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Some Wilker and Cusa type inequalities for generalized trigonometric and hyperbolic functions

Journal of Inequalities and Applications20182018:52

https://doi.org/10.1186/s13660-018-1644-8

Received: 6 December 2017

Accepted: 22 February 2018

Published: 2 March 2018

Abstract

The authors obtain some Wilker and Cusa type inequalities for generalized trigonometric and hyperbolic functions and generalize some known inequalities.

Keywords

Wilker type inequalityCusa type inequalityArithmetic–geometric means inequalityPower means inequalityGeneralized trigonometric functionGeneralized hyperbolic function

MSC

33B10

1 Introduction

It is well known from basic calculus that
$$ \arcsin x= \int_{0}^{x} \frac{1}{(1-t^{2})^{1/2}}\,dt $$
(1.1)
for \(0\leq x\leq1\) and
$$ \frac{\pi}{2}=\arcsin1= \int_{0}^{1} \frac{1}{(1-t^{2})^{1/2}}\,dt. $$
(1.2)
For \(1< p<\infty\) and \(0 \leq x \leq1\), the arc sine may be generalized as
$$ \arcsin_{p} x = \int_{0}^{x} {\frac{1}{(1-t^{p})^{1/p}}}\,dt $$
(1.3)
and
$$ \frac{\pi_{p}}{2} = \arcsin_{p} 1= \int_{0}^{1} {\frac{1}{ (1-t^{p})^{1/p}}}\,dt. $$
(1.4)

The inverse of \(\arcsin_{p}\) on \([0,\frac{\pi_{p}}{2} ]\) is called the generalized sine function, denoted by \(\sin_{p}\), and may be extended to \((-\infty, \infty)\). In the same way, we can define the generalized cosine function, the generalized tangent function, and their inverses, and also the corresponding hyperbolic functions. For their definitions and formulas, one may see recent references [13].

In [2], some classical inequalities for generalized trigonometric and hyperbolic functions, such as Mitrinović–Adamović inequality, Huygens’ inequality, and Wilker’s inequality, were generalized. In [3], some new second Wilker type inequalities for generalized trigonometric and hyperbolic functions were established. In [4], some Turán type inequalities for generalized trigonometric and hyperbolic functions were presented. Very recently, a conjecture posed in [5] was verified in [1]. For more about the Wilker type inequality and Huygens type inequalities, the reader may see [613].

In this paper, we establish some new Wilker and Cusa type inequalities for the generalized trigonometric and hyperbolic functions. Some known inequalities in [3] are the special cases of our results.

2 Lemmas

Lemma 2.1

([3, Lemma 2.7])

For \(p \in(1,\infty)\), we have
$$ \cos_{p}^{\alpha}x < \frac{\sin_{p} x}{x}< 1,\quad x\in \biggl(0,\frac{\pi _{p}}{2} \biggr) $$
(2.1)
and
$$ \cosh_{p}^{\alpha}x< \frac{\sinh_{p} x}{x}< \cosh_{p}^{\beta}x,\quad x>0, $$
(2.2)
where the constants \(\alpha=\frac{1}{p+1}\) and \(\beta=1\) are the best possible.

Lemma 2.2

([3, Theorem 3.5])

For \(p \in(1,2]\), then
$$ \biggl(\frac{x}{\sin_{p} x} \biggr)^{p} + \frac{x}{\tan_{p} x} > 2,\quad x\in \biggl(0,\frac{\pi_{p}}{2} \biggr). $$
(2.3)

Lemma 2.3

([14])

Let \(a>0, b>0\) and \(r\geq1\), then
$$ (a+b)^{r} \leq2^{r-1} \bigl(a^{r}+b^{r} \bigr). $$
(2.4)

Lemma 2.4

([15])

Let \(a_{k}>0, k=1,2,\ldots,n\), then
$$ \frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\geq \sqrt[n]{(1+a_{1}) (1+a_{2})\cdots(1+a_{n})} -1\geq\sqrt[n]{a_{1} a_{2} \cdots a_{n}}. $$
(2.5)

Lemma 2.5

([2, Theorem 3.4])

For \(p\in[2,\infty)\) and \(x\in (0,\frac{\pi_{p}}{2} )\), then
$$ \frac{\sin_{p} x}{x}< \frac{x}{\sinh_{p} x}. $$
(2.6)

Lemma 2.6

For \(p\in[2,\infty)\) and \(x\in (0,\frac{\pi_{p}}{2} )\), we have
$$ \biggl(\frac{\sin_{p} x}{x} \biggr)^{p}< \frac{x}{\sinh_{p} x}. $$
(2.7)

Proof

Using Lemma 2.5 and \(\frac{{\sin_{p} x}}{x} < 1\), we have
$$ \frac{x}{{\sinh_{p} x}} > \frac{{\sin_{p} x}}{x} > \biggl( { \frac{{\sin _{p} x}}{x}} \biggr)^{p}. $$
(2.8)
This implies inequality (2.7). □

Lemma 2.7

([2, Corollary 3.10])

For \(p\in[2,\infty)\) and \(x\in (0,\frac{\pi_{p}}{2} )\), then
$$ \biggl( \frac{x}{{\sinh_{p} x}} \biggr)^{p+1}< { \frac{{\sin_{p} x}}{x}}. $$
(2.9)

Lemma 2.8

([2, Theorem 3.22])

For \(p\in(1,2]\), the double inequality
$$ \frac{\sin_{p} x}{x}< \frac{\cos_{p} x+p}{1+p}\le\frac{\cos_{p} x+2}{3} $$
(2.10)
holds for all \(x\in (0,\frac{\pi_{p}}{2} ]\).

3 Main results

Theorem 3.1

For \(x\in (0,\frac{\pi_{p}}{2} )\), \(p\in(1,\infty)\), and \(\alpha-p\beta\leq0\), \(\beta>0\), we have
$$ \biggl(\frac{\sin_{p} x}{x} \biggr)^{\alpha}+ \biggl( \frac{\tan_{p} x}{x} \biggr)^{\beta}> 2. $$
(3.1)

Proof

From the arithmetic geometric means inequality and Lemma 2.1, it follows that
$$\begin{aligned} \biggl(\frac{\sin_{p} x}{x} \biggr)^{\alpha}+ \biggl(\frac{\tan_{p} x}{x} \biggr)^{\beta}&\geq2 \biggl(\frac{\sin_{p} x}{x} \biggr)^{\frac{\alpha}{2}} \biggl(\frac{\tan_{p} x}{x} \biggr)^{\frac{\beta}{2}} \\ & = 2 \biggl(\frac{\sin_{p} x}{x} \biggr)^{\frac{\alpha+\beta}{2}} \biggl(\frac{1}{\cos_{p} x} \biggr)^{\frac{\beta}{2}} \\ & > 2 \biggl(\frac{\sin_{p} x}{x} \biggr)^{\frac{\alpha+\beta}{2}} \biggl(\frac{\sin_{p} x}{x} \biggr)^{-\frac{(p+1)\beta}{2}} \\ &= 2 \biggl(\frac{\sin_{p} x}{x} \biggr)^{\frac{\alpha-p\beta}{2}} \\ &\geq2. \end{aligned}$$
 □

Remark 3.1

If \(p=\alpha=2, \beta=1\), inequality (3.1) turns into
$$ \biggl(\frac{\sin x}{x} \biggr)^{2}+ \frac{\tan x}{x} > 2. $$
(3.2)
Inequality (3.2) is called the first Wilker inequality in [16].

Remark 3.2

If \(\alpha=2p, \beta=p\), and \(p\geq2\), then \(\alpha-p\beta=2p-p^{2}\leq0\). So, inequality (3.1) reduces to
$$ \biggl(\frac{\sin_{p} x}{x} \biggr)^{2p}+ \biggl( \frac{\tan_{p} x}{x} \biggr)^{p} > 2. $$
(3.3)

Theorem 3.2

For \(p\in(1,2], x\in (0,\frac{\pi_{p}}{2} )\), and \(\alpha-p\beta\leq0, \beta\leq-1\), we have
$$ \biggl(\frac{\sin_{p} x}{x} \biggr)^{\alpha}+ \biggl( \frac{\tan_{p} x}{x} \biggr)^{\beta}>2. $$
(3.4)

Proof

Using \(\frac{x}{\sin_{p} x}\geq1\) and \(\alpha-p\beta\leq0\), we have
$$\begin{aligned} \biggl(\frac{\sin_{p} x}{x} \biggr)^{\alpha}+ \biggl(\frac{\tan_{p} x}{x} \biggr)^{\beta} &= \biggl(\frac{x}{\sin_{p} x} \biggr)^{-\alpha}+ \biggl(\frac{x}{\tan_{p} x} \biggr)^{-\beta} \\ &= \biggl(\frac{x}{\sin_{p} x} \biggr)^{-p\beta} \biggl(\frac{x}{\sin_{p} x} \biggr)^{p\beta-\alpha}+ \biggl(\frac{x}{\tan_{p} x} \biggr)^{-\beta} \\ &\geq \biggl[ \biggl(\frac{x}{\sin_{p} x} \biggr)^{p} \biggr]^{-\beta}+ \biggl(\frac{x}{\tan_{p} x} \biggr)^{-\beta}. \end{aligned}$$
Applying Lemmas 2.2 and 2.3, we obtain
$$\begin{aligned} \biggl(\frac{\sin_{p} x}{x} \biggr)^{\alpha}+ \biggl(\frac{\tan_{p} x}{x} \biggr)^{\beta} \geq2^{1+\beta} \biggl[ \biggl(\frac{x}{\sin_{p} x} \biggr)^{p}+\frac {x}{\tan_{p} x} \biggr]^{-\beta} >2. \end{aligned}$$
This completes the proof. □

Using the same method as that in Theorem 3.1, we can easily obtain the following Theorem 3.3 by Lemma 2.1 and the arithmetic and geometric means inequality. We omit the proof for the sake of simplicity.

Theorem 3.3

For \(p\in(1,\infty), x\in (0,\infty)\), and \(\alpha-p\beta\leq0, \beta>0\), then
$$ \biggl(\frac{\sinh_{p} x}{x} \biggr)^{\alpha}+ \biggl( \frac{\tanh_{p} x}{x} \biggr)^{\beta}>2. $$
(3.5)

Remark 3.3

Taking \(\alpha=2, \beta=1\) and \(p=2\) in inequality (3.5), we have
$$ \biggl(\frac{\sinh x}{x} \biggr)^{2}+ \frac{\tanh x}{x} > 2, $$
(3.6)
which is the (4) in Theorem 1 of [7]. Inequality (3.6) is called the first hyperbolic Wilker inequality.

Remark 3.4

Taking \(\alpha=2p, \beta=p\), and \(p\in[2,\infty)\), we have
$$ \biggl(\frac{\sinh_{p} x}{x} \biggr)^{2p}+ \biggl( \frac{\tanh_{p} x}{x} \biggr)^{p} > 2. $$
(3.7)

Theorem 3.4

For all \(x\in (0,\frac{\pi_{p}}{2} )\) and \(\alpha-p\beta\leq0, \beta>0\), we have
$$ \biggl[1+ \biggl(\frac{\sin_{p}x}{x} \biggr)^{\alpha}\biggr] \biggl[1+ \biggl(\frac {\tan_{p}x}{x} \biggr)^{\beta}\biggr] >4 $$
(3.8)
and
$$ \biggl(\frac{\sin_{p}x}{x} \biggr)^{\alpha}+ \biggl( \frac{\tan_{p}x}{x} \biggr)^{\beta}>2\sqrt{ \biggl[1+ \biggl( \frac{\sin_{p}x}{x} \biggr)^{\alpha}\biggr] \biggl[1+ \biggl( \frac{\tan_{p}x}{x} \biggr)^{\beta}\biggr]} -2>2. $$
(3.9)

Proof

Setting \(n=2, a_{1}= (\frac{\sin_{p}x}{x} )^{\alpha}\) and \(a_{2}= (\frac{\tan_{p}x}{x} )^{\beta}\) in Lemma 2.4, we have
$$\begin{aligned} & \biggl[1+ \biggl(\frac{\sin_{p}x}{x} \biggr)^{\alpha}\biggr] \biggl[1+ \biggl(\frac {\tan_{p}x}{x} \biggr)^{\beta}\biggr] \\ &\quad \geq \biggl[ \biggl(\frac{\sin_{p}x}{x} \biggr)^{\frac{\alpha}{2}} \biggl( \frac{\tan_{p}x}{x} \biggr)^{\frac{\beta}{2}}+1 \biggr]^{2} \\ &\quad > \biggl[ \biggl(\frac{\sin_{p}x}{x} \biggr)^{\frac{\alpha-p\beta }{2}}+1 \biggr]^{2} \\ &\quad > 4. \end{aligned}$$
Then it follows from Lemma 2.1 that
$$ \biggl(\frac{\sin_{p}x}{x} \biggr)^{\alpha}+ \biggl(\frac{\tan_{p}x}{x} \biggr)^{\beta}>2\sqrt{ \biggl[1+ \biggl(\frac{\sin_{p}x}{x} \biggr)^{\alpha}\biggr] \biggl[1+ \biggl(\frac{\tan_{p}x}{x} \biggr)^{\beta}\biggr]}-2>2. $$
 □

Remark 3.5

If \(n=3\) and \(a_{1}=a_{2}= (\frac{\sin_{p}x}{x} )^{\alpha}, a_{3}= (\frac{\tan_{p}x}{x} )^{\beta}\) in Lemma 2.4, it can be easily obtained that
$$ \biggl[1+ \biggl(\frac{\sin_{p}x}{x} \biggr)^{\alpha}\biggr]^{2} \biggl[1+ \biggl(\frac{\tan_{p}x}{x} \biggr)^{\beta}\biggr]>8 $$
(3.10)
and
$$ 2 \biggl(\frac{\sin_{p}x}{x} \biggr)^{\alpha}+ \biggl( \frac{\tan_{p}x}{x} \biggr)^{\beta}>3 \sqrt[3]{ \biggl[1+ \biggl( \frac{\sin_{p}x}{x} \biggr)^{\alpha}\biggr]^{2} \biggl[1+ \biggl( \frac{\tan_{p}x}{x} \biggr)^{\beta}\biggr]} -3>3, $$
(3.11)
by a similar method to that in Theorem 3.4 when changing the condition \(\alpha-p\beta\leq0\) to \(2\alpha-p\beta\leq0\).

Theorem 3.5

For \(p\in[2,\infty), t>0\), and \(x\in (0,\frac{\pi_{p}}{2} ]\), then
$$ \biggl( {\frac{x}{{\sin_{p} x}}} \biggr)^{pt} + \biggl( { \frac{x}{{\sinh _{p} x}}} \biggr)^{t} > 2. $$
(3.12)

Proof

Applying the AGM inequality \(a+b\geq2\sqrt{ab} \) and Lemma 2.6 for \(a = ( {\frac{x}{{\sin_{p} x}}} )^{pt}\) and \(b = ( {\frac {x}{{\sinh_{p} x}}} )^{t}\), we obtain
$$a + b \ge2\sqrt{ \biggl( {\frac{x}{{\sin_{p} x}}} \biggr)^{pt} \biggl( { \frac{x}{{\sinh_{p} x}}} \biggr)^{t} } > 2. $$
The proof is completed. □

Theorem 3.6

For \(p\in[2,\infty), t>0\) and \(x\in (0,\frac{\pi_{p}}{2} ]\), then
$$ (p+1) \biggl( {\frac{x}{{\sin_{p} x}}} \biggr)^{t} + \biggl( {\frac{x}{{\sinh _{p} x}}} \biggr)^{t} > p + 1. $$
(3.13)

Proof

From the AGM inequality \((n+1)a+b\geq(n+1)\sqrt[n+1]{a^{n}b} \) and Lemma 2.6, for \(a = ( {\frac{x}{{\sin_{p} x}}} )^{t}\) and \(b = ( {\frac {x}{{\sinh_{p} x}}} )^{t}\), inequality (3.13) follows readily. □

Applying AGM inequality and Lemma 2.7, Theorems 3.7 and 3.8 can be easily obtained by the similar method as before.

Theorem 3.7

For \(p\in[2,\infty), t>0\), and \(x\in (0,\frac{\pi_{p}}{2} ]\), then
$$ \biggl( {\frac{\sinh_{p} x}{{x}}} \biggr)^{(p+1)t} + \biggl( { \frac{\sin _{p} x}{{x}}} \biggr)^{t} > 2. $$
(3.14)

Theorem 3.8

For \(p\in[2,\infty), t>0\), and \(x\in (0,\frac{\pi_{p}}{2} ]\), then
$$ (p+2) \biggl( {\frac{\sinh_{p} x}{{x}}} \biggr)^{t} + \biggl( {\frac{\sin_{p} x}{{x}}} \biggr)^{t} > p + 2. $$
(3.15)

Finally, we give a Cusa type inequality.

Theorem 3.9

For \(p\in(1,2]\) and \(x\in(0,\frac{\pi_{p}}{2}]\), the function \(f(x) = \frac{{\ln {\frac{{\sin_{p} x}}{x}}}}{{\ln {\frac{{p + \cos_{p} x}}{p+1}} }} \) is strictly increasing. Consequently, we have the following inequality:
$$ \biggl( {\frac{{p + \cos_{p} x}}{p+1}} \biggr)^{\alpha}< \frac{{\sin_{p} x}}{x} < \biggl( {\frac{{p + \cos_{p} x}}{p+1}} \biggr)^{\beta}$$
(3.16)
with the best constants \(\alpha = \frac{{\ln {\frac{{2\sin_{p} \frac{{\pi_{p} }}{2}}}{{\pi_{p} }}} }}{{\ln{\frac{{p + \cos_{p} \frac{{\pi_{p} }}{2}}}{p+1}} }} \) and \(\beta=1\).

Proof

A simple computation yields
$$\begin{aligned} &f'(x)\ln^{2} {\frac{{p + \cos_{p} x}}{p+1}} \\ &\quad= \frac{{x\cos_{p} x - \sin_{p} x}}{{x\sin_{p} x}}\ln{\frac{{p + \cos_{p} x}}{p+1}} + \frac{{\cos_{p} x\tan_{p}^{p - 1} x}}{{p + \cos_{p} x}}\ln { \frac{{\sin_{p} x}}{x}} \\ &\quad> {\frac{{x\cos_{p} x - \sin_{p} x}}{{x\sin_{p} x}} + \frac{{\cos_{p} x\tan_{p}^{p - 1} x}}{{p + \cos_{p} x}}}\ln{\frac{{\sin_{p} x}}{x}} \\ &\quad= {\frac{{(x\cos_{p} x - \sin_{p} x)(p + \cos_{p} x) + x\sin_{p} x\cos_{p} x\tan_{p}^{p - 1} x}}{{x\sin_{p} x( p+ \cos_{p} x)}}} \ln{\frac{{\sin_{p} x}}{x}} \\ &\quad= \frac{{\ln {\frac{{\sin_{p} x}}{x}} }}{{x\sin_{p} x( p+ \cos_{p} x)}}g(x), \end{aligned}$$
where
$$g(x) = x\cos_{p}^{2} x\sec_{p}^{p} x + px\cos_{p} x - p\sin_{p} x - \sin_{p} x \cos_{p} x. $$
Since
$$g'(x) = \cos_{p} x\tan_{p}^{p - 1} xh(x), $$
where
$$h(x) = 2\sin_{p} x - px - (2 - p)x\sec_{p}^{p - 1} x, $$
with
$$h'(x) = 2\cos_{p} x - p - (2 - p) \sec_{p}^{p - 1} x- (2 - p) (p - 1)x\sec _{p}^{p - 1} x\tan_{p}^{p - 1} x $$
and
$$\begin{aligned} h''(x) ={}& {-} 2\cos_{p} x\tan_{p}^{p - 1} x - 2(2 - p) (p - 1)\sec_{p}^{p - 1} x\tan_{p}^{p - 1} x \\ & {}- (2 - p) (p - 1)^{2} x\sec_{p}^{p - 1} x \tan_{p}^{p - 1} x \bigl(\tan_{p}^{p-1} x + \csc_{p} x\sec_{p}^{p - 1}x\bigr) < 0. \end{aligned}$$
Hence \(h'(x)\) is decreasing on \((0,\frac{\pi_{p}}{2})\). It then follows that \(h'(x)< h'(0)=0\), which also implies that \(h(x)< h(0)=0\). Hence, \(g'(x)<0\), which shows that the function \(g(x)\) is also decreasing on \((0,\frac{\pi_{p}}{2})\). The inequality \(g(x)< g(0)=0\) indicates that \(f'(x)>0\). Hence, \(f(x)\) is strictly increasing for \(x\in(0,\frac{\pi_{p}}{2})\). As a result, we have \(f(0)< f(x)\leq f(\frac{\pi_{p}}{2})\).
Using L’Hôspital’s rule, we obtain that
$$\begin{aligned} f\bigl(0^{+} \bigr)&= \mathop{\lim} _{x \to0^{+} } \frac{{\ln {\frac{{\sin_{p} x}}{x}} }}{{\ln {\frac{{p + \cos_{p} x}}{p+1}} }} \\ &= \mathop{\lim} _{x \to0^{+} } - \frac{{x\cos_{p} x - \sin_{p} x}}{{x\sin_{p} x}}\frac{{p + \cos_{p} x}}{{\cos_{p} x\tan_{p}^{p - 1} x}} \\ &= - (p+1)\mathop{\lim} _{x \to0^{+} } \frac{{x\cos_{p} x - \sin_{p} x}}{{x^{p + 1} }} \\ &=1 \end{aligned}$$
and
$$f \biggl(\frac{\pi_{p}}{2} \biggr)=\frac{{\ln {\frac{{2\sin_{p} \frac{{\pi_{p} }}{2}}}{{\pi_{p} }}}}}{{\ln{\frac{{p + \cos_{p} \frac{{\pi_{p} }}{2}}}{p+1}} }}. $$
The proof is completed. □

4 A conjecture

Conjecture 4.1

For all \(x\in(0,\frac{\pi_{p}}{2}]\) and \(p\in(1,2]\), is the function \(\frac{{\ln\frac{x}{{\sin_{p} x}}}}{{\ln\cosh_{p} x}} \) strictly increasing?

Declarations

Acknowledgements

The work was supported by NSFC11401041, 51674038, NSF of Shandong Province under grant numbers ZR2017MA019, Science and Technology Project of Shandong Province under grant J16li52, and by the Science Foundation of Binzhou University under grant BZXYL1704. Also the authors would like to thank the editor and the anonymous referees for their valuable suggestions and comments which helped us to improve this paper greatly.

Authors’ contributions

All authors contributed equally to the manuscript and read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
College of Science, Binzhou University, Binzhou, China
(2)
College of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao, China
(3)
School of Mathematical Sciences, Qufu Normal University, Qufu, China

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