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 $\{1,(sinht)/t,exp(tcotht-1),cosht\}$. 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 [1–6])

$$1<\frac{sinht}{t}<e^{tcotht-1}<cosht$$

(1.1)

holds for all $t\ne 0$.

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

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

Let $p\ge 4/5$, and $t\ne 0$. Then the double inequality

$$(1-\lambda )+\lambda {(cosht)}^p<{\left(\frac{sinht}{t}\right)}^p<(1-\eta )+\eta {(cosht)}^p$$

(1.2)

holds if and only if $\eta \ge 1/3$ and $\lambda \le 0$.

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

Theorem 1.2 ([[8], Theorem 2])

Let $p>0$, and $t\ne 0$. Then

(1) if $0<p\le 6/5$, the double inequality

$$\alpha {(cosht)}^p+(1-\alpha )<e^{p(tcotht-1)}<\beta {(cosht)}^p+(1-\beta )$$

(1.3)

holds if and only if $\alpha \le 2/3$ and $\beta \ge {(2/e)}^p$;

(2) if $p\ge 2$, the double inequality

$$\alpha {(cosht)}^p+(1-\alpha )<e^{p(tcotht-1)}<\beta {(cosht)}^p+(1-\beta )$$

(1.4)

holds if and only if $\alpha \le {(2/e)}^p$ and $\beta \ge 2/3$.

Next, we do the work for each of the triples $\{(sinht)/t,exp(tcotht-1),cosht\}$ and $\{1,(sinht)/t,exp(tcotht-1)\}$, and obtain the following two new results.

Theorem 1.3 Let $0<p\le 8/5$, and $t\ne 0$. Then

$$\alpha {(cosht)}^p+(1-\alpha ){\left(\frac{sinht}{t}\right)}^p<e^{p(tcotht-1)}<\beta {(cosht)}^p+(1-\beta ){\left(\frac{sinht}{t}\right)}^p$$

(1.5)

holds if and only if $\alpha \le 1/2$ and $\beta \ge {(2/e)}^p$.

Theorem 1.4 Let $p\ge 286/693$, and $t\ne 0$. Then

$$\alpha +(1-\alpha )e^{p(tcotht-1)}<{\left(\frac{sinht}{t}\right)}^p<\beta +(1-\beta )e^{p(tcotht-1)}$$

(1.6)

holds if and only if $\beta \le 1/2$ and $\alpha \ge 1$.

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 ([9–11])

Let $f,g:[a,b]\to \mathbb{R}$ be two continuous functions which are differentiable on $(a,b)$. Further, let $g^{\prime }\ne 0$ on $(a,b)$. If $f^{\prime }/g^{\prime }$ is increasing (or decreasing) on $(a,b)$, then the functions $(f(x)-f(b^{-}))/(g(x)-g(b^{-}))$ and $(f(x)-f(a^{+}))/(g(x)-g(a^{+}))$ are also increasing (or decreasing) on $(a,b)$.

Lemma 2.2 Let $t\in (0,+\mathrm{\infty })$. Then the inequality

$$D(t)\triangleq t{sinh}^{5}t+2t{sinh}^{3}t+t^{4}cosht-{sinh}^{4}tcosht-t^3{sinh}^{3}t-2t^{3}sinht>0$$

holds.

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

$$\begin{array}{rcl}D(t)& =& \frac{1}{16}t(sinh5t-5sinh3t+10sinht)+\frac{1}{2}t(sinh3t-3sinht)+t^{4}cosht\\ -\frac{1}{16}(cosh5t-3cosh3t+2cosht)-\frac{1}{4}{t}^3(sinh3t-3sinht)-2t^{3}sinht\\ =& \frac{1}{16}\sum _{n=0}^{\mathrm{\infty }}\frac{5^{2n+1}-5\cdot 3^{2n+1}+10}{(2n+2)!}{t}^{2n+1}+\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{3^{2n+1}-3}{(2n+1)!}{t}^{2n+2}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n)!}{t}^{2n+4}\\ -\frac{1}{16}\sum _{n=0}^{\mathrm{\infty }}\frac{5^{2n}-3\cdot 3^{2n}+2}{(2n)!}{t}^{2n}-\frac{1}{4}\sum _{n=0}^{\mathrm{\infty }}\frac{3^{2n+1}-3}{(2n+1)!}{t}^{2n+4}-2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1)!}{t}^{2n+4}\\ =& \frac{1}{16}\sum _{n=3}^{\mathrm{\infty }}\frac{l_n}{(2n+4)!}{t}^{2n+4},\end{array}$$

where

$$\begin{array}{rcl}{l}_n& =& (2n+4)(5^{2n+3}-5\cdot 3^{2n+3}+10)+8(2n+4)(3^{2n+3}-3)\\ +16(2n+4)(2n+3)(2n+2)(2n+1)-(5^{2n+4}-3\cdot 3^{2n+4}+2)\\ -4(2n+4)(2n+3)(2n+2)(3^{2n+1}-3)-32(2n+4)(2n+3)(2n+2)\\ =& (250n-125){25}^n+(279-462n-432n^2-96n^3)9^n\\ +256n^4+1,120n^3+1,520n^2+532n-154,n=3,4,\dots .\end{array}$$

Using a basic differential method, we can easily prove

$$\begin{array}{rcl}f(x)& \triangleq & (250x-125){25}^x+(279-462x-432x^2-96x^3)9^x\\ +256x^4+1,120x^3+1,520x^2+532x-154>0\end{array}$$

on $[3,\mathrm{\infty })$. This leads to $l_n>0$ for $n=3,4,\dots$ , and $D(t)>0$. So, the proof of Lemma 2.2 is complete. □

3 Proof of Theorem 1.3

Let

$$F(t)\equiv \frac{{(\frac{t}{sinht}{e}^{tcotht-1})}^p-1}{{(tcotht)}^p-1}=\frac{f_1(t)-f_1(0^{+})}{g_1(t)-g_1(0^{+})},$$

where $f_1(t)={(\frac{t}{sinht}{e}^{tcotht-1})}^p$ and $g_1(t)={(tcotht)}^p$. Then

$$k_1(t)\triangleq \frac{f_1^{\prime }(t)}{g_1^{\prime }(t)}=\frac{e^{p(tcotht-1)}}{{(cosht)}^{p-1}}\cdot \frac{{sinh}^{2}t-t^2}{sinht(sinhtcosht-t)}.$$

We compute

$$k_1^{\prime }(t)=\frac{e^{p(tcotht-1)}}{{(cosht)}^p}\cdot \frac{u_1(t)}{{(sinht)}^3{(sinhtcosht-t)}^2},$$

where

$$\begin{array}{rcl}{u}_1(t)& =& 2t^2{sinh}^{4}tcosht+{sinh}^{4}tcosht-4t{sinh}^{5}t\\ -3t{sinh}^{3}t+3t^2{sinh}^{2}tcosht-t^{3}sinht\\ -p(t{sinh}^{5}t+2t{sinh}^{3}t+t^{4}cosht-{sinh}^{4}tcosht-t^3{sinh}^{3}t-2t^{3}sinht)\\ =& 2t^2{sinh}^{4}tcosht+{sinh}^{4}tcosht-4t{sinh}^{5}t\\ -3t{sinh}^{3}t+3t^2{sinh}^{2}tcosht-t^{3}sinht-pD(t).\end{array}$$

If $0<p\le 8/5$, by Lemma 2.2 we have

$$\begin{array}{rcl}5u_1(t)& \ge & 10t^2{sinh}^{4}tcosht+13{sinh}^{4}tcosht-28t{sinh}^{5}t\\ -46t{sinh}^{3}t+30t^2{sinh}^{2}tcosht+6t^{3}sinht-8t^{4}cosht+8t^3{sinh}^{3}t\\ =& \sum _{n=3}^{\mathrm{\infty }}\frac{h_n}{16(2n+4)!}{t}^{2n+4},\end{array}$$

where

$$\begin{array}{rcl}{h}_n& =& 10(2n+4)(2n+3)(5^{2n+2}-3\cdot 3^{2n+2}+2)+13(5^{2n+4}-3\cdot 3^{2n+4}+2)\\ -28(2n+4)(5^{2n+3}-5\cdot 3^{2n+3}+10)-184(2n+4)(3^{2n+3}-3)\\ +120(2n+4)(2n+3)(3^{2n+2}-1)+96(2n+4)(2n+3)(2n+2)(2n+1)2n\\ -128(2n+4)(2n+3)(2n+2)(2n+1)+96(2n+4)(2n+3)(2n+2)(3^{2n}-1)\\ =& (1,000n^2-3,500n-2,875){25}^n+(768n^3+6,696n^2+13,956n+4,113)9^n\\ +(2n+4)(2n+3)(2n+2)(2n+1)(192n-128)\\ -96(2n+4)(2n+3)(2n+2)-100(2n+4)(2n+3)+272(2n+4)+26\\ >& 0\end{array}$$

for $n=3,4,\dots$ .

We have $u_1(t)>0$ for $0<p\le 8/5$. So, $k_1^{\prime }(t)>0$ for $t>0$, and $f_1^{\prime }(t)/g_1^{\prime }(t)=k_1(t)$ is increasing on $(0,+\mathrm{\infty })$. Hence, $F(t)$ is increasing on $(0,+\mathrm{\infty })$ by Lemma 2.1. At the same time, ${lim}_{t\to 0^{+}}F(t)=1/2$ and ${lim}_{t\to +\mathrm{\infty }}F(t)={(2/e)}^p$. So, the proof of Theorem 1.3 is complete.

4 Proof of Theorem 1.4

Let

$$S(t)\equiv \frac{{(\frac{sinht}{t}{e}^{1-tcotht})}^p-1}{e^{p(1-tcotht)}-1}=\frac{f_2(t)-f_2(0^{+})}{g_2(t)-g_2(0^{+})},$$

where $f_2(t)={(\frac{sinht}{t}{e}^{1-tcotht})}^p$ and $g_2(t)=e^{p(1-tcotht)}$. Then

$$k_2(t)\triangleq \frac{f_2^{\prime }(t)}{g_2^{\prime }(t)}={\left(\frac{sinht}{t}\right)}^{p-1}\frac{{(sinht)}^3-t^{2}sinht}{t^2(sinhtcosht-t)},$$

and

$$k_2^{\prime }(t)={\left(\frac{sinht}{t}\right)}^{p-2}\frac{u_2(t)}{t^4{(sinhtcosht-t)}^2},$$

where

where $e_n=1-(d_n/c_n)$ and

Let

Then

$$e_n=1-\frac{d_n}{c_n}=\frac{j(n)}{i(n)}.$$

Let $\mathrm{\Delta }(n)=286i(n)-693j(n)$. Then

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

First, we check that $\mathrm{\Delta }(n)>0$ for $n=3,4,5,6,7$; second, we can easily obtain that $\mathrm{\Delta }(n)>0$ for $n\ge 8$. So, we have that $\mathrm{\Delta }(n)>0$ for $n=3,4,\dots$ .

So, we have $u_2(t)>0$ for $p\ge 286/693$. So, $k_2^{\prime }(t)>0$ for $t>0$, and $f_2^{\prime }(t)/g_2^{\prime }(t)=k_2(t)$ is increasing on $(0,+\mathrm{\infty })$. Hence, $S(t)$ is increasing on $(0,+\mathrm{\infty })$ by Lemma 2.1 when $p\ge 286/693$. At the same time, ${lim}_{t\to 0^{+}}S(t)=1/2$ and ${lim}_{t\to +\mathrm{\infty }}S(t)=1$. 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 $A(x,y)$, $G(x,y)$, $L(x,y)$, and $I(x,y)$ be the arithmetic, geometric, logarithmic, and identric means, respectively. Without loss of generality, we set $0<x<y$. By the transformation $t=(log(y/x))/2$, 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 $p\ge 4/5$, and x and y be positive real numbers with $x\ne y$. Then

$$\alpha A^p(x,y)+(1-\alpha )G^p(x,y)<L^p(x,y)<\beta A^p(x,y)+(1-\beta )G^p(x,y)$$

(5.1)

holds if and only if $\alpha \le 0$ and $\beta \ge 1/3$.

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

Corollary 5.2 ([[8], Theorem 1])

Let $p\ge 1$, and x and y be positive real numbers with $x\ne y$. Then

$$\alpha A^p(x,y)+(1-\alpha )G^p(x,y)<L^p(x,y)<\beta A^p(x,y)+(1-\beta )G^p(x,y)$$

(5.2)

holds if and only if $\alpha \le 0$ and $\beta \ge 1/3$.

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

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

$$\alpha A(x,y)+(1-\alpha )G(x,y)<L(x,y)<\beta A(x,y)+(1-\beta )G(x,y)$$

(5.3)

holds if and only if $\alpha \le 0$ and $\beta \ge 1/3$.

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

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

(5.4)

Theorem 5.4 Let $p>0$. Then

(1) if $0<p\le 6/5$, the double inequality

$$\alpha A^p(x,y)+(1-\alpha )G^p(x,y)<I^p(x,y)<\beta A^p(x,y)+(1-\beta )G^p(x,y)$$

(5.5)

holds if and only if $\alpha \le 2/3$ and $\beta \ge {(2/e)}^p$;

(2) if $p\ge 2$, the double inequality

$$\alpha A^p(x,y)+(1-\alpha )G^p(x,y)<I^p(x,y)<\beta A^p(x,y)+(1-\beta )G^p(x,y)$$

(5.6)

holds if and only if $\alpha \le {(2/e)}^p$ and $\beta \ge 2/3$.

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

When letting $p=2$ and $\beta =2/3$ in the right-hand inequality of (5.6), one can obtain the following result, which is from Sándor and Trif [18].

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

(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 $x\ne y$. Then

$$\alpha A(x,y)+(1-\alpha )G(x,y)<I(x,y)<\beta A(x,y)+(1-\beta )G(x,y)$$

(5.8)

holds if and only if $\alpha \le 2/3$ and $\beta \ge 2/e$.

When letting $\alpha =2/3$ in the left-hand inequality in (5.8), one can obtain the following result, which is from Sándor [19].

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

(5.9)

Theorem 5.6 Let $0<p\le 8/5$, x and y be positive real numbers with $x\ne y$. Then

$$\alpha A^p(x,y)+(1-\alpha )L^p(x,y)<I^p(x,y)<\beta A^p(x,y)+(1-\beta )L^p(x,y)$$

(5.10)

holds if and only if $\alpha \le 1/2$ and $\beta \ge {(2/e)}^p$.

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 $x\ne y$. Then

$$\alpha A(x,y)+(1-\alpha )L(x,y)<I(x,y)<\beta A(x,y)+(1-\beta )L(x,y)$$

(5.11)

holds if and only if $\alpha \le 1/2$ and $\beta \ge 2/e$.

When letting $\alpha =1/2$ in the left-hand inequality of (5.11), one can obtain the following result, which is from Sándor [4, 19].

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

(5.12)

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

Theorem 5.8 Let x and y be positive real numbers with $x\ne y$, and $p\ge 286/693$. Then

$$\alpha G^p(x,y)+(1-\alpha )I^p(x,y)<L^p(x,y)<\beta G^p(x,y)+(1-\beta )I^p(x,y)$$

(5.13)

holds if and only if $\beta \le 1/2$ and $\alpha \ge 1$.

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 $x\ne y$. Then

$$\alpha G(x,y)+(1-\alpha )I(x,y)<L(x,y)<\beta G(x,y)+(1-\beta )I(x,y)$$

(5.14)

holds if and only if $\beta \le 1/2$ and $\alpha \ge 1$.

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

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

(5.15)

which was given by Alzer [5].