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

1 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 [3] 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 [4] generalized inequality (1.2) to the Bessel functions.

In [5], 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 [6] presented the best possible parameter p such that the second inequality of (1.3) holds for $x\in(0, \pi/2)$.

Zhu [7] 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 [8], 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. [9] 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 $x\in(0, \pi/2)$.

2 Lemmas

Lemma 2.1

See [10], 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 [9], 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) The function $g(x)$ is strictly increasing from $(0, \pi/2)$ onto $(2\beta+12/5, 3-\pi^{2}/4)$ if $\beta=-1$.

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

• (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 [9], 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) 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) 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) 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) 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) 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) 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) 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) 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) 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) 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) 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). □

3 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) 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) 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) 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) 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) 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) 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) 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 $(\alpha ,\beta)\in\{(\alpha,\beta)|\alpha\leq-2\beta-12/5, \alpha\beta (\alpha +2\beta)>0\}\cup\{(\alpha,\beta)|\alpha\geq\alpha(\beta), \alpha\beta (\alpha+2\beta)<0\}$.