# Sharp One-Parameter Mean Bounds for Yang Mean

Authors: Wei-Mao Qian; Yu-Ming Chu; Xiao-Hui Zhang
Journal: Mathematical Problems in Engineering
 (2016)

---

## Abstract

We prove that the double inequalityJ
α
(
a
,
b
)
<
U
(
a
,
b
)
<
J
β
(
a
,
b
) holds for all a
,
b
>
0 with a
≠
b if and only if α
≤
2
/
(
π
-
2
)
=
0.8187
⋯ and β
≥
3
/
2, where U
(
a
,
b
)
=
(
a
-
b
)
/
[
2
arctan
⁡
(
(
a
-
b
)
/
2
a
b
)
], and J
p
(
a
,
b
)
=
p
(
a
p
+
1
-
b
p
+
1
)
/
[
(
p
+
1
)
(
a
p
-
b
p
)
]
(
p
≠
0
,
-
1
), J
0
(
a
,
b
)
=
(
a
-
b
)
/
(
log
⁡
a
-
log
⁡
b
), and J
-
1
(
a
,
b
)
=
a
b
(
log
⁡
a
-
log
⁡
b
)
/
(
a
-
b
) are the Yang and pth one-parameter means of a and b, respectively.


---

## Body

## 1. Introduction

Letp
∈
R and a
,
b
>
0 with a
≠
b. Then the pth one-parameter mean J
p
(
a
,
b
), pth power mean M
p
(
a
,
b
), harmonic mean H
(
a
,
b
), geometric mean G
(
a
,
b
), logarithmic mean L
(
a
,
b
), first Seiffert mean P
(
a
,
b
), identric mean I
(
a
,
b
), arithmetic mean A
(
a
,
b
), Yang mean U
(
a
,
b
), second Seiffert mean T
(
a
,
b
), and quadratic mean Q
(
a
,
b
) are, respectively, defined by (1)
J
p
a
,
b
=
p
a
p
+
1
-
b
p
+
1
p
+
1
a
p
-
b
p
,
p
≠
0
,
-
1
,
a
-
b
log
⁡
a
-
log
⁡
b
,
p
=
0
,
a
b
log
⁡
a
-
log
⁡
b
a
-
b
,
p
=
-
1
,
M
p
a
,
b
=
a
p
+
b
p
2
1
/
p
p
≠
0
,
M
0
a
,
b
=
a
b
,
H
a
,
b
=
2
a
b
a
+
b
,
G
a
,
b
=
a
b
,
L
a
,
b
=
b
-
a
log
⁡
b
-
log
⁡
a
,
P
a
,
b
=
a
-
b
2
arcsin
⁡
a
-
b
/
a
+
b
,
I
a
,
b
=
1
e
b
b
a
a
1
/
b
-
a
,
A
a
,
b
=
a
+
b
2
,
U
a
,
b
=
a
-
b
2
arctan
⁡
a
-
b
/
2
a
b
,
T
a
,
b
=
a
-
b
2
arctan
⁡
a
-
b
/
a
+
b
,
Q
a
,
b
=
a
2
+
b
2
2
.It is well known that both the meansJ
p
(
a
,
b
) and M
p
(
a
,
b
) are continuous and strictly increasing with respect to p
∈
R for fixed a
,
b
>
0 with a
≠
b. Recently, the one-parameter mean J
p
(
a
,
b
) and Yang mean U
(
a
,
b
) have attracted the attention of many researchers.Alzer [1] proved that the inequalities(2)
G
a
,
b
<
J
p
a
,
b
J
-
p
a
,
b
<
L
a
,
b
<
J
p
a
,
b
+
J
-
p
a
,
b
2
<
A
a
,
bhold for all a
,
b
>
0 with a
≠
b and p
≠
0.In [2, 3], the authors discussed the monotonicity and logarithmic convexity properties of the one-parameter mean J
p
(
a
,
b
).In [4, 5], the authors proved that the double inequalities (3)
J
p
1
a
,
b
<
α
A
a
,
b
+
1
-
α
L
a
,
b
<
J
q
1
a
,
b
J
p
2
a
,
b
<
α
A
a
,
b
+
1
-
α
H
a
,
b
<
J
q
2
a
,
b
,hold for all a
,
b
>
0 with a
≠
b and α
∈
(
0,1
) if and only if p
1
≤
α
/
(
2
-
α
), q
1
≥
α, p
2
≤
3
α
-
2, and q
2
≥
α
/
(
2
-
α
).Xia et al. [6] proved that the double inequality (4)
J
3
α
-
1
/
2
a
,
b
<
α
A
a
,
b
+
1
-
α
G
a
,
b
<
J
α
/
2
-
α
a
,
bholds for all a
,
b
>
0 with a
≠
b if α
∈
(
0,2
/
3
), and inequality (4) is reversed if α
∈
(
2
/
3,1
).Gao and Niu [7] presented the best possible parameters p and q such that the double inequality J
p
(
a
,
b
)
<
A
α
(
a
,
b
)
G
β
(
a
,
b
)
H
1
-
α
-
β
(
a
,
b
)
<
J
q
(
a
,
b
) holds for all a
,
b
>
0 with a
≠
b and α
+
β
∈
(
0,1
).In [8, 9], the authors proved that the double inequalities (5)
J
λ
1
a
,
b
<
T
a
,
b
<
J
μ
1
a
,
b
,
J
λ
2
a
,
b
<
I
a
,
b
<
J
μ
2
a
,
bhold for all a
,
b
>
0 with a
≠
b if and only if λ
1
≤
2
/
(
2
-
π
), μ
1
≥
2, λ
2
≤
1
/
2, and μ
2
≥
1
/
(
e
-
1
).Xia et al. [10] found that M
(
1
+
2
p
)
/
3
(
a
,
b
) is the best possible lower power mean bound for the one-parameter mean J
p
(
a
,
b
) if p
∈
(
-
2
,
-
1
/
2
)
∪
(
1
,
∞
) and M
(
1
+
2
p
)
/
3
(
a
,
b
) is the best possible upper power mean bound for the one-parameter mean J
p
(
a
,
b
) if p
∈
(
-
∞
,
-
2
)
∪
(
-
1
/
2,1
).For alla
,
b
>
0 with a
≠
b, Yang [11] provided the bounds for the Yang mean U
(
a
,
b
) in terms of other bivariate means as follows: (6)
P
a
,
b
<
U
a
,
b
<
T
a
,
b
,
G
a
,
b
T
a
,
b
A
a
,
b
<
U
a
,
b
<
P
a
,
b
Q
a
,
b
A
a
,
b
,
Q
1
/
2
a
,
b
2
G
a
,
b
+
Q
a
,
b
3
1
/
2
<
U
a
,
b
<
Q
2
/
3
a
,
b
G
a
,
b
+
Q
a
,
b
2
1
/
3
,
G
a
,
b
+
Q
a
,
b
2
<
U
a
,
b
<
2
3
G
a
,
b
+
Q
a
,
b
2
1
/
2
+
1
3
Q
1
/
2
a
,
b
2
.In [12, 13], the authors proved that the double inequalities (7)
2
3
G
a
,
b
+
Q
a
,
b
2
p
+
1
3
Q
p
a
,
b
1
/
p
<
U
a
,
b
<
2
3
G
a
,
b
+
Q
a
,
b
2
q
+
1
3
Q
q
a
,
b
1
/
q
2
1
-
λ
G
a
,
b
+
Q
a
,
b
λ
Q
a
,
b
+
G
a
,
b
Q
λ
a
,
b
2
1
-
λ
G
a
,
b
+
Q
a
,
b
λ
+
Q
λ
a
,
b
<
U
a
,
b
<
2
1
-
μ
G
a
,
b
+
Q
a
,
b
μ
Q
a
,
b
+
G
a
,
b
Q
μ
a
,
b
2
1
-
μ
G
a
,
b
+
Q
a
,
b
μ
+
Q
μ
a
,
b
,
M
α
a
,
b
<
U
a
,
b
<
M
β
a
,
b
,hold for all a
,
b
>
0 with a
≠
b if and only if p
≤
p
0, q
≥
1
/
5, λ
≥
1
/
5, μ
≤
p
1, α
≤
2
log
⁡
2
/
(
2
log
⁡
π
-
log
⁡
2
), and β
≥
4
/
3, where p
0
=
0.1941
⋯ is the unique solution of the equation p
log
⁡
(
2
/
π
)
-
log
⁡
1
+
2
1
-
p
+
log
⁡
3
=
0 on the interval (
1
/
10
,
∞
), and p
1
=
log
⁡
(
π
-
2
)
/
log
⁡
2
=
0.1910
⋯.Very recently, Zhou et al. [14] proved that α
=
1
/
2 and β
=
log
⁡
3
/
(
1
+
log
⁡
2
)
=
0.6488
⋯ are the best possible parameters such that the double inequality (8)
a
α
+
a
b
α
/
2
+
b
α
3
1
/
α
<
U
a
,
b
<
a
β
+
a
b
β
/
2
+
b
β
3
1
/
βholds for all a
,
b
>
0 with a
≠
b.The aim of this paper is to present the best possible parametersα and β such that the double inequality J
α
(
a
,
b
)
<
U
(
a
,
b
)
<
J
β
(
a
,
b
) holds for all a
,
b
>
0 with a
≠
b.


## 2. Main Result

In order to prove our main result we need a lemma, which we present in this section.Lemma 1.
Letp
∈
R, and (9)
f
x
,
p
=
p
x
4
p
+
6
-
p
+
1
x
4
p
+
5
+
p
x
4
p
+
4
-
p
+
1
x
4
p
+
1
-
p
p
+
1
x
2
p
+
7
+
2
p
+
1
2
x
2
p
+
5
-
2
p
x
2
p
+
4
-
2
p
p
+
1
x
2
p
+
3
-
2
p
x
2
p
+
2
+
2
p
+
1
2
x
2
p
+
1
-
p
p
+
1
x
2
p
-
1
-
p
+
1
x
5
+
p
x
2
-
p
+
1
x
+
p
.Then the following statements are true:(1)
ifp
=
3
/
2, then f
(
x
,
p
)
>
0 for all x
∈
(
1
,
∞
);
(2)
ifp
=
2
/
(
π
-
2
)
=
0.8187
⋯, then there exists λ
∈
(
1
,
∞
) such that f
(
x
,
p
)
<
0 for x
∈
(
1
,
λ
) and f
(
x
,
p
)
>
0 for x
∈
(
λ
,
∞
).Proof.
For part (1), ifp
=
3
/
2, then (9) becomes (10)
f
x
,
p
=
1
4
x
-
1
6
x
2
+
2
x
+
2
2
x
2
+
2
x
+
1
3
x
2
+
4
x
+
3
.
Therefore, part (1) follows from (10).
For part (2), letp
=
2
/
(
π
-
2
), f
1
(
x
,
p
)
=
∂
f
(
x
,
p
)
/
∂
x, f
2
(
x
,
p
)
=
1
/
2
∂
f
1
(
x
,
p
)
/
∂
x, f
3
(
x
,
p
)
=
1
/
(
p
+
1
)
x
2
∂
f
2
(
x
,
p
)
/
∂
x, f
4
(
x
,
p
)
=
x
7
-
2
p
/
2
p
∂
f
3
(
x
,
p
)
/
∂
x, f
5
(
x
,
p
)
=
1
/
2
x
∂
f
4
(
x
,
p
)
/
∂
x, f
6
(
x
,
p
)
=
∂
f
5
(
x
,
p
)
/
∂
x, f
7
(
x
,
p
)
=
1
/
2
∂
f
6
(
x
,
p
)
/
∂
x, f
8
(
x
,
p
)
=
∂
f
7
(
x
,
p
)
/
∂
x, f
9
(
x
,
p
)
=
1
/
2
(
p
+
1
)
∂
f
8
(
x
,
p
)
/
∂
x, and f
10
(
x
,
p
)
=
∂
f
9
(
x
,
p
)
/
∂
x. Then elaborated computations lead to (11)
lim
x
→
1
⁡
f
x
,
p
=
0
,
lim
x
→
+
∞
⁡
f
x
,
p
=
+
∞
,
(12)
lim
x
→
1
⁡
f
1
x
,
p
=
0
,
lim
x
→
+
∞
⁡
f
1
x
,
p
=
+
∞
,
(13)
lim
x
→
1
⁡
f
2
x
,
p
=
0
,
lim
x
→
+
∞
⁡
f
2
x
,
p
=
+
∞
,
(14)
lim
x
→
1
⁡
f
3
x
,
p
=
0
,
lim
x
→
+
∞
⁡
f
3
x
,
p
=
+
∞
,
(15)
lim
x
→
1
⁡
f
4
x
,
p
=
-
48
p
+
1
3
2
-
p
<
0
,
lim
x
→
+
∞
⁡
f
4
x
,
p
=
+
∞
,
(16)
lim
x
→
1
⁡
f
5
x
,
p
=
-
192
p
+
1
2
3
2
-
p
<
0
,
lim
x
→
+
∞
⁡
f
5
x
,
p
=
+
∞
,
(17)
lim
x
→
1
⁡
f
6
x
,
p
=
2
p
+
1
368
p
3
+
332
p
2
-
484
p
-
963
<
0
,
(18)
lim
x
→
+
∞
⁡
f
6
x
,
p
=
+
∞
,
(19)
lim
x
→
1
⁡
f
7
x
,
p
=
p
+
1
1024
p
4
+
2096
p
3
+
1844
p
2
-
2876
p
-
5193
<
0
,
(20)
lim
x
→
+
∞
⁡
f
7
x
,
p
=
+
∞
,
(21)
lim
x
→
1
⁡
f
8
x
,
p
=
p
+
1
2560
p
5
+
6336
p
4
+
14176
p
3
+
11028
p
2
-
12680
p
-
22005
<
0
,
(22)
l
i
m
x
→
+
∞
f
8
x
,
p
=
+
∞
,
(23)
l
i
m
x
→
1
f
9
x
,
p
=
6
512
p
6
+
1184
p
5
+
4064
p
4
+
6372
p
3
+
4068
p
2
-
3495
p
-
5775
=
15.2085
⋯
>
0
,
(24)
f
10
x
,
p
=
p
+
2
2
p
+
1
2
p
+
3
2
2
p
+
5
2
p
+
7
4
p
+
1
4
p
+
5
x
2
p
-
8
p
p
+
1
p
+
2
p
+
3
2
p
+
1
2
p
+
3
4
p
+
3
4
p
+
5
x
2
p
-
1
+
p
2
p
-
1
2
p
+
1
2
2
p
+
3
2
p
+
5
4
p
-
1
4
p
+
3
x
2
p
-
2
-
8
p
p
-
1
2
p
-
2
2
p
-
1
2
p
-
3
16
p
2
-
1
x
2
p
-
5
-
720
p
+
3
2
p
+
5
2
p
+
7
x
.
Note that(25)
2
p
>
1
>
2
p
-
1
>
0
>
2
p
-
2
>
2
p
-
5
,
1536
p
7
+
15040
p
6
+
59440
p
5
+
122280
p
4
+
137144
p
3
+
61850
p
2
-
49845
p
-
72450
=
85165.4405
⋯
>
0
.
It follows from (24) and (25) that (26)
f
10
x
,
p
>
p
+
2
2
p
+
1
2
p
+
3
2
2
p
+
5
2
p
+
7
4
p
+
1
4
p
+
5
-
8
p
p
+
1
p
+
2
p
+
3
2
p
+
1
2
p
+
3
4
p
+
3
4
p
+
5
-
720
p
+
3
2
p
+
5
2
p
+
7
x
+
p
2
p
-
1
2
p
+
1
2
2
p
+
3
2
p
+
5
4
p
-
1
4
p
+
3
-
8
p
p
-
1
2
p
-
2
2
p
-
1
2
p
-
3
16
p
2
-
1
x
2
p
-
2
=
1536
p
7
+
15040
p
6
+
59440
p
5
+
122280
p
4
+
137144
p
3
+
61850
p
2
-
49845
p
-
72450
x
+
p
2
p
-
1
4
p
-
1
704
p
4
+
136
p
3
+
1120
p
2
+
248
p
-
3
x
2
p
-
2
>
0
,for x
∈
(
1
,
∞
).
From (23) and (26) we clearly see that f
8
(
x
,
p
) is strictly increasing with respect to x on the interval (
1
,
∞
). Then (21) and (22) lead to the conclusion that there exists λ
1
>
1 such that the function x
→
f
7
(
x
,
p
) is strictly decreasing on (
1
,
λ
1
] and strictly increasing on [
λ
1
,
∞
).
It follows from (19) and (20) together with the piecewise monotonicity of the function x
→
f
7
(
x
,
p
) that there exists λ
2
>
1 such that the function x
→
f
6
(
x
,
p
) is strictly decreasing on (
1
,
λ
2
] and strictly increasing on [
λ
2
,
∞
).
Making use of (13)–(18) and the same method as the above we know that there exists λ
i
>
1
(
i
=
3,4
,
5,6
,
7
) such that the function x
→
f
8
-
i
(
x
,
p
) is strictly decreasing on (
1
,
λ
i
] and strictly increasing on [
λ
i
,
∞
).
It follows from (12) and the piecewise monotonicity of the function x
→
f
1
(
x
,
p
) that there exists λ
∗
>
1 such that the function x
→
f
(
x
,
p
) is strictly decreasing on (
1
,
λ
∗
] and strictly increasing on [
λ
∗
,
∞
).
Therefore, part (2) follows easily from (11) and the piecewise monotonicity of the function x
→
f
(
x
,
p
).Theorem 2.
The double inequality(27)
J
α
a
,
b
<
U
a
,
b
<
J
β
a
,
bholds for all a
,
b
>
0 with a
≠
b if and only if α
≤
2
/
(
π
-
2
)
=
0.8187
⋯ and β
≥
3
/
2.Proof.
SinceU
(
a
,
b
) and J
p
(
a
,
b
) are symmetric and homogeneous of degree one, without loss of generality, we assume that a
=
x
2
>
1 and b
=
1. Let p
∈
R and p
≠
0
,
-
1. Then (1) lead to (28)
J
p
a
,
b
-
U
a
,
b
=
J
p
x
2
,
1
-
U
x
2
,
1
=
p
x
2
p
+
2
-
1
p
+
1
x
2
p
-
1
arctan
⁡
x
2
-
1
/
2
x
F
x
,
p
,where (29)
F
x
,
p
=
arctan
⁡
x
2
-
1
2
x
-
p
+
1
x
2
-
1
x
2
p
-
1
2
p
x
2
p
+
2
-
1
,
(30)
lim
x
→
1
⁡
F
x
,
p
=
0
,
(31)
∂
F
x
,
p
∂
x
=
2
p
x
4
+
1
x
2
p
+
2
-
1
2
f
x
,
p
,where f
(
x
,
p
) is defined by (9).
We divide the proof into four cases.
Case
1 (
p
=
2
/
(
π
-
2
)
). Then it follows from Lemma 1(2), (29), and (31) that there exists λ
>
1 such that the function x
→
F
(
x
,
p
) is strictly decreasing on (
1
,
λ
] and strictly increasing on [
λ
,
∞
), and (32)
lim
x
→
∞
⁡
F
x
,
p
=
0
.
Therefore,(33)
J
2
/
π
-
2
a
,
b
<
U
a
,
bfollows easily from (28), (30), and (32) together with the piecewise monotonicity of the function x
→
F
(
x
,
p
).
Case
2 (
p
>
2
/
(
π
-
2
)
). Then (1) leads to (34)
lim
x
→
∞
⁡
J
p
x
,
1
U
x
,
1
=
2
p
2
p
+
1
π
>
1
.
Inequality (34) implies that there exists large enough X
=
X
(
p
)
>
1 such that U
(
a
,
b
)
<
J
p
(
a
,
b
) for all a
,
b
>
0 with a
/
b
∈
(
0,1
/
X
)
∪
(
X
,
∞
).
Case
3 (
p
=
3
/
2
). Then from Lemma 1(1) and (31) we know that the function x
→
F
(
x
,
p
) is strictly increasing on the interval (
1
,
∞
). Therefore, (35)
U
a
,
b
<
J
3
/
2
a
,
bfollows from (28) and (30) together with the monotonicity of the function x
→
F
(
x
,
p
).
Case
4 (
0
<
p
<
3
/
2
). Let x
>
0 and x
→
0; then making use of Taylor expansion we get (36)
U
1,1
+
x
-
J
p
1,1
+
x
=
x
2
arctan
⁡
x
/
2
1
+
x
-
p
1
-
1
+
x
p
+
1
p
+
1
1
-
1
+
x
p
=
3
-
2
p
24
x
2
+
o
x
2
.
Equation (36) implies that there exists small enough δ
>
0 such that U
(
1,1
+
x
)
>
J
p
(
1,1
+
x
) for all x
∈
(
0
,
δ
).



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*Source: 1579468-2016-02-29.xml*