# Sharp inequalities for hyperbolic functions and circular functions

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## Abstract

In this paper, we obtain some new sharp bounds for the exponential functions whose powers involve hyperbolic functions and circular functions, respectively.

## Introduction

In , Stolarsky obtained the bounds for the exponential function whose power involves a hyperbolic function.

### Theorem 1

Let $$x>0$$. Then

$$1< e^{x\coth x-1}< \cosh x.$$
(1.1)

On the other hand, Pittenger  and Stolarsky  got the lower bound for the function $$e^{x\coth x-1}$$ as follows.

### Theorem 2

Let $$x>0$$. Then

$$\biggl( \cosh \frac{2x}{3} \biggr) ^{\frac{3}{2}}< e^{x\coth x-1}.$$
(1.2)

In fact, Zhu  and Kouba  showed a new sharp lower bound for the function $$e^{x\coth x-1}$$ as follows.

### Theorem 3

Let $$x>0$$. Then

$$\biggl( \frac{2(\cosh x)^{\frac{6}{5}}+1}{3} \biggr) ^{\frac{5}{6}}< e ^{x\coth x-1}.$$
(1.3)

Zhu  proved

$$\biggl( \cosh \frac{2x}{3} \biggr) ^{\frac{3}{2}}< \biggl( \frac{2( \cosh x)^{\frac{6}{5}}+1}{3} \biggr) ^{\frac{5}{6}},\quad x>0$$
(1.4)

to illustrate that the inequality (1.3) is stronger than (1.2).

It should be pointed out that the paper of Yang et al.  has made a great deal of improvement on inequality (1.2). The subject of the present paper is to further study the inequality (1.3), and to obtain the following results.

### Theorem 4

Let $$p\neq 0$$, $$p_{1}=(\ln (3/2))/(\ln (e/2))=1. 321 4\ldots$$ , and $$x\in (0,+\infty )$$. Then we have

1. (i)

when $$p\in {}[ 2,+\infty )$$, the double inequality

$$e^{p(x\coth x-1)}< \frac{2\cosh ^{p}x+1}{3}< \frac{2}{3} \biggl( \frac{e}{2} \biggr) ^{p}e^{p(x\coth x-1)}$$
(1.5)

holds, where the constants 1 and $$(2/3)(e/2)^{p}$$ are the best possible;

2. (ii)

when $$p\in (-\infty ,6/5]$$, we have

1. (a)

if $$p\in (0,6/5]$$, the double inequality (1.5) reverses, that is,

$$\frac{2}{3} \biggl( \frac{e}{2} \biggr) ^{p}e^{p(x\coth x-1)}< \frac{2 \cosh ^{p}x+1}{3}< e^{p(x\coth x-1)}$$
(1.6)

holds, where the constants $$(2/3)(e/2)^{p}$$ and 1 are the best possible;

2. (b)

if $$p\in (-\infty ,0)$$, the left-hand side of the double inequality (1.5) holds too;

3. (iii)

when $$p\in (6/5,2)$$, we have

1. (c)

if $$p\in {}[ p_{1},2)$$, the left-hand side of inequality (1.5) holds;

2. (d)

if $$p\in (6/5,p_{1})$$, the left-hand side of inequality (1.6) holds.

As straightforward consequences of Theorem 4, Theorem 5 which is due to Kouba  may be derived immediately.

### Theorem 5

Let $$p\neq 0$$, and $$p_{1}=(\ln (3/2))/(\ln (e/2))=1. 321 4\ldots$$ . Then

1. (1)

the inequality

$$\frac{2\cosh ^{p}x+1}{3}< e^{p(x\coth x-1)}$$
(1.7)

holds for all $$x\in (0,+\infty )$$ if and only if $$p\in (0,6/5]$$;

2. (2)

the inequality (1.7) reverses for all $$x\in (0,+\infty )$$ if and only if $$p\in (-\infty ,0)\cup {}[ p_{1},+\infty )$$.

The analog of Theorem 4 for the circular functions is the following result.

### Theorem 6

Let $$p\neq 0$$, $$p_{2}=\ln 3=1. 0986\ldots$$ , and $$x\in (0,\pi /2)$$. Then we have

1. (i)

when $$p\in {}[ 6/5,+\infty )$$, the double inequality

$$e^{p(x\cot x-1)}< \frac{2\cos ^{p}x+1}{3}< \biggl( \frac{e^{p}}{3} \biggr) e^{p(x\cot x-1)}$$
(1.8)

holds, where the constants 1 and $$e^{p}/3$$ are the best possible;

2. (ii)

when $$p\in (-\infty ,1]$$, we have

1. (a)

if $$p\in (0,1]$$, the double inequality (1.8) reverses, that is,

$$\biggl( \frac{e^{p}}{3} \biggr) e^{p(x\cot x-1)}< \frac{2\cos ^{p}x+1}{3}< e^{p(x\cot x-1)}$$
(1.9)

holds, where the constants $$e^{p}/3$$ and 1 are the best possible;

2. (b)

if $$p\in (-\infty ,0)$$, the double inequality (1.8) holds too;

3. (iii)

when $$p\in (1,6/5)$$, we have

1. (c)

if $$p\in (p_{2},6/5)$$, the right-hand side of inequality (1.8) holds;

2. (d)

if $$p\in (1,p_{2}]$$, the right-hand side of inequality (1.9) holds.

The following result which is due to Yang et al.  is a straightforward consequence of Theorem 6.

### Theorem 7

Let $$p\neq 0$$, and $$p_{2}=\ln 3=1. 0986\ldots$$ . Then

1. (A)

the inequality

$$e^{p(x\cot x-1)}< \frac{2\cos ^{p}x+1}{3}$$
(1.10)

holds for all $$x\in (0,\pi /2)$$ if and only if $$p\in (-\infty ,0) \cup {}[ 6/5,+\infty )$$;

2. (B)

the inequality (1.10) reverses for all $$x\in (0,\pi/2 )$$ if and only if $$p\in (0,p_{2}]$$.

## Lemmas

### Lemma 1

()

For $$-\infty < a< b<\infty$$, let $$f,g:[a,b]\rightarrow \mathbb{R}$$ be continuous functions that are differentiable on $$( a,b )$$, with $$f ( a ) =g ( a ) =0$$ or $$f ( b ) =g ( b ) =0$$. Assume that $$g^{\prime }(x)\neq 0$$ for each x in $$(a,b)$$. If $$f^{\prime }/g^{ \prime }$$ is increasing (decreasing) on $$(a,b)$$, then so is $$f/g$$.

### Lemma 2

()

Let $$A ( x ) =\sum_{n=0}^{\infty }a_{n}x^{n}$$ and $$B ( x ) =\sum_{n=0}^{\infty }b_{n}x^{n}$$ be two real power series converging on $$( 0,r )$$ ($$r>0$$) with $$b_{n}>0$$ for all n. If the sequence $$\{a_{n}/b_{n}\}$$ is increasing (decreasing) for all n, then the function $$A ( x ) /B ( x )$$ is also increasing (decreasing) on $$( 0,r )$$.

### Lemma 3

Let $$l(x)$$ be defined by

$$l ( x ) =\frac{\ln \frac{\sinh 2x-2x}{4 ( x\cosh x- \sinh x ) }}{\ln \cosh x}.$$
(2.1)

Then $$l(x)$$ is strictly increasing from $$( 0,+\infty )$$ onto $$( 1/5,1 )$$.

### Proof

Let

$$l(x)=\frac{\ln \frac{\sinh 2x-2x}{4 ( x\cosh x-\sinh x ) }}{ \ln \cosh x}=:\frac{f(x)}{g(x)}=\frac{f(x)-f(0^{+})}{g(x)-g(0^{+})},$$

where

$$f(x)=\ln \frac{\sinh 2x-2x}{4 ( x\cosh x-\sinh x ) },\qquad g(x)=\ln \cosh x.$$

Then

\begin{aligned}& f^{\prime }(x) = \frac{2\cosh 2x-2}{\sinh 2x-2x}+\frac{x\sinh x}{\sinh x-x\cosh x} \\& \hphantom{f^{\prime }(x)}=\sinh x \biggl( \frac{4\sinh x}{\sinh 2x-2x}+\frac{x}{\sinh x-x\cosh x} \biggr) , \\& g^{\prime }(x) =\frac{\sinh x}{\cosh x}. \end{aligned}

We compute

\begin{aligned} \frac{f^{\prime }(x)}{g^{\prime }(x)} =&\frac{\sinh x ( \frac{4\sinh x}{\sinh 2x-2x}+\frac{x}{\sinh x-x\cosh x} ) }{\frac{\sinh x}{\cosh x}} \\ =& \biggl( \frac{x}{\sinh x-x\cosh x}+4\frac{\sinh x}{-2x+\sinh 2x} \biggr) \cosh x \\ =& ( \cosh x ) \frac{-4\sinh ^{2}x-x\sinh 2x+2x^{2}+4x\cosh x \sinh x}{ ( -2x+\sinh 2x ) ( -\sinh x+x\cosh x ) } \\ =& \frac{ ( 2x^{2}\cosh x+2x\cosh ^{2}x\sinh x-4\cosh x\sinh ^{2}x ) }{- ( 2x^{2}\cosh x-2x\cosh ^{2}x\sinh x-2x\sinh x+2\cosh x\sinh ^{2}x ) } \\ =&\frac{ ( \cosh x-\cosh 3x+\frac{1}{2}x\sinh x+\frac{1}{2}x \sinh 3x+ 2x^{2}\cosh x ) }{- ( \frac{1}{2}\cosh 3x-\frac{1}{2} \cosh x-\frac{5}{2}x\sinh x-\frac{1}{2}x\sinh 3x+ 2x^{2}\cosh x ) } \\ =&\frac{2\cosh x-2\cosh 3x+x\sinh x+x\sinh 3x+ 4x^{2}\cosh x}{-\cosh 3x+\cosh x+5x\sinh x+x\sinh 3x- 4x^{2}\cosh x} \\ :=&\frac{A(x)}{B(x)}, \end{aligned}

where

\begin{aligned}& A(x) =2\cosh x-2\cosh 3x+x\sinh x+x\sinh 3x+ 4x^{2}\cosh x, \\& B(x) =-\cosh 3x+\cosh x+5x\sinh x+x\sinh 3x- 4x^{2}\cosh x. \end{aligned}

Then

\begin{aligned} A(x) =&\sum_{n=2}^{+\infty } \frac{6 ( n-2 ) 3^{2n}+16n^{2}+26n+12}{(2n+2)!}x^{2n+2}=:\sum_{n=2} ^{+\infty }a_{n}x^{2n+2}, \\ B(x) =&\sum_{n=2}^{+\infty } \frac{3 ( 2n-1 ) 3^{2n}-16n^{2}-14n+3}{(2n+2)!}x^{2n+2}=:\sum_{n=2}^{+ \infty }b_{n}x^{2n+2}, \end{aligned}

where

\begin{aligned}& a_{n} =\frac{6 ( n-2 ) 3^{2n}+16n^{2}+26n+12}{(2n+2)!}, \\& b_{n} =\frac{3 ( 2n-1 ) 3^{2n}-16n^{2}-14n+3}{(2n+2)!}. \end{aligned}

We can obtain

$$\frac{a_{n}}{b_{n}}= 2\frac{3 ( n-2 ) 3^{2n}+8n^{2}+13n+6}{3 ( 2n-1 ) 3^{2n}-16n^{2}-14n+3}=:2s_{n},\quad n\geq 2.$$

Now we will prove that $$\{s_{n}\}_{n\geq 2}$$ is strictly increasing, which means

$$s_{n}< s_{n+1}\quad \Longleftrightarrow\quad h ( n ) 3^{2n}+32n^{2}+112n+81>0,$$

where

$$h ( n ) =:81\cdot 3^{2n}- \bigl( 256n^{3}+224n^{2}+112n+162 \bigr) >0$$

for $$n\geq 2$$ due to

\begin{aligned}& h ( n+1 ) -9h ( n ) = 2048n^{3}+1024n^{2}-320n+704 \\& \quad = 20{,}544+28{,}352 ( n-2 ) +13{,}312 ( n-2 ) ^{2}+2048 ( n-2 ) ^{3}>0 \end{aligned}

for $$n\geq 2$$ and $$h ( 2 ) = 3231>0$$. This leads to $$s_{n}< s_{n+1}$$ for $$n\geq 2$$. So $$\{a_{n}/b _{n}\}_{n\geq 2}$$ is strictly increasing. By Lemma 2, we know that $$A(x)/B(x)=f^{\prime }(x)/g^{\prime }(x)$$ is strictly increasing on $$( 0,+\infty )$$. Then $$l(x)=f(x)/g(x)$$ is strictly increasing on $$( 0,+\infty )$$ by Lemma 1.

Since

$$\lim_{x\rightarrow 0^{+}}\frac{\ln \frac{\sinh 2x-2x}{4 ( x \cosh x-\sinh x ) }}{\ln \cosh x}= \frac{1}{5}, \qquad \lim_{x\rightarrow +\infty }\frac{\ln \frac{ \sinh 2x-2x}{4 ( x\cosh x-\sinh x ) }}{\ln \cosh x}= 1,$$

this completes the proof of Lemma 3. □

### Lemma 4

Let $$x>0,B_{2n}$$ be the even-indexed Bernoulli numbers (see ). Then the following power series expansions:

\begin{aligned}& \tan x =\sum_{n=1}^{\infty } \frac{2^{2n}-1}{ ( 2n ) !}2^{2n} \vert B _{2n} \vert x^{2n-1}, \end{aligned}
(2.2)
\begin{aligned}& \sec ^{2}x =\sum_{n=1}^{\infty } \frac{2^{2n}(2^{2n}-1) ( 2n-1 ) }{ ( 2n ) !} \vert B_{2n} \vert x^{2n-2}, \end{aligned}
(2.3)
\begin{aligned}& \tan x\sec ^{2}x =\sum_{n=2}^{\infty } \frac{2^{2n}(2^{2n}-1) ( 2n-1 ) ( n-1 ) }{ ( 2n ) !} \vert B_{2n} \vert x^{2n-3} \end{aligned}
(2.4)

hold for all $$x\in (-\pi /2,\pi /2)$$.

### Proof

The power series expansion (2.2) can be found in [11, equations 1.3.1.4(3)]. By (2.2) we have

$$(\sec x)^{2}=(\tan x)^{\prime }=\sum _{n=1}^{\infty }\frac{2^{2n}(2^{2n}-1) ( 2n-1 ) }{ ( 2n ) !} \vert B_{2n} \vert x^{2n-2},\quad \vert x \vert < \frac{\pi }{2},$$

and

\begin{aligned} \tan x\sec ^{2}x =&\frac{1}{2} \bigl( \sec ^{2}x \bigr) ^{\prime } \\ =& \frac{1}{2}\sum _{n=2}^{\infty }\frac{2^{2n}(2^{2n}-1) ( 2n-1 ) ( 2n-2 ) }{ ( 2n ) !} \vert B_{2n} \vert x ^{2n-3} \\ =&\sum_{n=2}^{\infty }\frac{2^{2n}(2^{2n}-1) ( 2n-1 ) ( n-1 ) }{ ( 2n ) !} \vert B_{2n} \vert x^{2n-3},\quad \vert x \vert < \frac{\pi }{2}. \end{aligned}

□

### Lemma 5

([12, 13])

Let $$B_{2n}$$ be the even-indexed Bernoulli numbers. Then

$$\frac{(2n+2)(2n+1)(2^{2n-1}-1)}{\pi ^{2}(2^{2n+1}-1)}< \frac{ \vert B _{2n+2} \vert }{ \vert B_{2n} \vert }< \frac{(2n+2)(2n+1)(2^{2n}-1)}{ \pi ^{2}(2^{2n+2}-1)}$$
(2.5)

holds for $$n=1,2,\ldots$$ .

### Lemma 6

Let $$z(x)$$ be defined by

$$z ( x ) =\frac{\ln \frac{2x-\sin 2x}{4 ( \sin x-x \cos x ) }}{\ln \cos x}.$$
(2.6)

Then $$z(x)$$ is strictly decreasing from $$( 0,\pi /2 )$$ onto $$( 0,1/5 )$$.

### Proof

Let

$$z(x)=\frac{\ln \frac{2x-\sin 2x}{4 ( \sin x-x\cos x ) }}{ \ln \cos x}=\frac{p(x)}{q(x)}=\frac{p(x)-p(0^{+})}{q(x)-q(0^{+})},$$

where

$$p(x)=\ln \frac{2x-\sin 2x}{4 ( \sin x-x\cos x ) },\qquad q(x)= \ln \cos x.$$

Since

$$p^{\prime }(x)= ( \sin x ) \biggl[ \frac{4\sin x}{2x- \sin 2x}- \frac{x}{\sin x-x\cos x} \biggr] ,\qquad q^{\prime }(x)=-\frac{ \sin x}{\cos x},$$

we have

\begin{aligned} \frac{p^{\prime }(x)}{q^{\prime }(x)} =& \frac{x\cos x}{\sin x-x\cos x}-\frac{2\sin x\cos x}{x-\cos x\sin x} \\ =&\frac{x^{2}\cos x+x\cos ^{2}x\sin x-2\cos x\sin ^{2}x}{-x^{2}\cos x+x \cos ^{2}x\sin x+x\sin x-\cos x\sin ^{2}x} \\ =&\frac{ ( x^{2}\cos x+x\cos ^{2}x\sin x-2\cos x\sin ^{2}x ) /\cos ^{3}x}{ ( -x^{2}\cos x+x\cos ^{2}x\sin x+x\sin x-\cos x\sin ^{2}x ) /\cos ^{3}x} \\ =&\frac{x^{2}\sec ^{2}x+x\tan x-2\tan ^{2}x}{-x^{2}\sec ^{2}x +x\tan x+x\tan x\sec ^{2}x-\tan ^{2}x} \\ =:&\frac{C(x)}{D(x)}. \end{aligned}

From Lemma 4 we obtain

\begin{aligned}& C(x) =x^{2}\sec ^{2}x+x\tan x-2\tan ^{2}x \\& \hphantom{C(x)}=\sum_{n=3}^{\infty } \biggl[ \frac{ ( 2n ) ( 2^{2n}-1 ) \vert B_{2n} \vert }{ ( 2n ) !}-\frac{8(2^{2n+2}-1) ( 2n+1 ) \vert B_{2n+2} \vert }{ ( 2n+2 ) !} \biggr] 2^{2n}x ^{2n} \\& \hphantom{C(x)}=:\sum_{n=3}^{\infty }c_{n}x^{2n}, \\& D(x) =-x^{2}\sec ^{2}x +x\tan x+x\tan x\sec ^{2}x-\tan ^{2}x \\& \hphantom{D(x)}=\sum_{n=3}^{\infty } \biggl[ \frac{2 ( 2n+1 ) ( 2^{2n+2}-1 ) \vert B_{2n+2} \vert }{ ( 2n+2 ) !}-\frac{ ( 2^{2n}-1 ) \vert B_{2n} \vert }{ ( 2n ) !} \biggr] ( n-1 ) 2^{2n+1}x^{2n} \\& \hphantom{D(x)}=:\sum_{n=3}^{\infty }d_{n}x^{2n}. \end{aligned}

We consider the monotonicity of $$C(x)/D(x)$$, and compute that

$$\frac{c_{n}}{d_{n}}=\frac{1}{2 ( n-1 ) }\frac{ ( n+1 ) ( 2n ) ( 2^{2n}-1 ) \vert B_{2n} \vert -4(2^{2n+2}-1) \vert B _{2n+2} \vert }{ ( 2^{2n+2}-1 ) \vert B_{2n+2} \vert - ( n+1 ) ( 2^{2n}-1 ) \vert B_{2n} \vert }.$$

Then for $$n\geq 3$$

\begin{aligned}& \frac{c_{n}}{d_{n}}>\frac{c_{n+1}}{d_{n+1}}\quad \Longleftrightarrow \\& \frac{L}{M} =:\frac{ ( n+1 ) ( 2n ) ( 2^{2n}-1 ) \vert B_{2n} \vert -4(2^{2n+2}-1) \vert B _{2n+2} \vert }{ ( n-1 ) [ ( 2^{2n+2}-1 ) \vert B_{2n+2} \vert - ( n+1 ) ( 2^{2n}-1 ) \vert B_{2n} \vert ] } \\& \hphantom{\frac{L}{M}}>\frac{ ( n+2 ) ( 2n+2 ) ( 2^{2n+2}-1 ) \vert B_{2n+2} \vert -4(2^{2n+4}-1) \vert B_{2n+4} \vert }{n [ ( 2^{2n+4}-1 ) \vert B_{2n+4} \vert - ( n+2 ) ( 2^{2n+2}-1 ) \vert B_{2n+2} \vert ] } \\& \hphantom{\frac{L}{M}}:=\frac{X}{Y}. \end{aligned}

Since

\begin{aligned}& \frac{LY-MX}{2 \vert B_{2n+2} \vert ^{2}} \\& \quad =- ( n+2 ) \bigl( n^{2}-2n-1 \bigr) \bigl( 2^{2n+2}-1 \bigr) ^{2} \\& \qquad {}- \bigl( 2^{2n+2}-1 \bigr) \bigl( 2^{2n}-1 \bigr) ( n+2 ) ( n+1 ) \frac{ \vert B_{2n} \vert }{ \vert B_{2n+2} \vert } \\& \qquad {}+ ( n+1 ) \bigl( n^{2}-2n+2 \bigr) \bigl( 2^{2n+4}-1 \bigr) \bigl( 2^{2n}-1 \bigr) \frac{ \vert B_{2n} \vert }{ \vert B_{2n+2} \vert }\frac{ \vert B_{2n+4} \vert }{ \vert B_{2n+2} \vert } \\& \qquad {}-2 \bigl( 2^{2n+4}-1 \bigr) \bigl( 2^{2n+2}-1 \bigr) \frac{ \vert B _{2n+4} \vert }{ \vert B_{2n+2} \vert }, \end{aligned}

by Lemma 5 we have

\begin{aligned}& \frac{LY-MX}{2 \vert B_{2n+2} \vert ^{2}} \\& \quad >- ( n+2 ) \bigl( n^{2}-2n-1 \bigr) \bigl( 2^{2n+2}-1 \bigr) ^{2} \\& \qquad {}-\frac{ ( 2^{2n+2}-1 ) ( 2^{2n}-1 ) ( n+2 ) ( n+1 ) \pi ^{2}(2^{2n+1}-1)}{(2n+2)(2n+1)(2^{2n-1}-1)} \\& \qquad {}+ ( n+1 ) \bigl( n^{2}-2n+2 \bigr) \bigl( 2^{2n+4}-1 \bigr) \bigl( 2^{2n}-1 \bigr) \\& \qquad {}\times \frac{\pi ^{2}(2^{2n+2}-1)}{(2n+2)(2n+1)(2^{2n}-1)}\frac{(2n+4)(2n+3)(2^{2n+1}-1)}{ \pi ^{2}(2^{2n+3}-1)} \\& \qquad {}-2 \bigl( 2^{2n+4}-1 \bigr) \bigl( 2^{2n+2}-1 \bigr) \frac{(2n+4)(2n+3)(2^{2n+2}-1)}{ \pi ^{2}(2^{2n+4}-1)} \\& \quad =\frac{ ( 2^{2n+2}-1 ) ( n+2 ) }{\pi ^{2} ( 2^{2n+3}-1 ) ( 2^{2n}-2 ) ( 2n+1 ) } r(n) , \end{aligned}

where

$$r(n)= \bigl[ u_{1}(n)2^{2n}-v_{1}(n) \bigr] 2^{4n}+u_{2}(n)2^{2n}-v _{2}(n)$$

with

\begin{aligned}& u_{1}(n) = \bigl( 64\pi ^{2}-512 \bigr) n^{2}- \bigl( 1024-64\pi ^{2} \bigr) n+224\pi ^{2}-16\pi ^{4}-384, \\& v_{1}(n)=12\pi ^{2}n^{3}+ \bigl( 146\pi ^{2}-1216 \bigr) n^{2}- \bigl( 2432-140\pi ^{2} \bigr) n \\& \hphantom{v_{1}(n)=}{}+ \bigl( 568\pi ^{2}-26\pi ^{4}-912 \bigr) , \\& u_{2}(n) =24\pi ^{2}n^{3}- \bigl( 400-38 \pi ^{2} \bigr) n^{2}- \bigl( 800-26\pi ^{2} \bigr) n \\& \hphantom{u_{2}(n) =}{}+247\pi ^{2}-11\pi ^{4}-300, \\& v_{2}(n) = \bigl( 4\pi ^{2}-32 \bigr) n^{2}+ \bigl( 64-4\pi ^{2} \bigr) n- \bigl( 14\pi ^{2}-\pi ^{4}-24 \bigr) . \end{aligned}

Then we have $$c_{n}/d_{n}>c_{n+1}/d_{n+1}$$ for $$n\geq 3$$ when proving

\begin{aligned}& u_{1}(n)2^{2n}-v_{1}(n) >0 \quad \Longleftrightarrow\quad 2^{2n}>\frac{v_{1}(n)}{u _{1}(n)}, \end{aligned}
(2.7)
\begin{aligned}& u_{2}(n)2^{2n}-v_{2}(n) >0 \quad \Longleftrightarrow\quad 2^{2n}>\frac{v_{2}(n)}{u _{2}(n)}. \end{aligned}
(2.8)

Now we use mathematical induction to prove (2.7). When $$n=3$$, (2.7) clearly holds. Assuming that (2.7) holds for $$n=m$$, that is,

$$2^{2m}>\frac{v_{1}(m)}{u_{1}(m)}.$$
(2.9)

Next, we prove that (2.7) is valid for $$n=m+1$$. By (2.9) we have

$$2^{2 ( m+1 ) }=4\cdot 2^{2m}>4\frac{v_{1}(m)}{u_{1}(m)},$$

in order to complete the proof of (2.7) it suffices to show that

$$4\frac{v_{1}(m)}{u_{1}(m)}>\frac{v_{1}(m+1)}{u_{1}(m+1)}\quad \Longleftrightarrow\quad 4v_{1}(m)u_{1}(m+1)-v_{1}(m+1)u_{1}(m)>0.$$

In fact,

$$4v_{1}(m)u_{1}(m+1)-v_{1}(m+1)u_{1}(m)=k(m),$$

where

\begin{aligned} k(m) =& \bigl( 13{,}452{,}288\pi ^{4}-184{,}343{,}040\pi ^{2}-230{,}976\pi ^{6}+ 1248\pi ^{8}+728{,} 082{,} 432 \bigr) \\ &{}+ \bigl( 11{,} 200{,} 896\pi ^{4}-166{,} 279{,} 680\pi ^{2}-103 {,}872\pi ^{6}+ 663 {,}994 {,}368 \bigr) ( m-3 ) \\ &{}+ \bigl( 3{,}983{,} 424\pi ^{4}-59{,} 525{,} 376\pi ^{2}-16{,} 608\pi ^{6}+225{,} 067 {,}008 \bigr) ( m-3 ) ^{2} \\ &{}+ \bigl( 738{,} 048\pi ^{4}-10{,} 274{,} 304\pi ^{2}-576\pi ^{6}+33{,} 619{,} 968 \bigr) ( m-3 ) ^{3} \\ &{}+ \bigl( 68{,} 736\pi ^{4}-801{,} 792\pi ^{2}+1{,}867{,} 776 \bigr) ( m-3 ) ^{4} \\ &{}+ \bigl( 2304\pi ^{4}-18{,} 432\pi ^{2} \bigr) ( m-3 ) ^{5} \\ >&0 \end{aligned}

for $$m\geq 3$$.

Similarly, we can prove (2.8). By (2.7) and (2.8) we find that $$\{c_{n}/d_{n}\}_{n\geq 3}$$ is a monotonic decreasing sequence. Then we arrive at the conclusion that $$p^{\prime }(x)/q^{\prime }(x)=C(x)/D(x)$$ is decreasing on $$( 0,\pi /2 )$$ by Lemma 2. By Lemma 1 we see that $$z(x)$$ is decreasing on $$(0,\pi /2)$$.

Since

$$z\bigl(0^{+}\bigr)=\frac{1}{5},\qquad z \biggl( \biggl( \frac{\pi }{2} \biggr) ^{-} \biggr) =0,$$

this completes the proof of Lemma 6. □

## The proofs of main results

### Proof

Let

$$G(x)=\frac{2\cosh ^{p}x+1}{3e^{p(x\coth x-1)}},\quad x>0.$$

Then

$$G(+\infty )=\textstyle\begin{cases} \frac{2}{3} ( \frac{e}{2} ) ^{p},&p>0, \\ +\infty ,&p< 0, \end{cases}$$

and

$$G^{\prime }(x)=\frac{p}{3}\frac{Q(x)}{e^{p(x\coth x-1)}},$$
(3.1)

where

\begin{aligned} Q ( x ) =&\frac{2 ( x\cosh x-\sinh x ) }{\sinh ^{2}x}\cosh ^{p-1}x- \frac{\cosh x\sinh x-x}{\sinh ^{2}x} \\ =&\frac{2 ( x\cosh x-\sinh x ) }{\sinh ^{2}x} \biggl[ \cosh ^{p-1}x-\frac{\cosh x\sinh x-x}{2 ( x\cosh x-\sinh x ) } \biggr] \\ =&\frac{2 ( x\cosh x-\sinh x ) ( \ln \cosh x ) }{\sinh ^{2}x}\frac{\cosh ^{p-1}x-\frac{\cosh x\sinh x-x}{2 ( x \cosh x-\sinh x ) }}{\ln ( \cosh ^{p-1}x ) -\ln \frac{ \cosh x\sinh x-x}{2 ( x\cosh x-\sinh x ) }} \\ &{}\times \biggl( p-1-\frac{\ln \frac{\sinh 2x-2x}{4 ( x\cosh x- \sinh x ) }}{\ln \cosh x} \biggr) \\ =:&2\frac{ ( x\cosh x-\sinh x ) ( \ln \cosh x ) }{\sinh ^{2}x}\frac{\cosh ^{p-1}x-\frac{\cosh x\sinh x-x}{2 ( x \cosh x-\sinh x ) }}{\ln ( \cosh ^{p-1}x ) -\ln \frac{ \cosh x\sinh x-x}{2 ( x\cosh x-\sinh x ) }} \\ &{}\times \bigl( p-1-l(x) \bigr) \end{aligned}
(3.2)

with

$$l(x)=\frac{\ln \frac{\sinh 2x-2x}{4 ( x\cosh x-\sinh x ) }}{ \ln \cosh x}.$$

We consider the following three cases.

Case 1: $$p\geq 2$$.

From Lemma 3, we get $$\max_{x\in (0,+\infty )}l(x)=1$$. So $$p-1-l(x)>0$$. This leads to $$Q(x)>0$$ by (3.2), and $$G^{\prime }(x)>0$$ by (3.1). Then

$$G\bigl(0^{+}\bigr)< G(x)< G(+\infty ),$$

this is the double inequality (1.5).

Case 2: $$p\leq 6/5$$.

From Lemma 3, we get $$\min_{x\in (0,+\infty )}l(x)=1/5$$. So $$p-1-l(x)<0$$. This leads to $$Q(x)<0$$.

Subcase 2.1: $$0< p\leq 6/5$$. In this case, $$G^{\prime }(x)<0$$ by (3.1). Then

$$G(+\infty )< G(x)< G\bigl(0^{+}\bigr),$$

this is the double inequality (1.6).

Subcase 2.2: $$p<0$$. We have $$G^{\prime }(x)>0$$ by (3.1). In view of $$G(+\infty )=+\infty$$, the left-hand side of inequality (1.5) holds too.

Case 3: $$6/5< p<2$$.

Let $$r(x):=l(x)+1-p$$. Then

\begin{aligned}& r\bigl(0^{+}\bigr) =l\bigl(0^{+}\bigr)+1-p= \frac{6}{5}-p< 0, \\& r(+\infty ) =l(+\infty )+1-p=2-p>0, \end{aligned}

and there is the unique point $$\xi \in (0,+\infty )$$ such that $$r(x)<0$$ holds for all $$x\in (0,\xi )$$ and $$r(x)>0$$ holds for all for $$x\in (\xi ,+\infty )$$. That is, $$p-1-l(x)>0$$ holds for all $$x\in (0,\xi )$$ and $$p-1-l(x)>0$$ holds for all $$x\in (\xi ,+\infty )$$. By (3.2) and (3.1), we have $$G^{\prime }(x)>0$$ for all $$x\in (0, \xi )$$ and $$G^{\prime }(x)<0$$ holds for all $$x\in (\xi ,+\infty )$$. Then

$$\min (G\bigl(0^{+}\bigr),G(+\infty )< G(x)< G(\xi ).$$

Subcase 3.1: $$p_{1}=(\ln (3/2))/(\ln (e/2))< p<2$$. In this case, $$1<(2/3)(e/2)^{p}$$, that is, $$G(0^{+})< G(+\infty )$$ holds, so $$\min (G(0^{+}),G(+\infty ))=G(0^{+})$$. This leads to the left-hand side of inequality (1.5).

Subcase 3.2: $$6/5< p< p_{1}=(\ln (3/2))/(\ln (e/2))$$. In this case, $$1>(2/3)(e/2)^{p}$$, that is, $$G(0^{+})>G(+\infty )$$ holds, so $$\min (G(0^{+}),G(+\infty ))=G(+\infty )$$. This leads to the left-hand side of inequality (1.6).

The proof of Theorem 4 is complete. □

### Proof

Let

$$F(x)=\frac{2\cos ^{p}x+1}{3e^{p(x\cot x-1)}},\quad 0< x< \frac{\pi }{2}.$$

Then

$$F \biggl( \biggl( \frac{\pi }{2} \biggr) ^{-} \biggr) = \frac{e^{p}}{3}$$

and

\begin{aligned} F^{\prime } ( x ) =&\frac{p}{3} \frac{2 ( \sin x-x \cos x ) [ -\ln ( \cos x ) ] }{ ( \sin ^{2}x ) \exp (p(x\cot x-1))}\frac{\cos ^{p-1}x-\frac{x- \cos x\sin x}{2 ( \sin x-x\cos x ) }}{ ( p-1 ) \ln \cos x-\ln \frac{ ( x-\cos x\sin x ) }{2 ( \sin x-x \cos x ) }} \\ &{}\times \biggl( p-1-\frac{\ln \frac{2x-\sin 2x}{4 ( \sin x-x \cos x ) }}{\ln \cos x} \biggr) \\ =&\frac{p}{3}\frac{2 ( \sin x-x\cos x ) [ -\ln ( \cos x ) ] }{ ( \sin ^{2}x ) \exp (p(x \cot x-1))}\frac{\cos ^{p-1}x-\frac{x-\cos x\sin x}{2 ( \sin x-x \cos x ) }}{ ( p-1 ) \ln \cos x-\ln \frac{ ( x- \cos x\sin x ) }{2 ( \sin x-x\cos x ) }} \\ &{}\times \bigl[ p-1-z(x) \bigr] , \end{aligned}
(3.3)

where

$$z(x)=\frac{\ln \frac{2x-\sin 2x}{4 ( \sin x-x\cos x ) }}{ \ln \cos x}.$$

We consider the following three cases.

Case 1: $$p\geq 6/5$$.

From Lemma 6, we get $$\max_{x\in (0,+\infty )}z(x)=1/5$$. So $$p-1-z(x)>0$$ holds. This leads to $$F^{{\prime }}(x)<0$$ by (3.3). Then

$$F\bigl(0^{+}\bigr)< F(x)< F \biggl( \biggl( \frac{\pi }{2} \biggr) ^{-} \biggr) ,$$

which is the double inequality (1.8).

Case 2: $$p\leq 1$$.

From Lemma 6, we get $$\min_{x\in (0,+\infty )}z(x)=0$$. So $$p-1-z(x)<0$$ holds.

Subcase 2.1: $$0< p\leq 1$$. In this case, $$F^{\prime }(x)<0$$ by (3.3). Then

$$F \biggl( \biggl( \frac{\pi }{2} \biggr) ^{-} \biggr) < F(x)< F\bigl(0^{+}\bigr),$$

which is the double inequality (1.9).

Subcase 2.2: $$p<0$$. We have $$F^{\prime }(x)>0$$ by (3.3), the double inequality (1.8) holds too.

Case 3: $$1< p<6/5$$.

Let $$q(x):=z(x)-p+1$$. Then $$q(0^{+})=z(0^{+})-p+1=6/5-p>0$$, $$q( ( \pi /2 ) ^{-})=z( ( \pi /2 ) ^{-})-p+1=1-p<0$$. There is a unique point $$\eta \in (0,\pi /2)$$ such that $$q(x)>0$$ holds for all $$x\in (0,\eta )$$ and $$q(x)<0$$ holds for all for $$x\in (\eta ,\pi /2)$$. That is, $$p-1-z(x)<0$$ holds for all $$x\in (0,\eta )$$ and $$p-1-z(x)>0$$ holds for all $$x\in (\eta ,\pi /2)$$. By (3.3), we have $$F^{\prime }(x)<0$$ for all $$x\in (0,\eta )$$ and $$F^{\prime }(x)>0$$ for all $$x\in (\eta ,\pi /2)$$. Then

$$F(\eta )< F(x)< \max \bigl(F\bigl(0^{+}\bigr),F\bigl( ( \pi /2 ) ^{-}\bigr)\bigr).$$

Subcase 3.1: $$p_{2}=\ln 3=1. 0986< p<6/5$$. In this case, $$1< e^{p}/3$$, that is, $$F(0^{+})< F( ( \pi /2 ) ^{-})$$, so $$\max (F(0^{+}),F( ( \pi /2 ) ^{-}))=F( ( \pi /2 ) ^{-})$$. This leads to the right-hand side of inequality (1.8).

Subcase 3.2: $$1< p\leq p_{2}=\ln 3=1. 0986$$. In this case, $$1\geq e^{p}/3$$, that is, $$F(0^{+})\geq F( ( \pi /2 ) ^{-})$$, so $$\max (F(0^{+}),F( ( \pi /2 ) ^{-}))=F(0^{+})$$. This leads to the right-hand side of inequality (1.9).

The proof of Theorem 6 is complete. □

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## Acknowledgements

The author is thankful to reviewers for their careful corrections to and valuable comments on the original version of this paper.

### Funding

The author was supported by the National Natural Science Foundation of China (no. 61772025).

## Author information

The author provided the questions and gave the proof for all results. He read and approved this manuscript.

Correspondence to Ling Zhu.

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### Competing interests

The author declares that he has no competing interests. 