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Generalized Wilker-type inequalities with two parameters

Journal of Inequalities and Applications20162016:187

https://doi.org/10.1186/s13660-016-1127-8

  • Received: 26 April 2016
  • Accepted: 9 July 2016
  • Published:

Abstract

In the article, we present certain \(p, q\in\mathbb{R}\) such that the Wilker-type inequalities
$$\begin{aligned}& \frac{2q}{p+2q} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \frac{p}{p+2q} \biggl(\frac{\tan x}{x} \biggr)^{q}>(< )1\quad \mbox{and}\\& \biggl(\frac{\pi}{2} \biggr)^{p} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \biggl[1- \biggl(\frac{\pi}{2} \biggr)^{p} \biggr] \biggl(\frac{\tan x}{x} \biggr)^{q}>(< )1 \end{aligned}$$
hold for all \(x\in(0, \pi/2)\).

Keywords

  • Wilker inequality
  • trigonometric function
  • Bernoulli number
  • monotonicity

MSC

  • 26D05
  • 33B10

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 [310].

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 [1628].

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). □

Declarations

Acknowledgements

The research was supported by the Natural Science Foundation of China under Grants 11371125, 61374086, and 11401191 and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.

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

Authors’ Affiliations

(1)
School of Civil Engineering and Architecture, Changsha University of Sciences & Technology, Changsha, 410114, China
(2)
School of Mathematics and Computation Sciences, Hunan City University, Yiyang, 413000, China
(3)
Albert Einstein College of Medicine, Yeshiva University, New York, NY, 10033, United States

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