Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 348326, 12 pages
http://dx.doi.org/10.1155/2013/348326
Research Article
Best Possible Bounds for Neuman-Sándor Mean by the Identric,
Quadratic and Contraharmonic Means
Tie-Hong Zhao,1 Yu-Ming Chu,2 Yun-Liang Jiang,3 and Yong-Min Li4
1
Department of Mathematics, Hangzhou Normal University, Hangzhou 310036, China
School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China
3
School of Information & Engineering, Huzhou Teachers College, Huzhou 313000, China
4
School of Automation, Southeast University, Nanjing 210096, China
2
Correspondence should be addressed to Yu-Ming Chu; chuyuming2005@yahoo.com.cn
Received 19 January 2013; Accepted 1 February 2013
Academic Editor: Khalil Ezzinbi
Copyright © 2013 Tie-Hong Zhao et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We prove that the double inequalities 𝐼𝛼1 (𝑎, 𝑏)𝑄1−𝛼1 (𝑎, 𝑏) < 𝑀(𝑎, 𝑏) < 𝐼𝛽1 (𝑎, 𝑏)𝑄1−𝛽1 (𝑎, 𝑏), 𝐼𝛼2 (𝑎, 𝑏)𝐶1−𝛼2 (𝑎, 𝑏) < 𝑀(𝑎, 𝑏) <
𝐼𝛽2 (𝑎, 𝑏)𝐶1−𝛽2 (𝑎, 𝑏) hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛼1 ≥ 1/2, 𝛽1 ≤ log[√2 log(1 + √2)]/(1 − log √2), 𝛼2 ≥ 5/7, and
𝛽2 ≤ log[2 log(1+√2)], where 𝐼(𝑎, 𝑏), 𝑀(𝑎, 𝑏), 𝑄(𝑎, 𝑏), and 𝐶(𝑎, 𝑏) are the identric, Neuman-Sándor, quadratic, and contraharmonic
means of 𝑎 and 𝑏, respectively.
1. Introduction
For 𝑝 ∈ R and 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏, the identric mean 𝐼(𝑎, 𝑏),
Neuman-Sándor mean 𝑀(𝑎, 𝑏) [1], quadratic mean 𝑄(𝑎, 𝑏),
contraharmonic mean 𝐶(𝑎, 𝑏), and 𝑝th power mean 𝑀𝑝 (𝑎, 𝑏)
are defined by
1/(𝑏−𝑎)
1 𝑏𝑏
𝐼 (𝑎, 𝑏) = ( 𝑎 )
𝑒 𝑎
𝑀 (𝑎, 𝑏) =
𝑄 (𝑎, 𝑏) = √
,
𝑎−𝑏
,
2sinh [(𝑎 − 𝑏) / (𝑎 + 𝑏)]
−1
𝑎2 + 𝑏2
,
2
𝐶 (𝑎, 𝑏) =
𝑎2 + 𝑏2
,
𝑎+𝑏
(1)
1/𝑝
𝑎𝑝 + 𝑏𝑝
{
{
{(
)
2
𝑀𝑝 (𝑎, 𝑏) = {
{
{
√
{ 𝑎𝑏,
Recently, the identric, Neuman-Sándor, quadratic, and
contraharmonic means have attracted the interest of numerous eminent mathematicians. In particular, many remarkable
inequalities for these means can be found in the literature [1–
18].
Let 𝐻(𝑎, 𝑏) = 2𝑎𝑏/(𝑎 + 𝑏), 𝐺(𝑎, 𝑏) = √𝑎𝑏, 𝐿(𝑎, 𝑏) =
(𝑏 − 𝑎)/(log 𝑏 − log 𝑎), 𝑃(𝑎, 𝑏) = (𝑎 − 𝑏)/(4 arctan √𝑎/𝑏 − 𝜋),
𝐴(𝑎, 𝑏) = (𝑎+𝑏)/2, and 𝑇(𝑎, 𝑏) = (𝑎−𝑏)/[2 arctan((𝑎−𝑏)/(𝑎+
𝑏))] be the harmonic, geometric, logarithmic, first Seiffert,
arithmetic, and second Seiffert means of two distinct positive
numbers 𝑎 and 𝑏, respectively. Then it is well known that the
inequalities
, 𝑝 ≠ 0,
𝑝 = 0,
respectively, where sinh−1 (𝑥) = log(𝑥+√1 + 𝑥2 ) is the inverse
hyperbolic sine function.
𝐻 (𝑎, 𝑏) = 𝑀−1 (𝑎, 𝑏) < 𝐺 (𝑎, 𝑏) = 𝑀0 (𝑎, 𝑏) < 𝐿 (𝑎, 𝑏)
< 𝑃 (𝑎, 𝑏) < 𝐼 (𝑎, 𝑏) < 𝐴 (𝑎, 𝑏) < 𝑀1 (𝑎, 𝑏)
< 𝑀 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) < 𝑄 (𝑎, 𝑏) = 𝑀2 (𝑎, 𝑏)
< 𝐶 (𝑎, 𝑏)
hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏.
(2)
2
Abstract and Applied Analysis
Neuman and Sándor [1, 8] established that
𝐴 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) <
𝐴 (𝑎, 𝑏)
log (1 + √2)
which has been investigated in [23–25]. Alzer [26] proved that
the inequalities
,
√𝐴 (𝑎, 𝑏) 𝐺 (𝑎, 𝑏) < √𝐼 (𝑎, 𝑏) 𝐿 (𝑎, 𝑏)
𝜋
𝑇 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) ,
4
𝑀 (𝑎, 𝑏) <
𝐴2 (𝑎, 𝑏)
,
𝑃 (𝑎, 𝑏)
<
(3)
√𝐴 (𝑎, 𝑏) 𝑇 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏)
<√
𝑀 (𝑎, 𝑏) <
𝐴2 (𝑎, 𝑏) + 𝑇2 (𝑎, 𝑏)
,
2
𝑀0 (𝑎, 𝑏) < √𝐼 (𝑎, 𝑏) 𝐿 (𝑎, 𝑏) < 𝑀1/2 (𝑎, 𝑏) ,
for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏.
Let 0 < 𝑎, 𝑏 ≤ 1/2 with 𝑎 ≠ 𝑏, 𝑎󸀠 = 1 − 𝑎, and 𝑏󸀠 = 1 − 𝑏.
Then the Ky Fan inequalities
𝑇 (𝑎, 𝑏)
𝑀 (𝑎, 𝑏)
<
<
𝑀 (𝑎󸀠 , 𝑏󸀠 ) 𝑇 (𝑎󸀠 , 𝑏󸀠 )
hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏.
The following sharp bounds for 𝐼, (𝐼𝐿)1/2 , and (𝐼 + 𝐿)/2
in terms of the power mean and the convex combination of
arithmetic and geometric means are given in [27] as
𝑀2/3 (𝑎, 𝑏) < 𝐼 (𝑎, 𝑏) < 𝑀log 2 (𝑎, 𝑏) ,
2𝐴 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏)
3
𝐺 (𝑎, 𝑏)
𝐿 (𝑎, 𝑏)
𝑃 (𝑎, 𝑏)
𝐴 (𝑎, 𝑏)
<
<
<
𝐺 (𝑎󸀠 , 𝑏󸀠 ) 𝐿 (𝑎󸀠 , 𝑏󸀠 ) 𝑃 (𝑎󸀠 , 𝑏󸀠 ) 𝐴 (𝑎󸀠 , 𝑏󸀠 )
(4)
were presented in [1].
Li et al. [19] found the best possible bounds for the
Neuman-Sándor mean in terms of the generalized logarithmic mean 𝐿 𝑟 (𝑎, 𝑏). Neuman [20] and Zhao et al. [21] proved
that the inequalities
𝑀log 2/(1+log 2) (𝑎, 𝑏)
<
for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏.
Chu et al. [28] presented the optimal constants 𝛼1 , 𝛽1 , 𝛼2 ,
and 𝛽2 such that the double inequalities
𝛼1 𝑄 (𝑎, 𝑏) + (1 − 𝛼1 ) 𝐴 (𝑎, 𝑏)
<
2 𝜋/2 √ 2 2
𝑎 cos 𝜃 + 𝑏2 sin2 𝜃𝑑𝜃
∫
𝜋 0
< 𝛽1 𝑄 (𝑎, 𝑏) + (1 − 𝛽1 ) 𝐴 (𝑎, 𝑏) ,
𝜆𝐶 (𝑎, 𝑏) + (1 − 𝜆) 𝐴 (𝑎, 𝑏)
𝑄𝛼2 (𝑎, 𝑏) 𝐴1−𝛼2 (𝑎, 𝑏)
< 𝑀 (𝑎, 𝑏) < 𝜇𝐶 (𝑎, 𝑏) + (1 − 𝜇) 𝐴 (𝑎, 𝑏) ,
(5)
𝛼1 𝐻 (𝑎, 𝑏) + (1 − 𝛼1 ) 𝑄 (𝑎, 𝑏)
<
< 𝑀 (𝑎, 𝑏) < 𝛽1 𝐻 (𝑎, 𝑏) + (1 − 𝛽1 ) 𝑄 (𝑎, 𝑏) ,
< 𝑀 (𝑎, 𝑏) < 𝛽2 𝐶 (𝑎, 𝑏) + (1 − 𝛽2 ) 𝑄 (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)], 𝛽 ≥ 1/3, 𝜆 ≤ [1 − log(1 +
√2)]/ log(1 + √2), 𝜇 ≥ 1/6, 𝛼1 ≥ 2/9, 𝛽1 ≤ 1 − 1/[√2 log(1 +
√2)], 𝛼2 ≥ 1/3, and 𝛽2 ≤ 1 − 1/[√2 log(1 + √2)].
In [22], Chu and Long gave the best possible constants
𝑝, 𝑞, 𝛼, and 𝛽 such that the double inequalities 𝑀𝑝 (𝑎, 𝑏) <
𝑀(𝑎, 𝑏) < 𝑀𝑞 (𝑎, 𝑏) and 𝛼𝐼(𝑎, 𝑏) < 𝑀(𝑎, 𝑏) < 𝛽𝐼(𝑎, 𝑏) hold
for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏.
The ratio of identric means leads to the weighted geometric mean
1/(𝑎+𝑏)
,
(9)
2 𝜋/2 √ 2 2
𝑎 cos 𝜃 + 𝑏2 sin2 𝜃𝑑𝜃
∫
𝜋 0
< 𝑄𝛽2 (𝑎, 𝑏) 𝐴1−𝛽2 (𝑎, 𝑏)
𝛼2 𝐶 (𝑎, 𝑏) + (1 − 𝛼2 ) 𝑄 (𝑎, 𝑏)
= (𝑎𝑎 𝑏𝑏 )
(8)
2
2
< 𝐼 (𝑎, 𝑏) < 𝐴 (𝑎, 𝑏) + (1 − ) 𝐺 (𝑎, 𝑏)
𝑒
𝑒
< 𝑀 (𝑎, 𝑏) < 𝛽𝑄 (𝑎, 𝑏) + (1 − 𝛽) 𝐴 (𝑎, 𝑏) ,
𝐼 (𝑎, 𝑏)
𝐼 (𝑎, 𝑏) + 𝐿 (𝑎, 𝑏)
< 𝑀1/2 (𝑎, 𝑏) ,
2
2
1
𝐴 (𝑎, 𝑏) + 𝐺 (𝑎, 𝑏)
3
3
𝛼𝑄 (𝑎, 𝑏) + (1 − 𝛼) 𝐴 (𝑎, 𝑏)
𝐼 (𝑎2 , 𝑏2 )
𝐼 (𝑎, 𝑏) + 𝐿 (𝑎, 𝑏) 𝐴 (𝑎, 𝑏) + 𝐺 (𝑎, 𝑏)
<
2
2
(7)
(6)
hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏.
The aim of this paper is to find the best possible constants
𝛼1 , 𝛽1 , 𝛼2 and 𝛽2 such that the double inequalities
𝐼𝛼1 (𝑎, 𝑏) 𝑄1−𝛼1 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝐼𝛽1 (𝑎, 𝑏) 𝑄1−𝛽1 (𝑎, 𝑏) ,
𝐼𝛼2 (𝑎, 𝑏) 𝐶1−𝛼2 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝐼𝛽2 (𝑎, 𝑏) 𝐶1−𝛽2 (𝑎, 𝑏)
(10)
hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. All numerical computations
are carried out using mathematica software.
2. Lemmas
In order to prove our main results, we need several lemmas,
which we present in this section.
Abstract and Applied Analysis
3
Lemma 1. The double inequality
𝑥+
𝑥3 2𝑥5 √
𝑥3 2𝑥5 8𝑥7
−
< 1 + 𝑥2 sinh−1 (𝑥) < 𝑥 +
−
+
3
15
3
15
105
(11)
Proof. To prove inequalities (19) and (20), it suffices to show
that
𝜄1 (𝑥)
:= log [
holds for 𝑥 ∈ (0, 1).
−(
Proof. To prove Lemma 1, it suffices to prove that
𝜂1 (𝑥) = √1 + 𝑥2 sinh−1 (𝑥) − (𝑥 +
𝜂2 (𝑥) = √1 + 𝑥2 sinh−1 (𝑥) − (𝑥 +
𝑥3 2𝑥5
−
) > 0,
3
15
(12)
𝑥3 2𝑥5 8𝑥7
−
+
)<0
3
15
105
(13)
𝜂1 (0) = 𝜂2 (0) = 0,
𝜂1󸀠
𝑥𝜂1∗ (𝑥)
,
(𝑥) =
√1 + 𝑥2
𝜂2󸀠
𝑥𝜂2∗ (𝑥)
,
(𝑥) =
√1 + 𝑥2
(14)
2𝑥3 8𝑥5 √
+
) 1 + 𝑥2 ,
3
15
(15)
(1 + 𝑥)1+𝑥
] − 2𝑥 − 𝑥 log (1 − 𝑥2 )
(1 − 𝑥)1−𝑥
(22)
2𝑥3 2𝑥5 2𝑥7
+
+
+ 𝑥9 ) < 0
3
5
7
for 𝑥 ∈ (0, 3/4).
From (21) and (22), one has
𝜄1 (0+ ) = 𝜄2 (0+ ) = 0,
(23)
2𝑥8
>0
1 − 𝑥2
(24)
7
9𝑥8
( − 𝑥2 ) < 0
1 − 𝑥2 9
(25)
for 𝑥 ∈ (0, 3/4).
Therefore, inequality (21) follows from (23) and (24), and
inequality (22) follows from (23) and (25).
Lemma 3. Let
8𝑥4
󸀠
> 0,
𝜂1∗ (𝑥) =
3√1 + 𝑥2
(16)
16𝑥6
< 0,
5√1 + 𝑥2
(17)
󸀠
:= log [
𝜄2󸀠 (𝑥) = −
𝜂1∗ (0) = 𝜂2∗ (0) = 0,
𝜂2∗ (𝑥) = −
𝜄2 (𝑥)
for 𝑥 ∈ (0, 1), and
2𝑥3 √
𝜂1∗ (𝑥) = sinh−1 (𝑥) − (𝑥 −
) 1 + 𝑥2 ,
3
𝜂2∗ (𝑥) = sinh−1 (𝑥) − (𝑥 −
2𝑥3 2𝑥5 2𝑥7
+
+
)>0
3
5
7
𝜄1󸀠 (𝑥) =
where
(21)
for 𝑥 ∈ (0, 1), and
−(
for 𝑥 ∈ (0, 1).
From the expressions of 𝜂1 (𝑥) and 𝜂2 (𝑥), we get
(1 + 𝑥)1+𝑥
] − 2𝑥 − 𝑥 log (1 − 𝑥2 )
(1 − 𝑥)1−𝑥
Φ1 (𝑥) =
1
1
−
.
√1 + 𝑥2 sinh−1 (𝑥) 𝑥 (1 + 𝑥2 )
(26)
Then the double inequality
for 𝑥 ∈ (0, 1).
Therefore, inequality (12) follows from (14)–(16), and
inequality (13) follows from (14)–(17).
2𝑥 34𝑥3 𝑥5
2𝑥 34𝑥3 4𝑥5
−
+
< Φ1 (𝑥) <
−
+
3
45
2
3
45
5
holds for 𝑥 ∈ (0, 0.7).
(27)
Proof. To prove inequality (27), it suffices to show that
𝜙1 (𝑥) = 𝑥√1 + 𝑥2 − sinh−1 (𝑥)
Lemma 2. Let
𝐿 (𝑥) = log [
(1 + 𝑥)1+𝑥
] − 2𝑥 − 𝑥 log (1 − 𝑥2 ) .
(1 − 𝑥)1−𝑥
(18)
− 𝑥 (1 + 𝑥2 ) sinh−1 (𝑥) (
2𝑥 34𝑥3 𝑥5
−
+ ) (28)
3
45
2
> 0,
Then
𝐿 (𝑥) >
2𝑥3 2𝑥5 2𝑥7
+
+
3
5
7
(19)
− 𝑥 (1 + 𝑥2 ) sinh−1 (𝑥) (
for 𝑥 ∈ (0, 1), and
𝐿 (𝑥) <
for 𝑥 ∈ (0, 3/4).
𝜙2 (𝑥) = 𝑥√1 + 𝑥2 − sinh−1 (𝑥)
2𝑥3 2𝑥5 2𝑥7
+
+
+ 𝑥9
3
5
7
(20)
2𝑥 34𝑥3 4𝑥5
−
+
)
3
45
5
<0
(29)
for 𝑥 ∈ (0, 0.7).
4
Abstract and Applied Analysis
First, we prove inequality (28). From the expression of
𝜙1 (𝑥), we have
𝜙1 (0) = 0,
𝜙1 (0.7) = 0.0033 ⋅ ⋅ ⋅ ,
𝜙1󸀠 (𝑥) =
𝑥𝜙1∗
(𝑥)
90√1 + 𝑥2
,
for 𝑥 ∈ (0, 0.6], and
𝜙1∗∗ (𝑥) > − 56𝑥 − 309𝑥3 + 422𝑥5 + 675𝑥7
(30)
+ 2 (28 − 324𝑥2 + 735𝑥4 + 1260𝑥6 )
(31)
× (𝑥 +
where
𝑥3 2𝑥5
𝑥3
−
)=
3
15
15
×(−14075+25028𝑥2 +56571𝑥4 + 9660𝑥6 −5040𝑥8 )
𝜙1∗ (𝑥) = 120𝑥 + 8𝑥3 + 23𝑥5 − 45𝑥7
− 2 (60 − 16𝑥2 − 69𝑥4 + 180𝑥6 )
Equation (32) leads to
󸀠
𝜙1∗ (𝑥) = −
𝜙1∗ (0.7) = −3.551 ⋅ ⋅ ⋅ ,
(33)
𝑥𝜙1∗∗ (𝑥)
,
1 + 𝑥2
where
𝜙1∗∗ (𝑥) = − 56𝑥 − 309𝑥3 + 422𝑥5 + 675𝑥7
+ 2 (28 − 324𝑥2 + 735𝑥4 + 1260𝑥6 )
𝑥3
[−14075 + 25028 × (0.6)2 + 56571 × (0.6)4 ]
15
=
1416676𝑥3
>0
9375
(32)
× √1 + 𝑥2 sinh−1 (𝑥) .
𝜙1∗ (0.6) = 3.017 ⋅ ⋅ ⋅ ,
>
(38)
for 𝑥 ∈ [0.6, 0.7).
From (33), (37), and (38), we clearly see that there exists
𝑥1 ∈ (0.6, 0.7) such that 𝜙1∗ (𝑥) > 0 for 𝑥 ∈ (0, 𝑥1 ) and 𝜙1∗ (𝑥) <
0 for 𝑥 ∈ (𝑥1 , 0.7). Then (31) leads to the conclusion that
𝜙1 (𝑥) is strictly increasing on (0, 𝑥1 ] and strictly decreasing
on [𝑥1 , 0.7).
Therefore, inequality (28) follows from (30) and the
piecewise monotonicity of 𝜙1 (𝑥).
Next, we prove inequality (29). From the expression of
𝜙2 (𝑥), we get
𝜙2 (0) = 0,
(34)
× √1 + 𝑥2 sinh−1 (𝑥) .
𝜙2󸀠 (𝑥) = −
Note that
2𝑥𝜙2∗ (𝑥)
,
45√1 + 𝑥2
(39)
where
60 − 16𝑥2 − 69𝑥4 + 180𝑥6 > 0
𝜙2∗ (𝑥) = 𝑥 (18𝑥6 + 𝑥4 − 2𝑥2 − 30)
(35)
+ 2 (15 − 4𝑥2 + 3𝑥4 + 72𝑥6 )
for 𝑥 ∈ (0, 0.6], and
28 − 324𝑥2 + 735𝑥4 + 1260𝑥6 > 0
(40)
× √1 + 𝑥2 sinh−1 (𝑥) .
(36)
It follows from Lemma 1 and (40) that
for 𝑥 ∈ [0.6, 0.7).
It follows from (32) and (34)–(36) together with Lemma 1
that
𝜙1∗
3
5
(𝑥) > 120𝑥 + 8𝑥 + 23𝑥 − 45𝑥
+ 2 (15 − 4𝑥2 + 3𝑥4 + 72𝑥6 ) (𝑥 +
7
𝑥3
− 2 (60 − 16𝑥 − 69𝑥 + 180𝑥 ) (𝑥 + )
3
2
4
=
6
=
𝑥5
(515 − 1077𝑥2 − 360𝑥4 )
3
≥
𝑥5
9
81
[515 − 1077 ×
− 360 ×
]
3
25
625
10078𝑥5
=
>0
375
𝜙2∗ (𝑥) > 𝑥 (18𝑥6 + 𝑥4 − 2𝑥2 − 30)
(37)
𝑥3 2𝑥5
−
) (41)
3
15
𝑥5
(5 + 2476𝑥2 + 708𝑥4 − 288𝑥6 ) > 0
15
for 𝑥 ∈ (0, 0.7).
Therefore, inequality (29) follows from (39) together with
(41).
Lemma 4. Let
Φ2 (𝑥) =
1
√1 + 𝑥2 sinh−1 (𝑥)
−
1 − 𝑥2
.
𝑥 (1 + 𝑥2 )
(42)
Abstract and Applied Analysis
5
for 𝑥 ∈ (0, 0.66), and
Then the double inequality
5𝑥 79𝑥3 11𝑥5
5𝑥 79𝑥3 9𝑥5
−
+
< Φ2 (𝑥) <
−
+
3
45
10
3
45
5
4 − 108𝑥2 + 213𝑥4 + 396𝑥6
(43)
3 2
> 4 − 108 × ( ) + 213 × (0.66)4
4
holds for 𝑥 ∈ (0, 3/4).
+ 396 × (0.66)6 = 16.3972 > 0
Proof. To prove Lemma 4, it suffices to prove that
𝜑1 (𝑥) := 𝑥√1 + 𝑥2 − (1 − 𝑥2 ) sinh−1 (𝑥)
− 𝑥 (1 + 𝑥2 ) sinh−1 (𝑥) (
5𝑥 79𝑥3 11𝑥5
−
+
) > 0,
3
45
10
(44)
𝜑2 (𝑥) := 𝑥√1 + 𝑥2 − (1 − 𝑥2 ) sinh−1 (𝑥)
− 𝑥 (1 + 𝑥2 ) sinh−1 (𝑥) (
5𝑥 79𝑥3 9𝑥5
−
+
)<0
3
45
5
(45)
3
𝜑1 ( ) = 0.008457 ⋅ ⋅ ⋅ > 0,
4
𝜑1󸀠 (𝑥) =
𝑥𝜑1∗ (𝑥)
,
90√1 + 𝑥2
for 𝑥 ∈ [0.66, 3/4).
It follows from Lemma 1, (48), and (51)–(53) that
𝜑1∗ (𝑥)
> 120𝑥 + 8𝑥3 + 59𝑥5 − 99𝑥7
− 2 (60 − 16𝑥2 − 177𝑥4 + 396𝑥6 )
for 𝑥 ∈ (0, 3/4).
We first prove inequality (44). From the expression of
𝜑1 (𝑥), we obtain
𝜑1 (0) = 0,
× (𝑥 +
=
𝑥3 2𝑥5 8𝑥7
−
+
)
3
15
105
𝑥5
[46165 − 82573𝑥2 − 32420𝑥4
105
+7584𝑥6 + 6336𝑥6 (1 − 𝑥2 )]
(46)
>
𝑥5
[46165 − 82573 × (0.66)2 − 32420 × (0.66)4 ]
105
(47)
=
𝑥5
× 4044.5917 ⋅ ⋅ ⋅ > 0
105
where
𝜑1∗ (𝑥) = 120𝑥 + 8𝑥3 + 59𝑥5 − 99𝑥7 − 2
𝜑1∗∗ (𝑥) > −56𝑥 − 705𝑥3 + 836𝑥5 + 1485𝑥7
+ 14 (4 − 108𝑥2 + 213𝑥4 + 396𝑥6 ) (𝑥+
Equation (48) leads to
3
𝜑1∗ ( ) = −19.299 ⋅ ⋅ ⋅ < 0,
4
(49)
𝑥𝜙∗∗ (𝑥)
󸀠
𝜑1∗ (𝑥) = − 1 2 ,
1+𝑥
=
(50)
𝜑1∗∗ (𝑥) = − 56𝑥 − 705𝑥3 + 836𝑥5 + 1485𝑥7
(51)
× √1 + 𝑥2 sinh−1 (𝑥) .
Note that
60 − 16𝑥2 − 177𝑥4 + 396𝑥6
> 60 − 16 × (0.66)2 − 177 × (0.66)4
= 19.4451 > 0
(52)
𝑥3 2𝑥5
−
)
3
15
𝑥3
[−32975 + 49598𝑥2 + 123369𝑥4
15
+10668𝑥6 + 11088𝑥6 (1 − 𝑥2 )]
>
𝑥3
[−32975 + 49598 ×(0.66)2 +123369 × (0.66)4 ]
15
=
𝑥3
× 12038.83 ⋅ ⋅ ⋅ > 0
15
where
+ 14 (4 − 108𝑥2 + 213𝑥4 + 396𝑥6 )
(54)
for 𝑥 ∈ (0, 0.66), and
× (60 − 16𝑥2 − 177𝑥4 + 396𝑥6 ) √1 + 𝑥2 sinh−1 (𝑥) .
(48)
𝜑1∗ (0.66) = 6.02 ⋅ ⋅ ⋅ > 0,
(53)
(55)
for 𝑥 ∈ [0.66, 3/4).
From (50) and (55), we know that 𝜑1∗ (𝑥) is
strictly decreasing on [0.66, 3/4), and this in conjunction with (49) and (54) leads to the conclusion that
there exists 𝑥1 ∈ (0.66, 3/4) such that 𝜑1∗ (𝑥) > 0 for
𝑥 ∈ (0, 𝑥1 ) and 𝜑1∗ (𝑥) < 0 for 𝑥 ∈ (𝑥1 , 3/4). Then (47)
implies that 𝜑1 (𝑥) is strictly increasing on (0, 𝑥1 ] and strictly
decreasing on [𝑥1 , 3/4). Therefore, inequality (44) follows
from (46) and the piecewise monotonicity of 𝜑1 (𝑥).
6
Abstract and Applied Analysis
Next, we prove inequality (45). From the expression of
𝜑2 (𝑥) one has
12
23𝑥5
( − 𝑥2 ) > 0,
2
35 (1 + 𝑥 ) 23
Υ1 (𝑥) − (
𝜑2 (0) = 0,
𝜙2󸀠 (𝑥) = −
=
𝑥𝜙2∗ (𝑥)
,
45√1 + 𝑥2
(56)
<
where
1
2𝑥3 2𝑥5 2𝑥7
(
+
+
+ 𝑥9 )
2
2𝑥
3
5
7
𝑥
4𝑥 4𝑥3 8𝑥5
−
(
−
+
)
1 + 𝑥2
3
5
7
+
𝜙2∗ (𝑥) = − 60𝑥 − 4𝑥3 + 2𝑥5 + 81𝑥7
+ 4 (15 − 4𝑥2 + 3𝑥4 + 162𝑥6 )
(57)
4𝑥 4𝑥3 8𝑥5
−
+
)
3
5
7
=−
𝑥7 (1 − 𝑥2 )
2 (1 + 𝑥2 )
<0
(61)
× √1 + 𝑥2 sinh−1 (𝑥) .
for 𝑥 ∈ (0, 0.7).
Therefore, Lemma 5 follows easily from (61).
It follows from Lemma 1 and (52) that
Lemma 6. Let 𝐿(𝑥) be defined as in Lemma 2 and
𝜙2∗
3
5
(𝑥) > −60𝑥 − 4𝑥 + 2𝑥 + 81𝑥
7
3
𝑥
2𝑥
+ 4 (15 − 4𝑥 + 3𝑥 + 162𝑥 ) (𝑥 +
−
)
3
15
2
=
4
6
𝑥5
(10 + 11027𝑥2 + 3216𝑥4 − 1296𝑥6 ) > 0
15
(58)
for 𝑥 ∈ (0, 3/4).
Therefore, inequality (45) follows from (56) together with
(58).
Lemma 5. Let 𝐿(𝑥) be defined as in Lemma 2 and
Υ1 (𝑥) =
Υ2 (𝑥) =
5
7𝑥 9𝑥3 7𝑥5
7𝑥 9𝑥3 15𝑥5
−
+
< Υ2 (𝑥) <
−
+
3
5
5
3
5
7
Proof. It follows from Lemma 2 that
Υ2 (𝑥) − (
(59)
3
4𝑥 4𝑥
4𝑥 4𝑥
4𝑥
8𝑥
−
+
< Υ1 (𝑥) <
−
+
3
5
5
3
5
7
5
holds for 𝑥 ∈ (0, 0.7).
Proof. From Lemma 2, one has
4𝑥 4𝑥3 4𝑥5
Υ1 (𝑥) − (
−
+
)
3
5
5
>
2𝑥3 2𝑥5 2𝑥7
1
𝑥
(
+
+
)+
2
2𝑥
3
5
7
1 + 𝑥2
−(
4𝑥 4𝑥3 4𝑥5
−
+
)
3
5
5
=
(60)
7𝑥 9𝑥3 7𝑥5
−
+
)
3
5
5
1
2𝑥3 2𝑥5 2𝑥7
(
+
+
)
2𝑥2
3
5
7
Then the double inequality
5
(63)
holds for 𝑥 ∈ (0, 3/4).
+
3
(62)
Then the double inequality
>
𝐿 (𝑥)
𝑥
+
.
2𝑥2
1 + 𝑥2
𝐿 (𝑥)
2𝑥
+
.
2𝑥2
1 + 𝑥2
7𝑥 9𝑥3 7𝑥5
2𝑥
−
(
−
+
)
1 + 𝑥2
3
5
5
13
44𝑥5
( − 𝑥2 ) > 0,
35 (1 + 𝑥2 ) 22
7𝑥 9𝑥3 15𝑥5
Υ2 (𝑥) − (
−
+
)
3
5
7
<
2𝑥3 2𝑥5 2𝑥7
1
(
+
+
+ 𝑥9 )
2
2𝑥
3
5
7
+
=−
2𝑥
7𝑥 9𝑥3 15𝑥5
−
(
−
+
)
1 + 𝑥2
3
5
7
𝑥7 (3 − 𝑥2 )
2 (1 + 𝑥2 )
<0
for 𝑥 ∈ (0, 3/4).
Therefore, Lemma 6 follows from (64).
(64)
Abstract and Applied Analysis
7
Lemma 7 and 𝑥 > sinh−1 (𝑥) give 𝜔1 (𝑥) < 0 and
Lemma 7. The inequality
𝑥3
√1 + 𝑥2
> [sinh−1 (𝑥)]
3
(65)
𝜔1󸀠 (𝑥) =
2
−1
𝑥3 [sinh (𝑥)]
[
3
𝑥3
3
√1 +
− (sinh−1 (𝑥)) ] > 0
𝑥2
(73)
holds for 𝑥 ∈ (0, 1).
for 𝑥 ∈ (0, 1). This in turn implies that
Proof. Let
𝜁 (𝑥) =
𝑥3
3
− [sinh−1 (𝑥)] .
√1 + 𝑥2
(66)
󸀠
2
𝜔1 (𝑥) 󸀠 𝜔1 (𝑥) (1 + 𝑥 ) − 2𝑥𝜔1 (𝑥)
=
>0
]
2
1 + 𝑥2
(1 + 𝑥2 )
𝜁 (0) = 0,
𝜁1 (𝑥)
3/2
(1 + 𝑥2 )
𝜔2 (1) = 0.2796 ⋅ ⋅ ⋅ > 0,
(67)
,
𝜔2󸀠 (𝑥) = −
where
2𝑥
𝜔2∗ (𝑥) =
(68)
It follows from Lemma 1 and (68) that
(69)
6
37
208 36𝑥
64𝑥
+(
+
+
)
525
525
175
3675
6
32𝑥
]>0
735
[
𝜔2 (𝑥)
<0
2
−
1
(1 + 𝑥2 ) [sinh−1 (𝑥)]
𝑥
(1 + 𝑥2 )
3/2
sinh−1 (𝑥)
(75)
,
2
(70)
.
(76)
2
3/2
(1 + 𝑥2 )
<0
]
3/2
Lemma 8. Let
(𝑥 + 𝑥3 )
[sinh (𝑥)]
2
󸀠
3/2
(1 + 𝑥2 )
=
−
−1
for 𝑥 ∈ (0, 1).
From (75)–(76), we clearly see that 𝜔2󸀠 (𝑥) < 0 and 𝜔2 (𝑥) >
0 for 𝑥 ∈ (0, 1). This in turn implies that
for 𝑥 ∈ (0, 1).
Therefore, Lemma 7 follows from (67) together with (69).
1 + 3𝑥2
𝑥
𝜔2∗󸀠 (𝑥) = −
2
𝜇1 (𝑥) =
𝜔2∗ (𝑥)
𝜔2∗ (0) = 0,
𝑥3 2𝑥5 8𝑥7
> 𝑥2 (3 + 2𝑥2 ) − 3(𝑥 +
−
+
)
3
15
105
× (1 − 𝑥2 ) +
+
𝑥
− sinh−1 (𝑥) ,
√1 + 𝑥2
𝜁1 (𝑥)
2
3/2
𝑥2 )
(1 +
where
2
𝜁1 (𝑥) = 𝑥2 (3 + 2𝑥2 ) − 3[√1 + 𝑥2 sinh−1 (𝑥)] .
= 𝑥6 [
(74)
for 𝑥 ∈ (0, 1).
On the other hand, from the expression of 𝜔2 (𝑥), we get
Then
𝜁󸀠 (𝑥) =
[
𝜔2󸀠 (𝑥) (1 + 𝑥2 )
− 3𝑥√1 + 𝑥2 𝜔2 (𝑥)
(1 +
(77)
3
𝑥2 )
for 𝑥 ∈ (0, 1).
Equation (72) together with inequalities (74) and (77) lead
to the conclusion that
𝜇1 (𝑥)
≤
Then 𝜇1 (𝑥) < 0.2 for 𝑥 ∈ [0.7, 1).
𝜔1 (1)
𝜔2 (0.7)
+
3/2
2
[1 + (0.7)2 ]
(78)
= 0.167 ⋅ ⋅ ⋅ < 0.2
Proof. Let
𝜔1 (𝑥) =
𝜔2 (𝑥) =
1
1
−
,
2
2
𝑥
[sinh−1 (𝑥)]
2
√1 + 𝑥2
−
𝑥
.
sinh−1 (𝑥)
for 𝑥 ∈ [0.7, 1).
(71)
𝜇2 (𝑥) =
Then
𝜇1 (𝑥) =
𝜔2 (𝑥)
𝜔1 (𝑥)
+
.
1 + 𝑥2 (1 + 𝑥2 )3/2
Lemma 9. Let
1 + 4𝑥2 − 𝑥4
(𝑥 +
−
(72)
2
𝑥3 )
−
1
(1 +
𝑥2 ) [sinh−1
𝑥
(1 + 𝑥2 )
3/2
sinh−1 (𝑥)
Then 𝜇2 (𝑥) < 0.51 for 𝑥 ∈ [0.65, 1).
.
(𝑥)]
2
(79)
8
Abstract and Applied Analysis
Proof. Differentiating ]1 (𝑥) yields
Proof. Let
1
1
−
= 𝜇1 (𝑥) ,
2
2
𝑥
[sinh−1 (𝑥)]
𝜏1 (𝑥) =
]󸀠1 (𝑥) =
(80)
𝑥
3 − 𝑥2
−
,
𝜏2 (𝑥) =
2
√1 + 𝑥
sinh−1 (𝑥)
]󸀠1 (𝑥)
>
𝜏2 (𝑥)
𝜏1 (𝑥)
+
.
1 + 𝑥2 (1 + 𝑥2 )3/2
(81)
=
From (74), we clearly see that
[
𝜏1 (𝑥) 󸀠
𝜔1 (𝑥) 󸀠
]
=
[
] >0
1 + 𝑥2
1 + 𝑥2
>
(82)
>
𝜏2 (1) = 0.2796 ⋅ ⋅ ⋅ > 0,
1
−1
sinh (𝑥)
−
3/2
𝑥2 ) [sinh−1
(𝑥)]
2
2
𝜏2∗ (0.65) = 0.6033 ⋅ ⋅ ⋅ ,
+ 2[
(83)
𝜏2 (𝑥)
󸀠
] =
3/2
(1 + 𝑥2 )
− 3𝑥√1 + 𝑥2 𝜏2 (𝑥)
3
(1 + 𝑥2 )
<0
for 𝑥 ∈ [0.65, 1).
Equation (81) together with inequalities (82) and (84) lead
to the conclusion that
𝜏 (1)
𝜏2 (0.65)
𝜇2 (𝑥) ≤ 1
= 0.503 ⋅ ⋅ ⋅ < 0.51 (85)
+
3/2
2
[1 + (0.65)2 ]
]󸀠2
for 𝑥 ∈ [0.65, 1).
Lemma 10. Let 𝐿(𝑥) be defined as in Lemma 2 and
2
(1 − 𝑥2 ) (1 + 𝑥2 )
Then ]1 (𝑥) > 1.2 for x ∈ [0.7, 1).
349
5
2𝑥
2
3
35(1 − 𝑥2 ) (1 + 𝑥2 )
>0
−
𝐿 (𝑥)
.
𝑥3
3 − 2𝑥2 + 3𝑥4
(1 −
𝑥2 ) (1
+
2
𝑥2 )
𝐿 (𝑥)
.
𝑥3
−
2
4
6
8
3𝐿 (𝑥) 2 (1 + 7𝑥 − 17𝑥 + 5𝑥 − 4𝑥 )
−
.
(𝑥) =
2
3
𝑥4
𝑥(1 − 𝑥2 ) (1 + 𝑥2 )
(89)
(90)
It follows from (19) and (90) together with the monotonicity of the function 561𝑥2 − 272𝑥4 on [0.65, 1) that
󸀠
]2 (𝑥)
>
=
>
2 (1 + 𝑥4 )
+
[−84 + 316 × (0.7)2 −
Proof. Differentiating ]2 (𝑥) yields
(84)
]1 (𝑥) =
35(1 −
3
𝑥2 )
Then ]2 (𝑥) > 1.38 for 𝑥 ∈ [0.65, 1).
5 + 𝑥2 √
1 + 𝑥2 sinh−1 (𝑥) − 𝑥] > 0
1 + 𝑥2
3/2
2𝑥
2
𝑥2 ) (1
]2 (𝑥) =
2
𝜏2󸀠 (𝑥) (1 + 𝑥2 )
3
Lemma 11. Let 𝐿(𝑥) be defined as in Lemma 2 and
for 𝑥 ∈ (0, 1).
From (83), we clearly see that 𝜏2󸀠 (𝑥) < 0 and 𝜏2 (𝑥) > 0 for
𝑥 ∈ [0.65, 1). This in turn implies that
[
2
35(1 − 𝑥2 ) (1 + 𝑥2 )
for 𝑥 ∈ [0.7, 1).
Therefore, ]1 (𝑥) ≥ ]1 (0.7) = 1.214 ⋅ ⋅ ⋅ > 1.2 for 𝑥 ∈
[0.7, 1) follows from (88).
,
𝜏2∗ (𝑥) = (5 + 𝑥2 ) [sinh−1 (𝑥)] − (1 + 𝑥2 ) ,
𝜏2∗󸀠 (𝑥) = 2𝑥[sinh−1 (𝑥)]
2𝑥 (−84 + 316𝑥2 − 97𝑥4 + 68𝑥6 + 26𝑥8 + 36𝑥10 + 15𝑥12 )
(88)
𝑥𝜏2∗ (𝑥)
(1 +
2 2𝑥2 2𝑥4
2 + 8𝑥2 − 20𝑥4 − 6𝑥8
1
[3 ( +
+
)−
]
2
3
𝑥
3
5
7
(1 − 𝑥2 ) (1 + 𝑥2 )
2
+68𝑥4 (𝑥2 − )]
5
for 𝑥 ∈ (0, 1).
On the other hand, from the expression of 𝜏2 (𝑥) together
with Lemma 1, we get
𝜏2󸀠 (𝑥) = −
(87)
It follows from (19) and (87) that
then
𝜇2 (𝑥) =
3𝐿 (𝑥) 2 + 8𝑥2 − 20𝑥4 − 6𝑥8
−
.
2
3
𝑥4
𝑥(1 − 𝑥2 ) (1 + 𝑥2 )
=
2 (1 + 7𝑥2 − 17𝑥4 + 5𝑥6 − 4𝑥8 )
1
2 2𝑥2 2𝑥4
]
[3 ( +
+
)−
2
3
𝑥
3
5
7
(1 − 𝑥2 ) (1 + 𝑥2 )
2𝑥 (−189 + 561𝑥2 − 272𝑥4 + 103𝑥6 + 26𝑥8 + 36𝑥10 + 15𝑥12 )
2
3
35(1 − 𝑥2 ) (1 + 𝑥2 )
2𝑥 [−189 + 561 × (0.65)2 − 272 × (0.65)4 + 103 × (0.65)6 ]
2
3
35(1 − 𝑥2 ) (1 + 𝑥2 )
2𝑥 × 7.23 ⋅ ⋅ ⋅
2
3
35(1 − 𝑥2 ) (1 + 𝑥2 )
>0
(91)
(86)
for 𝑥 ∈ [0.65, 1).
Equation (91) leads to the conclusion that ]2 (𝑥)
]2 (0.65) = 1.389 ⋅ ⋅ ⋅ > 1.38 for 𝑥 ∈ [0.65, 1).
≥
Abstract and Applied Analysis
9
Lemma 12. Let Φ1 (𝑥) and Υ1 (𝑥) be defined, respectively, as
in Lemmas 3 and 5, and Θ1 (𝑥; 𝑝) = Φ1 (𝑥) − 𝑝Υ1 (𝑥). Then
Θ1 (𝑥; 𝑝) is strictly decreasing on [0.7, 1) if 𝑝 > 1/6.
Proof. Differentiating Θ1 (𝑥; 𝑝) with respect to 𝑥 and making
use of Lemmas 8 and 10, we get
we assume that 𝑎 > 𝑏. Let 𝑝 ∈ (0, 1), 𝑥 = (𝑎 − 𝑏)/(𝑎 + 𝑏), and
𝜆 1 = log[√2 log(1 + √2)]/(1 − log √2). Then 𝑥 ∈ (0, 1), and
1 (1 + 𝑥)1+𝑥
𝐼 (𝑎, 𝑏)
= [
]
𝐴 (𝑎, 𝑏) 𝑒 (1 − 𝑥)1−𝑥
𝑀 (𝑎, 𝑏)
𝑥
,
=
𝐴 (𝑎, 𝑏)
sinh−1 (𝑥)
𝑑Θ1 (𝑥; 𝑝)
= Φ󸀠1 (𝑥) − 𝑝Υ1󸀠 (𝑥) = 𝜇1 (𝑥) − 𝑝]1 (𝑥)
𝑑𝑥
1
< 0.2 − × 1.2 = 0
6
(92)
lim+
𝑥→0
𝑄 (𝑎, 𝑏) √
= 1 + 𝑥2 ,
𝐴 (𝑎, 𝑏)
(95)
log √1 + 𝑥2 − log 𝑥 + log [sinh−1 (𝑥)]
,
log √1 + 𝑥2 − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1
(96)
log √1 + 𝑥2 − log 𝑥 + log [sinh−1 (𝑥)]
log √1 + 𝑥2 − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1
1
= ,
2
Lemma 13. Let Φ2 (𝑥) and Υ2 (𝑥) be defined, respectively, as
in Lemmas 4 and 6, and Θ2 (𝑥; 𝑞) = Φ2 (𝑥) − 𝑞Υ2 (𝑥). Then
Θ2 (𝑥; 𝑞) is strictly decreasing on [0.65, 1) if 𝑞 > 2/5.
Proof. Differentiating Θ2 (𝑥; 𝑞) with respect to 𝑥 and making
use of Lemmas 9 and 11, we have
,
log [𝑄 (𝑎, 𝑏)] − log [𝑀 (𝑎, 𝑏)]
log [𝑄 (𝑎, 𝑏)] − log [𝐼 (𝑎, 𝑏)]
=
for 𝑥 ∈ [0.7, 1) and 𝑝 > 1/6. This in turn implies that Θ1 (𝑥; 𝑝)
is strictly decreasing on [0.7, 1) if 𝑝 > 1/6.
1/2𝑥
lim−
𝑥→1
(97)
log √1 + 𝑥2 − log 𝑥 + log [sinh−1 (𝑥)]
log √1 + 𝑥2 − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1
= 𝜆1.
𝑑Θ1 (𝑥; 𝑞)
= Φ󸀠2 (𝑥) − 𝑞Υ2󸀠 (𝑥) = 𝜇2 (𝑥) − 𝑞]2 (𝑥)
𝑑𝑥
< 0.51 −
2
× 1.38 = −0.042 < 0
5
(98)
(93)
The difference between the convex combination of
log[𝐼(𝑎, 𝑏)], log[𝑄(𝑎, 𝑏)] and log[𝑀(𝑎, 𝑏)] is as follows:
𝑝 log [𝐼 (𝑎, 𝑏)] + (1 − 𝑝) log [𝑄 (𝑎, 𝑏)] − log [𝑀 (𝑎, 𝑏)]
=
for 𝑥 ∈ [0.65, 1) and 𝑞 > 2/5. This in turn implies that
Θ2 (𝑥; 𝑞) is strictly decreasing on [0.65, 1) if 𝑞 > 2/5.
3. Main Results
Theorem 14. The double inequality
𝑝
(1 + 𝑥)1+𝑥
]−𝑝
log [
2𝑥
(1 − 𝑥)1−𝑥
+ (1 − 𝑝) log √1 + 𝑥2 − log [
𝑥
] := 𝐷𝑝 (𝑥) .
sinh−1 (𝑥)
(99)
Equation (99) leads to
𝐷𝑝 (0+ ) = 0,
𝐷𝑝 (1− ) = log [√2 log (1 + √2)] − 𝑝 (1 − log √2) ,
𝐼𝛼1 (𝑎, 𝑏) 𝑄1−𝛼1 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝐼𝛽1 (𝑎, 𝑏) 𝑄1−𝛽1 (𝑎, 𝑏)
(94)
𝐷𝜆 1 (1− ) = 0,
𝐷𝑝󸀠 (𝑥) = −
holds for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛽1 ≤
log[√2 log(1 + √2)]/(1 − log √2) = 0.337 ⋅ ⋅ ⋅ and 𝛼1 ≥ 1/2.
Proof. Since 𝐼(𝑎, 𝑏), 𝑀(𝑎, 𝑏), and 𝑄(𝑎, 𝑏) are symmetric and
homogeneous of degree one, then without loss of generality,
1 + 𝑝𝑥2
1
𝐿 (𝑥)
+
−
3
−1
2
√1 + 𝑥 sinh (𝑥)
𝑥+𝑥
2𝑥2
(100)
(101)
= Φ1 (𝑥) − 𝑝Υ1 (𝑥) = Θ1 (𝑥; 𝑝) ,
where 𝐿(𝑥), Φ1 (𝑥), Υ1 (𝑥), and Θ1 (𝑥; 𝑝) are defined as in
Lemmas 2, 3, 5, and 12, respectively.
10
Abstract and Applied Analysis
It follows from (101) together with Lemmas 3 and 5 that
󸀠
𝐷1/2
(𝑥)
<
2𝑥 34𝑥3 4𝑥5 1 4𝑥 4𝑥3 4𝑥5
−
+
− (
−
+
) (102)
3
45
5
2 3
5
5
2𝑥2 8
( − 𝑥2 ) < 0
5 9
for 𝑥 ∈ (0, 0.7). Moreover, we see clearly, from Lemma 12,
󸀠
󸀠
(𝑥) is strictly decreasing on [0.7, 1) and so 𝐷1/2
(𝑥) <
that 𝐷1/2
󸀠
𝐷1/2 (0.7) = −0.109 ⋅ ⋅ ⋅ < 0 for 𝑥 ∈ [0.7, 1). This in
conjunction with (100) and (102) implies that
(i) If 𝛼1 < 1/2, then (96) and (97) imply that there exists
𝛿1 ∈ (0, 1) such that 𝑀(𝑎, 𝑏) < 𝐼𝛼1 (𝑎, 𝑏)𝑄1−𝛼1 (𝑎, 𝑏) for
all 𝑎, 𝑏 > 0 with (𝑎 − 𝑏)/(𝑎 + 𝑏) ∈ (0, 𝛿1 ).
(ii) If 𝛽1 > 𝜆 1 , then (96) and (98) imply that there exists
𝛿2 ∈ (0, 1) such that 𝑀(𝑎, 𝑏) > 𝐼𝛽1 (𝑎, 𝑏)𝑄1−𝛽1 (𝑎, 𝑏) for
all 𝑎, 𝑏 > 0 with (𝑎 − b)/(𝑎 + 𝑏) ∈ (1 − 𝛿2 , 1).
=−
𝐷1/2 (𝑥) < 0
(103)
for 𝑥 ∈ (0, 1).
On the other hand, (101) and Lemmas 3 and 5 together
with the monotonicity of the function −2(17 − 18𝜆 1 )𝑥2 /45 +
(7 − 16𝜆 1 )𝑥4 /14 on (0, 0.7) lead to
𝐷𝜆󸀠 1 (𝑥)
>
2𝑥 34𝑥3 𝑥5
4𝑥 4𝑥3 8𝑥5
−
+
− 𝜆1 (
−
+
)
3
45
2
3
5
7
= 𝑥[
2 (1 − 2𝜆 1 ) 2 (17 − 18𝜆 1 ) 2 7 − 16𝜆 1 4
−
𝑥 +
𝑥]
3
45
14
> 𝑥[
2 (1 − 2𝜆 1 ) 2 (17 − 18𝜆 1 )
−
× (0.7)2
3
45
+
Theorem 15. The double inequality
𝐼𝛼2 (𝑎, 𝑏) 𝐶1−𝛼2 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝐼𝛽2 (𝑎, 𝑏) 𝐶1−𝛽2 (𝑎, 𝑏)
(108)
holds for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛼2 ≥ 5/7 and
𝛽2 ≤ log[2 log(1 + √2)] = 0.566 ⋅ ⋅ ⋅.
Proof. We will follow the same idea in the proof of
Theorem 14. Since 𝐼(𝑎, 𝑏), 𝑀(𝑎, 𝑏), and 𝐶(𝑎, 𝑏) are symmetric
and homogeneous of degree one. Without loss of generality,
we assume that 𝑎 > 𝑏. Let 𝑞 ∈ (0, 1), 𝜆 2 = log[2 log(1 + √2)],
and 𝑥 = (𝑎 − 𝑏)/(𝑎 + 𝑏). Then 𝑥 ∈ (0, 1).
Making use of (95) together with 𝐶(𝑎, 𝑏)/𝐴(𝑎, 𝑏) = 1 + 𝑥2
gives
log [𝐶 (𝑎, 𝑏)] − log [𝑀 (𝑎, 𝑏)]
log [𝐶 (𝑎, 𝑏)] − log [𝐼 (𝑎, 𝑏)]
=
7 − 16𝜆 1
× (0.7)4 ]
14
(74969 − 218832𝜆 1 ) 𝑥
=
>0
180000
lim+
𝑥→0
(104)
𝐷𝜆󸀠 1 (1− ) = −∞.
(105)
From (104) and (105) together with the monotonicity of
𝐷𝜆󸀠 1 (𝑥) on [0.7, 1), we clearly see that there exists 𝑐1 ∈ (0.7, 1)
such that 𝐷𝜆 1 (𝑥) is strictly increasing on (0, 𝑐1 ] and strictly
decreasing on [𝑐1 , 1). This in conjunction with (100) implies
that
𝐷𝜆 1 (𝑥) > 0
(106)
for 𝑥 ∈ (0, 1).
Equation (99) together with inequalities (103) and (106)
gives rise to
𝑀 (𝑎, 𝑏) > 𝐼
1/2
1/2
(𝑎, 𝑏) 𝑄
(𝑎, 𝑏) ,
𝑀 (𝑎, 𝑏) < 𝐼𝜆 1 (𝑎, 𝑏) 𝑄1−𝜆 1 (𝑎, 𝑏) .
,
log (1 + 𝑥2 ) − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1
(109)
log (1 + 𝑥2 ) − log 𝑥 + log [sinh−1 (𝑥)]
log (1 + 𝑥2 ) − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1
5
= ,
7
for 𝑥 ∈ (0, 0.7).
It follows from Lemma 12 that 𝐷𝜆󸀠 1 (𝑥) is strictly decreasing on [0.7, 1). Note that
𝐷𝜆󸀠 1 (0.7) = 0.0229 ⋅ ⋅ ⋅ > 0,
log (1 + 𝑥2 ) − log 𝑥 + log [sinh−1 (𝑥)]
log (1 + 𝑥2 ) − log 𝑥 + log [sinh−1 (𝑥)]
lim
𝑥 → 1− log (1
Therefore, Theorem 14 follows from (107) together with
the following statements.
+ 𝑥2 ) − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1
= 𝜆2.
(111)
The difference between the convex combination of
log[𝐼(𝑎, 𝑏)], log[𝐶(𝑎, 𝑏)] and log[𝑀(𝑎, 𝑏)] is as follows:
𝑞 log [𝐼 (𝑎, 𝑏)] + (1 − 𝑞) log [𝐶 (𝑎, 𝑏)] − log [𝑀 (𝑎, 𝑏)]
=
(107)
(110)
𝑞
(1 + 𝑥)1+𝑥
] − 𝑞 + (1 − 𝑞) log (1 + 𝑥2 )
log [
2𝑥
(1 − 𝑥)1−𝑥
− log [
𝑥
] := 𝐸𝑞 (𝑥) .
sinh−1 (𝑥)
(112)
Abstract and Applied Analysis
11
Equation (112) leads to
𝐸𝑞 (0+ ) = 0,
𝐸𝑞 (1− ) = log [2 log (1 + √2)] − 𝑞,
𝐸𝜆 2 (1− ) = 0,
(113)
𝐸𝑞󸀠 (𝑥)
=−
1 − 𝑥2 + 2𝑞𝑥2
1
𝐿 (𝑥)
(114)
+
−
√1 + 𝑥2 sinh−1 (𝑥)
𝑥 + 𝑥3
2𝑥2
such that 𝐸𝜆 2 (𝑥) is strictly increasing on (0, 𝑐2 ] and strictly
decreasing on [𝑐2 , 1). This in conjunction with (113) implies
that
𝐸𝜆 2 (𝑥) > 0
for 𝑥 ∈ (0, 1).
Equation (112) together with inequalities (116) and (119)
lead to the conclusion that
𝑀 (𝑎, 𝑏) > 𝐼5/7 (𝑎, 𝑏) 𝐶2/7 (𝑎, 𝑏) ,
= Φ2 (𝑥) − 𝑞Υ2 (𝑥) = Θ2 (𝑥; 𝑞) ,
where 𝐿(𝑥), Φ2 (𝑥), Υ2 (𝑥), and Θ2 (𝑥; 𝑞) are defined as in
Lemmas 2, 4, 6, and 13, respectively.
It follows from Lemmas 4, 6, and 13 together with (114)
that
󸀠
𝐸5/7
(𝑥)
<(
5𝑥 79𝑥3 9𝑥5
5 7𝑥 9𝑥3 7𝑥5
−
+
)− (
−
+
)
3
45
5
7 3
5
5
=−
4𝑥2 37
( − 𝑥2 ) < 0
5 63
(115)
󸀠
for 𝑥 ∈ (0, 0.65) and 𝐸5/7
(𝑥) is strictly decreasing on [0.65, 1).
󸀠
󸀠
Thus, we have 𝐸5/7 (𝑥) < 𝐸5/7
(0.65) = −0.117 ⋅ ⋅ ⋅ for 𝑥 ∈
[0.65, 1). This in conjunction with (113) and (115) implies that
𝐸5/7 (𝑥) < 0
(116)
for 𝑥 ∈ (0, 1).
On the other hand, Lemmas 4, 6, and 13 together with
(114) lead to
𝐸𝜆󸀠 2
(𝑥)
>(
5𝑥 79𝑥3 11𝑥5
7𝑥 9𝑥3 15𝑥5
−
+
) − 𝜆2 (
−
+
)
3
45
10
3
5
7
= 𝑥[
5 − 7𝜆 2 79 − 81𝜆 2 2 150𝜆 2 − 77 4
−
𝑥 −
𝑥]
3
45
70
> 𝑥[
5 − 7𝜆 2 79 − 81𝜆 2
−
× (0.65)2
3
45
−
=
150𝜆 2 − 77
× (0.65)4 ]
70
113027173 − 197098950𝜆 2
𝑥>0
100800000
(117)
for 𝑥 ∈ (0, 0.65) and 𝐸𝜆󸀠 2 (𝑥) is strictly decreasing on [0.65, 1).
Note that
𝐸𝜆󸀠 2 (0.65) = 0.0609 ⋅ ⋅ ⋅ ,
𝐸𝜆󸀠 2 (1− ) = −∞.
(118)
From (117) and (118) together with the monotonicity of 𝐸𝜆󸀠 2 (𝑥)
on [0.65, 1), we clearly see that there exists 𝑐2 ∈ (0.65, 1)
(119)
𝑀 (𝑎, 𝑏) < 𝐼𝜆 2 (𝑎, 𝑏) 𝐶1−𝜆 2 (𝑎, 𝑏) .
(120)
Therefore, Theorem 15 follows from (120) together with
the following statements.
(i) If 𝛼2 < 5/7, then (109) and (110) imply that there exists
𝛿3 ∈ (0, 1) such that 𝑀(𝑎, 𝑏) < 𝐼𝛼2 (𝑎, 𝑏)𝐶1−𝛼2 (𝑎, 𝑏) for
all 𝑎, 𝑏 > 0 with (𝑎 − 𝑏)/(𝑎 + 𝑏) ∈ (0, 𝛿3 ).
(ii) If 𝛽2 > 𝜆 2 , then (109) and (111) imply that there exists
𝛿4 ∈ (0, 1) such that 𝑀(𝑎, 𝑏) > 𝐼𝛽2 (𝑎, 𝑏)𝐶1−𝛽2 (𝑎, 𝑏) for
all 𝑎, 𝑏 > 0 with (𝑎 − 𝑏)/(𝑎 + 𝑏) ∈ (1 − 𝛿4 , 1).
Acknowledgments
This research was supported by the Natural Science Foundation of China under Grants 11171307, 61174076, and 61173123
and the Natural Science Foundation of Zhejiang Province
under Grants Z1110551 and LY12F02012.
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