# New inequalities for hyperbolic functions and their applications

## Abstract

In this paper, we obtain some new inequalities in the exponential form for the whole of the triples about the four functions $\left\{1,\left(sinht\right)/t,exp\left(tcothtâˆ’1\right),cosht\right\}$. Then we generalize some well-known inequalities for the arithmetic, geometric, logarithmic, and identric means to obtain analogous inequalities for their p th powers, where $p>0$.

MSC: 26E60, 26D07.

## 1 Introduction

Let sinht, cosht, and cotht be the hyperbolic sine, hyperbolic cosine, and hyperbolic cotangent, respectively. It is well known that (see [1â€“6])

$1<\frac{sinht}{t}<{e}^{tcothtâˆ’1}
(1.1)

holds for all .

In the recent paper [7], we have established the following Cusa-type inequalities of exponential type for the triple $\left\{1,\left(sinht\right)/t,cosht\right\}$ described as follows.

Theorem 1.1 (Cusa-type inequalities [[7], Part (i) of Theorem 1.1])

Let $pâ‰¥4/5$, and . Then the double inequality

$\left(1âˆ’\mathrm{Î»}\right)+\mathrm{Î»}{\left(cosht\right)}^{p}<{\left(\frac{sinht}{t}\right)}^{p}<\left(1âˆ’\mathrm{Î·}\right)+\mathrm{Î·}{\left(cosht\right)}^{p}$
(1.2)

holds if and only if $\mathrm{Î·}â‰¥1/3$ and $\mathrm{Î»}â‰¤0$.

On the other hand, the author of this paper [8] obtains the following inequalities of exponential type for the triple $\left\{1,exp\left(tcothtâˆ’1\right),cosht\right\}$.

Theorem 1.2 ([[8], Theorem 2])

Let $p>0$, and . Then

(1) if $0, the double inequality

$\mathrm{Î±}{\left(cosht\right)}^{p}+\left(1âˆ’\mathrm{Î±}\right)<{e}^{p\left(tcothtâˆ’1\right)}<\mathrm{Î²}{\left(cosht\right)}^{p}+\left(1âˆ’\mathrm{Î²}\right)$
(1.3)

holds if and only if $\mathrm{Î±}â‰¤2/3$ and $\mathrm{Î²}â‰¥{\left(2/e\right)}^{p}$;

(2) if $pâ‰¥2$, the double inequality

$\mathrm{Î±}{\left(cosht\right)}^{p}+\left(1âˆ’\mathrm{Î±}\right)<{e}^{p\left(tcothtâˆ’1\right)}<\mathrm{Î²}{\left(cosht\right)}^{p}+\left(1âˆ’\mathrm{Î²}\right)$
(1.4)

holds if and only if $\mathrm{Î±}â‰¤{\left(2/e\right)}^{p}$ and $\mathrm{Î²}â‰¥2/3$.

Next, we do the work for each of the triples $\left\{\left(sinht\right)/t,exp\left(tcothtâˆ’1\right),cosht\right\}$ and $\left\{1,\left(sinht\right)/t,exp\left(tcothtâˆ’1\right)\right\}$, and obtain the following two new results.

Theorem 1.3 Let $0, and . Then

$\mathrm{Î±}{\left(cosht\right)}^{p}+\left(1âˆ’\mathrm{Î±}\right){\left(\frac{sinht}{t}\right)}^{p}<{e}^{p\left(tcothtâˆ’1\right)}<\mathrm{Î²}{\left(cosht\right)}^{p}+\left(1âˆ’\mathrm{Î²}\right){\left(\frac{sinht}{t}\right)}^{p}$
(1.5)

holds if and only if $\mathrm{Î±}â‰¤1/2$ and $\mathrm{Î²}â‰¥{\left(2/e\right)}^{p}$.

Theorem 1.4 Let $pâ‰¥286/693$, and . Then

$\mathrm{Î±}+\left(1âˆ’\mathrm{Î±}\right){e}^{p\left(tcothtâˆ’1\right)}<{\left(\frac{sinht}{t}\right)}^{p}<\mathrm{Î²}+\left(1âˆ’\mathrm{Î²}\right){e}^{p\left(tcothtâˆ’1\right)}$
(1.6)

holds if and only if $\mathrm{Î²}â‰¤1/2$ and $\mathrm{Î±}â‰¥1$.

In this paper, we shall give the elementary proofs of Theorem 1.3 and Theorem 1.4. In the last section, we apply Theorems 1.1-1.4 to obtain some new results for four classical means.

## 2 Lemmas

Lemma 2.1 ([9â€“11])

Let $f,g:\left[a,b\right]â†’\mathbb{R}$ be two continuous functions which are differentiable on $\left(a,b\right)$. Further, let on $\left(a,b\right)$. If ${f}^{â€²}/{g}^{â€²}$ is increasing (or decreasing) on $\left(a,b\right)$, then the functions $\left(f\left(x\right)âˆ’f\left({b}^{âˆ’}\right)\right)/\left(g\left(x\right)âˆ’g\left({b}^{âˆ’}\right)\right)$ and $\left(f\left(x\right)âˆ’f\left({a}^{+}\right)\right)/\left(g\left(x\right)âˆ’g\left({a}^{+}\right)\right)$ are also increasing (or decreasing) on $\left(a,b\right)$.

Lemma 2.2 Let $tâˆˆ\left(0,+\mathrm{âˆž}\right)$. Then the inequality

$D\left(t\right)â‰œt{sinh}^{5}t+2t{sinh}^{3}t+{t}^{4}coshtâˆ’{sinh}^{4}tcoshtâˆ’{t}^{3}{sinh}^{3}tâˆ’2{t}^{3}sinht>0$

holds.

Proof Using the power series expansions of the functions sinh5t, sinh3t, cosht, ${sinh}^{4}tcosht$, and sinht, we have

$\begin{array}{rcl}D\left(t\right)& =& \frac{1}{16}t\left(sinh5tâˆ’5sinh3t+10sinht\right)+\frac{1}{2}t\left(sinh3tâˆ’3sinht\right)+{t}^{4}cosht\\ âˆ’\frac{1}{16}\left(cosh5tâˆ’3cosh3t+2cosht\right)âˆ’\frac{1}{4}{t}^{3}\left(sinh3tâˆ’3sinht\right)âˆ’2{t}^{3}sinht\\ =& \frac{1}{16}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{5}^{2n+1}âˆ’5â‹\dots {3}^{2n+1}+10}{\left(2n+2\right)!}{t}^{2n+1}+\frac{1}{2}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{3}^{2n+1}âˆ’3}{\left(2n+1\right)!}{t}^{2n+2}+\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{1}{\left(2n\right)!}{t}^{2n+4}\\ âˆ’\frac{1}{16}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{5}^{2n}âˆ’3â‹\dots {3}^{2n}+2}{\left(2n\right)!}{t}^{2n}âˆ’\frac{1}{4}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{3}^{2n+1}âˆ’3}{\left(2n+1\right)!}{t}^{2n+4}âˆ’2\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{1}{\left(2n+1\right)!}{t}^{2n+4}\\ =& \frac{1}{16}\underset{n=3}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{l}_{n}}{\left(2n+4\right)!}{t}^{2n+4},\end{array}$

where

$\begin{array}{rcl}{l}_{n}& =& \left(2n+4\right)\left({5}^{2n+3}âˆ’5â‹\dots {3}^{2n+3}+10\right)+8\left(2n+4\right)\left({3}^{2n+3}âˆ’3\right)\\ +16\left(2n+4\right)\left(2n+3\right)\left(2n+2\right)\left(2n+1\right)âˆ’\left({5}^{2n+4}âˆ’3â‹\dots {3}^{2n+4}+2\right)\\ âˆ’4\left(2n+4\right)\left(2n+3\right)\left(2n+2\right)\left({3}^{2n+1}âˆ’3\right)âˆ’32\left(2n+4\right)\left(2n+3\right)\left(2n+2\right)\\ =& \left(250nâˆ’125\right){25}^{n}+\left(279âˆ’462nâˆ’432{n}^{2}âˆ’96{n}^{3}\right){9}^{n}\\ +256{n}^{4}+1,120{n}^{3}+1,520{n}^{2}+532nâˆ’154,\phantom{\rule{1em}{0ex}}n=3,4,â€¦.\end{array}$

Using a basic differential method, we can easily prove

$\begin{array}{rcl}f\left(x\right)& â‰œ& \left(250xâˆ’125\right){25}^{x}+\left(279âˆ’462xâˆ’432{x}^{2}âˆ’96{x}^{3}\right){9}^{x}\\ +256{x}^{4}+1,120{x}^{3}+1,520{x}^{2}+532xâˆ’154>0\end{array}$

on $\left[3,\mathrm{âˆž}\right)$. This leads to ${l}_{n}>0$ for $n=3,4,â€¦$â€‰, and $D\left(t\right)>0$. So, the proof of Lemma 2.2 is complete.â€ƒâ–¡

## 3 Proof of Theorem 1.3

Let

$F\left(t\right)â‰¡\frac{{\left(\frac{t}{sinht}{e}^{tcothtâˆ’1}\right)}^{p}âˆ’1}{{\left(tcotht\right)}^{p}âˆ’1}=\frac{{f}_{1}\left(t\right)âˆ’{f}_{1}\left({0}^{+}\right)}{{g}_{1}\left(t\right)âˆ’{g}_{1}\left({0}^{+}\right)},$

where ${f}_{1}\left(t\right)={\left(\frac{t}{sinht}{e}^{tcothtâˆ’1}\right)}^{p}$ and ${g}_{1}\left(t\right)={\left(tcotht\right)}^{p}$. Then

${k}_{1}\left(t\right)â‰œ\frac{{f}_{1}^{â€²}\left(t\right)}{{g}_{1}^{â€²}\left(t\right)}=\frac{{e}^{p\left(tcothtâˆ’1\right)}}{{\left(cosht\right)}^{pâˆ’1}}â‹\dots \frac{{sinh}^{2}tâˆ’{t}^{2}}{sinht\left(sinhtcoshtâˆ’t\right)}.$

We compute

${k}_{1}^{â€²}\left(t\right)=\frac{{e}^{p\left(tcothtâˆ’1\right)}}{{\left(cosht\right)}^{p}}â‹\dots \frac{{u}_{1}\left(t\right)}{{\left(sinht\right)}^{3}{\left(sinhtcoshtâˆ’t\right)}^{2}},$

where

$\begin{array}{rcl}{u}_{1}\left(t\right)& =& 2{t}^{2}{sinh}^{4}tcosht+{sinh}^{4}tcoshtâˆ’4t{sinh}^{5}t\\ âˆ’3t{sinh}^{3}t+3{t}^{2}{sinh}^{2}tcoshtâˆ’{t}^{3}sinht\\ âˆ’p\left(t{sinh}^{5}t+2t{sinh}^{3}t+{t}^{4}coshtâˆ’{sinh}^{4}tcoshtâˆ’{t}^{3}{sinh}^{3}tâˆ’2{t}^{3}sinht\right)\\ =& 2{t}^{2}{sinh}^{4}tcosht+{sinh}^{4}tcoshtâˆ’4t{sinh}^{5}t\\ âˆ’3t{sinh}^{3}t+3{t}^{2}{sinh}^{2}tcoshtâˆ’{t}^{3}sinhtâˆ’pD\left(t\right).\end{array}$

If $0, by Lemma 2.2 we have

$\begin{array}{rcl}5{u}_{1}\left(t\right)& â‰¥& 10{t}^{2}{sinh}^{4}tcosht+13{sinh}^{4}tcoshtâˆ’28t{sinh}^{5}t\\ âˆ’46t{sinh}^{3}t+30{t}^{2}{sinh}^{2}tcosht+6{t}^{3}sinhtâˆ’8{t}^{4}cosht+8{t}^{3}{sinh}^{3}t\\ =& \underset{n=3}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{h}_{n}}{16\left(2n+4\right)!}{t}^{2n+4},\end{array}$

where

$\begin{array}{rcl}{h}_{n}& =& 10\left(2n+4\right)\left(2n+3\right)\left({5}^{2n+2}âˆ’3â‹\dots {3}^{2n+2}+2\right)+13\left({5}^{2n+4}âˆ’3â‹\dots {3}^{2n+4}+2\right)\\ âˆ’28\left(2n+4\right)\left({5}^{2n+3}âˆ’5â‹\dots {3}^{2n+3}+10\right)âˆ’184\left(2n+4\right)\left({3}^{2n+3}âˆ’3\right)\\ +120\left(2n+4\right)\left(2n+3\right)\left({3}^{2n+2}âˆ’1\right)+96\left(2n+4\right)\left(2n+3\right)\left(2n+2\right)\left(2n+1\right)2n\\ âˆ’128\left(2n+4\right)\left(2n+3\right)\left(2n+2\right)\left(2n+1\right)+96\left(2n+4\right)\left(2n+3\right)\left(2n+2\right)\left({3}^{2n}âˆ’1\right)\\ =& \left(1,000{n}^{2}âˆ’3,500nâˆ’2,875\right){25}^{n}+\left(768{n}^{3}+6,696{n}^{2}+13,956n+4,113\right){9}^{n}\\ +\left(2n+4\right)\left(2n+3\right)\left(2n+2\right)\left(2n+1\right)\left(192nâˆ’128\right)\\ âˆ’96\left(2n+4\right)\left(2n+3\right)\left(2n+2\right)âˆ’100\left(2n+4\right)\left(2n+3\right)+272\left(2n+4\right)+26\\ >& 0\end{array}$

for $n=3,4,â€¦$â€‰.

We have ${u}_{1}\left(t\right)>0$ for $0. So, ${k}_{1}^{â€²}\left(t\right)>0$ for $t>0$, and ${f}_{1}^{â€²}\left(t\right)/{g}_{1}^{â€²}\left(t\right)={k}_{1}\left(t\right)$ is increasing on $\left(0,+\mathrm{âˆž}\right)$. Hence, $F\left(t\right)$ is increasing on $\left(0,+\mathrm{âˆž}\right)$ by Lemma 2.1. At the same time, ${lim}_{tâ†’{0}^{+}}F\left(t\right)=1/2$ and ${lim}_{tâ†’+\mathrm{âˆž}}F\left(t\right)={\left(2/e\right)}^{p}$. So, the proof of Theorem 1.3 is complete.

## 4 Proof of Theorem 1.4

Let

$S\left(t\right)â‰¡\frac{{\left(\frac{sinht}{t}{e}^{1âˆ’tcotht}\right)}^{p}âˆ’1}{{e}^{p\left(1âˆ’tcotht\right)}âˆ’1}=\frac{{f}_{2}\left(t\right)âˆ’{f}_{2}\left({0}^{+}\right)}{{g}_{2}\left(t\right)âˆ’{g}_{2}\left({0}^{+}\right)},$

where ${f}_{2}\left(t\right)={\left(\frac{sinht}{t}{e}^{1âˆ’tcotht}\right)}^{p}$ and ${g}_{2}\left(t\right)={e}^{p\left(1âˆ’tcotht\right)}$. Then

${k}_{2}\left(t\right)â‰œ\frac{{f}_{2}^{â€²}\left(t\right)}{{g}_{2}^{â€²}\left(t\right)}={\left(\frac{sinht}{t}\right)}^{pâˆ’1}\frac{{\left(sinht\right)}^{3}âˆ’{t}^{2}sinht}{{t}^{2}\left(sinhtcoshtâˆ’t\right)},$

and

${k}_{2}^{â€²}\left(t\right)={\left(\frac{sinht}{t}\right)}^{pâˆ’2}\frac{{u}_{2}\left(t\right)}{{t}^{4}{\left(sinhtcoshtâˆ’t\right)}^{2}},$

where

where ${e}_{n}=1âˆ’\left({d}_{n}/{c}_{n}\right)$ and

Let

Then

${e}_{n}=1âˆ’\frac{{d}_{n}}{{c}_{n}}=\frac{j\left(n\right)}{i\left(n\right)}.$

Let $\mathrm{Î”}\left(n\right)=286i\left(n\right)âˆ’693j\left(n\right)$. Then

$\begin{array}{rcl}\mathrm{Î”}\left(n\right)& =& \left(741,313nâˆ’5,759,424\right){36}^{n}+{16}^{n}\left[2,275,328\left(2n+5\right)+4,009,984\\ +70,400\left(2n+5\right)\left(2n+4\right)\left(2n+3\right)âˆ’532,224\left(2n+5\right)\left(2n+4\right)\right]\\ +{4}^{n}\left[9,152\left(2n+5\right)\left(2n+4\right)\left(2n+3\right)\left(2n+2\right)âˆ’62,656\left(2n+5\right)\left(2n+4\right)\left(2n+3\right)\\ +133,056\left(2n+5\right)\left(2n+4\right)âˆ’610,016\left(2n+5\right)âˆ’156,640\right].\end{array}$

First, we check that $\mathrm{Î”}\left(n\right)>0$ for $n=3,4,5,6,7$; second, we can easily obtain that $\mathrm{Î”}\left(n\right)>0$ for $nâ‰¥8$. So, we have that $\mathrm{Î”}\left(n\right)>0$ for $n=3,4,â€¦$â€‰.

So, we have ${u}_{2}\left(t\right)>0$ for $pâ‰¥286/693$. So, ${k}_{2}^{â€²}\left(t\right)>0$ for $t>0$, and ${f}_{2}^{â€²}\left(t\right)/{g}_{2}^{â€²}\left(t\right)={k}_{2}\left(t\right)$ is increasing on $\left(0,+\mathrm{âˆž}\right)$. Hence, $S\left(t\right)$ is increasing on $\left(0,+\mathrm{âˆž}\right)$ by Lemma 2.1 when $pâ‰¥286/693$. At the same time, ${lim}_{tâ†’{0}^{+}}S\left(t\right)=1/2$ and ${lim}_{tâ†’+\mathrm{âˆž}}S\left(t\right)=1$. So, the proof of Theorem 1.4 is complete.

## 5 Applications of theorems

In this section, we assume that x and y are two different positive numbers. Let $A\left(x,y\right)$, $G\left(x,y\right)$, $L\left(x,y\right)$, and $I\left(x,y\right)$ be the arithmetic, geometric, logarithmic, and identric means, respectively. Without loss of generality, we set $0. By the transformation $t=\left(log\left(y/x\right)\right)/2$, we can compute and obtain

where $t>0$.

Now, the four results in Section 1 are equivalent to the following ones for four classical means.

Theorem 5.1 Let $pâ‰¥4/5$, and x and y be positive real numbers with . Then

$\mathrm{Î±}{A}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î±}\right){G}^{p}\left(x,y\right)<{L}^{p}\left(x,y\right)<\mathrm{Î²}{A}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î²}\right){G}^{p}\left(x,y\right)$
(5.1)

holds if and only if $\mathrm{Î±}â‰¤0$ and $\mathrm{Î²}â‰¥1/3$.

Theorem 5.1 can deduce the following one, which is from Zhu [8].

Corollary 5.2 ([[8], Theorem 1])

Let $pâ‰¥1$, and x and y be positive real numbers with . Then

$\mathrm{Î±}{A}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î±}\right){G}^{p}\left(x,y\right)<{L}^{p}\left(x,y\right)<\mathrm{Î²}{A}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î²}\right){G}^{p}\left(x,y\right)$
(5.2)

holds if and only if $\mathrm{Î±}â‰¤0$ and $\mathrm{Î²}â‰¥1/3$.

When letting $p=1$ in Theorem 5.1, one can obtain the result (see [12â€“14], [[15], Theorem 1]).

Corollary 5.3 Let x and y be positive real numbers with . Then

$\mathrm{Î±}A\left(x,y\right)+\left(1âˆ’\mathrm{Î±}\right)G\left(x,y\right)
(5.3)

holds if and only if $\mathrm{Î±}â‰¤0$ and $\mathrm{Î²}â‰¥1/3$.

When letting $\mathrm{Î²}=1/3$ in the right-hand inequality of (5.3), one can obtain the well-known inequality by Carlson [16]

$L\left(x,y\right)<\frac{1}{3}A\left(x,y\right)+\frac{2}{3}G\left(x,y\right).$
(5.4)

Theorem 5.4 Let $p>0$. Then

(1) if $0, the double inequality

$\mathrm{Î±}{A}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î±}\right){G}^{p}\left(x,y\right)<{I}^{p}\left(x,y\right)<\mathrm{Î²}{A}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î²}\right){G}^{p}\left(x,y\right)$
(5.5)

holds if and only if $\mathrm{Î±}â‰¤2/3$ and $\mathrm{Î²}â‰¥{\left(2/e\right)}^{p}$;

(2) if $pâ‰¥2$, the double inequality

$\mathrm{Î±}{A}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î±}\right){G}^{p}\left(x,y\right)<{I}^{p}\left(x,y\right)<\mathrm{Î²}{A}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î²}\right){G}^{p}\left(x,y\right)$
(5.6)

holds if and only if $\mathrm{Î±}â‰¤{\left(2/e\right)}^{p}$ and $\mathrm{Î²}â‰¥2/3$.

The part (2) of Theorem 5.4 is a result of Trif [17].

When letting $p=2$ and $\mathrm{Î²}=2/3$ in the right-hand inequality of (5.6), one can obtain the following result, which is from SÃ¡ndor and Trif [18].

${I}^{2}\left(x,y\right)<\frac{2}{3}{A}^{2}\left(x,y\right)+\frac{1}{3}{G}^{2}\left(x,y\right).$
(5.7)

When letting $p=1$ in the double inequality (5.5), one can obtain the following result (see [12], [[15], Theorem 2]).

Corollary 5.5 Let x and y be positive real numbers with . Then

$\mathrm{Î±}A\left(x,y\right)+\left(1âˆ’\mathrm{Î±}\right)G\left(x,y\right)
(5.8)

holds if and only if $\mathrm{Î±}â‰¤2/3$ and $\mathrm{Î²}â‰¥2/e$.

When letting $\mathrm{Î±}=2/3$ in the left-hand inequality in (5.8), one can obtain the following result, which is from SÃ¡ndor [19].

$\frac{2}{3}A\left(x,y\right)+\frac{1}{3}G\left(x,y\right)
(5.9)

Theorem 5.6 Let $0, x and y be positive real numbers with . Then

$\mathrm{Î±}{A}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î±}\right){L}^{p}\left(x,y\right)<{I}^{p}\left(x,y\right)<\mathrm{Î²}{A}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î²}\right){L}^{p}\left(x,y\right)$
(5.10)

holds if and only if $\mathrm{Î±}â‰¤1/2$ and $\mathrm{Î²}â‰¥{\left(2/e\right)}^{p}$.

Theorem 5.6 can deduce the following result (see Zhu [15]).

Corollary 5.7 ([[15], Theorem 3])

Let x and y be positive real numbers with . Then

$\mathrm{Î±}A\left(x,y\right)+\left(1âˆ’\mathrm{Î±}\right)L\left(x,y\right)
(5.11)

holds if and only if $\mathrm{Î±}â‰¤1/2$ and $\mathrm{Î²}â‰¥2/e$.

When letting $\mathrm{Î±}=1/2$ in the left-hand inequality of (5.11), one can obtain the following result, which is from SÃ¡ndor [4, 19].

$I\left(x,y\right)>\frac{A\left(x,y\right)+L\left(x,y\right)}{2}.$
(5.12)

Finally, we give the bounds for ${L}^{p}\left(x,y\right)$ in terms of ${G}^{p}\left(x,y\right)$ and ${I}^{p}\left(x,y\right)$, and obtain the following new result.

Theorem 5.8 Let x and y be positive real numbers with , and $pâ‰¥286/693$. Then

$\mathrm{Î±}{G}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î±}\right){I}^{p}\left(x,y\right)<{L}^{p}\left(x,y\right)<\mathrm{Î²}{G}^{p}\left(x,y\right)+\left(1âˆ’\mathrm{Î²}\right){I}^{p}\left(x,y\right)$
(5.13)

holds if and only if $\mathrm{Î²}â‰¤1/2$ and $\mathrm{Î±}â‰¥1$.

Theorem 5.8 can deduce a result of Zhu [15]:

Corollary 5.9 ([[15], Theorem 4])

Let x and y be positive real numbers with . Then

$\mathrm{Î±}G\left(x,y\right)+\left(1âˆ’\mathrm{Î±}\right)I\left(x,y\right)
(5.14)

holds if and only if $\mathrm{Î²}â‰¤1/2$ and $\mathrm{Î±}â‰¥1$.

Obviously, the right-hand side of (5.14) is an extension of the following inequality:

$L\left(x,y\right)<\frac{1}{2}\left(G\left(x,y\right)+I\left(x,y\right)\right),$
(5.15)

which was given by Alzer [5].

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Zhu, L. New inequalities for hyperbolic functions and their applications. J Inequal Appl 2012, 303 (2012). https://doi.org/10.1186/1029-242X-2012-303