1 Introduction

The well-known Wilker inequality \((\sin x/x)^{2}+\tan x/x>2\) for all \(x\in(0, \pi/2)\) was proposed by Wilker [1] and proved by Sumner et al. [2].

Recently, the Wilker inequality has attracted the attention of many researchers. Many generalizations, improvements, and refinements of the Wilker inequality can be found in the literature [3–10].

Pinelis [11] and Sun and Zhu [12] proved that the inequalities

$$ \biggl(\frac{\sin x}{x} \biggr)^{2}+\frac{\tan x}{x}-2>\lambda x^{3}\tan x\quad \mbox{and}\quad \biggl(\frac{y}{\sinh y} \biggr)^{2} + \frac{y}{\tanh y}-2< \mu y^{3}\sinh y $$

hold for all \(x\in(0, \pi/2)\) and \(y>0\) if and only if \(\lambda\leq 8/45\) and \(\mu\geq2/45\).

Wu and Srivastava [13] provided polynomials \(P_{1}(x)\) and \(P_{2}(x)\) of degree \(2n+3\) \((n\in\mathbb{N})\) with explicit expressions and coefficients concerning Bernoulli numbers such that the double inequality

$$ P_{1}(x)\tan x< \biggl(\frac{\sin x}{x} \biggr)^{2}+ \frac{\tan x}{x}-2< P_{2}(x)\tan x $$

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

Yang [14] proved that \(p=5/3\) and \(q=\log2/[2(\log\pi-\log2)]\) are the best possible parameters such that the double inequality

$$ \biggl(\frac{\sqrt{\cos^{2p}x+8}+\cos^{p}x}{4} \biggr)^{1/p}< \frac{\sin x}{x}< \biggl( \frac{\sqrt{\cos^{2q}x+8}+\cos^{q}x}{4} \biggr)^{1/q} $$

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

Very recently, Yang and Chu [15] proved that the Wilker-type inequality

$$ \frac{2}{k+2} \biggl(\frac{\sin x}{x} \biggr)^{kp}+ \frac{k}{k+2} \biggl(\frac{\tan x}{x} \biggr)^{p}>(< )1 $$

holds for any fixed \(k\geq1\) and all \(x\in(0, \pi/2)\) if and only if \(p>0\) or \(p\leq[\log2-\log(k+2)]/[k(\log\pi-\log2)]\) (\(-12/[5(k+2)]\leq p<0\)), and the hyperbolic version of Wilker-type inequality

$$ \frac{2}{k+2} \biggl(\frac{\sinh x}{x} \biggr)^{kp}+ \frac{k}{k+2} \biggl(\frac{\tanh x}{x} \biggr)^{p}>(< )1 $$

holds for any fixed \(k\geq1\) (\({<}-2\)) and all \(x\in(0, \infty)\) if and only if \(p>0\) or \(p\leq-12/[5(k+2)]\) (\(p<0\) or \(p\geq -12/[5(k+2)]\)).

More results of the Wilker-type inequalities for hyperbolic, Bessel, circular, inverse trigonometric, inverse hyperbolic, lemniscate, generalized trigonometric, generalized hyperbolic, Jacobian elliptic and theta, and hyperbolic Fibonacci functions can be found in the literature [16–28].

The main purpose of the article is to establish the Wilker-type inequalities

$$ \frac{2q}{p+2q} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \frac{p}{p+2q} \biggl(\frac{\sin x}{x} \biggr)^{q}>(< )1 $$

and

$$ \biggl(\frac{\pi}{2} \biggr)^{p} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \biggl[1- \biggl(\frac{\pi}{2} \biggr)^{p} \biggr] \biggl(\frac{\tan x}{x} \biggr)^{q}>(< )1 $$

for all \(x\in(0, \pi/2)\) and certain \(p, q\in\mathbb{R}\). Some complicated analytical computations are carried out using the computer algebra system Mathematica.

2 Lemmas

In order to prove our main results, we need several lemmas.

Lemma 2.1

(See [29, 30])

Let \(-\infty< a< b<\infty\), \(f, g: [a, b]\rightarrow\mathbb{R}\) be continuous on \([a, b]\) and differentiable on \((a, b)\), and \(g^{\prime}(x)\neq0\) on \((a, b)\). Then both of the functions

$$ \frac{f(x)-f(a)}{g(x)-g(a)} \quad\textit{and} \quad\frac{f(x)-f(b)}{g(x)-g(b)} $$

are increasing (decreasing) on \((a, b)\) if \(f^{\prime}(x)/g^{\prime}(x)\) is increasing (decreasing) on \((a, b)\). If \(f^{\prime}(x)/g^{\prime}(x)\) is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2.2

(See [31])

Let \(A(t)=\sum_{k=0}^{\infty }a_{k}t^{k}\) and \(B(t)=\sum_{k=0}^{\infty}b_{k}t^{k}\) be two real power series converging on \((-r,r)\) (\(r>0\)) with \(b_{k}>0\) for all k. If the nonconstant sequence \(\{a_{k}/b_{k}\}\) is increasing (decreasing) for all k, then the function \(t\mapsto A(t)/B(t)\) is strictly increasing (decreasing) on \((0,r)\).

Lemma 2.3

(See [32])

Let \(n\in\mathbb{N}\), and \(B_{n}\) be the Bernoulli numbers. Then the power series formulas

$$\begin{aligned}& \frac{1}{\sin x}=\frac{1}{x}+\sum_{n=1}^{\infty} \frac{2^{2n}-2}{(2n)!}|B_{2n}|x^{2n-1},\qquad \cot x=\frac{1}{x}- \sum_{n=1}^{\infty}\frac{2^{2n}}{(2n)!}|B_{2n}|x^{2n-1},\\& \frac{1}{\sin^{2} x}=\frac{1}{x^{2}}+\sum_{n=1}^{\infty} \frac {(2n-1)2^{2n}}{(2n)!}|B_{2n}|x^{2n-2}, \\& \frac{\cos x}{\sin^{3} x}= \frac{1}{x^{3}}-\sum_{n=1}^{\infty} \frac {n(2n+1)2^{2n+2}}{(2n+2)!}|B_{2n+2}|x^{2n-1} \end{aligned}$$

hold for \(x\in(-\pi, \pi)\), and the power series formulas

$$ \tan x=\sum_{n=1}^{\infty}\frac{ (2^{2n}-1 )2^{2n}}{(2n)!}|B_{2n}|x^{2n-1},\qquad \frac{1}{\cos^{2} x}=\sum_{n=1}^{\infty} \frac{(2n-1) (2^{2n}-1 )2^{2n}}{(2n)!}|B_{2n}|x^{2n-2} $$

hold for \(x\in(-\pi/2, \pi/2)\).

Lemma 2.4

(See [33])

Let \(B_{n}\) be the Bernoulli numbers. Then the double inequality

$$ \frac{2(2n)!}{(2\pi)^{2n}}< |B_{2n}|< \frac{2^{2n-1}}{2^{2n-1}-1}\frac {2(2n)!}{(2\pi)^{2n}} $$

holds for all \(n\in\mathbb{N}\).

From Lemma 2.4 we immediately get the following:

Remark 2.1

Let \(B_{n}\) be the Bernoulli numbers. Then the double inequality

$$ \frac{2^{2n-1}-1}{2^{2n-1}}\frac{(2\pi)^{2}}{2n(2n-1)}< \frac {|B_{2n-2}|}{|B_{2n}|}< \frac{2^{2n-3}}{2^{2n-3}-1} \frac{(2\pi)^{2}}{2n(2n-1)} $$

holds for all \(n\in\mathbb{N}\) and \(n\geq1\).

Lemma 2.5

Let \(n\in\mathbb{N}\), \(B_{n}\) be the Bernoulli numbers, and \(a_{n}\) and \(b_{n}\) be respectively defined by

$$\begin{aligned}& a_{n}=2^{2n}-2n^{2}-3n-2, \end{aligned}$$
(2.1)
$$\begin{aligned}& b_{n}=(2n-3)2^{2n}+2n^{2}+n+4-n(2n-1) \bigl(2^{2n-3}-1 \bigr)\frac {|B_{2n-2}|}{|B_{2n}|}. \end{aligned}$$
(2.2)

Then the sequence \(\{b_{n}/a_{n}\}\) is strictly increasing for \(n\geq3\).

Proof

Let \(n\geq3\) and

$$ u_{n}=\frac{b_{n+1}}{a_{n+1}}-\frac{b_{n}}{a_{n}}. $$
(2.3)

Then from (2.1)-(2.3) and Remark 2.1 we get

$$\begin{aligned} u_{n}={}&\frac{(2n-1)2^{2n+2}+2n^{2}+5n+7}{2^{2n+2}-2n^{2}-7n-7}-\frac {(n+1)(2n+1) (2^{2n-1}-1 )}{ 2^{2n+2}-2n^{2}-7n-7}\frac{|B_{2n}|}{|B_{2n+2}|} \\ &{}-\frac{(2n-3)2^{2n}+2n^{2}+n+4}{2^{2n}-2n^{2}-3n-2}+\frac{n(2n-1) (2^{2n-3}-1 )}{ 2^{2n}-2n^{2}-3n-2}\frac{|B_{2n-2}|}{|B_{2n}|} \\ >{}&\frac{2}{a_{n}a_{n+1}} \bigl[4\times 2^{4n}-\bigl(6n^{3}+7n^{2}+5n+11 \bigr)2^{2n}-\bigl(2n^{2}-2n-7\bigr) \bigr] \\ &{}+\frac{\pi^{2}}{2^{2n+2}}\frac {(6n^{2}+5n-39)2^{4n}+(20n^{2}+70n+134)2^{2n}-(32n^{2}+112n+112)}{a_{n}a_{n+1}}. \end{aligned}$$
(2.4)

Let

$$ u_{n}^{\ast}=4\times 2^{4n}-\bigl(6n^{3}+7n^{2}+5n+11 \bigr)2^{2n}-\bigl(2n^{2}-2n-7\bigr). $$
(2.5)

Then we clearly see that

$$\begin{aligned}& u_{3}^{\ast}=315>0, \end{aligned}$$
(2.6)
$$\begin{aligned}& u_{n+1}^{\ast}-16u_{n}^{\ast }= \bigl(18n^{3}+3n^{2}-17n+15\bigr)2^{2n+2}+ \bigl(30n^{2}-34n-105\bigr)>0 \end{aligned}$$
(2.7)

for \(n\geq3\).

It follows from (2.6) and (2.7) that

$$ u_{n}^{\ast}>0 $$
(2.8)

for all \(n\geq3\).

It is not difficult to verify that

$$ a_{n}>0 $$
(2.9)

and

$$ \bigl(6n^{2}+5n-39\bigr)2^{4n}+\bigl(20n^{2}+70n+134 \bigr)2^{2n}-\bigl(32n^{2}+112n+112\bigr)>0 $$
(2.10)

for all \(n\geq3\).

Therefore, Lemma 2.5 follows easily from (2.3)-(2.5) and (2.8)-(2.10). □

Lemma 2.6

Let \(n\in\mathbb{N}\), \(B_{n}\) be the Bernoulli numbers, \(u_{n}\) be defined by (2.3), and \(c_{n}\) and \(v_{n}\) be respectively defined by

$$\begin{aligned}& c_{n}=2n\bigl(2^{2n}-1\bigr)-2n(2n-1) \bigl(2^{2n-3}-1 \bigr)\frac{|B_{2n-2}|}{|B_{2n}|}, \end{aligned}$$
(2.11)
$$\begin{aligned}& v_{n}=\frac{c_{n+1}}{a_{n+1}}-\frac{c_{n}}{a_{n}}. \end{aligned}$$
(2.12)

Then \(v_{n}>u_{n}\) for all \(n\geq3\).

Proof

It follows from (2.1)-(2.3), (2.11), and (2.12) that

$$\begin{aligned} v_{n}-u_{n}={}&{-}\frac{2[(6n^{2}+5n-2)2^{2n}+4n+5]}{a_{n}a_{n+1}}+\frac {n(2n-1)(2^{2n-3}-1)}{2^{2n}-2n^{2}-3n-2} \frac {|B_{2n-2}|}{|B_{2n}|} \\ &{}-\frac{(n+1)(2n+1)(2^{2n-1}-1)}{2^{2n+2}-2n^{2}-7n-7}\frac {|B_{2n}|}{|B_{2n+2}|}. \end{aligned}$$
(2.13)

From (2.13), Remark 2.1, and the inequality \(\pi^{2}>9\) we get

$$\begin{aligned}& v_{3}-u_{3}=\frac{8}{105},\qquad v_{4}-u_{4}= \frac{104}{3{,}045}, \qquad v_{5}-u_{5}=\frac{15{,}496}{1{,}102{,}145}, \end{aligned}$$
(2.14)
$$\begin{aligned}& v_{6}-u_{6}=\frac{23{,}139{,}208}{4{,}326{,}527{,}205},\qquad v_{7}-u_{7}= \frac{2{,}511{,}041{,}224}{1{,}319{,}700{,}084{,}885}, \end{aligned}$$
(2.15)
$$\begin{aligned}& v_{n}-u_{n} \\& \quad>-\frac{2[(6n^{2}+5n-2)2^{2n}+4n+5]}{a_{n}a_{n+1}} \\& \qquad{}+\frac{n(2n-1)(2^{2n-3}-1)}{2^{2n}-2n^{2}-3n-2}\frac {2^{2n-1}-1}{2^{2n-1}}\frac{(2\pi)^{2}}{2n(2n-1)} \\& \qquad{}-\frac{(n+1)(2n+1)(2^{2n-1}-1)}{2^{2n+2}-2n^{2}-7n-7}\frac {2^{2n-1}}{2^{2n-1}-1}\frac{(2\pi)^{2}}{(2n+1)(2n+2)} \\& \quad=-\frac{2[(6n^{2}+5n-2)2^{2n}+4n+5]}{a_{n}a_{n+1}} \\& \qquad{}+\frac{\pi^{2}}{2^{2n+2}}\frac {(6n^{2}+5n-39)2^{4n}+(20n^{2}+70n+134)2^{2n}-(32n^{2}+112n+112)}{a_{n}a_{n+1}} \\& \quad>-\frac{2[(6n^{2}+5n-2)2^{2n}+4n+5]}{a_{n}a_{n+1}} \\& \qquad{}+\frac{81}{2^{2n+2}}\frac {(6n^{2}+5n-39)2^{4n}+(20n^{2}+70n+134)2^{2n}-(32n^{2}+112n+112)}{a_{n}a_{n+1}} \\& \quad=\frac {(6n^{2}+5n-335)2^{4n}+(180n^{2}+598n+1{,}166)2^{2n}-9(32n^{2}+112n+112)}{a_{n}a_{n+1}2^{2n+2}}. \end{aligned}$$
(2.16)

Note that

$$ a_{n}>0,\quad \bigl(180n^{2}+598n+1{,}166\bigr)2^{2n}-9 \bigl(32n^{2}+112n+112\bigr)>0, $$
(2.17)

and

$$ 6n^{2}+5n-335\geq6\times8^{2}+5\times8-335=89 $$
(2.18)

for all \(n\geq8\).

Therefore, Lemma 2.6 follows easily from (2.14)-(2.18). □

Lemma 2.7

Let \(n\in\mathbb{N}\), and \(w_{n}\) be defined by

$$\begin{aligned} w_{n}={}&32\times 2^{6n}- \bigl(48n^{3}+206n^{2}+165n+2{,}183 \bigr)2^{4n}\\ &{}+ \bigl(3{,}284n^{2}+5{,}526n+4{,}716 \bigr)2^{2n}- \bigl(1{,}320n^{2}+1{,}980n+1{,}320 \bigr). \end{aligned}$$

Then \(w_{n}>0\) for all \(n\geq5\).

Proof

Let

$$ w_{n}^{\ast}=32\times4^{n}- \bigl(48n^{3}+206n^{2}+165n+2{,}183 \bigr). $$

Then we clearly see that

$$\begin{aligned}& w_{n}=2^{4n}w_{n}^{\ast}+ \bigl(3{,}284n^{2}+5{,}526n+4{,}716 \bigr)2^{2n}- \bigl(1{,}320n^{2}+1{,}980n+1{,}320 \bigr), \end{aligned}$$
(2.19)
$$\begin{aligned}& w_{5}^{\ast}=18{,}160>0, \end{aligned}$$
(2.20)
$$\begin{aligned}& w_{n+1}^{\ast}-4w_{n}^{\ast}=144n^{3}+474n^{2}-61n+6{,}130>0 \end{aligned}$$
(2.21)

for all \(n\geq5\).

Inequalities (2.20) and (2.21) lead to the conclusion that

$$ w_{n}^{\ast}>0 $$
(2.22)

for all \(n\geq5\).

Note that

$$ \bigl(3{,}284n^{2}+5{,}526n+4{,}716 \bigr)2^{2n}- \bigl(1{,}320n^{2}+1{,}980n+1{,}320 \bigr)>0 $$
(2.23)

for all \(n\geq5\).

Therefore, Lemma 2.7 follows from (2.19), (2.22), and (2.23). □

Lemma 2.8

Let \(n\in\mathbb{N}\), and \(u_{n}\) and \(v_{n}\) be defined by (2.3) and (2.12), respectively. Then \(v_{3}=37u_{3}/35\) and \(v_{n}<37u_{n}/35\) for all \(n\geq4\).

Proof

It follows from (2.1)-(2.3), (2.11), and (2.12) that

$$\begin{aligned} &35v_{n}-37u_{n} \\ &\quad=-\frac{2 [8\times 2^{4n}- (12n^{3}-196n^{2}-165n+92 )2^{2n}- (4n^{2}-144n-189 ) ]}{ (2^{2n}-2n^{2}-3n-2 ) (2^{2n+2}-2n^{2}-7n-7 )} \\ &\qquad{}-\frac{33(2n+1)(n+1) (2^{2n-1}-1 )}{2^{2n+2}-2n^{2}-7n-7}\frac {|B_{2n}|}{|B_{2n+2}|} +\frac{33n(2n-1) (2^{2n-3}-1 )}{2^{2n}-2n^{2}-3n-2}\frac {|B_{2n-2}|}{|B_{2n}|}. \end{aligned}$$
(2.24)

From Remark 2.1, (2.24), and the inequality \(\pi^{2}<10\) we get

$$\begin{aligned}& 35v_{3}-37u_{3}=0,\qquad 35v_{4}-37u_{4}=- \frac{288}{145}, \end{aligned}$$
(2.25)
$$\begin{aligned}& 35v_{n}-37u_{n} \\& \quad< -\frac{2 [8\times 2^{4n}- (12n^{3}-196n^{2}-165n+92 )2^{2n}- (4n^{2}-144n-189 ) ]}{ (2^{2n}-2n^{2}-3n-2 ) (2^{2n+2}-2n^{2}-7n-7 )} \\& \qquad{}-\frac{33(2n+1)(n+1) (2^{2n-1}-1 )}{2^{2n+2}-2n^{2}-7n-7}\frac {2^{2n+1}-1}{2^{2n+1}}\frac{(2\pi)^{2}}{(2n+1)(2n+2)} \\& \qquad{}+\frac{33n(2n-1) (2^{2n-3}-1 )}{2^{2n}-2n^{2}-3n-2}\frac {2^{2n-3}}{2^{2n-3}-1}\frac{(2\pi)^{2}}{2n(2n+1)} \\& \quad=-\frac{2 [8\times 2^{4n}- (12n^{3}-196n^{2}-165n+92 )2^{2n}- (4n^{2}-144n-189 ) ]}{ a_{n}a_{n+1}} \\& \qquad{}+\frac{33\pi^{2}}{4}\frac {(6n^{2}+5n+11)2^{4n}-2(10n^{2}+15n+12)2^{2n}+8n^{2}+12n+8}{a_{n}a_{n+1}2^{2n}} \\& \quad< -\frac{2 [8\times 2^{4n}- (12n^{3}-196n^{2}-165n+92 )2^{2n}- (4n^{2}-144n-189 ) ]}{ a_{n}a_{n+1}} \\& \qquad{}+\frac{33\times 10}{4}\frac {(6n^{2}+5n+11)2^{4n}-2(10n^{2}+15n+12)2^{2n}+8n^{2}+12n+8}{a_{n}a_{n+1}2^{2n}} \\& \quad=-\frac{w_{n}}{a_{n}a_{n+1}2^{2n+1}}, \end{aligned}$$
(2.26)

where \(w_{n}\) is given in Lemma 2.7.

Therefore, Lemma 2.8 follows easily from Lemma 2.7, (2.25), and (2.26). □

Let

$$\begin{aligned}& A(x)=(x-\sin x\cos x) (\sin x-x\cos x)^{2}\cos x, \end{aligned}$$
(2.27)
$$\begin{aligned}& B(x)=(\sin x-x\cos x) (x-\sin x\cos x)^{2}, \end{aligned}$$
(2.28)
$$\begin{aligned}& C(x)=x \bigl(x\sin x-2x^{2}\cos x+\sin^{2}x\cos x \bigr) \sin ^{2}x \\& \hphantom{C(x)}=x^{3}\sin^{2}x\cos x \biggl(\frac{\sin^{2}x}{x^{2}}+ \frac{\tan x}{x}-2 \biggr). \end{aligned}$$
(2.29)

Then from the Wilker inequality and Lemma 2.3 we clearly see that

$$ A(x)>0,\qquad B(x)>0, \qquad C(x)>0 $$

for all \(x\in(0, \pi/2)\) and

$$\begin{aligned}& \frac{A(x)}{\sin^{3}x\cos^{2}x}=\sum_{n=3}^{\infty} \frac {2^{2n}|B_{2n}|a_{n}}{(2n)!}x^{2n},\qquad \frac{B(x)}{\sin^{3}x\cos^{2}x}=\sum _{n=3}^{\infty}\frac {2^{2n}|B_{2n}|b_{n}}{(2n)!}x^{2n}, \end{aligned}$$
(2.30)
$$\begin{aligned}& \frac{C(x)}{\sin^{3}x\cos^{2}x}=\sum_{n=3}^{\infty} \frac {2^{2n}|B_{2n}|c_{n}}{(2n)!}x^{2n}, \end{aligned}$$
(2.31)

where \(a_{n}\), \(b_{n}\), and \(c_{n}\) are respectively given by (2.1), (2.2), and (2.11).

Lemma 2.9

Let \(q\in\mathbb{R}\), \(A(x)\), \(B(x)\), and \(C(x)\) be respectively given by (2.27)-(2.29), and \(f(x): (0, \pi/2)\rightarrow\mathbb{R}\) be defined as

$$ f(x)=\frac{q B(x)+C(x)}{A(x)}. $$
(2.32)

Then the following statements are true:

  1. (1)

    if \(q=-1\), then \(f(x)\) is strictly increasing from \((0, \pi/2)\) onto \((2q+12/5, 3-\pi^{2}/4)\);

  2. (2)

    if \(q>-1\), then \(f(x)\) is strictly increasing from \((0, \pi/2)\) onto \((2q+12/5, \infty)\);

  3. (3)

    if \(q\leq-37/35\), then \(f(x)\) is strictly decreasing from \((0, \pi/2)\) onto \((-\infty, 2q+12/5)\).

Proof

Let \(a_{n}\), \(b_{n}\), \(c_{n}\), \(u_{n}\), and \(v_{n}\) be respectively defined by (2.1)-(2.3), (2.11), and (2.12). Then from (2.30)-(2.32) and Lemma 2.5 we have

$$\begin{aligned}& f(x)=\frac{\sum_{n=3}^{\infty}(qb_{n}+c_{n})x^{2n}}{\sum_{n=3}^{\infty }a_{n}x^{2n}}, \end{aligned}$$
(2.33)
$$\begin{aligned}& \frac{qb_{n+1}+c_{n+1}}{a_{n+1}}-\frac{qb_{n}+c_{n}}{a_{n}}=qu_{n}+v_{n}, \end{aligned}$$
(2.34)
$$\begin{aligned}& u_{n}=\frac{b_{n+1}}{a_{n+1}}-\frac{b_{n}}{a_{n}}>0 \end{aligned}$$
(2.35)

for all \(n\geq3\).

Note that

$$\begin{aligned}& f\bigl(0^{+}\bigr)=\frac{qb_{3}+c_{3}}{a_{3}}=2q+\frac{12}{5}, \end{aligned}$$
(2.36)
$$\begin{aligned}& \lim_{x\rightarrow{\frac{\pi}{2}}^{-}}f(x)=\lim_{x\rightarrow {\frac{\pi}{2}}^{-}} \frac{C(x)-B(x)}{A(x)}=3-\frac{\pi^{2}}{4}\quad (q=-1), \end{aligned}$$
(2.37)
$$\begin{aligned}& \lim_{x\rightarrow{\frac{\pi}{2}}^{-}}f(x)=\lim_{x\rightarrow {\frac{\pi}{2}}^{-}} \frac{qB(x)+C(x)}{A(x)}=+\infty \quad(q>-1), \end{aligned}$$
(2.38)
$$\begin{aligned}& \lim_{x\rightarrow{\frac{\pi}{2}}^{-}}f(x)=\lim_{x\rightarrow {\frac{\pi}{2}}^{-}} \frac{qB(x)+C(x)}{A(x)}=-\infty\quad (q< -1). \end{aligned}$$
(2.39)

We divide the proof into two cases.

Case 1 \(q\geq-1\). Then it follows from (2.34) and (2.35), together with Lemma 2.6, that

$$ \frac{qb_{n+1}+c_{n+1}}{a_{n+1}}-\frac{qb_{n}+c_{n}}{a_{n}}\geq v_{n}-u_{n}>0 $$
(2.40)

for \(n\geq3\).

Therefore, parts (1) and (2) follow from (2.33), (2.36)-(2.38), (2.40), and Lemma 2.2.

Case 2 \(q\leq-37/35\). Then (2.34) and (2.35), together with Lemma 2.8, lead to

$$\begin{aligned}& \frac{qb_{4}+c_{4}}{a_{4}}-\frac{qb_{3}+c_{3}}{a_{3}}\leq v_{3}-\frac{37}{35}u_{3}=0, \end{aligned}$$
(2.41)
$$\begin{aligned}& \frac{qb_{n+1}+c_{n+1}}{a_{n+1}}-\frac{qb_{n}+c_{n}}{a_{n}}\leq v_{n}-\frac{37}{35}u_{n}< 0 \end{aligned}$$
(2.42)

for \(n\geq4\).

Therefore, part (3) follows from (2.33), (2.36), (2.39), (2.41), (2.42), and Lemma 2.2. □

Let \(p, q\in\mathbb{R}\), \(x\in(0, \pi/2)\), and the functions \(x\rightarrow S_{p}(x)\), \(x\rightarrow T_{q}(x)\), and \(x\rightarrow W_{p,q}(x)\) be respectively defined by

$$\begin{aligned}& S_{p}(x)=\frac{1- (\frac{\sin x}{x} )^{p}}{p}\quad (p\neq 0),\qquad S_{0}(x)=\lim _{p\rightarrow0}S_{p}(x)=\log\frac{x}{\sin x}, \end{aligned}$$
(2.43)
$$\begin{aligned}& T_{q}(x)=\frac{ (\frac{\tan x}{x} )^{q}-1}{q} \quad(q\neq 0),\qquad T_{0}(x)=\lim _{q\rightarrow0}T_{q}(x)=\log\frac{\tan x}{x}, \end{aligned}$$
(2.44)

and

$$ W_{p, q}(x)=\frac{S_{p}(x)}{T_{q}(x)}. $$

Then we clearly see that

$$\begin{aligned} &S_{p}\bigl(0^{+}\bigr)=T_{q}\bigl(0^{+} \bigr)=0, \\ &W_{p,q}(x)=\frac{S_{p}(x)}{T_{q}(x)}=\frac {S_{p}(x)-S_{p}(0^{+})}{T_{q}(x)-T_{q}(0^{+})}= \textstyle\begin{cases} \frac{q}{p}\frac{1- (\frac{\sin x}{x} )^{p}}{ (\frac{\tan x}{x} )^{q}-1},& pq\neq0, \\ \frac{1}{p}\frac{1- (\frac{\sin x}{x} )^{p}}{\log\frac{\tan x}{x}},& p\neq0, q=0,\\ q \frac{\log\frac{x}{\sin x}}{ (\frac{\tan x}{x} )^{q}-1},& p=0, q\neq0,\\ \frac{\log (\frac{x}{\sin x} )}{\log (\frac{\tan x}{x} )},& p=q=0, \end{cases}\displaystyle \end{aligned}$$
(2.45)
$$\begin{aligned} &W_{p,q}\bigl(0^{+}\bigr)=\frac{1}{2}, \end{aligned}$$
(2.46)
$$\begin{aligned} &W_{p,q} \biggl({\frac{\pi}{2}}^{-} \biggr)= \frac{q}{p} \biggl[ \biggl(\frac {2}{\pi} \biggr)^{p}-1 \biggr] \quad(p\neq0, q< 0),\\ &W_{0,q} \biggl({\frac{\pi}{2}}^{-} \biggr)=\lim _{p\rightarrow 0}W_{p,q} \biggl({\frac{\pi}{2}}^{-} \biggr)=q\log\frac{2}{\pi}\quad (q< 0). \end{aligned}$$
(2.47)

Lemma 2.10

Let \(x\in(0, \pi/2)\), and \(W_{p, q}(x)\) be defined by (2.45). Then the following statements are true:

  1. (1)

    \(W_{p, q}(x)\) is strictly decreasing on \((0, \pi/2)\) if \(q\geq -1\) and \(p+2q+12/5\geq0\);

  2. (2)

    \(W_{p, q}(x)\) is strictly increasing on \((0, \pi/2)\) if \(-37/35< q\leq-1\) and \(p\leq\pi^{2}/4-3\);

  3. (3)

    \(W_{p, q}(x)\) is strictly increasing on \((0, \pi/2)\) if \(q\leq -37/35\) and \(p+2q+12/5\leq0\).

Proof

Let \(pq\neq0\) and \(x\in(0, \pi/2)\). Then (2.43) and (2.44) lead to

$$\begin{aligned} \biggl[\frac{S_{p}^{\prime}(x)}{T_{q}^{\prime}(x)} \biggr]^{\prime} &= \biggl[\frac{\sin x-x\cos x}{x-\sin x\cos x} \biggl(\frac{\sin x}{x} \biggr)^{p-q}\cos^{q+1}x \biggr]^{\prime} \\ &=-\frac{x^{q-p-1}\sin^{p-q-1}x\cos^{q}x}{(x-\sin x\cos x)^{2}}A(x)\bigl[f(x)+p\bigr], \end{aligned}$$
(2.48)

where \(A(x)\) and \(f(x)\) are respectively given by (2.27) and (2.32).

(1) If \(q\geq-1\) and \(p+2q+12/5\geq0\), then from Lemma 2.9(1) and (2) and from (2.48) we have

$$ \biggl[\frac{S_{p}^{\prime}(x)}{T_{q}^{\prime}(x)} \biggr]^{\prime}< -\frac {x^{q-p-1}\sin^{p-q-1}x\cos^{q}x}{(x-\sin x\cos x)^{2}}A(x) \biggl(p+2q+\frac{12}{5} \biggr)\leq0 $$
(2.49)

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

Therefore, Lemma 2.10(1) follows easily from (2.45) and (2.49) together with Lemma 2.1.

(2) If \(-37/35< q\leq-1\) and \(p\leq\pi^{2}/4-3\), then (2.48) and Lemma 2.9(1) lead to

$$\begin{aligned} \biggl[\frac{S_{p}^{\prime}(x)}{T_{q}^{\prime}(x)} \biggr]^{\prime} &\geq -\frac{x^{q-p-1}\sin^{p-q-1}x\cos^{q}x}{(x-\sin x\cos x)^{2}}A(x) \biggl[p+\frac{C(x)-B(x)}{A(x)} \biggr] \\ &>-\frac{x^{q-p-1}\sin^{p-q-1}x\cos^{q}x}{(x-\sin x\cos x)^{2}}A(x) \biggl(p+3-\frac{\pi^{2}}{4} \biggr)\geq0 \end{aligned}$$
(2.50)

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

Therefore, Lemma 2.10(2) follows from (2.45) and (2.50) together with Lemma 2.1.

(3) If \(q\leq-37/35\) and \(p+2q+12/5\leq0\), then Lemma 2.9(3) and (2.48) lead to the conclusion that

$$ \biggl[\frac{S_{p}^{\prime}(x)}{T_{q}^{\prime}(x)} \biggr]^{\prime}>-\frac {x^{q-p-1}\sin^{p-q-1}x\cos^{q}x}{(x-\sin x\cos x)^{2}}A(x) \biggl(p+2q+\frac{12}{5} \biggr)\geq0 $$
(2.51)

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

Therefore, Lemma 2.10(3) follows from (2.45) and (2.51) together with Lemma 2.1. □

Remark 2.2

It is not difficult to verify that (2.48) is also true if \(pq=0\).

3 Main results

Let

$$\begin{aligned}& E_{1}= \biggl\{ (p, q)\Big|q\geq-1, p+2q+\frac{12}{5}\geq0 \biggr\} , \end{aligned}$$
(3.1)
$$\begin{aligned}& E_{2}= \biggl\{ (p, q)\Big|-\frac{37}{35}< q\leq-1, p\leq \frac{\pi^{2}}{4}-3 \biggr\} , \end{aligned}$$
(3.2)
$$\begin{aligned}& E_{3}= \biggl\{ (p, q)\Big| q\leq-\frac{37}{35}, p+2q+ \frac{12}{5}\leq 0 \biggr\} , \end{aligned}$$
(3.3)
$$\begin{aligned}& D_{1}= \bigl\{ (p, q)\big| pq(p+2q)>0 \bigr\} ,\qquad D_{2}= \bigl\{ (p, q)| pq(p+2q)< 0 \bigr\} , \end{aligned}$$
(3.4)
$$\begin{aligned}& D_{3}= \bigl\{ (p, q)\big| p>0, q< 0 \bigr\} ,\qquad D_{4}= \bigl\{ (p, q)|p< 0, q< 0 \bigr\} , \end{aligned}$$
(3.5)
$$\begin{aligned}& G_{1}=E_{1}\cap D_{1},\qquad G_{2}=E_{2} \cup E_{3}\cap D_{2}, \end{aligned}$$
(3.6)
$$\begin{aligned}& G_{3}=E_{1}\cap D_{2},\qquad G_{4}=E_{2} \cup E_{3}\cap D_{1}, \end{aligned}$$
(3.7)
$$\begin{aligned}& G_{5}=E_{1}\cap D_{3},\qquad G_{6}=E_{2} \cup E_{3}\cap D_{4}, \end{aligned}$$
(3.8)
$$\begin{aligned}& G_{7}=E_{1}\cap D_{4},\qquad G_{8}=E_{2} \cup E_{3}\cap D_{3}. \end{aligned}$$
(3.9)

Then (3.1)-(3.9) lead to

$$\begin{aligned} G_{1}={}& \bigl\{ (p,q)|p>0, q>0 \bigr\} \cup \bigl\{ (p,q)|0< p< -2q, q\geq -1 \bigr\} \\ &{}\cup \biggl\{ (p,q)\Big|q>0, -\frac{12}{5}\leq p+2q< 0 \biggr\} , \end{aligned}$$
(3.10)
$$\begin{aligned} G_{2}={}&G_{6}= \biggl\{ (p,q)\Big|p\leq\frac{\pi^{2}}{4}-3, q \leq -1 \biggr\} \\ &{}\cup \biggl\{ (p,q)\Big|\frac{\pi^{2}}{4}-3< p< 0, q\leq-\frac{37}{35}, p+2q+ \frac{12}{5}\leq0 \biggr\} , \end{aligned}$$
(3.11)
$$\begin{aligned} G_{3}={}& \bigl\{ (p,q)|p< 0, p+2q>0 \bigr\} \cup \bigl\{ (p,q)|-1\leq q< 0, p+2q>0 \bigr\} \\ &{}\cup \biggl\{ (p,q)|-1\leq q< 0, -2q-\frac{12}{5}\leq p< 0 \biggr\} , \end{aligned}$$
(3.12)
$$\begin{aligned} G_{4}={}&G_{8}= \biggl\{ (p,q)\Big|0< p\leq-2q-\frac{12}{5} \biggr\} , \end{aligned}$$
(3.13)
$$\begin{aligned} G_{5}={}& \bigl\{ (p,q)|p>0, -1\leq q< 0 \bigr\} , \end{aligned}$$
(3.14)
$$\begin{aligned} G_{7}={}& \biggl\{ (p,q)\Big|-1\leq q< 0,-2q-\frac{12}{5}\leq p< 0 \biggr\} . \end{aligned}$$
(3.15)

Theorem 3.1

Let \(G_{1}\), \(G_{2}\), \(G_{3}\), and \(G_{4}\) be respectively defined by (3.10)-(3.13). Then the Wilker-type inequality

$$ \frac{2q}{p+2q} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \frac{p}{p+2q} \biggl(\frac{\tan x}{x} \biggr)^{q}>1 $$
(3.16)

holds for all \(x\in(0, \pi/2)\) if \((p, q)\in G_{1}\cup G_{2}\), and inequality (3.16) is reversed if \((p, q)\in G_{3}\cup G_{4}\).

Proof

Let \(W_{p,q}(x)\) be defined by (2.45). We only prove that inequality (3.16) holds for all \(x\in(0, \pi/2)\) if \((p, q)\in G_{1}\cup G_{2}\); the reversed inequality for \((p, q)\in G_{3}\cup G_{4}\) can be proved by a completely similar method.

We divide the proof into two cases.

Case 1 \((p, q)\in G_{1}\). Then (3.1), (3.4), and (3.6) lead to

$$\begin{aligned}& q\geq-1,\quad p+2q+\frac{12}{5}\geq0, \end{aligned}$$
(3.17)
$$\begin{aligned}& pq(p+2q)>0. \end{aligned}$$
(3.18)

It follows from (2.45), (2.46), Lemma 2.10(1), and (3.17) that

$$ w_{p, q}(x)=\frac{q}{p}\frac{1-(\frac{\sin x}{x})^{p}}{ (\frac{\tan x}{x} )^{q}-1}< \frac{1}{2} $$
(3.19)

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

Therefore, inequality (3.16) follows easily from (3.18) and (3.19).

Case 2 \((p, q)\in G_{2}\). Then from (2.45), (2.46), Lemma 2.10(2) and (3), (3.2)-(3.4), and (3.6) we clearly see that

$$ w_{p, q}(x)=\frac{q}{p}\frac{1-(\frac{\sin x}{x})^{p}}{ (\frac{\tan x}{x} )^{q}-1}>\frac{1}{2} $$
(3.20)

and

$$ pq(p+2q)< 0. $$
(3.21)

Therefore, inequality (3.16) follows from (3.20) and (3.21). □

Theorem 3.2

Let \(G_{5}\), \(G_{6}\), \(G_{7}\), and \(G_{8}\) be respectively defined by (3.11) and (3.13)-(3.15). Then the Wilker-type inequality

$$ \biggl(\frac{\pi}{2} \biggr)^{p} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \biggl[1- \biggl(\frac{\pi}{2} \biggr)^{p} \biggr] \biggl(\frac{\tan x}{x} \biggr)^{q}< 1 $$
(3.22)

holds for all \(x\in(0, \pi/2)\) if \((p, q)\in G_{5}\cup G_{6}\), and inequality (3.22) is reversed if \((p, q)\in G_{7}\cup G_{8}\).

Proof

Let \(W_{p,q}(x)\) be defined by (2.45). We only prove that inequality (3.22) holds for all \(x\in(0, \pi/2)\) if \((p, q)\in G_{5}\cup G_{6}\); the reversed inequality for \((p, q)\in G_{7}\cup G_{8}\) can be proved by a completely similar method.

We divide the proof into two cases.

Case 1 \((p, q)\in G_{5}\). Then from (2.45), (2.47), Lemma 2.10(1), (3.1), (3.5), and (3.8) we clearly see that

$$ w_{p, q}(x)=\frac{q}{p}\frac{1-(\frac{\sin x}{x})^{p}}{ (\frac{\tan x}{x} )^{q}-1}>\frac{q}{p} \biggl[ \biggl(\frac{2}{\pi} \biggr)^{p}-1 \biggr] $$
(3.23)

and

$$ p>0. $$
(3.24)

Therefore, inequality (3.22) follows easily from (3.23) and (3.24).

Case 2 \((p, q)\in G_{6}\). Then (2.45), (2.47), Lemma 2.10(2) and (3), (3.2), (3.3), (3.5), and (3.8) lead to the conclusion that

$$ w_{p, q}(x)=\frac{q}{p}\frac{1-(\frac{\sin x}{x})^{p}}{ (\frac{\tan x}{x} )^{q}-1}< \frac{q}{p} \biggl[ \biggl(\frac{2}{\pi} \biggr)^{p}-1 \biggr] $$
(3.25)

and

$$ p< 0. $$
(3.26)

Therefore, inequality (3.22) follows easily from (3.25) and (3.26). □