Sharp inequalities for hyperbolic functions and circular functions

Abstract

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

1 Introduction

In [1], 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 [2] and Stolarsky [3] 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 [4] and Kouba [5] 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 [6] 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. [7] 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

• (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;

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

  • (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;

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

• (iii) when $p\in (6/5,2)$, we have

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

  • (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 [5] 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) 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) 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

• (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;

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

  • (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;

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

• (iii) when $p\in (1,6/5)$, we have

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

  • (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. [7] is a straightforward consequence of Theorem 6.

Theorem 7

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

• (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 $p\in (-\infty ,0) \cup {}[ 6/5,+\infty )$;

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

2 Lemmas

Lemma 1

([8])

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

([9])

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 [10]). 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. □

3 The proofs of main results

3.1 The proof of Theorem 4

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. □

3.2 The proof of Theorem 6

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. □