# Refinements of Bounds for Neuman Means

Authors: Yu-Ming Chu; Wei-Mao Qian
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We present the sharp bounds for the Neuman meansS
H
A, S
A
H, S
C
A and S
A
C in terms of the arithmetic, harmonic, and contraharmonic means. Our results are the refinements or improvements of the results given by Neuman.


---

## Body

## 1. Introduction

Fora
,
b
>
0 with a
≠
b, the Schwab-Borchardt mean SB
(
a
,
b
) of a and b is given by
(1)
SB
(
a
,
b
)
=
{
b
2
-
a
2
cos
⁡
-
1
(
a
/
b
)
,
a
<
b
,
a
2
-
b
2
cosh
⁡
-
1
(
a
/
b
)
,
a
>
b
,
where cos
⁡
-
1
(
x
) and cosh
⁡
-
1
(
x
)
=
log
⁡
⁡
(
x
+
x
2
-
1
) are the inverse cosine and inverse hyperbolic cosine functions, respectively.It is well-known that the meanSB
(
a
,
b
) is strictly increasing in both a and b, nonsymmetric and homogeneous of degree 1 with respect to a and b. Many symmetric bivariate means are special cases of the Schwab-Borchardt mean; for example,
(2)
P
(
a
,
b
)
=
a
-
b
2
sin
-
1
⁡
[
(
a
-
b
)
/
(
a
+
b
)
]
=
SB
(
G
,
A
)
is
the
first
Seiffert
mean
,
T
(
a
,
b
)
=
a
-
b
2
tan
-
1
[
(
a
-
b
)
/
(
a
+
b
)
]
=
SB
(
A
,
Q
)
is
the
second
Seiffert
mean
,
M
(
a
,
b
)
=
a
-
b
2
sinh
⁡
-
1
[
(
a
-
b
)
/
(
a
+
b
)
]
=
SB
(
Q
,
A
)
is
the
Neuman-Sándor
mean
,
L
(
a
,
b
)
=
a
-
b
2
tanh
-
1
[
(
a
-
b
)
/
(
a
+
b
)
]
=
SB
(
A
,
G
)
is
the
logarithmic
mean
,
where G
(
a
,
b
)
=
a
b, A
(
a
,
b
)
=
(
a
+
b
)
/
2, and Q
(
a
,
b
)
=
(
a
2
+
b
2
)
/
2 denote the classical geometric mean, arithmetic mean, and quadratic mean of a and b, respectively. The Schwab-Borchardt mean SB
(
a
,
b
) was investigated in [1, 2].LetH
(
a
,
b
)
=
2
a
b
/
(
a
+
b
) and C
(
a
,
b
)
=
(
a
2
+
b
2
)
/
(
a
+
b
) be the harmonic and contraharmonic means of two positive numbers a and b, respectively. Then, it is well-known that
(3)
H
(
a
,
b
)
<
G
(
a
,
b
)
<
L
(
a
,
b
)
<
P
(
a
,
b
)
<
A
(
a
,
b
)
<
M
(
a
,
b
)
H
(
a
,
b
)
<
T
(
a
,
b
)
<
Q
(
a
,
b
)
<
C
(
a
,
b
)
,
for a
,
b
>
0 with a
≠
b.Recently, the Schwab-Borchardt mean and its special cases have been the subject of intensive research. Neuman and Sándor [3, 4] proved that the inequalities
(4)
P
(
a
,
b
)
>
2
π
A
(
a
,
b
)
,
A
(
a
,
b
)
log
⁡
(
1
+
2
)
>
M
(
a
,
b
)
>
π
4
log
⁡
(
1
+
2
)
T
(
a
,
b
)
,
T
(
A
(
a
,
b
)
,
G
(
a
,
b
)
)
<
P
(
a
,
b
)
,
T
(
a
,
b
)
>
T
(
A
(
a
,
b
)
,
Q
(
a
,
b
)
)
,
L
(
a
,
b
)
<
L
(
A
(
a
,
b
)
,
G
(
a
,
b
)
)
,
M
(
a
,
b
)
<
L
(
A
(
a
,
b
)
,
Q
(
a
,
b
)
)
,
L
(
a
,
b
)
>
H
(
P
(
a
,
b
)
,
G
(
a
,
b
)
)
,
P
(
a
,
b
)
>
H
(
L
(
a
,
b
)
,
A
(
a
,
b
)
)
,
M
(
a
,
b
)
>
H
(
T
(
a
,
b
)
,
A
(
a
,
b
)
)
,
T
(
a
,
b
)
>
H
(
M
(
a
,
b
)
,
Q
(
a
,
b
)
)
,
G
(
a
,
b
)
P
(
a
,
b
)
<
L
2
(
a
,
b
)
<
G
2
(
a
,
b
)
+
P
2
(
a
,
b
)
2
,
L
(
a
,
b
)
A
(
a
,
b
)
<
P
2
(
a
,
b
)
<
L
2
(
a
,
b
)
+
A
2
(
a
,
b
)
2
,
A
(
a
,
b
)
T
(
a
,
b
)
<
M
2
(
a
,
b
)
<
A
2
(
a
,
b
)
+
T
2
(
a
,
b
)
2
,
M
(
a
,
b
)
Q
(
a
,
b
)
<
T
2
(
a
,
b
)
<
M
2
(
a
,
b
)
+
Q
2
(
a
,
b
)
2
,
Q
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
<
M
(
a
,
b
)
<
1
3
Q
(
a
,
b
)
+
2
3
A
(
a
,
b
)
hold for all a, b
>
0 with a
≠
b. In [5], the author proved that the double inequalities
(5)
α
Q
(
a
,
b
)
+
(
1
-
α
)
A
(
a
,
b
)
<
M
(
a
,
b
)
<
β
Q
(
a
,
b
)
+
(
1
-
β
)
A
(
a
,
b
)
,
λ
C
(
a
,
b
)
+
(
1
-
λ
)
A
(
a
,
b
)
<
M
(
a
,
b
)
<
μ
C
(
a
,
b
)
+
(
1
-
μ
)
A
(
a
,
b
)
hold for all a, b
>
0 with a
≠
b if and only if α
≤
[
1
-
log
⁡
⁡
(
1
+
2
)
]
/
[
(
2
-
1
)
log
⁡
⁡
(
1
+
2
)
]
=
0.3249
⋯, β
≥
1
/
3, λ
≤
[
1
-
log
⁡
⁡
(
1
+
2
)
]
/
log
⁡
⁡
(
1
+
2
)
=
0.1345
⋯, and μ
≥
1
/
6. Chu and Long [6] found that the double inequality
(6)
M
p
(
a
,
b
)
<
M
(
a
,
b
)
<
q
I
(
a
,
b
)
holds for all a
,
b
>
0 with a
≠
b if and only if p
≤
log
⁡
⁡
2
/
log
⁡
⁡
[
2
log
⁡
⁡
(
1
+
2
)
]
=
1.224
⋯ and q
≥
e
/
[
2
log
⁡
⁡
(
1
+
2
)
], where M
p
(
a
,
b
)
=
[
(
a
p
+
b
p
)
/
2
]
1
/
p
(
p
≠
0
) and M
0
(
a
,
b
)
=
a
b is the pth power mean of a and b. Zhao et al. [7] presented the least values α
1, α
2, and α
3 and the greatest values β
1, β
2, and β
3 such that the double inequalities
(7)
α
1
H
(
a
,
b
)
+
(
1
-
α
1
)
Q
(
a
,
b
)
<
M
(
a
,
b
)
<
β
1
H
(
a
,
b
)
+
(
1
-
β
1
)
Q
(
a
,
b
)
,
α
2
G
(
a
,
b
)
+
(
1
-
α
2
)
Q
(
a
,
b
)
<
M
(
a
,
b
)
<
β
2
G
(
a
,
b
)
+
(
1
-
β
2
)
Q
(
a
,
b
)
,
α
3
H
(
a
,
b
)
+
(
1
-
α
3
)
C
(
a
,
b
)
<
M
(
a
,
b
)
<
β
3
H
(
a
,
b
)
+
(
1
-
β
3
)
C
(
a
,
b
)
hold for all a, b
>
0 with a
≠
b.Very recently, the bivariate meansS
A
H, S
H
A, S
C
A, and S
A
C derived from the Schwab-Borchardt mean are defined by Neuman [8, 9] as follows:
(8)
S
A
H
=
SB
(
A
,
H
)
,
S
H
A
=
SB
(
H
,
A
)
,
S
C
A
=
SB
(
C
,
A
)
,
S
A
C
=
SB
(
A
,
C
)
.We call the meansS
A
H, S
H
A, S
C
A, and S
A
C given in (8) the Neuman means. Moreover, let v
=
(
a
-
b
)
/
(
a
+
b
)
∈
(
-
1,1
); then the following explicit formulas for S
A
H, S
H
A, S
A
C, and S
C
A are found by Neuman [8]:
(9)
S
A
H
=
A
tanh
⁡
(
p
)
p
,
S
H
A
=
A
sin
⁡
(
q
)
q
,
(10)
S
C
A
=
A
sinh
⁡
⁡
(
r
)
r
,
S
A
C
=
A
tan
⁡
(
s
)
s
,
where p, q, r, and s are defined implicitly as sech
⁡
(
p
)
=
1
-
v
2, cos
⁡
⁡
(
q
)
=
1
-
v
2, cosh
⁡
⁡
(
r
)
=
1
+
v
2, and sec
⁡
(
s
)
=
1
+
v
2, respectively. Clearly, p
∈
(
0
,
∞
), q
∈
(
0
,
π
/
2
), r
∈
(
0
,
log
⁡
(
2
+
3
)
), and s
∈
(
0
,
π
/
3
).In [8, 9], Neuman proved that the inequalities
(11)
H
(
a
,
b
)
<
S
A
H
(
a
,
b
)
<
L
(
a
,
b
)
<
S
H
A
(
a
,
b
)
<
P
(
a
,
b
)
,
T
(
a
,
b
)
<
S
C
A
(
a
,
b
)
<
Q
(
a
,
b
)
<
S
A
C
(
a
,
b
)
<
C
(
a
,
b
)
,
(12)
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
<
S
H
A
(
a
,
b
)
<
1
3
H
(
a
,
b
)
+
2
3
A
(
a
,
b
)
,
C
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
<
S
C
A
(
a
,
b
)
<
1
3
C
(
a
,
b
)
+
2
3
A
(
a
,
b
)
,
A
1
/
3
(
a
,
b
)
H
2
/
3
(
a
,
b
)
<
S
A
H
(
a
,
b
)
<
1
3
A
(
a
,
b
)
+
2
3
H
(
a
,
b
)
,
A
1
/
3
(
a
,
b
)
C
2
/
3
(
a
,
b
)
<
S
A
C
(
a
,
b
)
<
1
3
A
(
a
,
b
)
+
2
3
C
(
a
,
b
)
hold for a, b
>
0 with a
≠
b.He et al. [10] found the greatest values α
1, α
2
∈
[
0,1
/
2
], α
3, α
4
∈
[
1
/
2,1
], and the least values β
1, β
2
∈
[
0,1
/
2
], β
3, β
4
∈
[
1
/
2,1
] such that the double inequalities
(13)
H
(
α
1
a
+
(
1
-
α
1
)
b
,
α
1
b
+
(
1
-
α
1
)
a
)
<
S
A
H
(
a
,
b
)
<
H
(
β
1
a
+
(
1
-
β
1
)
b
,
β
1
b
+
(
1
-
β
1
)
a
)
,
H
(
α
2
a
+
(
1
-
α
2
)
b
,
α
2
b
+
(
1
-
α
2
)
a
)
<
S
H
A
(
a
,
b
)
<
H
(
β
2
a
+
(
1
-
β
2
)
b
,
β
2
b
+
(
1
-
β
2
)
a
)
,
C
(
α
3
a
+
(
1
-
α
3
)
b
,
α
3
b
+
(
1
-
α
3
)
a
)
<
S
C
A
(
a
,
b
)
<
C
(
β
3
a
+
(
1
-
β
3
)
b
,
β
3
b
+
(
1
-
β
3
)
a
)
,
C
(
α
4
a
+
(
1
-
α
4
)
b
,
α
4
b
+
(
1
-
α
4
)
a
)
<
S
A
C
(
a
,
b
)
<
C
(
β
4
a
+
(
1
-
β
4
)
b
,
β
4
b
+
(
1
-
β
4
)
a
)
hold for all a
,
b
>
0 with a
≠
b.Motivated by inequalities (12), it is natural to ask what the greatest values α
1, α
2, α
3, and α
4 and the least values β
1, β
2, β
3, and β
4 are such that the double inequalities
(14)
α
1
[
H
(
a
,
b
)
3
+
2
A
(
a
,
b
)
3
]
+
(
1
-
α
1
)
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
<
S
H
A
(
a
,
b
)
<
β
1
[
H
(
a
,
b
)
3
+
2
A
(
a
,
b
)
3
]
+
(
1
-
β
1
)
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
,
α
2
[
C
(
a
,
b
)
3
+
2
A
(
a
,
b
)
3
]
+
(
1
-
α
2
)
C
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
<
S
C
A
(
a
,
b
)
<
β
2
[
C
(
a
,
b
)
3
+
2
A
(
a
,
b
)
3
]
+
(
1
-
β
2
)
C
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
,
α
3
[
A
(
a
,
b
)
3
+
2
H
(
a
,
b
)
3
]
+
(
1
-
α
3
)
A
1
/
3
(
a
,
b
)
H
2
/
3
(
a
,
b
)
<
S
A
H
(
a
,
b
)
<
β
3
[
A
(
a
,
b
)
3
+
2
H
(
a
,
b
)
3
]
+
(
1
-
β
3
)
A
1
/
3
(
a
,
b
)
H
2
/
3
(
a
,
b
)
,
α
4
[
A
(
a
,
b
)
3
+
2
C
(
a
,
b
)
3
]
+
(
1
-
α
4
)
A
1
/
3
(
a
,
b
)
C
2
/
3
(
a
,
b
)
<
S
A
C
(
a
,
b
)
<
β
4
[
A
(
a
,
b
)
3
+
2
C
(
a
,
b
)
3
]
+
(
1
-
β
4
)
A
1
/
3
(
a
,
b
)
C
2
/
3
(
a
,
b
)
hold for all a, b
>
0 with a
≠
b.The purpose of this paper is to answer these questions. All numerical computations are carried out using MATHEMATICA software. Our main results are the following Theorems1–4.Theorem 1.
The double inequality(15)
α
1
[
H
(
a
,
b
)
3
+
2
A
(
a
,
b
)
3
]
+
(
1
-
α
1
)
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
<
S
H
A
(
a
,
b
)
<
β
1
[
H
(
a
,
b
)
3
+
2
A
(
a
,
b
)
3
]
+
(
1
-
β
1
)
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
holds for all a
,
b
>
0 with a
≠
b if and only if α
1
≤
4
/
5 and β
1
≥
3
/
π.Theorem 2.
The two-sided inequality(16)
α
2
[
C
(
a
,
b
)
3
+
2
A
(
a
,
b
)
3
]
+
(
1
-
α
2
)
C
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
<
S
C
A
(
a
,
b
)
<
β
2
[
C
(
a
,
b
)
3
+
2
A
(
a
,
b
)
3
]
+
(
1
-
β
2
)
C
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
holds true for all a
,
b
>
0 with a
≠
b if and only if α
2
≤
3
[
2
3
log
⁡
⁡
(
2
+
3
)
-
3
]
/
[
(
3
2
3
-
4
)
log
⁡
⁡
(
2
+
3
)
]
=
0.7528
⋯ and β
2
≥
4
/
5.Theorem 3.
The double inequality(17)
α
3
[
A
(
a
,
b
)
3
+
2
H
(
a
,
b
)
3
]
+
(
1
-
α
3
)
A
1
/
3
(
a
,
b
)
H
2
/
3
(
a
,
b
)
<
S
A
H
(
a
,
b
)
<
β
3
[
A
(
a
,
b
)
3
+
2
H
(
a
,
b
)
3
]
+
(
1
-
β
3
)
A
1
/
3
(
a
,
b
)
H
2
/
3
(
a
,
b
)
holds for all a
,
b
>
0 with a
≠
b if and only if α
3
≤
0 and β
3
≥
4
/
5.Theorem 4.
The two-sided inequality(18)
α
4
[
A
(
a
,
b
)
3
+
2
C
(
a
,
b
)
3
]
+
(
1
-
α
4
)
A
1
/
3
(
a
,
b
)
C
2
/
3
(
a
,
b
)
<
S
A
C
(
a
,
b
)
<
β
4
[
A
(
a
,
b
)
3
+
2
C
(
a
,
b
)
3
]
+
(
1
-
β
4
)
A
1
/
3
(
a
,
b
)
C
2
/
3
(
a
,
b
)
holds true for all a
,
b
>
0 with a
≠
b if and only if α
4
≤
4
/
5 and β
2
≥
3
(
3
3
-
4
3
π
)
/
[
(
5
-
3
4
3
)
π
]
=
0.8400
⋯.


## 2. Two Lemmas

In order to prove our main results, we need two lemmas, which we present in this section.Lemma 5.
Letp
∈
ℝ and
(19)
f
(
x
)
=
p
2
x
6
+
2
p
2
x
5
+
3
(
-
p
2
+
4
p
-
2
)
x
4
f
(
x
)
=
+
2
(
-
2
p
2
+
9
p
-
6
)
x
3
+
(
4
p
2
+
6
p
-
9
)
x
2
f
(
x
)
=
+
6
(
p
-
1
)
x
+
3
(
p
-
1
)
.
Then, the following statements are true.(1)
Ifp
=
4
/
5, then f
(
x
)
<
0 for all x
∈
(
0,1
) and f
(
x
)
>
0 for all x
∈
(
1
,
2
3
).
(2)
Ifp
=
3
/
π, then there exists λ
1
∈
(
0,1
) such that f
(
x
)
<
0 for x
∈
(
0
,
λ
1
) and f
(
x
)
>
0 for x
∈
(
λ
1
,
1
).
(3)
Ifp
=
3
[
2
3
log
⁡
⁡
(
2
+
3
)
-
3
]
/
[
(
3
2
3
-
4
)
log
⁡
⁡
(
2
+
3
)
]
=
0.7528
⋯, then there exists λ
2
∈
(
1
,
2
3
) such that f
(
x
)
<
0 for x
∈
(
1
,
λ
2
) and f
(
x
)
>
0 for x
∈
(
λ
2
,
2
3
).Proof.
For part (1), ifp
=
4
/
5, then (19) becomes
(20)
f
(
x
)
=
1
25
(
x
-
1
)
(
16
x
5
+
48
x
4
+
90
x
3
+
86
x
2
w
f
(
x
)
=
1
25
(
x
-
10
)
+
45
x
+
15
16
x
5
+
48
x
4
)
.
Therefore, part (1) follows easily from (20).
For part (2), ifp
=
3
/
π, then simple computations lead to
(21)
-
p
2
+
4
p
-
2
=
-
2
π
2
+
12
π
-
9
π
2
>
0
,
(22)
-
2
p
2
+
9
p
-
6
=
-
6
π
2
+
27
π
-
18
π
2
>
0
,
(23)
4
p
2
+
6
p
-
9
=
-
9
π
2
+
18
π
+
36
π
2
>
0
,
(24)
f
(
0
)
=
-
3
(
π
-
3
)
π
<
0
,
(25)
f
(
1
)
=
9
(
15
-
4
π
)
π
>
0
,
(26)
f
′
(
x
)
=
6
p
2
x
5
+
10
p
2
x
4
+
12
(
-
p
2
+
4
p
-
2
)
x
3
+
6
(
-
2
p
2
+
9
p
-
6
)
x
2
+
2
(
4
p
2
+
6
p
-
9
)
x
+
6
(
p
-
1
)
,
(27)
f
′
(
0
)
=
6
(
3
-
π
)
π
<
0
,
(28)
f
′
(
1
)
=
12
(
30
-
7
π
)
π
>
0
,
(29)
f
′′
(
x
)
=
30
p
2
x
4
+
40
p
2
x
3
+
36
(
-
p
2
+
4
p
-
2
)
x
2
+
12
(
-
2
p
2
+
9
p
-
6
)
x
+
2
(
4
p
2
+
6
p
-
9
)
.
It follows from (21)–(23) and (29) that f
′
(
x
) is strictly increasing on (
0,1
). Then, (27) and (28) lead to the conclusion that there exists x
0
∈
(
0,1
) such that f
(
x
) is strictly decreasing in (
0
,
x
0
] and strictly increasing in [
x
0
,
1
).
Therefore, part (2) follows from (24) and (25) together with the piecewise monotonicity of f
(
x
).
For part (3), ifp
=
3
[
2
3
log
⁡
⁡
(
2
+
3
)
-
3
]
/
[
(
3
2
3
-
4
)
log
⁡
⁡
(
2
+
3
)
]
=
0.7528
⋯, then numerical computations lead to
(30)
-
p
2
+
4
p
-
2
=
0.444
⋯
>
0
,
(31)
4
p
2
+
6
p
-
9
=
-
2.215
⋯
<
0
,
(32)
6
(
p
-
1
)
=
-
1.483
⋯
<
0
,
(33)
f
(
1
)
=
9
(
5
p
-
4
)
=
-
2.120
⋯
<
0
,
(34)
f
(
2
3
)
=
1.669
⋯
>
0
.
It follows from (26) and (30)–(32) that
(35)
f
′
(
x
)
>
6
p
2
x
2
+
10
p
2
x
2
+
12
(
-
p
2
+
4
p
-
2
)
x
2
+
6
(
-
2
p
2
+
9
p
-
6
)
x
2
+
2
(
4
p
2
+
6
p
-
9
)
x
2
+
6
(
p
-
1
)
x
2
=
12
(
10
p
-
7
)
x
2
>
0
for x
∈
(
1
,
2
3
).
Therefore, part (3) follows easily from (33)–(35).Lemma 6.
Letp
∈
ℝ and
(36)
g
(
x
)
=
3
(
1
-
p
)
x
6
+
6
(
1
-
p
)
x
5
+
(
-
4
p
2
-
6
p
+
9
)
x
4
+
2
(
2
p
2
-
9
p
+
6
)
x
3
+
3
(
p
2
-
4
p
+
2
)
x
2
-
2
p
2
x
-
p
2
.
Then, the following statements are true.(1)
Ifp
=
4
/
5, then g
(
x
)
<
0 for all x
∈
(
0,1
) and g
(
x
)
>
0 for all x
∈
(
1
,
2
3
).
(2)
Ifp
=
3
(
3
3
-
4
3
π
)
/
[
(
5
-
3
4
3
)
π
]
=
0.8400
⋯, then there exists λ
3
∈
(
1
,
2
3
) such that g
(
x
)
<
0 for x
∈
(
1
,
λ
3
) and g
(
x
)
>
0 for x
∈
(
λ
3
,
2
3
).Proof.
For part (1), ifp
=
4
/
5, then (36) becomes
(37)
g
(
x
)
=
1
25
(
x
-
1
)
(
15
x
5
+
45
x
4
+
86
x
3
+
90
x
2
w
+
48
x
+
16
15
x
5
+
45
x
4
)
.
Therefore, part (1) follows from (37).
For part (2), ifp
=
3
(
3
3
-
4
3
π
)
/
[
(
5
-
3
4
3
)
π
]
=
0.8400
⋯, then numerical computations lead to
(38)
-
4
p
2
-
6
p
+
9
=
1.137
⋯
>
0
,
(39)
p
2
-
4
p
+
2
=
-
0.654
⋯
<
0
,
(40)
g
(
1
)
=
9
(
4
-
5
p
)
=
-
1.801
⋯
<
0
,
(41)
g
(
2
3
)
=
1.635
⋯
>
0
,
(42)
g
′
(
x
)
=
18
(
1
-
p
)
x
5
+
30
(
1
-
p
)
x
4
+
4
(
-
4
p
2
-
6
p
+
9
)
x
3
+
6
(
2
p
2
-
9
p
+
6
)
x
2
+
6
(
p
2
-
4
p
+
2
)
x
-
2
p
2
.
From (38) and (39) together with (42), we clearly see that
(43)
g
′
(
x
)
>
18
(
1
-
p
)
x
2
+
30
(
1
-
p
)
x
2
+
4
(
-
4
p
2
-
6
p
+
9
)
x
2
+
6
(
2
p
2
-
9
p
+
6
)
x
2
+
6
(
p
2
-
4
p
+
2
)
x
2
-
2
p
2
x
2
=
6
(
22
-
25
p
)
x
2
>
0
for x
∈
(
1
,
2
3
).
Therefore, part (2) follows from (40) and (41) together with (43).


## 3. Proofs of Theorems1–4

Proof of Theorem1.
Without loss of generality, we assume thata
>
b. Let v
=
(
a
-
b
)
/
(
a
+
b
),  λ
=
v
2
-
v
2,  x
=
1
-
λ
2
6, and p
∈
{
4
/
5,3
/
π
}. Then, v, λ, x
∈
(
0,1
),
(44)
S
H
A
(
a
,
b
)
-
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
H
(
a
,
b
)
/
3
+
2
A
(
a
,
b
)
/
3
-
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
=
λ
/
sin
-
1
(
λ
)
-
(
1
-
λ
2
)
1
/
6
2
/
3
+
(
1
-
λ
2
)
1
/
2
/
3
-
(
1
-
λ
2
)
1
/
6
,
(45)
S
H
A
(
a
,
b
)
-
[
p
(
1
3
H
(
a
,
b
)
+
2
A
(
a
,
b
)
3
)
S
H
A
(
a
,
b
)
-
e
+
(
1
-
p
)
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
(
1
3
H
(
a
,
b
)
+
2
A
(
a
,
b
)
3
)
]
=
A
(
a
,
b
)
[
λ
sin
-
1
(
λ
)
-
p
(
(
1
-
λ
2
)
1
/
2
3
+
2
3
)
S
H
A
(
a
,
)
-
-
(
1
-
p
)
(
1
-
λ
2
)
1
/
6
(
(
1
-
λ
2
)
1
/
2
3
+
2
3
)
]
=
(
A
(
a
,
b
)
[
p
(
(
1

-

λ
2
)
1
/
2

+

2
)

+

3
(
1
-
p
)
(
1
-
λ
2
)
1
/
6
]
)
×
(
3
sin
-
1
(
λ
)
)
-
1
F
(
x
)
,
where
(46)
F
(
x
)
=
3
1
-
x
6
p
x
3
+
3
(
1
-
p
)
x
+
2
p
-
sin
-
1
(
1
-
x
6
)
,
(47)
F
(
0
)
=
3
2
p
-
π
2
,
(48)
F
(
1
)
=
0
,
(49)
F
′
(
x
)
=
3
(
x
-
1
)
2
1
-
x
6
[
p
x
3
+
3
(
1
-
p
)
x
+
2
p
]
2
f
(
x
)
,
where f
(
x
) is defined as in Lemma 5.
We divide the proof into two cases.
Case  1  (
p
=
4
/
5
). Then, from Lemma 5(1) and (49), we clearly see that F
(
x
) is strictly decreasing in (
0,1
). Therefore,
(50)
S
H
A
(
a
,
b
)
>
4
5
[
1
3
H
(
a
,
b
)
+
2
3
A
(
a
,
b
)
]
+
1
5
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
for all a
,
b
>
0 with a
≠
b follows from (45) and (48) together with the monotonicity of F
(
x
).
Case  2  (
p
=
3
/
π
). Then, from (47) and (49) and Lemma 5(2), we know that
(51)
F
(
0
)
=
0
and there exists λ
1
∈
(
0,1
) such that F
(
x
) is strictly decreasing in (
0
,
λ
1
] and strictly increasing in [
λ
1
,
1
). Therefore,
(52)
S
H
A
(
a
,
b
)
<
3
π
[
1
3
H
(
a
,
b
)
+
2
3
A
(
a
,
b
)
]
+
(
1
-
3
π
)
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
for all a
,
b
>
0 with a
≠
b follows from (45) and (48) together with (51) and the piecewise monotonicity of F
(
x
).
Note that(53)
lim
⁡
λ
→
0
+
λ
/
sin
-
1
(
λ
)
-
(
1
-
λ
2
)
1
/
6
2
/
3
+
(
1
-
λ
2
)
1
/
2
/
3
-
(
1
-
λ
2
)
1
/
6
=
4
5
,
(54)
lim
⁡
λ
→
1
-
λ
/
sin
-
1
(
λ
)
-
(
1
-
λ
2
)
1
/
6
2
/
3
+
(
1
-
λ
2
)
1
/
2
/
3
-
(
1
-
λ
2
)
1
/
6
=
3
π
.
Therefore, Theorem1 follows from (50) and (52)–(54) together with the following statements.(i)
Ifα
>
4
/
5, then (44) and (53) imply that there exists small enough δ
>
0 such that S
H
A
(
a
,
b
)
<
α
(
H
(
a
,
b
)
/
3
+
2
A
(
a
,
b
)
/
3
)
+
(
1
-
α
)
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
) for all a
>
b
>
0 with b
/
a
∈
(
0
,
δ
).
(ii)
Ifβ
<
3
/
π, then (44) and (54) imply that there exists large enough M
>
1 such that S
H
A
(
a
,
b
)
>
β
(
H
(
a
,
b
)
/
3
+
2
A
(
a
,
b
)
/
3
)
+
(
1
-
β
)
H
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
) for all a
>
b
>
0 with a
/
b
∈
(
M
,
+
∞
).Proof of Theorem2.
Without loss of generality, we assume thata
>
b. Let v
=
(
a
-
b
)
/
(
a
+
b
),  μ
=
v
2
+
v
2,  x
=
1
+
μ
2
6, and p
∈
{
3
[
2
3
log
⁡
(
2
+
3
)
-
3
]
/
[
(
3
2
3
-
4
)
log
⁡
(
2
+
3
)
]
,
4
/
5
}. Then, v
∈
(
0,1
),  μ
∈
(
0
,
3
),  x
∈
(
1
,
2
3
),
(55)
S
C
A
(
a
,
b
)
-
C
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
C
(
a
,
b
)
/
3
+
2
A
(
a
,
b
)
/
3
-
C
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
=
μ
/
sinh
⁡
-
1
(
μ
)
-
(
1
+
μ
2
)
1
/
6
2
/
3
+
(
1
+
μ
2
)
1
/
2
/
3
-
(
1
+
μ
2
)
1
/
6
,
(56)
S
C
A
(
a
,
b
)
-
[
p
(
1
3
C
(
a
,
b
)
+
2
A
(
a
,
b
)
3
)
w
S
C
A
(
a
,
b
)
-
+
(
1
-
p
)
C
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
(
1
3
C
(
a
,
b
)
+
2
A
(
a
,
b
)
3
)
]
=
A
(
a
,
b
)
[
μ
sinh
⁡
-
1
(
μ
)
-
p
(
(
1
+
μ
2
)
1
/
2
3
+
2
3
)
S
C
A
(
a
,
b
)
-
ww
-
(
1
-
p
)
(
1
+
μ
2
)
1
/
6
(
(
1
+
μ
2
)
1
/
2
3
+
2
3
)
]
=
(
A
(
a
,
b
)
[
p
(
(
1
+
μ
2
)
1
/
2

+

2
)

+

3
(
1

-

p
)
(
1

+

μ
2
)
1
/
6
]
)
×
(
3
sinh
⁡
-
1
(
μ
)
)
-
1
G
(
x
)
,
where
(57)
G
(
x
)
=
3
x
6
-
1
p
x
3
+
3
(
1
-
p
)
x
+
2
p
-
sinh
⁡
-
1
(
x
6
-
1
)
,
(58)
G
(
1
)
=
0
,
(59)
G
(
2
3
)
=
3
3
(
4
-
3
2
3
)
p
+
3
2
3
-
log
⁡
⁡
(
1
+
3
)
,
(60)
G
′
(
x
)
=
-
3
(
x
-
1
)
2
x
6
-
1
[
p
x
3
+
3
(
1
-
p
)
x
+
2
p
]
2
f
(
x
)
,
where f
(
x
) is defined as in Lemma 5.
We divide the proof into two cases.
Case  1 (
p
=
3
[
2
3
log
⁡
(
2
+
3
)
-
3
]
/
[
(
3
2
3
-
4
)
log
⁡
(
2
+
3
)
]
=
0.7528
⋯
). Then, from (59) and (60) together with Lemma 5(3), we clearly see that there exists λ
2
∈
(
1
,
2
3
) such that G
(
x
) is strictly increasing in (
1
,
λ
2
] and strictly decreasing in [
λ
2
,
3
2
), and
(61)
G
(
2
3
)
=
0
.
Therefore,(62)
S
C
A
(
a
,
b
)
>
3
(
2
3
log
⁡
(
2
+
3
)
-
3
)
(
3
2
3
-
4
)
log
⁡
(
2
+
3
)
[
1
3
C
(
a
,
b
)
+
2
3
A
(
a
,
b
)
]
+
(
1
-
3
(
2
3
log
⁡
⁡
(
2
+
3
)
-
3
)
(
3
2
3
-
4
)
log
⁡
⁡
(
2
+
3
)
)
×
C
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
for all a
,
b
>
0 with a
≠
b follows easily from (56) and (58) together with (61) and the piecewise monotonicity of G
(
x
).
Case  2 (
p
=
4
/
5
). Then, Lemma 5(1) and (60) lead to the conclusion that G
(
x
) is strictly decreasing in (
1
,
2
3
). Therefore,
(63)
S
C
A
(
a
,
b
)
<
4
5
[
1
3
C
(
a
,
b
)
+
2
3
A
(
a
,
b
)
]
+
1
5
C
1
/
3
(
a
,
b
)
A
2
/
3
(
a
,
b
)
for all a
,
b
>
0 with a
≠
b follows from (56) and (58) together with the monotonicity of G
(
x
).
Note that(64)
lim
⁡
μ
→
0
+
μ
/
sinh
⁡
-
1
(
μ
)
-
(
1
+
μ
2
)
1
/
6
2
/
3
+
(
1
+
μ
2
)
1
/
2
/
3
-
(
1
+
μ
2
)
1
/
6
=
4
5
,
(65)
lim
⁡
μ
→
3
-
μ
/
sinh
⁡
-
1
(
μ
)
-
(
1
+
μ
2
)
1
/
6
2
/
3
+
(
1
+
μ
2
)
1
/
2
/
3
-
(
1
+
μ
2
)
1
/
6
=
3
(
2
3
log
⁡
(
2
+
3
)
-
3
)
(
3
2
3
-
4
)
log
⁡
(
2
+
3
)
.
Therefore, Theorem2 follows from (55) and (62)–(65).Proof of Theorem3.
Without loss of generality, we assume thata
>
b. Let v
=
(
a
-
b
)
/
(
a
+
b
), λ
=
v
2
-
v
2, x
=
1
-
λ
2
6, and p
∈
{
4
/
5,0
}. Then, v
,
λ
,
x
∈
(
0,1
) and (9) leads to
(66)
S
A
H
(
a
,
b
)
=
A
(
a
,
b
)
λ
tanh
-
1
(
λ
)
.
It follows from (66) that
(67)
S
A
H
(
a
,
b
)
-
A
1
/
3
(
a
,
b
)
H
2
/
3
(
a
,
b
)
A
(
a
,
b
)
/
3
+
2
H
(
a
,
b
)
/
3
-
A
1
/
3
(
a
,
b
)
H
2
/
3
(
a
,
b
)
=
λ
/
tanh
-
1
(
λ
)
-
(
1
-
λ
2
)
1
/
3
1
/
3
+
2
(
1
-
λ
2
)
1
/
2
/
3
-
(
1
-
λ
2
)
1
/
3
,
(68)
S
A
H
(
a
,
b
)
-
[
p
(
1
3
A
(
a
,
b
)
+
2
H
(
a
,
b
)
3
)
S
A
H
(
a
,
b
)
-
w
+
(
1
-
p
)
A
1
/
3
(
a
,
b
)
H
2
/
3
(
a
,
b
)
(
1
3
A
(
a
,
b
)
+
2
H
(
a
,
b
)
3
)
]
=
A
(
a
,
b
)
[
λ
tanh
-
1
(
λ
)
-
p
(
2
(
1
-
λ
2
)
1
/
2
3
+
1
3
)
S
A
H
(
a
,
b
)
-
w
-
(
1
-
p
)
(
1
-
λ
2
)
1
/
3
(
2
(
1
-
λ
2
)
1
/
2
3
+
1
3
)
]
=
A
(
a
,
b
)
[
p
(
2
(
1
-
λ
2
)
1
/
2

+

1
)

+

3
(
1

-

p
)
(
1

-

λ
2
)
1
/
3
]
3
tanh
-
1
(
λ
)
×
H
(
x
)
,
where
(69)
H
(
x
)
=
3
1
-
x
6
2
p
x
3
+
3
(
1
-
p
)
x
2
+
p
-
tanh
-
1
(
1
-
x
6
)
(70)
H
(
1
)
=
0
,
(71)
H
′
(
x
)
=
-
3
(
1
-
x
)
2
x
1
-
x
6
[
2
p
x
3
+
3
(
1
-
p
)
x
2
+
p
]
2
g
(
x
)
,
where g
(
x
) is defined as in Lemma 6.
Ifp
=
4
/
5, then Lemma 6(1) and (71) lead to the conclusion that H
(
x
) is strictly increasing in (
0,1
). Therefore,
(72)
S
A
H
(
a
,
b
)
<
4
5
(
1
3
A
(
a
,
b
)
+
2
H
(
a
,
b
)
3
)
S
A
H
(
a
,
b
)
<
+
1
5
A
1
/
3
(
a
,
b
)
H
2
/
3
(
a
,
b
)
for all a, b
>
0 with a
≠
b follows from (68) and (70) together with the monotonicity of H
(
x
).
Note that(73)
lim
⁡
λ
→
0
+
λ
/
tanh
-
1
(
λ
)
-
(
1
-
λ
2
)
1
/
3
1
/
3
+
2
(
1
-
λ
2
)
1
/
2
/
3
-
(
1
-
λ
2
)
1
/
3
=
4
5
,
(74)
lim
⁡
λ
→
1
-
λ
/
tanh
-
1
(
λ
)
-
(
1
-
λ
2
)
1
/
3
1
/
3
+
2
(
1
-
λ
2
)
1
/
2
/
3
-
(
1
-
λ
2
)
1
/
3
=
0
.
Therefore, Theorem3 follows from (12) and (67) together with (72)–(74).Proof of Theorem4.
Without loss of generality, we assume thata
>
b. Let v
=
(
a
-
b
)
/
(
a
+
b
),  μ
=
v
2
+
v
2, x
=
1
+
μ
2
6, and p
∈
{
3
(
3
3
-
4
3
π
)
/
[
(
5
-
3
4
3
)
π
]
,
4
/
5
}. Then, v
∈
(
0,1
),  μ
∈
(
0
,
3
), and  x
∈
(
1
,
2
3
) and (10) leads to
(75)
S
A
C
(
a
,
b
)
=
A
(
a
,
b
)
μ
tan
-
1
(
μ
)
.
It follows from (75) that
(76)
S
A
C
(
a
,
b
)
-
A
1
/
3
(
a
,
b
)
C
2
/
3
(
a
,
b
)
A
(
a
,
b
)
/
3
+
2
C
(
a
,
b
)
/
3
-
A
1
/
3
(
a
,
b
)
C
2
/
3
(
a
,
b
)
=
μ
/
tan
-
1
(
μ
)
-
(
1
+
μ
2
)
1
/
3
1
/
3
+
2
(
1
+
μ
2
)
1
/
2
/
3
-
(
1
+
μ
2
)
1
/
3
,
(77)
S
A
C
(
a
,
b
)
-
[
p
(
1
3
A
(
a
,
b
)
+
2
C
(
a
,
b
)
3
)
S
A
C
(
a
,
b
)
-
e
+
(
1
-
p
)
A
1
/
3
(
a
,
b
)
C
2
/
3
(
a
,
b
)
(
1
3
A
(
a
,
b
)
+
2
C
(
a
,
b
)
3
)
]
=
A
(
a
,
b
)
[
μ
tan
-
1
(
μ
)
-
p
(
2
(
1
+
μ
2
)
1
/
2
3
+
1
3
)
S
A
C
(
a
,
b
)
-
w
-
(
1
-
p
)
(
1
+
μ
2
)
1
/
3
(
2
(
1
+
μ
2
)
1
/
2
3
+
1
3
)
]
=
A
(
a
,
b
)
[
p
(
2
(
1

+

μ
2
)
1
/
2

+

1
)

+

3
(
1

-

p
)
(
1

+

μ
2
)
1
/
3
]
3
tan
-
1
(
μ
)
×
J
(
x
)
,
where
(78)
J
(
x
)
=
3
x
6
-
1
2
p
x
3
+
3
(
1
-
p
)
x
2
+
p
-
tan
-
1
(
x
6
-
1
)
,
(79)
J
(
1
)
=
0
,
(80)
J
(
2
3
)
=
3
3
(
5
-
3
4
3
)
p
+
3
4
3
-
π
3
,
(81)
J
′
(
x
)
=
3
(
x
-
1
)
2
x
6
-
1
[
2
p
x
3
+
3
(
1
-
p
)
x
2
+
p
]
2
g
(
x
)
,
where g
(
x
) is defined as in Lemma 6.
We divide the proof into two cases.
Case  1 (
p
=
4
/
5
). Then, (81) and Lemma 6(1) lead to the conclusion that J
(
x
) is strictly increasing in (
1
,
2
3
). Therefore,
(82)
S
A
C
(
a
,
b
)
>
4
5
(
1
3
A
(
a
,
b
)
+
2
C
(
a
,
b
)
3
)
S
A
C
(
a
,
b
)
>
+
1
5
A
1
/
3
(
a
,
b
)
C
2
/
3
(
a
,
b
)
for all a
,
b
>
0 with a
≠
b follows easily from (77) and (79) together with the monotonicity of J
(
x
).
Case  2 (
p
=
3
(
3
3
-
4
3
π
)
/
(
5
-
3
4
3
)
π
). Then, (80) and (81) together with Lemma 6(2) lead to the conclusion that there exists λ
3
∈
(
1
,
2
3
) such that J
(
x
) is strictly decreasing in (
1
,
λ
3
] and strictly increasing in [
λ
3
,
2
3
), and
(83)
J
(
2
3
)
=
0
.
Therefore,
(84)
S
A
C
(
a
,
b
)
<
3
(
3
3
-
4
3
π
)
(
5
-
3
4
3
)
π
(
1
3
A
(
a
,
b
)
+
2
C
(
a
,
b
)
3
)
S
A
C
(
a
,
b
)
<
+
(
1
-
3
(
3
3
-
4
3
π
)
(
5
-
3
4
3
)
π
)
A
1
/
3
(
a
,
b
)
C
2
/
3
(
a
,
b
)
for all a
,
b
>
0 with a
≠
b follows easily from (77) and (79) together with (83) and the piecewise monotonicity of J
(
x
).
Note that(85)
lim
⁡
μ
→
0
+
μ
/
tan
-
1
(
μ
)
-
(
1
+
μ
2
)
1
/
3
1
/
3
+
2
(
1
+
μ
2
)
1
/
2
/
3
-
(
1
+
μ
2
)
1
/
3
=
4
5
,
(86)
lim
⁡
μ
→
3
-
μ
/
tan
-
1
(
μ
)
-
(
1
+
μ
2
)
1
/
3
1
/
3
+
2
(
1
+
μ
2
)
1
/
2
/
3
-
(
1
+
μ
2
)
1
/
3
=
3
(
3
3
-
4
3
π
)
(
5
-
3
4
3
)
π
.
Therefore, Theorem4 follows from (76) and (82) together with (84)–(86).



---

*Source: 354132-2014-03-18.xml*