# Best Possible Bounds for Yang Mean Using Generalized Logarithmic Mean

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

---

## Abstract

We prove that the double inequalityL
p
(
a
,
b
)
<
U
(
a
,
b
)
<
L
q
(
a
,
b
) holds for all a
,
b
>
0 with a
≠
b if and only if p
≤
p
0 and q
≥
2 and find several sharp inequalities involving the trigonometric, hyperbolic, and inverse trigonometric functions, where p
0
=
0.5451
⋯ is the unique solution of the equation (
p
+
1
)
1
/
p
=
2
π
/
2 on the interval (
0
,
∞
), U
(
a
,
b
)
=
(
a
-
b
)
/
[
2
arctan
⁡
(
(
a
-
b
)
/
2
a
b
)
], and L
p
(
a
,
b
)
=
[
(
a
p
+
1
-
b
p
+
1
)
/
(
(
p
+
1
)
(
a
-
b
)
)
]
1
/
p
(
p
≠
-
1,0
), L
-
1
(
a
,
b
)
=
(
a
-
b
)
/
(
log
⁡
a
-
log
⁡
b
) and L
0
(
a
,
b
)
=
(
a
a
/
b
b
)
1
/
(
a
-
b
)
/
e are the Yang, and pth generalized logarithmic means of a and b, respectively.


---

## Body

## 1. Introduction

Forp
∈
R and a
,
b
>
0 with a
≠
b, the pth generalized logarithmic mean L
p
(
a
,
b
) is defined by(1)
L
p
a
,
b
=
a
p
+
1
-
b
p
+
1
p
+
1
a
-
b
1
/
p
,
p
≠
0
,
-
1
,
1
e
a
a
b
b
1
/
a
-
b
,
p
=
0
,
a
-
b
log
⁡
a
-
log
⁡
b
,
p
=
-
1
.It is well known thatL
p
(
a
,
b
) is continuous and strictly increasing with respect to p
∈
R for fixed a
,
b
>
0 with a
≠
b. Many classical bivariate means are the special case of the generalized logarithmic mean. For example, G
(
a
,
b
)
=
a
b
=
L
-
2
(
a
,
b
) is the geometric mean, L
(
a
,
b
)
=
(
a
-
b
)
/
(
log
⁡
a
-
log
⁡
b
)
=
L
-
1
(
a
,
b
) is the logarithmic mean, I
(
a
,
b
)
=
(
a
a
/
b
b
)
1
/
(
a
-
b
)
/
e
=
L
0
(
a
,
b
) is the identric mean, and A
(
a
,
b
)
=
(
a
+
b
)
/
2
=
L
1
(
a
,
b
) is the arithmetic mean. Recently, the generalized logarithmic mean has been the subject of intensive research.Stolarsky [1] proved that the inequality(2)
L
p
a
,
b
<
M
2
+
p
/
3
a
,
bholds for all a
,
b
>
0 with a
≠
b and p
∈
(
-
2
,
-
1
/
2
)
∪
(
1
,
∞
), and inequality (2) is reversed for p
∈
(
-
∞
,
-
2
)
∪
(
-
1
/
2,1
), where M
r
(
a
,
b
)
=
[
(
a
r
+
b
r
)
/
2
]
1
/
r
(
r
≠
0
) and M
0
(
a
,
b
)
=
a
b is the rth power mean of a and b.Yang [2] proved that the double inequality(3)
A
a
,
b
<
L
p
a
,
b
<
M
p
a
,
bholds for all a
,
b
>
0 with a
≠
b if p
>
1, and inequality (3) is reversed if p
<
0.In [3], the authors proved that the inequality (4)
L
p
a
,
b
<
a
+
b
p
+
1
1
/
pholds for all a
,
b
>
0 with a
≠
b and p
>
1.Li et al. [4] proved that the function p
↦
L
p
(
a
,
b
)
/
L
p
(
1
-
a
,
1
-
b
) is strictly increasing (decreasing) on R if 0
<
a
<
b
≤
1
/
2 (1
/
2
≤
a
<
b
<
1). In [5, 6], the authors proved that the function q
↦
L
q
(
a
,
b
)
/
L
q
(
a
,
c
) is strictly decreasing on R if 0
<
a
<
b
<
c and the function r
↦
L
r
(
d
,
d
+
ε
)
/
L
r
(
d
+
δ
,
d
+
ε
+
δ
) is strictly increasing on R for all d
,
ε
,
δ
>
0.Shi and Wu [7] proved that the double inequality(5)
λ
b
+
1
-
λ
/
2
a
+
b
p
+
1
-
λ
a
+
1
-
λ
/
2
a
+
b
p
+
1
λ
p
+
1
b
-
a
1
/
p
<
L
p
a
,
b
<
a
+
b
-
c
p
+
1
-
c
p
+
1
p
+
1
a
+
b
-
2
c
1
/
pfor all b
>
a
>
c
>
0 and 0
<
λ
<
1 if p
>
1, and inequality (5) is reversed if p
∈
(
-
1,0
)
∪
(
0,1
).Long and Chu [8] and Matejíčka [9] presented the best possible parameters p
=
p
(
α
) and q
=
q
(
α
) such that the double inequality (6)
L
p
a
,
b
<
α
A
a
,
b
+
1
-
α
G
a
,
b
<
L
q
a
,
bholds for all a
,
b
>
0 with a
≠
b and α
∈
(
0,1
/
2
)
∪
(
1
/
2,1
).In [10], Qian and Long answered the question: what are the greatest value p and the least value q such that the double inequality(7)
L
p
a
,
b
<
G
α
a
,
b
H
1
-
α
a
,
b
<
L
q
a
,
bholds for all a
,
b
>
0 with a
≠
b and α
∈
(
0,1
), where H
(
a
,
b
)
=
2
a
b
/
(
a
+
b
) is the harmonic mean of a and b.In [11, 12], the authors proved that the double inequalities (8)
L
p
1
a
,
b
<
M
a
,
b
<
L
q
1
a
,
b
,
L
p
2
a
,
b
<
T
a
,
b
<
L
q
2
a
,
bhold for all a
,
b
>
0 with a
≠
b if and only if p
1
≤
p
1
∗, q
1
≥
2, p
2
≤
3, q
2
≥
q
2
∗, where p
1
∗
=
1.843
⋯ is the unique solution of the equation (
p
+
1
)
1
/
p
=
2
log
⁡
(
1
+
2
) on the interval (
0
,
∞
), q
2
∗
=
3.152
⋯ is the unique solution of the equation (
q
+
1
)
1
/
q
=
π
/
2 on the interval (
0
,
∞
), M
(
a
,
b
)
=
(
a
-
b
)
/
[
2
s
i
n
h
-
1
(
(
a
-
b
)
/
(
a
+
b
)
)
] is the Neuman-Sándor mean, and T
(
a
,
b
)
=
(
a
-
b
)
/
[
2
arctan
⁡
(
(
a
-
b
)
/
(
a
+
b
)
)
] is the second Seiffert mean.In [13, 14], the authors presented the best possible parameters p
1
=
p
1
(
q
), p
2
=
p
2
(
q
), λ
=
λ
(
α
), and μ
=
μ
(
α
) such that the double inequalities(9)
L
p
1
a
,
b
<
L
a
q
,
b
q
1
/
q
<
L
p
2
a
,
b
,
L
λ
a
,
b
<
G
α
a
,
b
A
a
,
b
+
G
a
,
b
2
1
-
α
<
L
μ
a
,
bhold for all a
,
b
>
0 with a
≠
b, q
>
0 with q
≠
1 and α
∈
(
0,2
/
3
)
∪
(
2
/
3,1
).Gao et al. [15] provided the greatest value α and the least value β such that the double inequality(10)
L
α
a
,
b
<
P
a
,
b
<
L
β
a
,
bholds for all a
,
b
>
0 with a
≠
b, where P
(
a
,
b
)
=
(
a
-
b
)
/
[
2
arcsin
⁡
(
(
a
-
b
)
/
(
a
+
b
)
)
] is the first Seiffert mean of a and b.Very recently, Yang [16] introduced the Yang mean(11)
U
a
,
b
=
a
-
b
2
arctan
⁡
a
-
b
/
2
a
bof two distinct positive real numbers a and b and proved that the inequalities (12)
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
2hold for all a
,
b
>
0 with a
≠
b, where Q
(
a
,
b
)
=
(
a
2
+
b
2
)
/
2 is the quadratic mean of a and b.The Yang meanU
(
a
,
b
) is the special case of the Seiffert type mean T
M
,
q
(
a
,
b
)
=
(
a
-
b
)
/
[
q
arctan
⁡
(
(
a
-
b
)
/
(
q
M
(
a
,
b
)
)
)
] defined by Toader in [17], where M
(
a
,
b
) is a bivariate mean and q is a positive real number. Indeed, U
(
a
,
b
)
=
T
G
,
2
(
a
,
b
).In [18, 19], the authors proved that the double inequalities (13)
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
,
bhold 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
⋯.Zhou et al. [20] proved that α
=
1
/
2 and β
=
log
⁡
3
/
(
1
+
log
⁡
2
)
=
0.6488
⋯ are the best possible parameters such that the double inequality(14)
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 main purpose of this paper is to present the best possible parametersp and q such that the double inequality L
p
(
a
,
b
)
<
U
(
a
,
b
)
<
L
q
(
a
,
b
) holds for all a
,
b
>
0 with a
≠
b. As application, we derive several sharp inequalities involving the trigonometric, hyperbolic, and inverse trigonometric functions. Some complicated computations are carried out using Mathematica computer algebra system.


## 2. Lemmas

In order to prove our main result we need two lemmas, which we present in this section.Lemma 1.
Letn
≥
4, -
∞
≤
a
<
b
≤
+
∞, f
0
,
f
i
,
f
i
∗
:
(
a
,
b
)
→
R be n-times differentiable functions such that f
i
(
x
)
=
f
i
∗
(
x
)
f
i
-
1
′
(
x
) and f
i
∗
(
x
)
>
0 for 1
≤
i
≤
n and x
∈
(
a
,
b
). If(15)
f
n
-
2
a
-
<
0
,
f
n
-
2
b
+
>
0
,
f
n
-
1
a
-
≥
0
,
(16)
f
i
a
-
≤
0
,
f
i
b
+
>
0for 0
≤
i
≤
n
-
3 and(17)
f
n
x
>
0for x
∈
(
a
,
b
), then there exists x
0
∈
(
a
,
b
) such that f
0
(
x
)
<
0 for x
∈
(
a
,
x
0
) and f
0
(
x
)
>
0 for x
∈
(
x
0
,
b
).Proof.
From (15) and (17) we clearly see that there exists x
n
-
2
∈
(
a
,
b
) such that f
n
-
2
(
x
)
<
0 for x
∈
(
a
,
x
n
-
2
) and f
n
-
2
(
x
)
>
0 for x
∈
(
x
n
-
2
,
b
), which implies that f
n
-
3
(
x
) is strictly decreasing on (
a
,
x
n
-
2
] and strictly increasing on [
x
n
-
2
,
b
). Then (16) leads to the conclusion that there exists x
n
-
3
∈
(
a
,
b
) such that f
n
-
3
(
x
)
<
0 for x
∈
(
a
,
x
n
-
3
) and f
n
-
3
(
x
)
>
0 for x
∈
(
x
n
-
3
,
b
).
Making use of (16) and the same method as above we know that for 0
≤
i
≤
n
-
4 there exists x
i
∈
(
a
,
b
) such that f
i
(
x
)
<
0 for x
∈
(
a
,
x
i
) and f
i
(
x
)
>
0 for x
∈
(
x
i
,
b
).Lemma 2.
Letp
∈
R, and(18)
f
x
,
p
=
p
x
4
p
+
10
-
2
p
+
1
x
4
p
+
8
+
2
3
p
-
1
x
4
p
+
6
-
2
p
+
1
x
4
p
+
4
-
3
p
+
2
x
4
p
+
2
-
p
2
p
+
1
x
2
p
+
10
+
2
p
2
+
5
p
+
4
x
2
p
+
8
+
4
1
-
p
x
2
p
+
6
+
4
1
-
p
x
2
p
+
4
+
2
p
2
+
5
p
+
4
x
2
p
+
2
-
p
2
p
+
1
x
2
p
-
3
p
+
2
x
8
-
2
p
+
1
x
6
+
2
3
p
-
1
x
4
-
2
p
+
1
x
2
+
p
.Then the following statements are true:(1)
ifp
=
2, then f
(
x
,
p
)
>
0 for all x
∈
(
1
,
∞
);
(2)
ifp
0
=
0.5451
⋯ is the unique solution of the equation (
p
+
1
)
1
/
p
=
2
π
/
2 on the interval (
0
,
∞
) and p
=
p
0, 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
), if p
=
2, then (18) becomes(19)
f
x
,
p
=
2
x
-
1
6
x
+
1
6
x
2
+
1
3
.
Therefore, part(
1
) follows from (19).
For part(
2
), let p
=
p
0
=
0.5451
⋯ be the unique solution of the equation (
p
+
1
)
1
/
p
=
2
π
/
2 on the interval (
0
,
∞
), f
1
(
x
,
p
)
=
1
/
2
x
∂
f
(
x
,
p
)
/
∂
x, f
2
(
x
,
p
)
=
1
/
2
x
∂
f
1
(
x
,
p
)
/
∂
x, f
3
(
x
,
p
)
=
1
/
2
x
∂
f
2
(
x
,
p
)
/
∂
x, f
4
(
x
,
p
)
=
1
/
2
x
∂
f
3
(
x
,
p
)
/
∂
x, f
5
(
x
,
p
)
=
x
9
-
2
p
/
2
p
∂
f
4
(
x
,
p
)
/
∂
x, f
6
(
x
,
p
)
=
1
/
2
(
p
+
1
)
x
∂
f
5
(
x
,
p
)
/
∂
x, f
7
(
x
,
p
)
=
1
/
8
x
∂
f
6
(
x
,
p
)
/
∂
x, f
8
(
x
,
p
)
=
1
/
2
x
∂
f
7
(
x
,
p
)
/
∂
x, and f
9
(
x
,
p
)
=
1
/
2
x
∂
f
8
(
x
,
p
)
/
∂
x. Then elaborated computations lead to(20)
f
1
-
,
p
=
0
,
lim
x
→
+
∞
⁡
f
x
,
p
=
+
∞
,
f
1
1
-
,
p
=
0
,
lim
x
→
+
∞
⁡
f
1
x
,
p
=
+
∞
,
f
2
1
-
,
p
=
0
,
lim
x
→
+
∞
⁡
f
2
x
,
p
=
+
∞
,
f
3
1
-
,
p
=
0
,
lim
x
→
+
∞
⁡
f
3
x
,
p
=
+
∞
,
f
4
1
-
,
p
=
-
48
p
2
2
-
p
p
+
1
<
0
,
lim
x
→
+
∞
⁡
f
4
x
,
p
=
+
∞
,
f
5
1
-
,
p
=
-
120
p
p
+
1
2
-
p
2
p
+
1
<
0
,
lim
x
→
+
∞
⁡
f
5
x
,
p
=
+
∞
,
f
6
1
-
,
p
=
-
80
p
p
+
1
-
7
p
2
+
8
p
+
8
<
0
,
lim
x
→
+
∞
⁡
f
6
x
,
p
=
+
∞
,
f
7
1
-
,
p
=
5
p
p
+
1
44
p
3
+
16
p
2
-
47
p
-
58
<
0
,
lim
x
→
+
∞
⁡
f
7
x
,
p
=
+
∞
,
f
8
1
-
,
p
=
p
288
p
5
+
606
p
4
+
1009
p
3
+
1676
p
2
+
1571
p
+
250
>
0
,
f
9
x
,
p
=
p
+
2
2
p
+
3
p
+
4
p
+
5
2
p
+
1
2
p
+
3
2
p
+
5
x
2
p
+
2
-
4
p
+
1
2
p
+
2
2
p
+
3
p
+
4
2
p
+
1
2
p
+
3
x
2
p
+
2
p
p
+
1
p
+
2
p
+
3
4
p
2
-
1
2
p
+
3
3
p
-
1
x
2
p
-
2
-
4
p
p
2
-
1
2
p
+
2
4
p
2
-
1
x
2
p
-
4
-
p
p
-
1
2
p
-
2
2
p
-
3
4
p
2
-
1
3
p
+
2
x
2
p
-
6
-
30
p
+
2
p
+
3
p
+
4
p
+
5
2
p
+
1
x
2
+
6
p
+
2
p
+
3
p
+
4
2
p
2
+
5
p
+
4
>
p
+
2
p
+
3
p
+
4
2
p
+
1
p
+
2
p
+
5
2
p
+
3
2
p
+
5
-
4
p
+
1
2
p
+
2
2
p
+
3
-
30
p
+
5
×
x
2
p
+
2
+
2
p
p
+
1
p
+
2
p
+
3
4
p
2
-
1
2
p
+
3
3
p
-
1
-
4
p
p
2
-
1
2
p
+
2
4
p
2
-
1
-
p
p
-
1
2
p
-
2
2
p
-
3
4
p
2
-
1
3
p
+
2
+
6
p
+
2
p
+
3
p
+
4
2
p
2
+
5
p
+
4
x
2
p
-
4
=
p
+
2
p
+
3
p
+
4
2
p
+
1
-
4
p
4
+
79
p
2
+
159
p
-
24
x
2
p
+
2
+
8
p
8
+
428
p
7
+
694
p
6
+
781
p
5
+
100
p
4
+
163
p
3
+
1250
p
2
+
1400
p
+
576
x
2
p
-
4
>
0for x
∈
(
1
,
∞
).
Therefore, part(
2
) follows easily from Lemma 1 and (20).


## 3. Main Result

Theorem 3.
The double inequality(21)
L
p
a
,
b
<
U
a
,
b
<
L
q
a
,
bholds for all a
,
b
>
0 with a
≠
b if and only if p
≤
p
0 and q
≥
2, where p
0
=
0.5451
⋯ is the unique solution of the equation (
p
+
1
)
1
/
p
=
2
π
/
2 on the interval (
0
,
∞
).Proof.
SinceU
(
a
,
b
) and L
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) and (11) lead to(22)
log
⁡
U
a
,
b
-
log
⁡
L
p
a
,
b
=
log
⁡
U
x
2
,
1
-
log
⁡
L
p
x
2
,
1
=
log
⁡
x
2
-
1
2
arctan
⁡
x
2
-
1
/
2
x
-
1
p
log
⁡
x
2
p
+
2
-
1
p
+
1
x
2
-
1
≕
F
x
,
p
,
(23)
lim
x
→
1
⁡
F
x
,
p
=
0
,
(24)
lim
x
→
∞
⁡
F
x
,
p
=
1
p
log
⁡
p
+
1
+
1
2
log
⁡
2
-
log
⁡
π
,
(25)
∂
F
x
,
p
∂
x
=
2
p
+
1
x
x
2
p
-
1
p
x
2
-
1
x
2
p
+
2
-
1
arctan
⁡
x
2
-
1
/
2
x
F
1
x
,
p
,where (26)
F
1
x
,
p
=
arctan
⁡
x
2
-
1
2
x
-
p
x
4
-
1
x
2
p
+
2
-
1
2
p
+
1
x
x
4
+
1
x
2
p
-
1
,
(27)
lim
x
→
1
⁡
F
1
x
,
p
=
0
,
(28)
lim
x
→
∞
⁡
F
1
x
,
p
=
-
∞
,
(29)
∂
F
1
x
,
p
∂
x
=
-
2
2
p
+
1
x
2
x
4
+
1
2
x
2
p
-
1
2
f
x
,
p
,where f
(
x
,
p
) is defined by (15).
We divide the proof into four cases.
Case 1
(
p
=
2
). Then from Lemma 2
(
1
) and (29) we clearly see that the function x
→
F
1
(
x
,
p
) is strictly decreasing on (
1
,
∞
). Then (27) leads to the conclusion that(30)
F
1
x
,
p
<
0for all x
∈
(
1
,
∞
). Therefore, (31)
U
a
,
b
<
L
2
a
,
bfollows easily from (22), (23), (25), and (30).
Case 2
(
p
=
p
0
). Then from Lemma 2
(
2
) and (29) we know that there exists λ
∈
(
1
,
∞
) such that the function x
→
F
1
(
x
,
p
) is strictly increasing on (
1
,
λ
] and strictly decreasing on λ
,
∞
).
It follows from (25)–(28) 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 increasing on (
1
,
λ
∗
] and strictly decreasing on [
λ
∗
,
∞
).
Note that (24) becomes(32)
lim
x
→
∞
⁡
F
x
,
p
=
0
.
Therefore,(33)
U
a
,
b
>
L
p
0
a
,
bfollows easily from (22), (23), and (32) together with the piecewise monotonicity of the function x
→
F
(
x
,
p
).
Case 3
(
p
<
2
). Let x
>
0 and x
→
0; then making use of Taylor expansion we get(34)
U
1
+
x
,
1
-
L
p
1
+
x
,
1
=
x
2
arctan
⁡
x
/
2
1
+
x
-
1
+
x
p
+
1
-
1
p
+
1
x
1
/
p
=
2
-
p
24
x
2
+
o
x
2
.
Equation (34) implies that there exists small enough δ
∈
(
0,1
) such that (35)
U
1
+
x
,
1
>
L
p
1
+
x
,
1for all x
∈
(
0
,
δ
).
Case 4
(
p
>
p
0
). Then from (24) and the fact that the function p
→
log
⁡
(
p
+
1
)
/
p is strictly decreasing on (
0
,
∞
) we get(36)
lim
x
→
∞
⁡
F
x
,
p
<
1
p
0
log
⁡
1
+
p
0
+
1
2
log
⁡
2
-
log
⁡
π
=
0
.
Equation (22) and inequality (36) imply that there exists large enough X
>
1 such that (37)
U
x
2
,
1
<
L
p
x
2
,
1for all x
∈
(
X
,
∞
).


## 4. Applications

As applications of Theorem3 in engineering problems, we present several sharp inequalities involving the trigonometric, hyperbolic, and inverse trigonometric functions in this section.From (1) and (11) together with Theorem 3 we get Theorem 4 immediately.Theorem 4.
Letp
0
=
0.5451
⋯ be the unique solution of the equation (
p
+
1
)
1
/
p
=
2
π
/
2 on the interval (
0
,
∞
). Then the double inequality (38)
6
a
-
b
2
a
2
+
a
b
+
b
2
<
arctan
⁡
a
-
b
2
a
b
<
π
a
-
b
1
+
1
/
p
0
2
a
p
0
+
1
-
b
p
0
+
1
1
/
p
0holds for all a
,
b
>
0 with a
≠
b.Lett
>
0, b
=
1, and a
=
t
2
+
t
t
2
+
2
+
1. Then Theorem 4 leads to the following.Theorem 5.
Letp
0
=
0.5451
⋯ be the unique solution of the equation (
p
+
1
)
1
/
p
=
2
π
/
2 on the interval (
0
,
∞
). Then the double inequality (39)
6
t
2
+
t
t
2
+
2
2
2
t
4
+
5
t
2
+
2
t
3
+
3
t
t
2
+
2
+
3
<
arctan
⁡
t
<
π
t
2
+
t
t
2
+
2
1
+
1
/
p
0
2
t
2
+
t
t
2
+
2
1
+
p
0
-
1
1
/
p
0holds for all t
>
0.Leta
>
b
>
0, x
=
l
o
g
a
/
b
∈
(
0
,
∞
). Then (1) and (11) lead to(40)
L
p
a
,
b
a
b
=
sinh
⁡
p
+
1
x
p
+
1
sinh
⁡
x
1
/
p
,
U
a
,
b
a
b
=
2
sinh
⁡
x
arctan
⁡
2
sinh
⁡
x
.It follows from Theorem3 and (40) that one has the following theorem.Theorem 6.
Letp
0
=
0.5451
⋯ be the unique solution of the equation (
p
+
1
)
1
/
p
=
2
π
/
2 on the interval (
0
,
∞
). Then the double inequality (41)
6
sinh
⁡
x
3
+
4
s
i
n
h
2
x
<
arctan
⁡
2
sinh
⁡
x
<
π
sinh
⁡
x
sinh
⁡
p
0
+
1
x
1
/
p
0holds for all x
>
0.Leta
>
b
>
0, x
=
arcsin
⁡
[
(
a
-
b
)
/
(
a
+
b
)
]
∈
(
0
,
π
/
2
). Then (1) and (11) lead to(42)
L
p
a
,
b
a
b
=
1
+
sin
⁡
x
/
cos
⁡
x
p
+
1
-
cos
⁡
x
/
1
+
sin
⁡
x
p
+
1
2
p
+
1
tan
⁡
x
1
/
p
,
U
a
,
b
a
b
=
2
tan
⁡
x
arctan
⁡
2
tan
⁡
x
.Theorem3 and (42) lead to the following.Theorem 7.
Letp
0
=
0.5451
⋯ be the unique solution of the equation (
p
+
1
)
1
/
p
=
2
π
/
2 on the interval (
0
,
∞
). Then the double inequality (43)
6
sin
⁡
x
3
+
s
i
n
2
x
<
arctan
⁡
2
tan
⁡
x
<
2
1
/
p
0
π
s
i
n
1
+
1
/
p
0
x
1
+
sin
⁡
x
1
+
p
0
-
1
-
sin
⁡
x
1
+
p
0
1
/
p
0holds for all x
∈
(
0
,
π
/
2
).Leta
>
b
>
0, x
=
arctan
⁡
(
(
a
-
b
)
/
(
a
+
b
)
)
∈
(
0
,
π
/
4
), y
=
s
i
n
h
-
1
(
(
a
-
b
)
/
(
a
+
b
)
)
∈
(
0
,
log
⁡
(
1
+
2
)
). Then from (1) and (11) we have(44)
L
p
a
,
b
a
b
=
1
2
p
+
1
1
/
p
1
cos
⁡
2
x
cos
⁡
x
+
sin
⁡
x
p
+
1
-
cos
⁡
x
-
sin
⁡
x
p
+
1
sin
⁡
x
1
/
p
=
1
2
p
+
1
1
/
p
1
1
-
s
i
n
h
2
y
1
+
sinh
⁡
y
p
+
1
-
1
-
sinh
⁡
y
p
+
1
sinh
⁡
y
1
/
p
,
(45)
U
a
,
b
a
b
=
2
sin
⁡
x
/
cos
⁡
2
x
arctan
⁡
2
sin
⁡
x
/
cos
⁡
2
x
=
2
sinh
⁡
y
/
1
-
s
i
n
h
2
y
arctan
⁡
2
sinh
⁡
y
/
1
-
s
i
n
h
2
y
.From (44), (45), and Theorem 3 one has the following.Theorem 8.
Letp
0
=
0.5451
⋯ be the unique solution of the equation (
p
+
1
)
1
/
p
=
2
π
/
2 on the interval (
0
,
∞
). Then the double inequalities (46)
6
sin
⁡
x
2
+
cos
⁡
2
x
<
arctan
⁡
2
sin
⁡
x
cos
⁡
2
x
<
2
1
/
p
0
π
s
i
n
1
+
1
/
p
0
x
cos
⁡
x
+
sin
⁡
x
1
+
1
/
p
0
-
cos
⁡
x
-
sin
⁡
x
1
+
1
/
p
0
1
/
p
0
,
6
sinh
⁡
y
3
+
s
i
n
h
2
y
<
arctan
⁡
2
sinh
⁡
y
3
-
cosh
⁡
2
y
<
2
1
/
p
0
π
s
i
n
h
1
+
1
/
p
0
y
1
+
sinh
⁡
y
1
+
1
/
p
0
-
1
-
sinh
⁡
y
1
+
1
/
p
0
1
/
p
0hold for all x
∈
(
0
,
π
/
4
) and y
∈
(
0
,
log
⁡
(
1
+
2
)
).



---

*Source: 8901258-2016-04-10.xml*