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# Necessary and sufficient conditions for the two parameter generalized Wilker-type inequalities

## Abstract

In the article, we provide the necessary and sufficient conditions for the parameters α and β such that the generalized Wilker-type inequality

$$\frac{2\beta}{\alpha+2\beta} \biggl(\frac{\sin x}{x} \biggr)^{\alpha}+ \frac{\alpha}{\alpha+2\beta} \biggl(\frac{\tan x}{x} \biggr)^{\beta}-1>(< )0$$

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

## Introduction

The Wilker inequality [1, 2] for sine and tangent functions states that the inequality

$$\biggl(\frac{\sin x}{x} \biggr)^{2}+ \frac{\tan x}{x}-2>0$$
(1.1)

holds for all $$x\in(0, \pi/2)$$. The generalizations and improvements for the Wilker inequality (1.1) have been the subject of intensive research in the recent years. Wu and Srivastava  proved that the inequality

$$\frac{\lambda}{\lambda+\mu} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \frac{\mu }{\lambda+\mu} \biggl(\frac{\tan x}{x} \biggr)^{q}>1$$
(1.2)

holds for all $$x\in(0, \pi/2)$$ if $$\lambda>0$$, $$\mu>0$$, $$q>0$$ or $$q\leq \min\{-1, -\lambda/\mu\}$$, and $$p\leq2q\mu/\lambda$$. Baricz and Sándor  generalized inequality (1.2) to the Bessel functions.

In , Zhu proved that the inequalities

$$\biggl(\frac{\sin x}{x} \biggr)^{2p}+ \biggl( \frac{\tan x}{x} \biggr)^{p}> \biggl(\frac{x}{\sin x} \biggr)^{2p}+ \biggl(\frac{x}{\tan x} \biggr)^{p}>2$$
(1.3)

hold for $$x\in(0, \pi/2)$$ and $$p\geq1$$. Matejíčka  presented the best possible parameter p such that the second inequality of (1.3) holds for $$x\in(0, \pi/2)$$.

Zhu  proved that the inequalities

$$(1-\lambda) \biggl(\frac{x}{\sin x} \biggr)^{p}+\lambda \biggl( \frac {x}{\tan x} \biggr)^{p}< 1< (1-\eta) \biggl(\frac{x}{\sin x} \biggr)^{p}+\eta \biggl(\frac {x}{\tan x} \biggr)^{p}$$

are valid for all $$x\in(0, \pi/2)$$ if $$(p, \lambda, \eta)\in\{(p, \lambda, \eta)| p\geq1, \lambda\geq1-(2/\pi)^{p}, \eta\leq1/3\} \cup\{(p, \lambda, \eta)| 0\leq p\leq4/5, \lambda\geq1/3, \eta \leq 1-(2/\pi)^{p}\}$$.

In , Yang and Chu provided the necessary and sufficient condition for the parameter μ such that the generalized Wilker-type inequality

$$\frac{2}{\lambda+2} \biggl(\frac{\sin x}{x} \biggr)^{\lambda\mu }+ \frac {\lambda}{\lambda+2} \biggl(\frac{\tan x}{x} \biggr)^{\mu}-1>(< )0$$

holds for any fixed $$\lambda\geq1$$ and all $$x\in(0, \pi/2)$$.

Very recently, Chu et al.  proved that the two parameter generalized Wilker-type inequality

$$\frac{2\beta}{\alpha+2\beta} \biggl(\frac{\sin x}{x} \biggr)^{\alpha}+ \frac {\alpha}{\alpha+2\beta} \biggl(\frac{\tan x}{x} \biggr)^{\beta}-1>0$$
(1.4)

holds for all $$x\in(0, \pi/2)$$ if $$(\alpha, \beta)\in E_{0}$$, and inequality (1.4) is reversed if $$(\alpha, \beta)\in E_{1}$$, where

\begin{aligned}& \begin{aligned}[b] E_{0}={}& \bigl\{ (\alpha,\beta)|\alpha>0, \beta>0 \bigr\} \cup \bigl\{ (\alpha ,\beta)| 0< \alpha< -2\beta, \beta\geq-1 \bigr\} \\ &{}\cup \biggl\{ (\alpha ,\beta )\Big|\beta>0, -\frac{12}{5}\leq\alpha+2\beta< 0 \biggr\} \\ &{}\cup \biggl\{ (\alpha,\beta)\Big|\alpha\leq\frac{\pi^{2}}{4}-3, \beta \leq -1 \biggr\} \\ &{}\cup \biggl\{ (\alpha,\beta)\Big|\frac{\pi^{2}}{4}-3< \alpha < 0, \beta \leq- \frac{37}{35}, \alpha+2\beta+\frac{12}{5}\leq0 \biggr\} , \end{aligned} \\& \begin{aligned}[b] E_{1}={}& \bigl\{ (\alpha,\beta)|\alpha< 0, \alpha+2 \beta>0 \bigr\} \cup \bigl\{ (\alpha,\beta)|-1\leq\beta< 0, \alpha+2\beta>0 \bigr\} \\ &{}\cup \biggl\{ (\alpha,\beta)\Big|-1\leq\beta< 0, -2\beta-\frac {12}{5}\leq \alpha< 0 \biggr\} \cup \biggl\{ (\alpha,\beta)\Big|0< \alpha\leq-2\beta - \frac {12}{5} \biggr\} . \end{aligned} \end{aligned}

The main purpose of this paper is to provide the necessary and sufficient conditions for the parameters α and β such that the generalized Wilker-type inequality (1.4) and its reversed inequality hold for all $$x\in(0, \pi/2)$$.

## Lemmas

### Lemma 2.1

See , Lemma 2.3

Let $$-\infty<\alpha<\beta <\infty$$, $$f_{1}, f_{2}: [\alpha,\beta] \rightarrow\mathbb{R}$$ be continuous on $$[\alpha, \beta]$$ and differentiable on $$(\alpha, \beta)$$, and $$f^{\prime}_{2}(x)\neq0$$ on $$(\alpha,\beta)$$. Then the inequality

$$\frac{f_{1} (x )-f_{1} (\alpha )}{f_{2} (x )-f_{2} (\alpha )}> (< )\frac{f^{\prime }_{1}(\alpha^{+})}{ f^{\prime}_{2}(\alpha^{+})}$$

holds for all $$x\in (\alpha,\beta )$$ if there exists $$\eta\in (\alpha, \beta)$$ such that $$f^{\prime}_{1}(x)/f^{\prime}_{2}(x)$$ is strictly increasing (decreasing) on $$(\alpha, \eta)$$ and strictly decreasing (increasing) on $$(\eta, \beta)$$, and

$$\frac{f_{1}(\beta)-f_{1}(\alpha)}{f_{2}(\beta)-g_{2}(\alpha)}\geq (\leq )\frac{f^{\prime}_{1}(\alpha^{+})}{f^{\prime}_{2}(\alpha^{+})}\neq \infty.$$

### Lemma 2.2

See , Lemma 2.9

Let $$\beta\in\mathbb{R}$$, $$x\in (0, \pi/2)$$, and $$\mathsf{F}(x)$$, $$\mathsf{G}(x)$$, $$\mathsf{H}(x)$$ and $$g(x)$$ be defined by

\begin{aligned}& \mathsf{F}(x)=\cos x(\sin x-x\cos x)^{2}(x-\sin x\cos x), \end{aligned}
(2.1)
\begin{aligned}& \mathsf{G}(x)=(x-\sin x\cos x)^{2}(\sin x-x\cos x), \end{aligned}
(2.2)
\begin{aligned}& \mathsf{H}(x)=x^{3} \biggl(\frac{\sin^{2}x}{x^{2}}+\frac{\tan x}{x}-2 \biggr)\sin^{2}x\cos x, \end{aligned}
(2.3)

and

$$g(x)=\frac{\beta\mathsf{G}(x)+\mathsf{H}(x)}{\mathsf{F}(x)},$$
(2.4)

respectively. Then the following statements are true:

1. (1)

The function $$g(x)$$ is strictly increasing from $$(0, \pi/2)$$ onto $$(2\beta+12/5, 3-\pi^{2}/4)$$ if $$\beta=-1$$.

2. (2)

The function $$g(x)$$ is strictly increasing from $$(0, \pi/2)$$ onto $$(2\beta+12/5, \infty)$$ if $$\beta>-1$$.

3. (3)

The function $$g(x)$$ is strictly decreasing from $$(0, \pi/2)$$ onto $$(-\infty, 2\beta+12/5)$$ if $$\beta\leq-37/35$$.

Let $$\alpha, \beta\in\mathbb{R}$$, $$x\in(0, \pi/2)$$ and the functions $$\mathsf{I}_{\alpha}(x)$$, $$\mathsf{J}_{\beta}(x)$$ and $$\mathsf{Q}_{\alpha,\beta}(x)$$ be defined by

\begin{aligned}& \mathsf{I}_{\alpha}(x)=\frac{1- (\frac{\sin x}{x} )^{\alpha }}{\alpha} \quad (\alpha\neq0), \qquad \mathsf{I}_{0}(x)=\log x-\log (\sin x), \end{aligned}
(2.5)
\begin{aligned}& \mathsf{J}_{\beta}(x)=\frac{ (\frac{\tan x}{x} )^{\beta }-1}{\beta} \quad (\beta\neq0), \qquad \mathsf{J}_{0}(x)=\log(\tan x)-\log x, \end{aligned}
(2.6)

and

$$\mathsf{Q}_{\alpha, \beta}(x)=\frac{\mathsf{I}_{\alpha }(x)}{\mathsf {J}_{\beta}(x)},$$

respectively.

Then it is not difficult to verify that

\begin{aligned} &\mathsf{I}_{\alpha}\bigl(0^{+}\bigr)=\mathsf{J}_{\beta} \bigl(0^{+}\bigr)=0, \\ &\mathsf{Q}_{\alpha, \beta}(x)=\frac{\mathsf{I}_{\alpha }(x)}{\mathsf {J}_{\beta}(x)}= \frac{\mathsf{I}_{\alpha}(x)- \mathsf{I}_{\alpha}(0^{+})}{\mathsf{J}_{\beta}(x)-\mathsf {J}_{\beta}(0^{+})}, \end{aligned}
(2.7)
\begin{aligned} &\mathsf{Q}_{\alpha,\beta}\bigl(0^{+}\bigr)= \frac{1}{2}, \end{aligned}
(2.8)
\begin{aligned} &\mathsf{Q}_{\alpha,\beta} \biggl({\frac{\pi}{2}}^{-} \biggr)=\frac {\beta }{\alpha} \biggl[ \biggl(\frac{2}{\pi} \biggr)^{\alpha}-1 \biggr] \quad(\alpha\neq0, \beta< 0), \end{aligned}
(2.9)
\begin{aligned} &\mathsf{Q}_{0,\beta} \biggl({\frac{\pi}{2}}^{-} \biggr) =\lim_{\alpha\rightarrow0}Q_{\alpha,\beta} \biggl({ \frac{\pi }{2}}^{-} \biggr)=\beta\log\frac{2}{\pi} \quad ( \beta< 0). \end{aligned}
(2.10)

### Lemma 2.3

See , Lemma 2.10

Let $$x\in(0, \pi/2)$$ and $$\mathsf{Q}_{\alpha, \beta}(x)$$ be defined by (2.7). Then the following statements are true:

1. (1)

If $$\alpha+2\beta+12/5\geq0$$ and $$\beta\geq-1$$, then $$\mathsf {Q}_{\alpha, \beta}(x)$$ is strictly decreasing on $$(0, \pi/2)$$.

2. (2)

If $$\alpha\leq\pi^{2}/4-3$$ and $$-37/35<\beta\leq-1$$, then $$\mathsf {Q}_{\alpha, \beta}(x)$$ is strictly increasing on $$(0, \pi/2)$$.

3. (3)

If $$\alpha+2\beta+12/5\leq0$$ and $$\beta\leq-37/35$$, then $$\mathsf {Q}_{\alpha, \beta}(x)$$ is strictly increasing on $$(0, \pi/2)$$.

### Lemma 2.4

Let $$x\in(0, \pi/2)$$, $$\mathsf{Q}_{\alpha, \beta}(x)$$ be defined by (2.7) and the function $$x\rightarrow\mathsf{D}(\alpha, \beta; x)$$ be defined by

$$\mathsf{D}(\alpha, \beta; x)=\mathsf{Q}_{\alpha, \beta}(x)- \frac{1}{2}.$$
(2.11)

Then the following statements are true:

(1) If $$\alpha\in\mathbb{R}$$ is fixed and $$\beta<0$$, then there exists a unique solution $$\beta=\beta(\alpha)$$ given by

$$\beta(\alpha)=\frac{\alpha}{2 [ (\frac{2}{\pi} )^{\alpha }-1 ]} \quad(\alpha\neq0), \qquad \beta(0)=\frac{1}{2\log \frac {2}{\pi}}$$
(2.12)

satisfies the equation $$\mathsf{D} (\alpha, \beta; {\frac{\pi }{2}}^{-} )=0$$ such that $$\mathsf{D} (\alpha, \beta; {\frac{\pi }{2}}^{-} )>0$$ for $$\beta<\beta(\alpha)$$ and $$\mathsf{D} (\alpha, \beta; {\frac{\pi}{2}}^{-} )<0$$ for $$\beta>\beta (\alpha)$$.

(2) If $$\beta<0$$ is fixed, then there exists a unique solution $$\alpha =\alpha(\beta)$$ satisfies the equation $$\mathsf{D} (\alpha, \beta; {\frac{\pi}{2}}^{-} )=0$$ such that $$\mathsf{D} (\alpha, \beta; {\frac{\pi}{2}}^{-} )>0$$ for $$\alpha<\alpha(\beta)$$ and $$\mathsf {D} (\alpha, \beta; {\frac{\pi}{2}}^{-} )<0$$ for $$\alpha >\alpha (\beta)$$. In particular, one has

$$\alpha_{0}=\alpha(-1)=-0.44367302\cdots, \qquad \alpha^{\ast }_{0}=\alpha \biggl(-\frac{37}{35} \biggr)=-0.20340978\cdots.$$
(2.13)

(3) The two functions $$\alpha\rightarrow\beta(\alpha)$$ and $$\beta \rightarrow\alpha(\beta)$$ are strictly decreasing.

### Proof

Part (1) follows easily from (2.9)-(2.11) and the fact that $$[(2/\pi)^{\alpha}-1]/\alpha<0$$.

(2) It follows from (2.9) and (2.11) that

$$\lim_{\alpha\rightarrow-\infty}\mathsf{D} \biggl(\alpha, \beta; { \frac {\pi}{2}}^{-} \biggr)=\infty,\qquad \lim_{\alpha\rightarrow\infty } \mathsf{D} \biggl(\alpha, \beta; {\frac{\pi}{2}}^{-} \biggr)=- \frac{1}{2}.$$
(2.14)

Note that

$$\frac{d}{d\alpha} \biggl[\frac{ (\frac{2}{\pi} )^{\alpha }-1}{\alpha} \biggr] = \frac{ (\frac{2}{\pi} )^{\alpha}}{\alpha^{2}} \biggl[\log \biggl(\frac{2}{\pi} \biggr)^{\alpha}+ \biggl(\frac{2}{\pi} \biggr)^{-\alpha }-1 \biggr]>0$$
(2.15)

for $$\alpha\neq0$$.

From (2.9), (2.11), and (2.15) we clearly see that the function $$\alpha \rightarrow\mathsf{D} (\alpha, \beta; {\frac{\pi }{2}}^{-} )$$ is strictly decreasing. Therefore, there exists a unique solution $$\alpha=\alpha(\beta)$$ that satisfies the equation $$\mathsf{D} (\alpha, \beta; {\frac{\pi}{2}}^{-} )=0$$ such that $$\mathsf {D} (\alpha, \beta; {\frac{\pi}{2}}^{-} )>0$$ for $$\alpha<\alpha (\beta )$$ and $$\mathsf{D} (\alpha, \beta; {\frac{\pi}{2}}^{-} )<0$$ for $$\alpha>\alpha(\beta)$$ follows from (2.14) and the monotonicity of the function $$\alpha\rightarrow\mathsf{D} (\alpha, \beta; {\frac {\pi}{2}}^{-} )$$. Numerical computations show that

$$\alpha(-1)=-0.44367302\cdots, \qquad\alpha \biggl(-\frac {37}{35} \biggr)=-0.20340978\cdots.$$

(3) The function $$\alpha\rightarrow\beta(\alpha)$$ is strictly decreasing follows easily from (2.12) and (2.15). The function $$\beta\rightarrow\alpha(\beta)$$ is strictly decreasing due to it is the inverse function of $$\alpha\rightarrow\beta(\alpha)$$. □

### Lemma 2.5

Let $$\beta(\alpha)$$ be defined by (2.12). Then

$$\alpha_{1}=-0.36131140\cdots$$
(2.16)

is the unique solution of the equation $$\beta(\alpha)=-\alpha/2-6/5$$ such that $$\beta(\alpha)<-\alpha/2-6/5$$ for $$\alpha<\alpha_{1}$$ and $$\beta(\alpha)>-\alpha/2-6/5$$ for $$\alpha>\alpha_{1}$$.

### Proof

Let $$P(\alpha)=\beta(\alpha)+\alpha/2+6/5$$. Then from (2.12) we clearly see that

\begin{aligned}& P(\alpha)=\frac{ (\frac{2}{\pi} )^{\alpha}}{2}\frac {\alpha }{ (\frac{2}{\pi} )^{\alpha}-1}+\frac{6}{5}, \\& \lim_{\alpha\rightarrow-\infty}P(\alpha)=-\infty, \qquad \lim_{\alpha \rightarrow\infty}P( \alpha)=\frac{6}{5}, \end{aligned}
(2.17)
\begin{aligned}& \frac{dP(\alpha)}{d\alpha}=-\frac{ (\frac{2}{\pi} )^{\alpha }}{2}\frac{\log (\frac{2}{\pi} )^{\alpha} - (\frac{2}{\pi} )^{\alpha}+1}{ [ (\frac {2}{\pi} )^{\alpha}-1 ]^{2}}>0 \end{aligned}
(2.18)

for $$\alpha\neq0$$, where the last of (2.18) due to $$\log x-x+1<0$$ for all $$x>0$$ with $$x\neq1$$.

Inequality (2.18) implies that the function $$\alpha\rightarrow P(\alpha )$$ is strictly increasing on $$(0, \infty)$$. Therefore, there exists a unique $$\alpha=\alpha_{1}$$ that satisfies the equation $$\beta(\alpha )=-\alpha/2-6/5$$ such that $$\beta(\alpha)<-\alpha/2-6/5$$ for $$\alpha <\alpha_{1}$$ and $$\beta(\alpha)>-\alpha/2-6/5$$ for $$\alpha>\alpha_{1}$$ follows from (2.17) and the monotonicity of the function $$\alpha \rightarrow P(\alpha)$$. Numerical computations show that $$\alpha _{1}=-0.36131140\cdots$$. □

### Lemma 2.6

Let $$\mathsf{Q}_{\alpha, \beta}(x)$$, $$\beta(\alpha)$$, $$\alpha_{0}$$ and $$\alpha^{\ast}_{0}$$ be defined by (2.7), (2.12), and (2.13), respectively. Then the following statements are true:

1. (1)

If $$\alpha\geq-2/7=-0.28571428\cdots$$, then the inequality $$\mathsf {Q}_{\alpha, \beta}(x)>1/2$$ holds for all $$x\in(0, \pi/2)$$ if and only if $$\beta\leq-\alpha/2-6/5$$.

2. (2)

If $$\alpha\geq\alpha^{\ast}_{0}$$, then the inequality $$\mathsf {Q}_{\alpha, \beta}(x)<1/2$$ holds for all $$x\in(0, \pi/2)$$ if and only if $$\beta\geq\beta(\alpha)$$.

3. (3)

If $$\alpha\leq-2/5$$, then the inequality $$\mathsf{Q}_{\alpha, \beta }(x)<1/2$$ holds for all $$x\in(0, \pi/2)$$ if and only if $$\beta\geq -\alpha/2-6/5$$.

4. (4)

If $$\alpha\leq\alpha_{0}$$, then the inequality $$\mathsf {Q}_{\alpha , \beta}(x)>1/2$$ holds for all $$x\in(0, \pi/2)$$ if and only if $$\beta \leq\beta(\alpha)$$.

### Proof

(1) If $$\alpha\geq-2/7$$ and $$\mathsf{Q}_{\alpha, \beta }(x)>1/2$$ for all $$x\in(0, \pi/2)$$, then from (2.5)-(2.7) one has

$$\lim_{x\rightarrow0^{+}}x^{-2} \biggl[\mathsf{Q}_{\alpha, \beta }(x)- \frac {1}{2} \biggr]=\lim_{x\rightarrow0^{+}}x^{-2} \biggl[- \frac{5\alpha +10\beta+12}{120}x^{2}+o \bigl(x^{2} \bigr) \biggr]=- \frac{5\alpha +10\beta +12}{120}\geq0,$$

which implies that $$\beta\leq-\alpha/2-6/5$$.

If $$\alpha\geq-2/7$$ and $$\beta\leq-\alpha/2-6/5$$, then we clearly see

$$\alpha+2\beta+\frac{12}{5}\leq0, \quad \beta\leq- \frac{37}{35}.$$
(2.19)

Therefore, $$\mathsf{Q}_{\alpha, \beta}(x)>1/2$$ for all $$x\in(0, \pi /2)$$ follows from Lemma 2.3(3) and (2.8) together with (2.19).

(2) If $$\alpha\geq\alpha^{\ast}_{0}$$ and $$\mathsf{Q}_{\alpha, \beta }(x)<1/2$$ for all $$x\in(0, \pi/2)$$, then from (2.11) and Lemma 2.4(1) we clearly see that $$\mathsf{D} (\alpha, \beta; {\frac{\pi }{2}}^{-} )\leq0$$ and $$\beta\geq\beta(\alpha)$$.

Next, we prove that $$\mathsf{Q}_{\alpha, \beta}(x)<1/2$$ for all $$x\in (0, \pi/2)$$ if $$\alpha\geq\alpha^{\ast}_{0}$$ and $$\beta\geq\beta (\alpha)$$. It follows from (2.6) and (2.7) together with the fact that

$$\frac{\partial\mathsf{J}_{\beta}(x)}{\partial\beta}=\frac{ (\frac {\tan x}{x} )^{\beta}}{\beta^{2}} \biggl[\log \biggl(\frac{\tan x}{x} \biggr)^{\beta}+ \biggl(\frac {x}{\tan x} \biggr)^{\beta}-1 \biggr]>0$$

for $$x\in(0, \pi/2)$$ and $$\beta\neq0$$ that the function $$\beta \rightarrow\mathsf{Q}_{\alpha, \beta}(x)$$ is strictly decreasing. Therefore, it suffices to prove that $$\mathsf{Q}_{\alpha, \beta }(x)<1/2$$ for all $$x\in(0, \pi/2)$$ if $$\alpha\geq\alpha^{\ast}_{0}$$ and $$\beta=\beta(\alpha)$$.

From (2.13) and Lemma 2.4(3) we get

$$\beta=\beta(\alpha)\leq\beta\bigl(\alpha^{\ast}_{0} \bigr)=-\frac{37}{35}.$$
(2.20)

Let $$\alpha\beta\neq0$$, $$\mathsf{F}(x)$$, $$\mathsf{G}(x)$$, $$\mathsf {H}(x)$$, $$g(x)$$, $$\mathsf{I}_{\alpha}(x)$$ and $$\mathsf{J}_{\beta}(x)$$ be defined by (2.1)-(2.6), respectively. Then simple computations lead to

\begin{aligned}& \biggl[\frac{\mathsf{I}^{\prime}_{\alpha}(x)}{\mathsf{J}^{\prime }_{\beta }(x)} \biggr]^{\prime}=-\frac{\cos^{\beta}x\sin^{\alpha-\beta -1}x}{(x-\sin x\cos x)^{2}}\bigl[g(x)+ \alpha\bigr]\mathsf{F}(x)x^{\beta-\alpha-1}, \end{aligned}
(2.21)
\begin{aligned}& \mathsf{J}^{\prime}_{\beta}(x)=\frac{2x-\sin(2x)}{2x^{2}\cos ^{2}x} \biggl( \frac{\tan x}{x} \biggr)^{\beta-1}>0 \end{aligned}
(2.22)

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

Let $$\alpha_{1}=-0.36131140\cdots$$ be defined by (2.16). Then it follows from Lemma 2.2(3), Lemma 2.5, and (2.20) together with $$\alpha \geq\alpha^{\ast}_{0}=-0.20340978\cdots>\alpha_{1}$$ that the function $$x\rightarrow g(x)+\alpha$$ is strictly decreasing on $$(0, \pi/2)$$ and

$$\lim_{x\rightarrow0^{+}}\bigl[g(x)+\alpha\bigr]=\alpha+2\beta+ \frac{12}{5}>0,\qquad \lim_{x\rightarrow{\frac{\pi}{2}}^{-}}\bigl[g(x)+\alpha\bigr]=- \infty.$$
(2.23)

From (2.21) and (2.23) together with the monotonicity of the function $$x\rightarrow g(x)+\alpha$$ on the interval $$(0, \pi/2)$$ we clearly see that there exists $$x_{0}\in(0, \pi/2)$$ such that the function $$x\rightarrow\mathsf{I}^{\prime}_{\alpha}(x)/\mathsf{J}^{\prime }_{\beta }(x)$$ is strictly decreasing on $$(0, x_{0})$$ and strictly increasing on $$(x_{0}, \pi/2)$$.

Note that

$$\frac{\mathsf{I}_{\alpha} ({\frac{\pi}{2}}^{-} )-\mathsf {I}_{\alpha}(0^{+})}{\mathsf{J}_{\beta(\alpha)} ({\frac{\pi}{2}}^{-} )-\mathsf{J}_{\beta(\alpha)}(0^{+})} =\mathsf{Q}_{\alpha, \beta(\alpha)} \biggl({ \frac{\pi }{2}}^{-} \biggr)=\mathsf{D} \biggl(\alpha, \beta( \alpha); {\frac{\pi }{2}}^{-} \biggr)+\frac{1}{2}= \frac{1}{2}.$$
(2.24)

Therefore, $$\mathsf{Q}_{\alpha, \beta}(x)<1/2$$ for all $$x\in(0, \pi /2)$$ follows from Lemma 2.1, (2.7), (2.22), (2.24), and the piecewise monotonicity of the function $$x\rightarrow\mathsf{I}^{\prime }_{\alpha }(x)/\mathsf{J}^{\prime}_{\beta}(x)$$ on the interval $$(0, \pi/2)$$.

(3) If $$\alpha\leq-2/5$$ and $$\mathsf{Q}_{\alpha, \beta}(x)<1/2$$ for all $$x\in(0, \pi/2)$$, then from (2.5)-(2.7) we have

$$\lim_{x\rightarrow0^{+}}x^{-2} \biggl[\mathsf{Q}_{\alpha, \beta }(x)- \frac {1}{2} \biggr]=\lim_{x\rightarrow0^{+}}x^{-2} \biggl[- \frac{5\alpha +10\beta+12}{120}x^{2}+o \bigl(x^{2} \bigr) \biggr]=- \frac{5\alpha +10\beta +12}{120}\leq0,$$

which implies that $$\beta\geq-\alpha/2-6/5$$.

If $$\alpha\leq-2/5$$ and $$\beta\geq-\alpha/2-6/5$$, then we clearly see that

$$\alpha+2\beta+\frac{12}{5}\geq0, \quad \beta\geq-1.$$
(2.25)

Therefore, $$\mathsf{Q}_{\alpha, \beta}(x)<1/2$$ for $$x\in(0, \pi/2)$$ follows easily from Lemma 2.3(1), (2.8), and (2.25).

(4) If $$\alpha\leq\alpha_{0}$$ and $$\mathsf{Q}_{\alpha, \beta}(x)>1/2$$ for all $$x\in(0, \pi/2)$$, then (2.11) and Lemma 2.4(1) lead to the conclusion that $$\mathsf{D} (\alpha, \beta; {\frac{\pi }{2}}^{-} )\geq0$$ and $$\beta\leq\beta(\alpha)$$.

Next, we prove that $$\mathsf{Q}_{\alpha, \beta}(x)>1/2$$ for all $$x\in (0, \pi/2)$$ if $$\alpha\leq\alpha_{0}$$ and $$\beta\leq\beta(\alpha)$$. Since the function $$\beta\rightarrow\mathsf{Q}_{\alpha, \beta}(x)$$ is strictly decreasing which was proved in part (2), we only need to prove that $$\mathsf{Q}_{\alpha, \beta}(x)>1/2$$ for all $$x\in(0, \pi/2)$$ if $$\alpha\leq\alpha_{0}$$ and $$\beta=\beta(\alpha)$$. It follows from Lemma 2.2(1) and (2), Lemma 2.4(3), Lemma 2.5, and $$\alpha\leq\alpha _{0}<\alpha_{1}$$ that $$\beta\geq\beta(\alpha_{0})=-1$$ and the function $$g(x)+\alpha$$ is strictly increasing on $$(0, \pi/2)$$ such that

\begin{aligned}& \lim_{x\rightarrow0^{+}}\bigl[g(x)+\alpha\bigr]=\alpha+2\beta+ \frac{12}{5}< 0, \end{aligned}
(2.26)
\begin{aligned}& \begin{aligned}[b] \lim_{x\rightarrow{\frac{\pi}{2}}^{-}}\bigl[g(x)+\alpha\bigr]&= \textstyle\begin{cases} \alpha+\infty, & \beta(\alpha)>-1, \\ \alpha+3-\frac{\pi^{2}}{4}, & \beta(\alpha)=-1, \end{cases}\displaystyle \\ &= \textstyle\begin{cases} \infty, & \beta>-1, \\ \alpha_{0}+3-\frac{\pi^{2}}{4}>0, & \beta=-1. \end{cases}\displaystyle \end{aligned} \end{aligned}
(2.27)

From (2.21), (2.26), and (2.27) we clearly see that there exists $$x^{\ast}\in(0, \pi/2)$$ such that the function $$x\rightarrow\mathsf {I}^{\prime}_{\alpha}(x)/\mathsf{J}^{\prime}_{\beta}(x)$$ is strictly increasing on $$(0, x^{\ast})$$ and strictly decreasing on $$(x^{\ast}, \pi /2)$$. Therefore, $$\mathsf{Q}_{\alpha, \beta}(x)>1/2$$ for all $$x\in(0, \pi/2)$$ follows from Lemma 2.1, (2.7), (2.22), (2.24), and the piecewise monotonicity of the function $$x\rightarrow\mathsf {I}^{\prime }_{\alpha}(x)/\mathsf{J}^{\prime}_{\beta}(x)$$ on the interval $$(0, \pi /2)$$. □

### Lemma 2.7

Let $$\mathsf{Q}_{\alpha, \beta}(x)$$, $$\alpha_{0}$$, $$\alpha^{\ast}_{0}$$ and $$\alpha(\beta)$$ be defined by (2.7) and Lemma  2.4, respectively. Then the following statements are true:

1. (1)

If $$\beta\geq-1$$, then the inequality $$\mathsf{Q}_{\alpha, \beta }(x)<1/2$$ holds for all $$x\in(0, \pi/2)$$ if and only if $$\alpha\geq -2\beta-12/5$$.

2. (2)

If $$-1\leq\beta<0$$, then the inequality $$\mathsf{Q}_{\alpha, \beta }(x)>1/2$$ holds for all $$x\in(0, \pi/2)$$ if and only if $$\alpha\leq \alpha(\beta)$$.

3. (3)

If $$\beta\leq-37/35$$, then the inequality $$\mathsf{Q}_{\alpha, \beta}(x)>1/2$$ holds for all $$x\in(0, \pi/2)$$ if and only if $$\alpha \leq-2\beta-12/5$$.

4. (4)

If $$\beta\leq-37/35$$, then the inequality $$\mathsf{Q}_{\alpha, \beta}(x)<1/2$$ holds for all $$x\in(0, \pi/2)$$ if and only if $$\alpha \geq\alpha(\beta)$$.

### Proof

(1) If $$\beta\geq-1$$ and $$\mathsf{Q}_{\alpha, \beta }(x)<1/2$$ for all $$x\in(0, \pi/2)$$, then from (2.5)-(2.7) we get

$$\lim_{x\rightarrow0^{+}}x^{-2} \biggl[\mathsf{Q}_{\alpha, \beta }(x)- \frac {1}{2} \biggr]=\lim_{x\rightarrow0^{+}}x^{-2} \biggl[- \frac{5\alpha +10\beta+12}{120}x^{2}+o \bigl(x^{2} \bigr) \biggr]=- \frac{5\alpha +10\beta +12}{120}\leq0,$$

which implies that $$\alpha\geq-2\beta-12/5$$.

If $$\beta\geq-1$$ and $$\alpha\geq-2\beta-12/5$$, then $$\mathsf {Q}_{\alpha, \beta}(x)<1/2$$ for all $$x\in(0, \pi/2)$$ follows from (2.8) and Lemma 2.3(1).

(2) If $$-1\leq\beta<0$$ and $$\mathsf{Q}_{\alpha, \beta}(x)>1/2$$ for all $$x\in(0, \pi/2)$$, then (2.11) and Lemma 2.4(2) lead to the conclusion that $$\mathsf{D} (\alpha,\beta;{\frac{\pi}{2}}^{-} )\geq0$$ and $$\alpha\leq\alpha(\beta)$$.

Next, we prove that $$\mathsf{Q}_{\alpha, \beta}(x)>1/2$$ for all $$x\in (0, \pi/2)$$ if $$-1\leq\beta<0$$ and $$\alpha\leq\alpha(\beta)$$. It follows from $$-1\leq\beta<0$$ and $$\alpha\leq\alpha(\beta)$$ together with Lemma 2.4(3) that

$$\alpha\leq\alpha(-1)=\alpha_{0}, \quad\beta\leq\beta( \alpha).$$
(2.28)

Therefore, $$\mathsf{Q}_{\alpha, \beta}(x)>1/2$$ for all $$x\in(0, \pi /2)$$ follows from Lemma 2.6(4) and (2.28).

(3) If $$\beta\leq-37/35$$ and $$\mathsf{Q}_{\alpha, \beta}(x)>1/2$$ for all $$x\in(0, \pi/2)$$, then from (2.5)-(2.7) we have

$$\lim_{x\rightarrow0^{+}}x^{-2} \biggl[\mathsf{Q}_{\alpha, \beta }(x)- \frac {1}{2} \biggr]=\lim_{x\rightarrow0^{+}}x^{-2} \biggl[- \frac{5\alpha +10\beta+12}{120}x^{2}+o \bigl(x^{2} \bigr) \biggr]=- \frac{5\alpha +10\beta +12}{120}\geq0,$$

which implies that $$\alpha\leq-2\beta-12/5$$.

If $$\beta\leq-37/35$$ and $$\alpha\leq-2\beta-12/5$$, then $$\mathsf {Q}_{\alpha, \beta}(x)>1/2$$ for all $$x\in(0, \pi/2)$$ follows from (2.8) and Lemma 2.3(3).

(4) If $$\beta\leq-37/35$$ and $$\mathsf{Q}_{\alpha, \beta}(x)<1/2$$ for all $$x\in(0, \pi/2)$$, then (2.11) and Lemma 2.4(2) lead to the conclusion that $$\mathsf{D} (\alpha, \beta; {\frac{\pi}{2}}^{-} )\leq 0$$ and $$\alpha\geq\alpha(\beta)$$.

Next, we prove that $$\mathsf{Q}_{\alpha, \beta}(x)<1/2$$ for all $$x\in (0, \pi/2)$$ if $$\beta\leq-37/35$$ and $$\alpha\geq\alpha(\beta)$$. It follows from $$\beta\leq-37/35$$ and $$\alpha\geq\alpha(\beta)$$ together with Lemma 2.4(3) that

$$\alpha\geq\alpha \biggl(-\frac{37}{35} \biggr)= \alpha^{\ast}_{0},\quad \beta\geq\beta(\alpha).$$
(2.29)

Therefore, the desired result follows from Lemma 2.6(2) and (2.29). □

## Main results

Let $$\alpha, \beta\in\mathbb{R}$$ with $$\alpha\beta(\alpha+2\beta )\neq 0$$ and $$\mathsf{Q}_{\alpha, \beta}(x)$$ be defined by (2.7), then we clearly see that the generalized Wilker-type inequality

$$\frac{2\beta}{\alpha+2\beta} \biggl(\frac{\sin x}{x} \biggr)^{\alpha}+ \frac {\alpha}{\alpha+2\beta} \biggl(\frac{\tan x}{x} \biggr)^{\beta}-1>0$$
(3.1)

holds for all $$x\in(0, \pi/2)$$ if and only if $$\mathsf{Q}_{\alpha, \beta}(x)<1/2$$ and $$\alpha\beta(\alpha+2\beta)> 0$$ or $$\mathsf {Q}_{\alpha, \beta}(x)>1/2$$ and $$\alpha\beta(\alpha+2\beta)<0$$, while the generalized Wilker-type inequality

$$\frac{2\beta}{\alpha+2\beta} \biggl(\frac{\sin x}{x} \biggr)^{\alpha}+ \frac {\alpha}{\alpha+2\beta} \biggl(\frac{\tan x}{x} \biggr)^{\beta}-1< 0$$
(3.2)

holds for all $$x\in(0, \pi/2)$$ if and only if $$\mathsf{Q}_{\alpha, \beta}(x)<1/2$$ and $$\alpha\beta(\alpha+2\beta)<0$$ or $$\mathsf {Q}_{\alpha , \beta}(x)>1/2$$ and $$\alpha\beta(\alpha+2\beta)>0$$.

From Lemmas 2.6 and 2.7 together with inequalities (3.1) and (3.2) we get Theorems 3.1 and 3.2 immediately.

### Theorem 3.1

Let $$\alpha, \beta\in\mathbb{R}$$ with $$\alpha \beta (\alpha+2\beta)\neq0$$, $$\beta(\alpha)$$, $$\alpha_{0}$$ and $$\alpha ^{\ast }_{0}$$ be defined by (2.12) and (2.13), respectively. Then the following statements are true:

1. (1)

If $$\alpha\geq-2/7$$, then inequality (3.1) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha, \beta)\in\{(\alpha, \beta)|\beta \leq -\alpha/2-6/5, \alpha\beta(\alpha+2\beta)<0\}$$ and inequality (3.2) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha, \beta)\in\{ (\alpha, \beta)|\beta\leq-\alpha/2-6/5, \alpha\beta(\alpha +2\beta)>0\}$$.

2. (2)

If $$\alpha\geq\alpha^{\ast}_{0}$$, then inequality (3.1) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha, \beta)\in\{(\alpha, \beta)|\beta\geq\beta(\alpha), \alpha\beta(\alpha+2\beta)>0\}$$ and inequality (3.2) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha , \beta)\in\{(\alpha, \beta)|\beta\geq\beta(\alpha), \alpha \beta (\alpha+2\beta)<0\}$$.

3. (3)

If $$\alpha\leq-2/5$$, then inequality (3.1) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha, \beta)\in\{(\alpha, \beta)|\beta \geq -\alpha/2-6/5, \alpha\beta(\alpha+2\beta)>0\}$$ and inequality (3.2) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha, \beta)\in\{ (\alpha, \beta)|\beta\geq-\alpha/2-6/5, \alpha\beta(\alpha +2\beta)<0\}$$.

4. (4)

If $$\alpha\leq\alpha_{0}$$, then inequality (3.1) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha,\beta)\in\{(\alpha,\beta )|\beta\leq\beta(\alpha),\alpha\beta(\alpha+2\beta)<0\}$$ and inequality (3.2) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha ,\beta)\in\{(\alpha,\beta)|\beta\leq\beta(\alpha), \alpha\beta (\alpha +2\beta)>0\}$$.

### Theorem 3.2

Let $$\alpha,\beta\in\mathbb{R}$$ with $$\alpha \beta (\alpha+2\beta)\neq0$$, $$\alpha_{0}$$, $$\alpha^{\ast}_{0}$$, and $$\alpha (\beta)$$ be defined by Lemma  2.4. Then the following statements are true:

1. (1)

If $$\beta\geq-1$$, then inequality (3.1) holds for all $$x\in(0, \pi /2)$$ if and only if $$(\alpha,\beta)\in\{(\alpha,\beta)|\alpha\geq -2\beta-12/5, \alpha\beta(\alpha+2\beta)>0\}$$ and inequality (3.2) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha,\beta)\in\{ (\alpha,\beta)|\alpha\geq-2\beta-12/5, \alpha\beta(\alpha +2\beta)<0\}$$.

2. (2)

If $$-1\leq\beta<0$$, then inequality (3.1) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha,\beta)\in\{(\alpha,\beta)|\alpha \leq \alpha(\beta),\alpha\beta(\alpha+2\beta)<0\}$$ and inequality (3.2) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha,\beta)\in\{ (\alpha,\beta)|\alpha\leq\alpha(\beta), \alpha\beta(\alpha +2\beta)>0\}$$.

3. (3)

If $$\beta\leq-37/35$$, then inequality (3.1) holds for all $$x\in (0, \pi/2)$$ if and only if $$(\alpha,\beta)\in\{(\alpha,\beta )|\alpha \leq-2\beta-12/5, \alpha\beta(\alpha+2\beta)<0\}\cup\{(\alpha ,\beta )|\alpha\geq\alpha(\beta), \alpha\beta(\alpha+2\beta)>0\}$$ and inequality (3.2) holds for all $$x\in(0, \pi/2)$$ if and only if $$(\alpha ,\beta)\in\{(\alpha,\beta)|\alpha\leq-2\beta-12/5, \alpha\beta (\alpha +2\beta)>0\}\cup\{(\alpha,\beta)|\alpha\geq\alpha(\beta), \alpha\beta (\alpha+2\beta)<0\}$$.

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## Acknowledgements

The research was supported by the Natural Science Foundation of China under Grants 11371125, 61374086, and 11401191.

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Correspondence to Yu-Ming Chu.

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All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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