1
/
β
j
+
∏
j
=
1
m
‍
x
j
≤
1
-
1
ξ
(
m
)
×
∑
j
=
1
ξ
(
m
)
‍
[
1
max
⁡
⁡
{
β
2
j
,
1
}
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
.Proof.
From the assumptions we have that(14)
1
β
1
≥
1
β
2
≥
⋯
≥
1
β
m
-
1
≥
1
β
m
>
0
,
1
β
j
-
1
β
j
+
1
≥
0
(
j
=
1,2
,
…
,
m
-
1
)
.
Case (I) (let m be even). In view of (
1
/
β
1
-
1
/
β
2
)
+
1
/
β
2
+
1
/
β
2
+
(
1
/
β
3
-
1
/
β
4
)
+
1
/
β
4
+
1
/
β
4
+
⋯
+
(
1
/
β
m
-
1
-
1
/
β
m
)
+
1
/
β
m
+
1
/
β
m
=
1
/
β
1
+
1
/
β
2
+
⋯
+
1
/
β
m
≥
1 by using inequality (9),
we get(15)
∏
j
=
1
m
/
2
[
1
-
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
1
/
β
2
j
=
∏
j
=
1
m
/
2
{
[
(
1
-
x
2
j
-
1
β
2
j
-
1
)
+
x
2
j
-
1
β
2
j
-
1
]
(
1
/
β
2
j
-
1
)
-
(
1
/
β
2
j
)
[
(
1
-
x
2
j
-
1
β
2
j
-
1
)
+
x
2
j
β
2
j
]
1
/
β
2
j
00000000
×
[
(
1
-
x
2
j
β
2
j
)
+
x
2
j
-
1
β
2
j
-
1
]
1
/
β
2
j
00000000
×
[
(
1
-
x
2
j
-
1
β
2
j
-
1
)
+
x
2
j
-
1
β
2
j
-
1
]
1
/
β
2
j
-
1
-
1
/
β
2
j
[
(
1
-
x
2
j
-
1
β
2
j
-
1
)
+
x
2
j
-
1
β
2
j
-
1
]
(
1
/
β
2
j
-
1
)
-
(
1
/
β
2
j
)
[
(
1
-
x
2
j
-
1
β
2
j
-
1
)
+
x
2
j
β
2
j
]
1
/
β
2
j
}
=
[
(
1
-
x
1
β
1
)
+
x
2
β
2
]
1
/
β
2
[
(
1
-
x
2
β
2
)
+
x
1
β
1
]
1
/
β
2
×
[
(
1
-
x
1
β
1
)
+
x
1
β
1
]
1
/
β
1
-
1
/
β
2
×
[
(
1
-
x
3
β
3
)
+
x
4
β
4
]
1
/
β
4
[
(
1
-
x
4
β
4
)
+
x
3
β
3
]
1
/
β
4
×
[
(
1
-
x
3
β
3
)
+
x
3
β
3
]
1
/
β
3
-
1
/
β
4
⋮
×
[
(
1
-
x
m
-
1
β
m
-
1
)
+
x
m
β
m
]
1
/
β
m
[
(
1
-
x
m
β
m
)
+
x
m
-
1
β
m
-
1
]
1
/
β
m
×
[
(
1
-
x
m
-
1
β
m
-
1
)
+
x
m
-
1
β
m
-
1
]
1
/
β
m
-
1
-
1
/
β
m
≥
∏
j
=
1
m
/
2
[
[
(
1
-
x
2
j
-
1
β
2
j
-
1
)
+
x
2
j
-
1
β
2
j
-
1
]
(
1
/
β
2
j
-
1
)
-
(
1
/
β
2
j
)
(
1
-
x
2
j
-
1
β
2
j
-
1
)
1
/
β
2
j
(
1
-
x
2
j
β
2
j
)
1
/
β
2
j
00000000
×
(
1
-
x
2
j
-
1
β
2
j
-
1
)
1
/
β
2
j
-
1
-
1
/
β
2
j
]
+
∏
j
=
1
m
/
2
[
[
(
1
-
x
2
j
-
1
β
2
j
-
1
)
+
x
2
j
-
1
β
2
j
-
1
]
(
1
/
β
2
j
-
1
)
-
(
1
/
β
2
j
)
(
x
2
j
β
2
j
)
1
/
β
2
j
(
x
2
j
-
1
β
2
j
-
1
)
1
/
β
2
j
0000000000
×
(
x
2
j
-
1
β
2
j
-
1
)
1
/
β
2
j
-
1
-
1
/
β
2
j
]
=
∏
j
=
1
m
/
2
(
1
-
x
j
β
j
)
1
/
β
j
+
∏
j
=
1
m
x
j
.
On the other hand, applying Lemma4 and the arithmetic-geometric means inequality we obtain(16)
∏
j
=
1
m
/
2
‍
[
1
-
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
1
/
β
2
j
≤
∏
j
=
1
m
/
2
‍
[
1
-
1
max
⁡
⁡
{
β
2
j
,
1
}
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
≤
{
2
m
∑
j
=
1
m
/
2
‍
[
1
-
1
max
⁡
⁡
{
β
2
j
,
1
}
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
}
m
/
2
=
{
1
-
2
m
∑
j
=
1
m
/
2
‍
[
1
max
⁡
⁡
{
β
2
j
,
1
}
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
}
m
/
2
.
Applying Lemma 4 again, we get
(17)
{
1
-
2
m
∑
j
=
1
m
/
2
‍
[
1
max
⁡
⁡
{
β
2
j
,
1
}
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
}
m
/
2
≤
1
-
2
m
∑
j
=
1
m
/
2
‍
[
1
max
⁡
⁡
{
β
2
j
,
1
}
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
.
Combining (15), (16), and (17) yields immediately inequality (13).
Case (II) (let m be odd). In view of (
1
/
β
1
-
1
/
β
2
)+1
/
β
2+1
/
β
2+(
1
/
β
3
-
1
/
β
4
)
+
1
/
β
4
+
1
/
β
4
+
⋯+(
1
/
β
m
-
2
-
1
/
β
m
-
1
)+1
/
β
m
-
1
+
1
/
β
m
-
1+1
/
β
m=1
/
β
1
+
1
/
β
2+⋯+1
/
β
m
≥
1, by using inequality (9), we have
(18)
∏
j
=
1
(
m
-
1
)
/
2
‍
[
1
-
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
1
/
β
2
j
=
{
∏
j
=
1
(
m
-
1
)
/
2
‍
[
1
-
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
1
/
β
2
j
}
×
[
(
1
-
x
m
β
m
)
+
x
m
β
m
]
1
/
β
m
=
{
∏
j
=
1
(
m
-
1
)
/
2
{
[
(
1
-
x
2
j
-
1
β
2
j
-
1
)
+
x
2
j
β
2
j
]
1
/
β
2
j
‍
0000000000000
×
[
(
1
-
x
2
j
β
2
j
)
+
x
2
j
-
1
β
2
j
-
1
]
1
/
β
2
j
0000000000000
×
[
(
1
-
x
2
j
-
1
β
2
j
-
1
)
+
x
2
j
-
1
β
2
j
-
1
]
1
/
β
2
j
-
1
-
1
/
β
2
j
}
∏
j
=
1
(
m
-
1
)
/
2
‍
}
×
[
(
1
-
x
m
β
m
)
+
x
m
β
m
]
1
/
β
m
≥
{
∏
j
=
1
(
m
-
1
)
/
2
‍
[
(
1
-
x
2
j
-
1
β
2
j
-
1
)
1
/
β
2
j
(
1
-
x
2
j
β
2
j
)
1
/
β
2
j
0000000000000
(
1
-
x
2
j
-
1
β
2
j
-
1
)
1
/
β
2
j
-
1
-
1
/
β
2
j
]
∏
j
=
1
(
m
-
1
)
/
2
‍
}
(
1
-
x
m
β
m
)
1
/
β
m
+
{
∏
j
=
1
(
m
-
1
)
/
2
‍
[
(
x
2
j
β
2
j
)
1
/
β
2
j
(
x
2
j
-
1
β
2
j
-
1
)
1
/
β
2
j
00000000000000
×
(
x
2
j
-
1
β
2
j
-
1
)
1
/
β
2
j
-
1
-
1
/
β
2
j
]
∏
j
=
1
(
m
-
1
)
/
2
‍
}
(
x
m
β
m
)
1
/
β
m
=
∏
j
=
1
m
‍
(
1
-
x
j
β
j
)
1
/
β
j
+
∏
j
=
1
m
‍
x
j
.
On the other hand, applying Lemma 4 and the arithmetic-geometric means inequality we obtain
(19)
∏
j
=
1
(
m
-
1
)
/
2
‍
[
1
-
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
1
/
β
2
j
≤
∏
j
=
1
(
m
-
1
)
/
2
‍
[
1
-
1
max
⁡
⁡
{
β
2
j
,
1
}
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
≤
{
2
m
-
1
∑
j
=
1
(
m
-
1
)
/
2
‍
[
1
-
1
max
⁡
⁡
{
β
2
j
,
1
}
000000000000000000
×
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
1
max
⁡
⁡
{
β
2
j
,
1
}
]
∑
j
=
1
(
m
-
1
)
/
2
‍
}
(
m
-
1
)
/
2
=
{
1
-
2
m
-
1
∑
j
=
1
(
m
-
1
)
/
2
‍
[
1
max
⁡
⁡
{
β
2
j
,
1
}
000000000000000000000
×
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
1
max
⁡
⁡
{
β
2
j
,
1
}
]
∑
j
=
1
(
m
-
1
)
/
2
‍
}
(
m
-
1
)
/
2
.
Applying Lemma 4 again, we have
(20)
{
1
-
2
m
-
1
∑
j
=
1
(
m
-
1
)
/
2
‍
[
1
max
⁡
⁡
{
β
2
j
,
1
}
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
}
(
m
-
1
)
/
2
≤
1
-
2
m
-
1
∑
j
=
1
(
m
-
1
)
/
2
‍
[
1
max
⁡
⁡
{
β
2
j
,
1
}
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
]
.
Hence, combining (18), (19), and (20) yields immediately inequality (13).Similar to the proof of Lemma5 but using Lemma 2 in place of Lemma 4, we immediately obtain the following result.Lemma 6.
Letβ
1
>
0, 0
>
β
2
≥
β
3
≥
⋯
≥
β
m, ∑
j
=
1
m
‍
(
1
/
β
j
)
≤
1, m
≥
2, let 0
<
x
1
<
1, x
j
>
1
(
j
=
2,3
,
…
,
m
), and let ξ
(
m
)
=
{
m
/
2
i
f
m
e
v
e
n
(
m
-
1
)
/
2
i
f
m
o
d
d.
Then(21)
∏
j
=
1
m
‍
(
1
-
x
j
β
j
)
1
/
β
j
+
∏
j
=
1
m
‍
x
j
≥
1
-
∑
j
=
1
ξ
(
m
)
‍
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
β
2
j
.Using the same methods as in Lemma6, we get the following Lemma.Lemma 7.
Let0
>
β
1
≥
β
2
≥
⋯
≥
β
m, m
≥
2, let x
j
>
1
(
j
=
1,2
,
…
,
m
), and let ξ
(
m
)
=
{
m
/
2
i
f
m
e
v
e
n
(
m
-
1
)
/
2
i
f
m
o
d
d.
Then(22)
∏
j
=
1
m
‍
(
1
-
x
j
β
j
)
1
/
β
j
+
∏
j
=
1
m
‍
x
j
≥
1
-
∑
j
=
1
ξ
(
m
)
‍
(
x
2
j
β
2
j
-
x
2
j
-
1
β
2
j
-
1
)
2
β
2
j
.Now, we present some new refinements of inequalities (4) and (5).Theorem 8.
Leta
r
j
>
0, r
=
1,2
,
…
,
n, j
=
1,2
,
…
,
m, m
≥
2
,
n
≥
2, let 0
<
β
1
≤
β
2
≤
⋯
≤
β
m, ∑
j
=
1
m
‍
(
1
/
β
j
)
≥
1, a
1
j
β
j
-
∑
r
=
2
n
‍
a
r
j
β
j
>
0, and let ξ
(
m
)
=
{
m
/
2
i
f
m
e
v
e
n
(
m
-
1
)
/
2
i
f
m
o
d
d.
Then(23)
∏
j
=
1
m
‍
(
a
1
j
β
j
-
∑
r
=
2
n
‍
a
r
j
β
j
)
1
/
β
j
≤
∏
j
=
1
m
‍
a
1
j
-
∑
r
=
2
n
‍
∏
j
=
1
m
‍
a
r
j
-
a
11
a
12
…
a
1
m
ξ
(
m
)
×
∑
j
=
1
ξ
(
m
)
‍
{
[
∑
r
=
2
n
‍
(
a
r
(
2
j
)
β
2
j
a
1
(
2
j
)
β
2
j
-
a
r
(
2
j
-
1
)
β
2
j
-
1
a
1
(
2
j
-
1
)
β
2
j
-
1
)
]
2
1
max
⁡
⁡
{
β
2
j
,
1
}
000000000000
×
[
∑
r
=
2
n
‍
(
a
r
(
2
j
)
β
2
j
a
1
(
2
j
)
β
2
j
-
a
r
(
2
j
-
1
)
β
2
j
-
1
a
1
(
2
j
-
1
)
β
2
j
-
1
)
]
2
}
.Proof.
From the assumptions we find that(24)
0
<
(
a
1
j
β
j
-
∑
r
=
2
n
a
r
j
β
j
)
1
/
β
j
(
a
1
j
β
j
)
1
/
β
j
<
1
(
j
=
1,2
,
…
,
m
)
.
Thus, by using Lemma5 with a substitution x
j
→
(
(
a
1
j
β
j
-
∑
r
=
2
n
‍
a
r
j
β
j
)
/
a
1
j
β
j
)
1
/
β
j
(
j
=
1,2
,
…
,
m
) in (13), we obtain
(25)
∏
j
=
1
m
‍
(
∑
r
=
2
n
a
r
j
β
j
a
1
j
β
j
)
1
/
β
j
+
∏
j
=
1
m
‍
(
a
1
j
β
j
-
∑
r
=
2
n
a
r
j
β
j
a
1
j
β
j
)
1
/
β
j
≤
1
-
1
ξ
(
m
)
∑
j
=
1
ξ
(
m
)
‍
{
-
(
1
-
∑
r
=
2
n
‍
a
r
(
2
j
-
1
)
β
2
j
-
1
a
1
(
2
j
-
1
)
β
2
j
-
1
)
]
2
1
max
⁡
⁡
{
β
2
j
,
1
}
00000000000000000
×
[
(
1
-
∑
r
=
2
n
a
r
(
2
j
)
β
2
j
a
1
(
2
j
)
β
2
j
)
00000000000000000000
-
(
1
-
∑
r
=
2
n
a
r
(
2
j
-
1
)
β
2
j
-
1
a
1
(
2
j
-
1
)
β
2
j
-
1
)
]
2
}
=
1
-
1
ξ
(
m
)
∑
j
=
1
ξ
(
m
)
‍
{
[
∑
r
=
2
n
‍
(
a
r
(
2
j
)
β
2
j
a
1
(
2
j
)
β
2
j
-
a
r
(
2
j
-
1
)
β
2
j
-
1
a
1
(
2
j
-
1
)
β
2
j
-
1
)
]
2
1
max
⁡
⁡
{
β
2
j
,
1
}
0000000000000000
×
[
∑
r
=
2
n
‍
(
a
r
(
2
j
)
β
2
j
a
1
(
2
j
)
β
2
j
-
a
r
(
2
j
-
1
)
β
2
j
-
1
a
1
(
2
j
-
1
)
β
2
j
-
1
)
]
2
}
,
which implies
(26)
∏
j
=
1
m
‍
(
a
1
j
β
j
-
∑
r
=
2
n
‍
a
r
j
β
j
)
1
/
β
j
≤
∏
j
=
1
m
‍
a
1
j
-
∏
j
=
1
m
‍
(
∑
r
=
2
n
‍
a
r
j
β
j
)
1
/
β
j
-
a
11
a
12
…
a
1
m
ξ
(
m
)
×
∑
j
=
1
ξ
(
m
)
‍
{
[
∑
r
=
2
n
‍
(
a
r
(
2
j
)
β
2
j
a
1
(
2
j
)
β
2
j
-
a
r
(
2
j
-
1
)
β
2
j
-
1
a
1
(
2
j
-
1
)
β
2
j
-
1
)
]
2
1
max
⁡
⁡
{
β
2
j
,
1
}
000000000000
×
[
∑
r
=
2
n
‍
(
a
r
(
2
j
)
β
2
j
a
1
(
2
j
)
β
2
j
-
a
r
(
2
j
-
1
)
β
2
j
-
1
a
1
(
2
j
-
1
)
β
2
j
-
1
)
]
2
}
.
On the other hand, we get from Lemma3 that
(27)
∏
j
=
1
m
‍
(
∑
r
=
2
n
‍
a
r
j
β
j
)
1
/
β
j
≥
∑
r
=
2
n
∏
j
=
1
m
a
r
j
.
Combining (26) and (27) yields immediately the desired inequality (23).Theorem 9.
Leta
r
j
>
0, 0
>
β
1
≥
β
2
≥
⋯
≥
β
m, a
1
j
β
j
-
∑
r
=
2
n
‍
a
r
j
β
j
>
0, r
=
1,2
,
…
,
n, j
=
1,2
,
…
,
m, let m
≥
2
,
n
≥
2, and let ξ
(
m
)
=
{
m
/
2
i
f
m
e
v
e
n
(
m
-
1
)
/
2
i
f
m
o
d
d.
Then(28)
∏
j
=
1
m
‍
(
a
1
j
β
j
-
∑
r
=
2
n
‍
a
r
j
β
j
)
1
/
β
j
≥
∏
j
=
1
m
‍
a
1
j
-
∑
r
=
2
n
‍
∏
j
=
1
m
‍
a
r
j
-
a
11
a
12
,
…
,
a
1
m
×
∑
j
=
1
ξ
(
m
)
{
1
β
2
j
[
∑
r
=
2
n
‍
(
a
r
(
2
j
)
β
2
j
a
1
(
2
j
)
β
2
j
-
a
r
(
2
j
-
1
)
β
2
j
-
1
a
1
(
2
j
-
1
)
β
2
j
-
1
)
]
2
}
.
Inequality (28) is also valid for β
1
>
0, 0
>
β
2
≥
β
3
≥
⋯
≥
β
m, ∑
j
=
1
m
‍
(
1
/
β
j
)
≤
1.Proof.
The proof of Theorem9 is similar to the one of Theorem 8, and we omit it.


## 3. Applications

In this section, we show two applications of the inequalities newly obtained in Section2.Firstly, we present a new refinement of inequality (6) by using Theorem 9.Theorem 10.
Lett
j
>
0
(
j
=
1,2
,
…
,
m
), β
1
>
0, 0
>
β
2
≥
β
3
≥
⋯
≥
β
m, ∑
j
=
1
m
‍
(
1
/
β
j
)
=
1, let f
j
(
x
)
(
j
=
1,2
,
…
,
m
) be positive integrable functions defined on [
a
,
b
] with t
j
β
j
-
∫
a
b
‍
f
j
β
j
(
x
)
d
x
>
0, and let ξ
(
m
)
=
{
m
/
2
if
m
even
(
m
-
1
)
/
2
if
m
odd.
Then(29)
∏
j
=
1
m
‍
(
t
j
β
j
-
∫
a
b
‍
f
j
β
j
(
x
)
d
x
)
1
/
β
j
≥
∏
j
=
1
m
‍
t
j
-
∫
a
b
‍
∏
j
=
1
m
‍
f
j
(
x
)
d
x
-
t
1
t
2
,
…
,
t
m
×
∑
j
=
1
ξ
(
m
)
‍
[
1
β
2
j
∫
a
b
‍
(
f
2
j
β
2
j
(
x
)
t
2
j
β
2
j
-
f
2
j
-
1
β
2
j
-
1
(
x
)
t
2
j
-
1
β
2
j
-
1
)
d
x
]
2
.Proof.
For any positive integern, we choose an equidistant partition of [
a
,
b
] as
(30)
a
<
a
+
b
-
a
n
<
⋯
<
a
+
b
-
a
n
k
<
⋯
<
a
+
b
-
a
n
(
n
-
1
)
<
b
,
x
i
=
a
+
b
-
a
n
i
,
i
=
0,1
,
…
,
n
,
(31)
Δ
x
k
=
b
-
a
n
,
k
=
1,2
,
…
,
n
.
Sincet
j
β
j
-
∫
a
b
‍
f
j
β
j
(
x
)
d
x
>
0
(
j
=
1,2
,
…
,
m
), it follows that
(32)
t
j
β
j
-
lim
⁡
n
→
∞
⁡
∑
k
=
1
n
f
j
β
j
(
a
+
k
(
b
-
a
)
n
)
b
-
a
n
>
0
(
j
=
1
,
2
,
…
,
m
)
.
Therefore, there exists a positive integer N such that
(33)
t
j
β
j
-
∑
k
=
1
n
‍
f
j
β
j
(
a
+
k
(
b
-
a
)
n
)
b
-
a
n
>
0
,
for all n
>
N and j
=
1,2
,
…
,
m.
Moreover, for anyn
>
N, it follows from Theorem 9 that
(34)
∏
j
=
1
m
‍
[
t
j
β
j
-
∑
k
=
1
n
‍
f
j
β
j
(
a
+
k
(
b
-
a
)
n
)
b
-
a
n
]
1
/
β
j
≥
∏
j
=
1
m
‍
t
j
β
j
-
∑
k
=
1
n
‍
(
∏
j
=
1
m
‍
f
j
(
a
+
k
(
b
-
a
)
n
)
)
×
(
b
-
a
n
)
1
/
β
1
+
1
/
β
2
+
⋯
+
1
/
β
m
-
t
1
t
2
,
…
,
t
m
∑
j
=
1
ξ
(
m
)
‍
1
β
2
j
×
[
(
a
+
k
(
b
-
a
)
n
)
b
-
a
n
∑
k
=
1
n
‍
(
1
t
2
j
β
2
j
f
2
j
β
2
j
(
a
+
k
(
b
-
a
)
n
)
b
-
a
n
00000000000
-
1
t
2
j
-
1
β
2
j
-
1
f
2
j
-
1
β
2
j
-
1
(
a
+
k
(
b
-
a
)
n
)
b
-
a
n
)
]
2
.
Noting that(35)
∑
j
=
1
m
‍
1
β
j
=
1
,
we get
(36)
∏
j
=
1
m
‍
[
t
j
β
j
-
∑
k
=
1
n
‍
f
j
β
j
(
a
+
k
(
b
-
a
)
n
)
b
-
a
n
]
1
/
β
j
≥
∏
j
=
1
m
‍
t
j
β
j
-
∑
k
=
1
n
‍
(
∏
j
=
1
m
‍
f
j
(
a
+
k
(
b
-
a
)
n
)
)
(
b
-
a
n
)
-
t
1
t
2
,
…
,
t
m
∑
j
=
1
ξ
(
m
)
‍
1
β
2
j
×
[
(
a
+
k
(
b
-
a
)
n
)
b
-
a
n
∑
k
=
1
n
‍
(
1
t
2
j
β
2
j
f
2
j
β
2
j
(
a
+
k
(
b
-
a
)
n
)
b
-
a
n
000000000000
-
1
t
2
j
-
1
β
2
j
-
1
f
2
j
-
1
β
2
j
-
1
(
a
+
k
(
b
-
a
)
n
)
b
-
a
n
)
]
2
.
In view of the assumption that f
j
(
x
)
(
j
=
1,2
,
…
,
m
) are positive Riemann integrable functions on [
a
,
b
], we find that ∏
j
=
1
m
‍
f
j
(
x
) and f
j
λ
j
(
x
) are also integrable on [
a
,
b
]. Letting n
→
∞ on both sides of inequality (36), we get the desired inequality (29).Next, we give a new refinement of inequality (7) by using Theorem 8.Theorem 11.
Lett
j
>
0
(
j
=
1,2
,
…
,
m
), 0
<
β
1
≤
β
2
≤
⋯
≤
β
m, ∑
j
=
1
m
‍
(
1
/
β
j
)
=
1, m
≥
2, and let f
j
(
x
)
(
j
=
1,2
,
…
,
m
) be positive integrable functions defined on [
a
,
b
] with t
j
β
j
-
∫
a
b
‍
f
j
β
j
(
x
)
d
x
>
0, and let ξ
(
m
)
=
{
m
/
2
if
m
even
(
m
-
1
)
/
2
if
m
odd.
Then(37)
∏
j
=
1
m
‍
(
t
j
β
j
-
∫
a
b
‍
f
j
β
j
(
x
)
d
x
)
1
/
β
j
≤
∏
j
=
1
m
‍
t
j
-
∫
a
b
‍
∏
j
=
1
m
‍
f
j
(
x
)
d
x
-
t
1
t
2
,
…
,
t
m
ξ
(
m
)
×
∑
j
=
1
ξ
(
m
)
‍
{
[
∫
a
b
‍
(
f
2
j
β
2
j
(
x
)
t
2
j
β
2
j
-
f
2
j
-
1
β
2
j
-
1
(
x
)
t
2
j
-
1
β
2
j
-
1
)
d
x
]
2
1
β
2
j
00000000000
×
[
∫
a
b
‍
(
f
2
j
β
2
j
(
x
)
t
2
j
β
2
j
-
f
2
j
-
1
β
2
j
-
1
(
x
)
t
2
j
-
1
β
2
j
-
1
)
d
x
]
2
}
.Proof.
The proof of Theorem11 is similar to the one of Theorem 10, and we omit it.



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*Source: 581354-2014-06-26.xml*