1. Introduction
The well known Jordan inequality [1] is given by
(1)
2
π
x
<
sin
(
x
)
<
x
,
x
∈
0
,
π
2
.
During the past few years, the improvements, refinements, and generalizations for inequality (1) have attracted the attention of many researchers [2–13]. Recently, the hyperbolic counterpart and its generalizations have been the subject of intensive research.
Zhu [14] proved that the inequality
(2)
sinh
(
x
)
x
q
>
p
+
(
1
-
p
)
cosh
(
x
)
holds for all
x
>
0
if and only if
q
≥
3
(
1
-
p
)
if
p
∈
(
-
∞
,
8
/
15
]
∪
(
1
,
∞
)
.
In [3, 15], Neuman and Sándor proved that
(3)
cosh
4
/
3
x
2
<
sinh
(
x
)
x
<
cosh
3
(
x
)
for all
x
>
0
.
Klén et al. [5] proved that the double inequality
(4)
cosh
1
/
4
(
x
)
<
sinh
(
x
)
x
<
cosh
1
/
2
(
x
)
holds for all
x
∈
(
0,1
)
.
In [4], the authors proved that the double inequality
(5)
cosh
p
(
x
)
<
sinh
(
x
)
x
<
cosh
q
(
x
)
holds for all
x
∈
(
0,1
)
if and only if
p
≤
1
/
3
and
q
≥
[
log
(
sinh
(
1
)
)
]
/
[
log
(
cosh
(
1
)
)
]
=
0.3721
⋯
.
Zhu [16, 17] proved that the inequalities
(6)
1
-
λ
+
λ
cosh
p
x
<
sinh
(
x
)
x
p
<
(
1
-
μ
)
+
μ
cosh
p
(
x
)
,
sinh
x
x
q
<
(
1
-
η
)
+
η
cosh
q
(
x
)
,
α
+
1
-
α
e
r
t
coth
t
-
1
<
sinh
(
x
)
x
r
<
β
+
(
1
-
β
)
e
r
[
t
coth
(
t
)
-
1
]
hold for all
x
>
0
if and only if
λ
≤
0
,
μ
≥
1
/
3
,
η
≤
1
/
3
,
α
≥
1
, and
β
≤
1
/
2
if
p
≥
4
/
5
,
q
<
0
, and
r
≥
286
/
693
.
Very recently, Yang [18] proved that the double inequality
(7)
cosh
p
x
1
/
3
p
2
<
sinh
(
x
)
x
<
cosh
q
x
1
/
3
q
2
holds for all
x
>
0
if and only if
p
≥
5
/
5
and
q
≤
1
/
3
.
The main purpose of this paper is to find the best possible parameters
p
,
q
∈
(
0
,
∞
)
such that the double inequality
1
/
3
p
2
cosh
(
p
x
)
+
1
-
1
/
3
p
2
<
sinh
(
x
)
/
x
<
1
/
3
q
2
cosh
(
q
x
)
+
1
-
1
/
3
q
2
holds for all
x
>
0
and present several new inequalities for certain special function and bivariate means.
2. Main Result
Theorem 1.
Let
p
,
q
∈
(
0
,
∞
)
. Then the double inequality
(8)
1
3
p
2
cosh
p
x
+
1
-
1
3
p
2
<
sinh
(
x
)
x
<
1
3
q
2
cosh
(
q
x
)
+
1
-
1
3
q
2
holds for all
x
>
0
if and only if
p
≤
15
/
5
and
q
≥
1
.
Proof.
Let
λ
>
0
and let the function
f
λ
be defined on
(
0
,
∞
)
by
(9)
f
λ
(
x
)
=
sinh
(
x
)
x
-
1
3
λ
2
cosh
(
λ
x
)
+
1
-
1
3
λ
2
.
Then making use of power series expansions and (9) we get
(10)
f
λ
x
=
∑
n
=
0
∞
x
2
n
(
2
n
+
1
)
!
-
1
3
λ
2
∑
n
=
0
∞
λ
x
2
n
(
2
n
)
!
+
1
-
1
3
λ
2
=
∑
n
=
2
∞
3
-
(
2
n
+
1
)
λ
2
n
-
2
3
(
2
n
+
1
)
!
x
2
n
.
Let
(11)
a
n
(
λ
)
=
3
-
(
2
n
+
1
)
λ
2
n
-
2
.
Then
(12)
a
n
(
1
)
=
3
-
(
2
n
+
1
)
=
-
2
(
n
-
1
)
<
0
for all
n
≥
2
. Consider
(13)
a
2
15
5
=
0
,
a
3
15
5
=
12
25
>
0
,
a
n
+
1
15
5
-
a
n
15
5
=
4
×
3
n
-
1
5
n
(
n
-
1
)
>
0
for all
n
≥
2
.
It follows from (13) that
(14)
a
2
15
5
=
0
,
a
n
15
5
>
0
for all
n
≥
3
.
Therefore, inequality (8) holds for all
x
>
0
with
p
=
15
/
5
and
q
=
1
follows from (9)–(12) and (14).
Next, we prove that
p
≤
15
/
5
and
q
≥
1
if inequality (8) holds for all
x
>
0
.
If the first inequality of (8) holds for all
x
>
0
, then from (9) and (10) we have
(15)
lim
x
→
0
+
f
p
(
x
)
x
4
=
3
-
5
p
2
360
≥
0
and
p
≤
15
/
5
.
If the second inequality of (8) holds for all
x
>
0
, then it follows from (9) that
(16)
lim
x
→
+
∞
f
q
(
x
)
e
q
x
=
1
-
e
-
2
x
2
x
e
(
1
-
q
)
x
-
1
+
e
-
2
q
x
6
q
2
-
1
-
1
3
q
2
e
-
q
x
≤
0
.
We clearly see that
lim
x
→
+
∞
f
q
(
x
)
/
e
q
x
=
+
∞
if
q
<
1
. Therefore,
q
≥
1
follows from (16).
Remark 2.
It is not difficult to verify that the bound
(17)
g
p
(
x
)
=
1
3
p
2
cosh
(
p
x
)
+
1
-
1
3
p
2
given in Theorem 1 is strictly increasing with respect to
p
on
(
0
,
∞
)
for fixed
x
∈
(
0
,
∞
)
.
Remark 3.
Let
p
=
15
/
5
>
3
/
4
>
2
/
2
>
2
/
3
>
3
/
3
and
q
=
1
<
2
3
/
3
. Then Theorem 1 and Remark 2 lead to
(18)
cosh
3
x
3
<
3
4
cosh
2
x
3
+
1
4
<
2
3
cosh
2
x
2
+
1
3
<
16
27
cosh
3
x
4
+
11
27
<
5
9
cosh
15
x
5
+
4
9
<
sinh
(
x
)
x
<
1
3
cosh
(
x
)
+
2
3
<
1
2
cosh
2
3
x
3
+
1
2
for all
x
>
0
.
3. Applications
It is well known that
(19)
∫
0
∞
x
sinh
(
x
)
=
1
2
ψ
′
1
2
,
where
ψ
′
is the trigamma function defined by
(20)
ψ
′
(
x
)
=
∫
0
∞
t
e
-
x
t
1
-
e
-
t
d
t
.
Let
(21)
Sh
(
x
)
=
∫
0
x
t
sinh
(
t
)
d
t
.
Then Remark 3 leads to
(22)
∫
0
x
3
cosh
(
t
)
+
2
d
t
<
Sh
(
x
)
<
∫
0
x
9
5
cosh
15
t
/
5
+
4
d
t
for all
x
>
0
.
From (19) and (22) we get the following.
Remark 4.
For all
x
>
0
one has
(23)
3
log
e
x
+
2
-
3
e
x
+
2
+
3
+
3
log
(
2
+
3
)
<
Sh
(
x
)
<
2
15
h
h
h
h
h
h
h
h
h
h
h
h
×
arctan
5
3
e
15
x
/
5
+
4
3
-
arctan
(
3
)
.
In particular, we have
(24)
4.5620
⋯
=
2
3
log
(
2
+
3
)
<
ψ
′
1
2
<
2
15
[
π
-
2
arctan
(
3
)
]
=
4.9845
⋯
.
For
a
,
b
>
0
, the Schwab-Borchardt mean
SB
(
a
,
b
)
[19–21] is given by
(25)
SB
(
a
,
b
)
=
b
2
-
a
2
arccos
(
a
/
b
)
(
a
<
b
)
,
SB
(
a
,
b
)
=
a
(
a
=
b
)
,
(26)
SB
(
a
,
b
)
=
a
2
-
b
2
cosh
-
1
(
a
/
b
)
(
a
>
b
)
.
Let
a
>
b
and let
x
=
cosh
-
1
(
a
/
b
)
. Then
cosh
(
x
)
=
a
/
b
and
a
2
-
b
2
/
cosh
-
1
(
a
/
b
)
=
b
sinh
(
x
)
/
x
. It follows from Remark 3 and (26) that
(27)
16
27
cosh
3
x
4
+
11
27
b
<
SB
(
a
,
b
)
<
1
3
cosh
(
x
)
+
2
3
b
.
Note that
(28)
1
3
cosh
(
x
)
+
2
3
b
=
a
+
2
b
3
,
16
27
cosh
3
x
4
+
11
27
b
=
8
b
1
/
4
27
2
(
a
+
b
)
(
2
a
-
b
)
2
+
2
b
3
/
2
1
/
2
+
11
b
27
.
From (27) and (28) we get the following.
Remark 5.
Let
a
>
b
>
0
; then the Schwab-Borchardt mean
SB
(
a
,
b
)
satisfies the double inequality
(29)
8
b
1
/
4
27
2
(
a
+
b
)
2
a
-
b
2
+
2
b
3
/
2
1
/
2
+
11
b
27
<
SB
(
a
,
b
)
<
a
+
2
b
3
.
Let
p
∈
R
and let
a
,
b
>
0
with
a
≠
b
. Then the arithmetic mean
A
(
a
,
b
)
, logarithmic mean
L
(
a
,
b
)
, geometric mean
G
(
a
,
b
)
, and
p
th power mean
M
p
(
a
,
b
)
are defined by
(30)
A
(
a
,
b
)
=
a
+
b
2
,
L
(
a
,
b
)
=
b
-
a
log
b
-
log
a
,
G
(
a
,
b
)
=
a
b
,
M
p
(
a
,
b
)
=
a
p
+
b
p
2
1
/
p
(
p
≠
0
)
,
M
0
(
a
,
b
)
=
a
b
=
G
(
a
,
b
)
.
It is well known that
M
p
(
a
,
b
)
is continuous and strictly increasing with respect to
p
∈
R
for fixed
a
,
b
>
0
with
a
≠
b
; the main properties for the power mean are given in [22]. Recently, the arithmetic, logarithmic, geometric, and power means have been the subject of intensive research. In particular, many remarkable inequalities can be found in the literature [23–35].
Let
x
=
1
/
2
log
(
a
/
b
)
; then (30) leads to
(31)
sinh
(
x
)
x
=
L
(
a
,
b
)
G
(
a
,
b
)
,
cosh
(
p
x
)
=
M
p
(
a
,
b
)
G
(
a
,
b
)
p
.
From Theorem 1 and (31) we get the following.
Remark 6.
Let
p
,
q
∈
(
0
,
∞
)
; then the double inequality
(32)
1
3
p
2
M
p
p
(
a
,
b
)
G
1
-
p
(
a
,
b
)
+
1
-
1
3
p
2
G
(
a
,
b
)
<
L
(
a
,
b
)
<
1
3
q
2
M
q
q
(
a
,
b
)
G
1
-
q
(
a
,
b
)
hhhhhhhhhhhhhl
+
1
-
1
3
q
2
G
(
a
,
b
)
holds for all
a
,
b
>
0
with
a
≠
b
if and only if
p
≤
15
/
5
and
q
≥
1
. In particular, the double inequality
(33)
5
9
M
15
/
5
15
/
5
(
a
,
b
)
G
1
-
15
/
5
(
a
,
b
)
+
4
9
G
(
a
,
b
)
<
L
(
a
,
b
)
<
1
3
A
(
a
,
b
)
+
2
3
G
(
a
,
b
)
holds for all
a
,
b
>
0
with
a
≠
b
.