Sharp Wilker-type inequalities with applications

Abstract

In this paper, we prove that the Wilker-type inequality

$$\frac{2}{k+2}{\left(\frac{sinx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanx}{x}\right)}^p>(<)1$$

holds for any fixed $k\ge 1$ and all $x\in (0,\pi /2)$ if and only if $p>0$ or $p\le -\frac{ln(k+2)-ln2}{k(ln\pi -ln2)}$ ($-\frac{12}{5(k+2)}\le p<0$), and the hyperbolic version of Wilker-type inequality

$$\frac{2}{k+2}{\left(\frac{sinhx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanhx}{x}\right)}^p>(<)1$$

holds for any fixed $k\ge 1$ ($<-2$) and all $x\in (0,\mathrm{\infty })$ if and only if $p>0$ or $p\le -\frac{12}{5(k+2)}$ ($p<0$ or $p\ge -\frac{12}{5(k+2)}$). As applications, several new analytic inequalities are presented.

MSC:26D05, 33B10.

1 Introduction

Wilker [1] proposed two open problems, the first of which states that the inequality

$${\left(\frac{sinx}{x}\right)}^2+\frac{tanx}{x}>2$$

(1.1)

holds for all $x\in (0,\pi /2)$. Inequality (1.1) was proved by Sumner et al. in [2].

Recently, the Wilker inequality (1.1) and its generalizations, improvements, refinements and applications have attracted the attention of many mathematicians (see [3–17] and related references therein).

In [9], Wu and Srivastava established the following Wilker-type inequality:

$${\left(\frac{x}{sinx}\right)}^2+\frac{x}{tanx}>2\text{for }x\in (0,\pi /2)$$

(1.2)

and its weighted and exponential generalization.

Theorem Wu ([[9], Theorem 1])

Let $\lambda >0$, $\mu >0$ and $p\le 2q\mu /\lambda$. If $q>0$ or $q\le min(-1,-\lambda /\mu )$, then the inequality

$$\frac{\lambda }{\lambda +\mu }{\left(\frac{sinx}{x}\right)}^p+\frac{\mu }{\lambda +\mu }{\left(\frac{tanx}{x}\right)}^q>1$$

(1.3)

holds for $x\in (0,\pi /2)$.

As an application of inequality (1.3), an open problem was proposed, answered and improved by Sándor and Bencze in [18]. Recently, inequality (1.3) and its related inequalities in [9] were extended to Bessel functions [3], and the hyperbolic version of Theorem Wu was presented in [12].

In 2009, Zhu [16] gave another exponential generalization of Wilker inequality (1.1) as follows.

Theorem Zh1 ([[16], Theorems 1.1 and 1.2])

Let $0<x<\pi /2$. Then the inequalities

$${\left(\frac{sinx}{x}\right)}^{2p}+{\left(\frac{tanx}{x}\right)}^p>{\left(\frac{x}{sinx}\right)}^{2p}+{\left(\frac{x}{tanx}\right)}^p>2$$

(1.4)

hold if $p\ge 1$, while the first one in (1.4) holds if and only if $p>0$.

Theorem Zh2 ([[16], Theorems 1.3 and 1.4])

Let $x>0$. Then the inequalities

$${\left(\frac{sinhx}{x}\right)}^{2p}+{\left(\frac{tanhx}{x}\right)}^p>{\left(\frac{x}{sinhx}\right)}^{2p}+{\left(\frac{x}{tanhx}\right)}^p>2$$

(1.5)

hold if $p\ge 1$, while the first one in (1.5) holds if and only if $p>0$.

In [16], Zhu also proposed an open problem: find the respectively largest range of p such that inequalities (1.4) and (1.5) hold. It was solved by Matejička in [19].

Another inequality associated with the Wilker inequality is the following:

$$2\frac{sinx}{x}+\frac{tanx}{x}>3$$

(1.6)

for $x\in (0,\pi /2)$, which is known as the Huygens inequality [20]. The following refinement of Huygens inequality is due to Neuman and Sándor [7]:

$$2\frac{sinx}{x}+\frac{tanx}{x}>2\frac{x}{sinx}+\frac{x}{tanx}>3$$

(1.7)

for $x\in (0,\pi /2)$. Very recently, the generalizations of (1.7) were given by Neuman in [8]. In [21], Zhu proved that the inequalities

$$(1-{\xi }_1)\frac{sinx}{x}+{\xi }_1\frac{tanx}{x}>1>(1-{\eta }_1)\frac{sinx}{x}+{\eta }_1\frac{tanx}{x},$$

(1.8)

$$(1-{\xi }_2)\frac{x}{sinx}+{\xi }_2\frac{x}{tanx}>1>(1-{\eta }_2)\frac{x}{sinx}+{\eta }_2\frac{x}{tanx}$$

(1.9)

hold for all $x\in (0,\pi /2)$ with the best constants ${\xi }_1=1/3$, ${\eta }_1=0$, ${\xi }_2=1/3$, ${\eta }_2=1-2/\pi$. Later, Zhu [15] generalized inequalities (1.8) and (1.9) to the exponential form as follows.

Theorem Zh3 ([[15], Theorems 1.1 and 1.2])

Let $0<x<\pi /2$. Then we have

• (i) If $p\ge 1$, then the double inequality

  $$(1-\lambda ){\left(\frac{x}{sinx}\right)}^p+\lambda {\left(\frac{x}{tanx}\right)}^p<1<(1-\eta ){\left(\frac{x}{sinx}\right)}^p+\eta {\left(\frac{x}{tanx}\right)}^p$$

  (1.10)

holds if and only if $\eta \le 1/3$ and $\lambda \ge 1-{(2/\pi )}^p$.

• (ii) If $0\le p\le 4/5$, then double inequality (1.10) holds if and only if $\lambda \ge 1/3$ and $\eta \le 1-{(2/\pi )}^p$.

• (iii) If $p<0$, then the second inequality in (1.10) holds if and only if $\eta \ge 1/3$.

The hyperbolic version of inequalities (1.7) was given in [7] by Neuman and Sándor. Later, Zhu showed the following.

Theorem Zh4 ([[17], Theorem 4.1])

Let $x>0$. Then one has

• (i) If $p\ge 4/5$, then the double inequality

  $$(1-\lambda ){\left(\frac{x}{sinhx}\right)}^p+\lambda {\left(\frac{x}{tanhx}\right)}^p<1<(1-\eta ){\left(\frac{x}{sinhx}\right)}^p+\eta {\left(\frac{x}{tanhx}\right)}^p$$

  (1.11)

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

• (ii) If $p<0$, then the inequality

  $$(1-\eta ){\left(\frac{x}{sinhx}\right)}^p+\eta {\left(\frac{x}{tanhx}\right)}^p>1$$

  (1.12)

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

The main aim of this paper is to present the best possible parameter p such that the inequalities

$$\frac{2}{k+2}{\left(\frac{sinx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanx}{x}\right)}^p>1\text{for }x\in (0,\pi /2),$$

(1.13)

$$\frac{2}{k+2}{\left(\frac{sinhx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanhx}{x}\right)}^p>1\text{for }x\in (0,\mathrm{\infty })$$

(1.14)

or their reversed inequalities hold for certain fixed k with $k(k+2)\ne 0$. As applications, we also present several new analytic inequalities.

2 Lemmas

In order to establish our main results, we need several lemmas, which we present in this section.

Lemma 1 Let A, B and C be defined on $(0,\pi /2)$ by

$$A=A(x)=cosx{(sinx-xcosx)}^2(x-cosxsinx),$$

(2.1)

$$B=B(x)={(x-cosxsinx)}^2(sinx-xcosx),$$

(2.2)

$$C=C(x)={sin}^{2}x(-2x^{2}cosx+xsinx+cosx{sin}^{2}x).$$

(2.3)

Then, for fixed $k\ge 1$, the function $x\mapsto C(x)/(kA(x)+B(x))$ is increasing on $(0,\pi /2)$. Moreover, we have

$$\frac{5}{12(k+2)}<\frac{C(x)}{kA(x)+B(x)}<1.$$

(2.4)

Proof We clearly see that $A,B>0$ for $x\in (0,\pi /2)$ because of $sinx-xcosx>0$ and $x-cosxsinx=(2x-sin2x)/2>0$, and $C>0$ because of

$$(-2x^{2}cosx+xsinx+cosx{sin}^{2}x)=x^{2}cosx({\left(\frac{sinx}{x}\right)}^2+\frac{tanx}{x}-2)>0$$

by Wilker inequality (1.1).

Let $D=(kA+B)/C$, then simple computations lead to

$$\begin{array}{rcl}D(x)& =& \frac{x{sin}^{2}x(-2x^{2}cosx+xsinx+cosx{sin}^{2}x)}{(sinx-xcosx)(x-cosxsinx)((1-k{cos}^{2}x)x+(k-1)cosxsinx)}\\ =& \frac{-2x^{2}cosx+xsinx+cosx{sin}^{2}x}{(sinx-xcosx)(x-cosxsinx)}\times \frac{x{sin}^{2}x}{k(sinx-xcosx)cosx+(x-cosxsinx)}\\ :=& D_1(x)\times D_2(x).\end{array}$$

It follows from [[16], Lemma 2.9] that the function $D_1$ is positive and increasing on $(0,\pi /2)$. Hence it remains to prove that the function $D_2$ is also positive and increasing. Clearly, $D_2(x)>0$, we only need to show that $D_2^{\prime }(x)>0$ for $x\in (0,\pi /2)$. Indeed,

$$\begin{array}{rcl}{D}_2^{\prime }(x)& =& (k-1)sinx\frac{(-2x^{2}cosx+cosx{sin}^{2}x+xsinx)}{{(k(sinx-xcosx)cosx+(x-cosxsinx))}^2}\\ =& \frac{(k-1)x^{2}sinxcosx}{{(k(sinx-xcosx)cosx+(x-cosxsinx))}^2}({\left(\frac{sinx}{x}\right)}^2+\frac{tanx}{x}-2),\end{array}$$

which is clearly positive due to Wilker inequality (1.1). Therefore, $C/(kA+B)$ is increasing on $(0,\pi /2)$, and

$$\frac{5}{12(k+2)}=\underset{x\to 0}{lim}\frac{C(x)}{kA(x)+B(x)}<D(x)<\underset{x\to \pi /2^{-}}{lim}\frac{C(x)}{kA(x)+B(x)}=1.$$

This completes the proof. □

Lemma 2 Let E, F and G be defined on $(0,\mathrm{\infty })$ by

$$E=E(x)=coshx{(sinhx-xcoshx)}^2(x-coshxsinhx),$$

(2.5)

$$F=F(x)=(sinhx-xcoshx){(x-coshxsinhx)}^2,$$

(2.6)

$$G=G(x)=x{sinh}^{2}x(2x^{2}coshx-xsinhx-coshx{sinh}^{2}x).$$

(2.7)

Then, for fixed $k\ge 1$ ($k<-2$), the function $x\mapsto G(x)/(kE(x)+F(x))$ is decreasing (increasing) on $(0,\mathrm{\infty })$. Moreover, we have

$$min(0,\frac{12}{5(k+2)})<\frac{G(x)}{kE(x)+F(x)}<max(0,\frac{12}{5(k+2)}).$$

(2.8)

Proof It is easy to verify that $E,F<0$ for $x\in (0,\mathrm{\infty })$ due to

$$\begin{array}{c}(x-coshxsinhx)=(2x-sinh2x)/2<0,\hfill \\ (sinhx-xcoshx)=x(\frac{sinhx}{x}-cosx)<0.\hfill \end{array}$$

While $G<0$ because of

$$(2x^{2}coshx-xsinhx-coshx{sinh}^{2}x)=-x^{2}coshx({\left(\frac{sinhx}{x}\right)}^2+\frac{tanhx}{x}-2)<0$$

by Wilker inequality (1.5).

Denote $G/(kE+F)$ by H and simple computations give

$$\begin{array}{rcl}H(x)& =& \frac{x{sinh}^{2}x(2x^{2}coshx-xsinhx-coshx{sinh}^{2}x)}{coshx{(sinhx-xcoshx)}^2(x-sinhxcoshx)k+(sinhx-xcoshx){(x-sinhxcoshx)}^2}\\ =& \frac{-2x^{2}coshx+xsinhx+coshx{sinh}^{2}x}{(xcoshx-sinhx)(sinhxcoshx-x)}\times \frac{x{sinh}^{2}x}{(k(xcoshx-sinhx)coshx+sinhxcoshx-x)}\\ :=& H_1(x)\times H_2(x).\end{array}$$

Clearly, $H_1(x)>0$, and it was proved in [[19], Proof of Lemma 2.2] that $H_1$ is decreasing on $(0,\mathrm{\infty })$. In order to prove the monotonicity of H, we only need to deal with the sign and monotonicity of $H_2$.

• (i) Clearly, $H_2(x)>0$ for $k\ge 1$. And we claim that $H_2$ is also decreasing on $(0,\mathrm{\infty })$. Indeed,

  $$\begin{array}{rcl}{H}_2^{\prime }(x)& =& -(k-1)sinhx\frac{(-2x^{2}coshx+coshx{sinh}^{2}x+xsinhx)}{{(xcoshx-sinhx)}^2{(coshxsinhx-x)}^2}\\ =& -\frac{(k-1)x^{2}sinhxcoshx}{{(xcoshx-sinhx)}^2{(coshxsinhx-x)}^2}({\left(\frac{sinhx}{x}\right)}^2+\frac{tanhx}{x}-2)<0.\end{array}$$

Consequently, $H=H_1\times H_2$ is positive and decreasing on $(0,\mathrm{\infty })$, and so

$$0=\underset{x\to \mathrm{\infty }}{lim}\frac{G(x)}{kE(x)+F(x)}<\frac{G(x)}{kE(x)+F(x)}<\underset{x\to 0}{lim}\frac{G(x)}{kE(x)+F(x)}=\frac{12}{5(k+2)}.$$

• (ii) For $k<-2$, by the previous proof we clearly see that $-H_2^{\prime }$ is decreasing on $(0,\mathrm{\infty })$, and so

  $$0<-\frac{1}{k}=\underset{x\to \mathrm{\infty }}{lim}(-H_2(x))<-H_2(x)<\underset{x\to 0}{lim}(-H_2(x))=-\frac{3}{k+2},$$

which implies that $-H_2$ is positive and decreasing on $(0,\mathrm{\infty })$, and so is the function $-H=H_1\times (-H_2)$. That is, H is negative and increasing on $(0,\mathrm{\infty })$, and inequality (2.8) holds true.

This completes the proof. □

Remark 1 It should be noted that $kE(x)+F(x)<0$ for $k\ge 1$ and $kE(x)+F(x)>0$ for $k<-2$. In fact, it suffices to notice (2.8) and $G(x)<0$ for $x\in (0,\mathrm{\infty })$.

Lemma 3 For $k\ge 1$, we have

$$1>\frac{ln(k+2)-ln2}{k(ln\pi -ln2)}>\frac{12}{5(k+2)}.$$

Proof It suffices to show that

$$\begin{array}{c}{\delta }_1(k)=\frac{ln(k+2)-ln2}{ln\pi -ln2}-k<0,\hfill \\ {\delta }_2(k)=\frac{ln(k+2)-ln2}{ln\pi -ln2}-\frac{12k}{5(k+2)}>0\hfill \end{array}$$

for $k\ge 1$.

Differentiation gives

$$\begin{array}{c}{\delta }_1^{\prime }(k)=\frac{1}{(ln\pi -ln2)(k+2)}-1<0,\hfill \\ {\delta }_2^{\prime }(k)=\frac{1}{5}\frac{5k+24ln2-24ln\pi +10}{{(k+2)}^2(ln\pi -ln2)}>0\hfill \end{array}$$

for $k\ge 1$. Therefore, Lemma 3 follows from ${\delta }_1(k)\le {\delta }_1(1)=(ln3-ln2)/(ln3-ln\pi )<0$ and ${\delta }_2(k)\ge {\delta }_2(1)=(ln3-ln2)/(ln\pi -ln2)-4/5>0$. □

3 Main results

Theorem 1 For fixed $k\ge 1$, inequality (1.13) holds for $x\in (0,\pi /2)$ if and only if $p>0$ or $p\le -\frac{ln(k+2)-ln2}{k(ln\pi -ln2)}$.

Proof Inequality (1.13) is equivalent to

$$f(x)=\frac{2}{k+2}{\left(\frac{sinx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanx}{x}\right)}^p-1>0$$

(3.1)

for $x\in (0,\pi /2)$. Differentiation yields

$$\begin{array}{rcl}{f}^{\prime }(x)& =& -\frac{2kp}{k+2}\frac{sinx-xcosx}{x^2}{\left(\frac{sinx}{x}\right)}^{kp-1}+\frac{kp}{k+2}\frac{x-sinxcosx}{x^2{cos}^{2}x}{\left(\frac{tanx}{x}\right)}^{p-1}\\ =& \frac{kp}{k+2}\frac{x-sinxcosx}{x^2{cos}^{2}x}{\left(\frac{tanx}{x}\right)}^{p-1}g(x),\end{array}$$

(3.2)

where

$$g(x)=1-4\frac{sinx-xcosx}{2x-sin2x}{\left(\frac{sinx}{x}\right)}^{(k-1)p}{(cosx)}^{p+1}.$$

(3.3)

A simple computation leads to $g(0^{+})=0$.

Differentiation again and simplifying give

$$g^{\prime }(x)=8\frac{{(\frac{sinx}{x})}^{(k-1)p}{(cosx)}^p}{xsinx{(2x-sin2x)}^2}h(x),$$

(3.4)

where

$$\begin{array}{rcl}h(x)& =& cosx{(sinx-xcosx)}^2(x-cosxsinx)kp\\ +{(x-cosxsinx)}^2(sinx-xcosx)p\\ +x{sin}^{2}x(-2x^{2}cosx+xsinx+cosx{sin}^{2}x)\\ =& kpA(x)+pB(x)+C(x)\\ =& (kA+B)(p+\frac{C}{kA+B}),\end{array}$$

(3.5)

where $A(x)$, $B(x)$ and $C(x)$ are defined as in (2.1), (2.2) and (2.3), respectively.

By (3.2), (3.4) we easily get

$$sgn{f}^{\prime }(x)=sgnpsgng(x),$$

(3.6)

$$sgn{g}^{\prime }(x)=sgnh(x).$$

(3.7)

Necessity. We first present two limit relations:

$$\underset{x\to 0^{+}}{lim}{x}^{4}f(x)=\frac{kp}{36}(p+\frac{12}{5(k+2)}),$$

(3.8)

$$\underset{x\to {(\pi /2)}^{-}}{lim}f(x)=\{\begin{array}{ll}\mathrm{\infty }& \text{if }p>0,\\ \frac{2}{k+2}{(\frac{2}{\pi })}^{kp}-1& \text{if }p<0.\end{array}$$

(3.9)

In fact, using power series extension yields

$$f(x)=\frac{kp}{36}\frac{kp+2p+12/5}{k+2}{x}^4+o\left(x^4\right),$$

which implies the first limit relation (3.8). From the fact that ${lim}_{x\to \pi /2^{-}}tanx=\mathrm{\infty }$, the second one (3.9) easily follows.

Now we can derive that the necessary condition of (1.13) holds for $x\in (0,\pi /2)$ from the simultaneous inequalities ${lim}_{x\to 0^{+}}{x}^{4}f(x)\ge 0$ and ${lim}_{x\to {(\pi /2)}^{-}}f(x)\ge 0$. Solving for p yields $p>0$ or

$$p\le min(-\frac{12}{5(k+2)},-\frac{ln(k+2)-ln2}{k(ln\pi -ln2)})=-\frac{ln(k+2)-ln2}{k(ln\pi -ln2)},$$

where the equality holds due to Lemma 3.

Sufficiency. We prove that the condition $p>0$ or $p\le -\frac{ln(k+2)-ln2}{k(ln\pi -ln2)}$ is sufficient. We divide the proof into three cases.

Case 1 $p>0$. Clearly, $h(x)>0$, then $g^{\prime }(x)>0$ and $g(x)>g(0^{+})=0$, which together with $sgnp=1$ yields $f^{\prime }(x)>0$ and $f(x)>f(0^{+})=0$.

Case 2 $p\le -1$. By Lemma 1 it is easy to get

$$p+\frac{C}{kA+B}<p+1\le 0,$$

which reveals that $h(x)<0$, $g^{\prime }(x)<0$ and $g(x)<g(0^{+})=0$, which in combination with $sgnp=-1$ implies $f^{\prime }(x)>0$ and $f(x)>f(0^{+})=0$.

Case 3 $-1<p\le -\frac{ln(k+2)-ln2}{k(ln\pi -ln2)}$. Lemma 1 reveals that $\frac{C}{kA+B}$ is increasing on $(0,\pi /2)$, so is the function $x\mapsto p+\frac{C}{kA+B}:=\lambda (x)$. Since

$$\lambda \left(0^{+}\right)=p+\frac{12}{5(k+2)}<0,\lambda \left({\frac{\pi }{2}}^{-}\right)=p+1>0,$$

there exists $x_1\in (0,\pi /2)$ such that $\lambda (x)<0$ for $x\in (0,x_1)$ and $\lambda (x)>0$ for $x\in (x_1,\pi /2)$, and so is $g^{\prime }(x)$. Therefore, $g(x)<g(0^{+})=0$ for $x\in (0,x_1)$ but $g(\pi /2^{-})=1$, which implies that there exists $x_0\in (x_1,\pi /2)$ such that $g(x)<0$ for $x\in (0,x_0)$ and $g(x)>0$ for $x\in (x_0,\pi /2)$. Due to $sgnp=-1$, it is deduced that $f^{\prime }(x)>0$ for $x\in (0,x_0)$ and $f^{\prime }(x)<0$ for $x\in (x_0,\pi /2)$, which reveals that f is increasing on $(0,x_0)$ and decreasing on $(x_0,\pi /2)$. It follows that

$$\begin{array}{c}0=f\left(0^{+}\right)<f(x)<f(x_0)=0\text{for }x\in (0,x_0),\hfill \\ f(x_0)>f(x)>f(\pi /2^{-})=\frac{2}{k+2}{\left(\frac{2}{\pi }\right)}^{kp}-1\ge 0\text{for }x\in (x_0,\pi /2),\hfill \end{array}$$

that is, $f(x)>0$ for $x\in (0,\pi /2)$.

This completes the proof. □

Theorem 2 For fixed $k\ge 1$, the reversed inequality of (1.13), that is,

$$\frac{2}{k+2}{\left(\frac{sinx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanx}{x}\right)}^p<1,$$

(3.10)

holds for $x\in (0,\pi /2)$ if and only if $-\frac{12}{5(k+2)}\le p<0$.

Proof Necessity. If inequality (3.10) holds for $x\in (0,\pi /2)$, then we have

$$\underset{x\to 0^{+}}{lim}\frac{f(x)}{x^4}=\frac{kp}{36}(p+\frac{12}{5(k+2)})\le 0.$$

Solving the inequality for p yields $-\frac{12}{5(k+2)}\le p<0$.

Sufficiency. We prove that the condition $-\frac{12}{5(k+2)}\le p<0$ is sufficient. It suffices to show that $f(x)<0$ for $x\in (0,\pi /2)$. By Lemma 1 it is easy to get

$$p+\frac{C}{kA+B}\ge p+\frac{12}{5(k+2)}\ge 0,$$

which reveals that $h(x)>0$, $g^{\prime }(x)>0$ and $g(x)>g(0^{+})=0$. In combination with $sgnp=-1$, it implies $f^{\prime }(x)<0$. Thus, $f(x)<f(0^{+})=0$, which proves the sufficiency and the proof is completed. □

Theorem 3 For fixed $k\ge 1$, inequality (1.14) holds for $x\in (0,\mathrm{\infty })$ if and only if $p>0$ or $p\le -\frac{12}{5(k+2)}$.

Proof Let

$$u(x)=\frac{2}{k+2}{\left(\frac{sinhx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanhx}{x}\right)}^p-1.$$

(3.11)

Then inequality (1.14) is equivalent to $u(x)>0$. Differentiation leads to

$$u^{\prime }(x)=-\frac{kp}{2(k+2)}\frac{sinh2x-2x}{x^2{cosh}^{2}x}{\left(\frac{tanhx}{x}\right)}^{p-1}v(x),$$

(3.12)

where

$$v(x)=1-4\frac{sinhx-xcoshx}{2x-sinh2x}{\left(\frac{sinhx}{x}\right)}^{kp-p}{(coshx)}^{p+1}.$$

(3.13)

Differentiation again gives

$$v^{\prime }(x)=\frac{2{cosh}^{p}x{(\frac{sinhx}{x})}^{kp-p}}{xsinhx{(x-coshxsinhx)}^2}w(x),$$

(3.14)

where

$$\begin{array}{rcl}w(x)& =& coshx{(sinhx-xcoshx)}^2(x-coshxsinhx)kp\\ +(sinhx-xcoshx){(x-coshxsinhx)}^{2}p\\ +x{sinh}^{2}x(2x^{2}coshx-xsinhx-coshx{sinh}^{2}x)\\ =& kpE(x)+pF(x)+G(x)=(kE+F)(p+\frac{G}{kE+F}),\end{array}$$

(3.15)

where $E(x)$, $F(x)$ and $G(x)$ are defined as in (2.5), (2.6) and (2.7), respectively.

By (3.12) and (3.14) we easily get

$$sgn{u}^{\prime }(x)=-sgn\frac{k}{k+2}sgnpsgnv(x),$$

(3.16)

$$sgn{v}^{\prime }(x)=sgnw(x).$$

(3.17)

Necessity. If inequality (1.14) holds for $x\in (0,\mathrm{\infty })$, then we have ${lim}_{x\to 0^{+}}{x}^{-4}u(x)\ge 0$. Expanding $u(x)$ in power series gives

$$u(x)=\frac{k}{36}p(p+\frac{12}{5p(k+2)})x^4+o\left(x^4\right).$$

Hence we get

$$\underset{x\to 0^{+}}{lim}{x}^{-4}u(x)=\frac{k}{36}p(p+\frac{12}{5(k+2)})\ge 0.$$

Solving the inequality for p yields $p>0$ or $p\le -\frac{12}{5(k+2)}$.

Sufficiency. We prove that the condition $p>0$ or $p\le -\frac{12}{5(k+2)}$ is sufficient for (1.14) to hold.

If $p>0$, then $w(x)<0$ due to $E,F,G<0$. Hence, from (3.17) we have $v^{\prime }(x)<0$ and $v(x)<{lim}_{x\to 0^{+}}v(x)=0$. It is derived by (3.16) that $u^{\prime }(x)>0$, and so $u(x)>{lim}_{x\to 0^{+}}u(x)=0$.

If $p\le -\frac{12}{5(k+2)}$, then by Lemma 2 we have

$$p+\frac{G}{kE+F}\le -\frac{12}{5(k+2)}+\frac{G}{kE+F}<0$$

and

$$w(x)=(kE+F)(p+\frac{G}{kE+F})>0.$$

From (3.17) we have $v^{\prime }(x)>0$ and $v(x)>{lim}_{x\to 0^{+}}v(x)=0$. It follows by (3.16) that $u^{\prime }(x)>0$, which implies that $u(x)>{lim}_{x\to 0^{+}}u(x)=0$.

This completes the proof. □

Remark 2 For $k\ge 1$, since ${lim}_{x\to \mathrm{\infty }}u(x)=\mathrm{\infty }$ for $p\ne 0$ and ${lim}_{x\to \mathrm{\infty }}u(x)=0$ for $p=0$, there does not exist p such that the reverse inequality of (1.14) holds for all $x>0$. But we can show that there exists $x_0\in (0,\mathrm{\infty })$ such that $u(x)<0$, that is, the reverse inequality of (1.14) holds for $-\frac{12}{5(k+2)}<p<0$. The details of the proof are omitted.

Theorem 4 For fixed $k<-2$, the reverse of (1.14), that is,

$$\frac{2}{k+2}{\left(\frac{sinhx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanhx}{x}\right)}^p<1,$$

(3.18)

holds for $x\in (0,\mathrm{\infty })$ if and only if $p<0$ or $p\ge -\frac{12}{5(k+2)}$.

Proof Necessity. If inequality (3.18) holds for $x\in (0,\mathrm{\infty })$, then we have

$$\underset{x\to 0^{+}}{lim}\frac{u(x)}{x^4}=\frac{k}{36}p(p+\frac{12}{5(k+2)})\le 0.$$

Solving the inequality for p yields $p<0$ or $p\ge -\frac{12}{5(k+2)}$.

Sufficiency. We prove that the condition $p<0$ or $p\ge -\frac{12}{5(k+2)}$ is sufficient for (3.18) to hold.

If $p<0$, then $w(x)=(kE+F)(p+\frac{G}{kE+F})<0$ due to $kE+F>0$ and $G<0$. Hence, from (3.17) we have $v^{\prime }(x)<0$ and $v(x)<{lim}_{x\to 0^{+}}v(x)=0$. It is derived by (3.16) that $u^{\prime }(x)<0$, and so $u(x)<{lim}_{x\to 0^{+}}u(x)=0$.

If $p\ge -\frac{12}{5(k+2)}$, then by Lemma 2 we have

$$p+\frac{G}{kE+F}\ge p+\frac{12}{5(k+2)}>0$$

and

$$w(x)=(kE+F)(p+\frac{G}{kE+F})>0.$$

From (3.17) we have $v^{\prime }(x)>0$ and $v(x)>{lim}_{x\to 0^{+}}v(x)=0$. It follows by (3.16) that $u^{\prime }(x)<0$, which implies that $u(x)<{lim}_{x\to 0^{+}}u(x)=0$.

This completes the proof. □

4 Applications

4.1 Huygens-type inequalities

Letting $k=1$ in Theorems 1 and 2, we have the following proposition.

Proposition 1 For $x\in (0,\pi /2)$, the double inequality

$$\frac{2}{3}{\left(\frac{sinx}{x}\right)}^p+\frac{1}{3}{\left(\frac{tanx}{x}\right)}^p>1>\frac{2}{3}{\left(\frac{sinx}{x}\right)}^q+\frac{1}{3}{\left(\frac{tanx}{x}\right)}^q$$

(4.1)

holds if and only if $p>0$ or $p\le -\frac{ln3-ln2}{ln\pi -ln2}\approx -0.898$ and $-4/5\le q<0$.

Let $M_r(a,b;w)$ denote the r th weighted power mean of positive numbers $a,b>0$ defined by

$$M_r(a,b;w):={(w{a}^r+(1-w)b^r)}^{1/r}\text{if }r\ne 0\text{ and }{M}_0(a,b;w)=a^w{b}^{1-w},$$

(4.2)

where $w\in (0,1)$.

Since

$$\frac{2}{3}{\left(\frac{sinx}{x}\right)}^p+\frac{1}{3}{\left(\frac{tanx}{x}\right)}^p=\frac{\frac{2}{3}+\frac{1}{3}{(cosx)}^{-p}}{{(\frac{sinx}{x})}^{-p}},$$

by Proposition 1 the inequality

$$\frac{sinx}{x}>{(\frac{2}{3}+\frac{1}{3}{(cosx)}^{-p})}^{-1/p}=M_{-p}(1,cosx;\frac{2}{3})$$

holds for $x\in (0,\pi /2)$ if and only if $-p\le 4/5$. Similarly, its reversed inequality holds if and only if $-p\ge \frac{ln3-ln2}{ln\pi -ln2}$. The facts can be stated as a corollary.

Corollary 1 Let $M_r(a,b;w)$ be defined by (4.2). Then, for $x\in (0,\pi /2)$, the inequalities

$$M_{\alpha }(1,cosx;\frac{2}{3})<\frac{sinx}{x}<M_{\beta }(1,cosx;\frac{2}{3})$$

(4.3)

hold if and only if $\alpha \le 4/5$ and $\beta \ge \frac{ln3-ln2}{ln\pi -ln2}\approx -0.898$.

Remark 3 The Cusa-Huygens inequality [20] refers to

$$\frac{sinx}{x}<\frac{2}{3}+\frac{1}{3}cosx$$

(4.4)

holds for $x\in (0,\pi /2)$, which is equivalent to the second inequality in (1.7). As an improvement and generalization, Corollary 1 was proved in [22] by Yang. Here we provide a new proof.

Remark 4 Let $a>b>0$ and let $x=arcsin\frac{a-b}{a+b}\in (0,\pi /2)$. Then $sinx/x=P/A$, $cosx=G/A$ and inequalities (4.3) can be rewritten as

$$M_{\alpha }(A,G;\frac{2}{3})<P<M_{\beta }(A,G;\frac{2}{3}),$$

(4.5)

where P is the first Seiffert mean [23] defined by

$$P=P(a,b)=\frac{a-b}{2arcsin\frac{a-b}{a+b}},$$

A and G denote the arithmetic and geometric means of a and b, respectively.

Let $x=arctan\frac{a-b}{a+b}$. Then $sinx/x=T/Q$, $cosx=A/Q$, and inequalities (4.3) can be rewritten as

$$M_{\alpha }(Q,A;\frac{2}{3})<T<M_{\beta }(Q,A;\frac{2}{3}),$$

(4.6)

where T is the second Seiffert mean [24] defined by

$$T=T(a,b)=\frac{a-b}{2arctan\frac{a-b}{a+b}},$$

Q denotes the quadratic mean of a and b.

Obviously, by Corollary 2, the two double inequalities (4.5) (see [22]) and (4.6) hold if and only if $\alpha \le 4/5$ and $\beta \ge \frac{ln3-ln2}{ln\pi -ln2}\approx -0.898$, (4.6) seems to be a new inequality.

In the same way, taking $k=1$ in Theorem 3, we get the following.

Proposition 2 For $x\in (0,\mathrm{\infty })$, the inequality

$$\frac{2}{3}{\left(\frac{sinhx}{x}\right)}^p+\frac{1}{3}{\left(\frac{tanhx}{x}\right)}^p>1$$

(4.7)

holds if and only if $p>0$ or $p\le -\frac{4}{5}$.

Similar to Corollary 1, we have the following.

Corollary 2 Let $M_r(a,b;w)$ be defined by (4.2). Then, for $x\in (0,\mathrm{\infty })$, the inequalities

$$M_{\alpha }(1,coshx;\frac{2}{3})<\frac{sinhx}{x}<M_{\beta }(1,coshx;\frac{2}{3})$$

(4.8)

hold if and only if $\alpha \le 0$ and $\beta \ge 4/5$.

Remark 5 Let $a>b>0$ and $x=ln\sqrt{a/b}$. Then $sinhx/x=L/G$, $coshx=A/G$, and (4.8) can be rewritten as

$$M_{\alpha }(G,A;\frac{2}{3})<L<M_{\beta }(G,A;\frac{2}{3}),$$

(4.9)

where L is the logarithmic means of a and b defined by

$$L=L(a,b)=\frac{a-b}{lna-lnb}.$$

Making use of $x=arcsinh\frac{b-a}{a+b}$ yields $sinhx/x=NS/A$ and $coshx=Q/A$, where NS is the Nueman-Sándor mean defined by

$$NS=NS(a,b)=\frac{a-b}{2arcsinh\frac{a-b}{a+b}}.$$

Thus, (4.8) is equivalent to

$$M_{\alpha }(A,Q;\frac{2}{3})<NS<M_{\beta }(A,Q;\frac{2}{3}).$$

(4.10)

Corollary 2 implies that inequalities (4.9) and (4.10) hold if and only if $\alpha \le 0$ and $\beta \ge 4/5$. The second inequality in (4.10) is a new inequality.

Remark 6 It should be pointed out that all inequalities involving $sinx/x$and cosx or $sinhx/x$and coshx in this paper can be rewritten as the equivalent inequalities for bivariate means mentioned previously. In what follows we no longer mention this.

4.2 Wilker-Zhu-type inequalities

Letting $k=2$ in Theorems 1 and 2, we have the following.

Proposition 3 For $x\in (0,\pi /2)$, the double inequality

$${\left(\frac{sinx}{x}\right)}^{2p}+{\left(\frac{tanx}{x}\right)}^p>2>{\left(\frac{sinx}{x}\right)}^{2q}+{\left(\frac{tanx}{x}\right)}^q$$

(4.11)

holds if and only if $p>0$ or $p\le -\frac{ln2}{2(ln\pi -ln2)}\approx -0.767$ and $-3/5\le q<0$.

Note that

$$\frac{{(\frac{sinx}{x})}^{2p}+{(\frac{tanx}{x})}^p-2}{{(\frac{sinx}{x})}^p+\frac{\sqrt{8+{cos}^{-2p}x}+{cos}^{-p}x}{2}}={\left(\frac{x}{sinx}\right)}^{-p}-\frac{\sqrt{8+{cos}^{-2p}x}-{cos}^{-p}x}{2}.$$

By Proposition 3 the inequality

$$\frac{x}{sinx}>{\left(\frac{\sqrt{8+{cos}^{-2p}x}-{cos}^{-p}x}{2}\right)}^{-1/p}$$

or

$$\frac{sinx}{x}<{\left(\frac{\sqrt{8+{cos}^{-2p}x}+{cos}^{-p}x}{4}\right)}^{-1/p}:=H_{-p}(cosx)$$

holds for $x\in (0,\pi /2)$ if and only if $-p\ge \frac{ln2}{2(ln\pi -ln2)}$, where $H_r$ is defined on $(0,\mathrm{\infty })$ by

$$H_r(t)={\left(\frac{\sqrt{8+t^{2r}}+t^r}{4}\right)}^{1/r}\text{if }r\ne 0\text{ and }{H}_0(t)=\sqrt[3]{t}.$$

(4.12)

Likewise, its reversed inequality holds if and only if $-p\le 3/5$. This result can be stated as a corollary.

Corollary 3 Let $H_r(t)$ be defined by (4.12). Then, for $x\in (0,\pi /2)$, the inequalities

$$H_{\alpha }(cosx)<\frac{sinx}{x}<H_{\beta }(cosx)$$

(4.13)

are true if and only if $\alpha \le 3/5$ and $\beta \ge \frac{ln2}{2(ln\pi -ln2)}\approx 0.767$.

Taking $k=2$ in Theorem 3, we have the following.

Proposition 4 For $x\in (0,\mathrm{\infty })$, the inequality

$${\left(\frac{sinhx}{x}\right)}^{2p}+{\left(\frac{tanhx}{x}\right)}^p>2$$

holds if and only if $p>0$ or $p\le -3/5$.

In a similar way, we get Corollary 4.

Corollary 4 Let $H_r(t)$ be defined by (4.12). Then, for $x\in (0,\mathrm{\infty })$, the inequalities

$$H_{\alpha }(coshx)<\frac{sinhx}{x}<H_{\beta }(coshx)$$

(4.14)

are true if and only if $\alpha \le 0$ and $\beta \ge 3/5$.

Now we give a generalization of inequalities (1.4) given by Zhu [15].

Proposition 5 For fixed $k\ge 1$, both chains of inequalities

$$\begin{array}{rl}\frac{2}{k+2}{\left(\frac{sinx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanx}{x}\right)}^p& \ge \frac{k}{k+2}{\left(\frac{sinx}{x}\right)}^{kp}+\frac{2}{k+2}{\left(\frac{tanx}{x}\right)}^p\\ >\frac{2}{k+2}{\left(\frac{x}{sinx}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{x}{tanx}\right)}^p>1,\end{array}$$

(4.15)

$$\begin{array}{rl}\frac{2}{k+2}{\left(\frac{sinx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanx}{x}\right)}^p& >\frac{2}{k+2}{\left(\frac{x}{tanx}\right)}^p+\frac{k}{k+2}{\left(\frac{x}{sinx}\right)}^{kp}\\ \ge \frac{2}{k+2}{\left(\frac{x}{sinx}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{x}{tanx}\right)}^p>1\end{array}$$

(4.16)

hold for $x\in (0,\pi /2)$ if and only if $k\ge 2$ and $p\ge \frac{ln(k+2)-ln2}{k(ln\pi -ln2)}$.

Proof The first inequality in (4.15) is equivalent to

$$\begin{array}{c}\frac{2}{k+2}{\left(\frac{sinx}{x}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{tanx}{x}\right)}^p-\frac{k}{k+2}{\left(\frac{sinx}{x}\right)}^{kp}-\frac{2}{k+2}{\left(\frac{tanx}{x}\right)}^p\hfill \\ =\frac{k-2}{k+2}({\left(\frac{tanx}{x}\right)}^p-{\left(\frac{sinx}{x}\right)}^{kp})>0.\hfill \end{array}$$

Due to $\frac{tanx}{x}>1$ and $\frac{sinx}{x}<1$, it holds for $x\in (0,\pi /2)$ if and only if

$$(k,p)\in \{k\ge 2,p>0\}\cup \{1\le k\le 2,p<0\}:={\mathrm{\Omega }}_1.$$

The second one is equivalent to

$$\frac{\frac{k}{k+2}{(\frac{sinx}{x})}^{kp}+\frac{2}{k+2}{(\frac{tanx}{x})}^p}{\frac{2}{k+2}{(\frac{x}{sinx})}^{kp}+\frac{k}{k+2}{(\frac{x}{tanx})}^p}>1,$$

which can be simplified to

$${\left(\frac{sinx}{x}\right)}^{kp}{\left(\frac{tanx}{x}\right)}^p={\left({\left(\frac{sinx}{x}\right)}^{k+1}\frac{1}{cosx}\right)}^p>1.$$

It is true for $x\in (0,\pi /2)$ if and only if $(k,p)\in \{k+1\ge 3,p\ge 0\}:={\mathrm{\Omega }}_2$.

By Theorem 1, the third one in (4.15) holds for $x\in (0,\pi /2)$ if and only if

$$(k,p)\in \{k\ge 1,-p>0\}\cup \{k\ge 1,-p\le -\frac{ln(k+2)-ln2}{k(ln\pi -ln2)}\}:={\mathrm{\Omega }}_3.$$

Hence, inequalities (4.15) hold for $x\in (0,\pi /2)$ if and only if

$$(k,p)\in {\mathrm{\Omega }}_1\cap {\mathrm{\Omega }}_2\cap {\mathrm{\Omega }}_3=\{k\ge 2,p\ge \frac{ln(k+2)-ln2}{k(ln\pi -ln2)}\},$$

which proves (4.15).

In the same way, we can prove (4.16), the details are omitted. □

Letting $k=2$ in Proposition 5, we have the following.

Corollary 5 For $x\in (0,\pi /2)$, inequality (1.4) holds if and only if $p\ge \frac{ln2}{2(ln\pi -ln2)}\approx 0.767$.

Similarly, using Theorem 3 we easily prove the following proposition.

Proposition 6 For fixed $k\ge 1$, the inequalities

$$\frac{k}{k+2}{\left(\frac{sinhx}{x}\right)}^{kp}+\frac{2}{k+2}{\left(\frac{tanhx}{x}\right)}^p>\frac{2}{k+2}{\left(\frac{x}{sinhx}\right)}^{kp}+\frac{k}{k+2}{\left(\frac{x}{tanhx}\right)}^p>1$$

(4.17)

hold for $x\in (0,\mathrm{\infty })$ if and only if $k\ge 2$ and $p\ge \frac{12}{5(k+2)}$.

Letting $k=2$ in Proposition 6, we have the following.

Corollary 6 For $x\in (0,\mathrm{\infty })$, inequality (1.5) holds if and only if $p\ge 3/5$.

Remark 7 Clearly, Corollaries 5 and 6 offer another method for solving the problems posed by Zhu in [16].

4.3 Other Wilker-type inequalities

Taking $k=3,4$ in Theorems 1 and 2, we obtain the following.

Proposition 7 For $x\in (0,\pi /2)$, the inequality

$$\frac{2}{5}{\left(\frac{sinx}{x}\right)}^{3p}+\frac{3}{5}{\left(\frac{tanx}{x}\right)}^p>1$$

(4.18)

holds if and only if $p>0$ or $p\le -\frac{ln5-ln2}{3(ln\pi -ln2)}\approx -0.676$. It is reversed if and only if $-12/25\le p<0$.

Proposition 8 For $x\in (0,\pi /2)$, the inequality

$$\frac{1}{3}{\left(\frac{sinx}{x}\right)}^{4p}+\frac{2}{3}{\left(\frac{tanx}{x}\right)}^p>1$$

(4.19)

holds if and only if $p>0$ or $p\le -\frac{ln3}{4(ln\pi -ln2)}\approx -0.608$. It is reversed if and only if $-2/5\le p<0$.

Putting $k=-3,-4$ in Theorem 3, we get the following.

Proposition 9 For $x\in (0,\mathrm{\infty })$, the inequality

$${\left(\frac{tanhx}{x}\right)}^p<\frac{2}{3}{\left(\frac{x}{sinhx}\right)}^{3p}+\frac{1}{3}$$

(4.20)

holds if and only if $p<0$ or $p\ge 12/5$.

Proposition 10 For $x\in (0,\pi /2)$, the inequality

$$2{\left(\frac{tanhx}{x}\right)}^p<{\left(\frac{x}{sinhx}\right)}^{4p}+1$$

(4.21)

holds if and only if $p<0$ or $p\ge 6/5$.