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].
Anzeige
Pinelis [11] and Sun and Zhu [12] proved that the inequalities hold for all \(x\in(0, \pi/2)\) and \(y>0\) if and only if \(\lambda\leq 8/45\) and \(\mu\geq2/45\).
$$ \biggl(\frac{\sin x}{x} \biggr)^{2}+\frac{\tan x}{x}-2>\lambda x^{3}\tan x\quad \mbox{and}\quad \biggl(\frac{y}{\sinh y} \biggr)^{2} + \frac{y}{\tanh y}-2< \mu y^{3}\sinh y $$
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 holds for all \(x\in(0, \pi/2)\).
$$ P_{1}(x)\tan x< \biggl(\frac{\sin x}{x} \biggr)^{2}+ \frac{\tan x}{x}-2< P_{2}(x)\tan x $$
Yang [14] proved that \(p=5/3\) and \(q=\log2/[2(\log\pi-\log2)]\) are the best possible parameters such that the double inequality holds for all \(x\in(0, \pi/2)\).
$$ \biggl(\frac{\sqrt{\cos^{2p}x+8}+\cos^{p}x}{4} \biggr)^{1/p}< \frac{\sin x}{x}< \biggl( \frac{\sqrt{\cos^{2q}x+8}+\cos^{q}x}{4} \biggr)^{1/q} $$
Very recently, Yang and Chu [15] proved that the Wilker-type inequality 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 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 \(p\geq -12/[5(k+2)]\)).
$$ \frac{2}{k+2} \biggl(\frac{\sin x}{x} \biggr)^{kp}+ \frac{k}{k+2} \biggl(\frac{\tan x}{x} \biggr)^{p}>(< )1 $$
$$ \frac{2}{k+2} \biggl(\frac{\sinh x}{x} \biggr)^{kp}+ \frac{k}{k+2} \biggl(\frac{\tanh x}{x} \biggr)^{p}>(< )1 $$
Anzeige
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 and 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.
$$ \frac{2q}{p+2q} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \frac{p}{p+2q} \biggl(\frac{\sin x}{x} \biggr)^{q}>(< )1 $$
$$ \biggl(\frac{\pi}{2} \biggr)^{p} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \biggl[1- \biggl(\frac{\pi}{2} \biggr)^{p} \biggr] \biggl(\frac{\tan x}{x} \biggr)^{q}>(< )1 $$
2 Lemmas
In order to prove our main results, we need several lemmas.
Lemma 2.1
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
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.
$$ \frac{f(x)-f(a)}{g(x)-g(a)} \quad\textit{and} \quad\frac{f(x)-f(b)}{g(x)-g(b)} $$
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
hold for
\(x\in(-\pi, \pi)\), and the power series formulas
hold for
\(x\in(-\pi/2, \pi/2)\).
$$\begin{aligned}& \frac{1}{\sin x}=\frac{1}{x}+\sum_{n=1}^{\infty} \frac{2^{2n}-2}{(2n)!}|B_{2n}|x^{2n-1},\qquad \cot x=\frac{1}{x}- \sum_{n=1}^{\infty}\frac{2^{2n}}{(2n)!}|B_{2n}|x^{2n-1},\\& \frac{1}{\sin^{2} x}=\frac{1}{x^{2}}+\sum_{n=1}^{\infty} \frac {(2n-1)2^{2n}}{(2n)!}|B_{2n}|x^{2n-2}, \\& \frac{\cos x}{\sin^{3} x}= \frac{1}{x^{3}}-\sum_{n=1}^{\infty} \frac {n(2n+1)2^{2n+2}}{(2n+2)!}|B_{2n+2}|x^{2n-1} \end{aligned}$$
$$ \tan x=\sum_{n=1}^{\infty}\frac{ (2^{2n}-1 )2^{2n}}{(2n)!}|B_{2n}|x^{2n-1},\qquad \frac{1}{\cos^{2} x}=\sum_{n=1}^{\infty} \frac{(2n-1) (2^{2n}-1 )2^{2n}}{(2n)!}|B_{2n}|x^{2n-2} $$
Lemma 2.4
(See [33])
Let
\(B_{n}\)
be the Bernoulli numbers. Then the double inequality
holds for all
\(n\in\mathbb{N}\).
$$ \frac{2(2n)!}{(2\pi)^{2n}}< |B_{2n}|< \frac{2^{2n-1}}{2^{2n-1}-1}\frac {2(2n)!}{(2\pi)^{2n}} $$
From Lemma 2.4 we immediately get the following:
Remark 2.1
Let \(B_{n}\) be the Bernoulli numbers. Then the double inequality holds for all \(n\in\mathbb{N}\) and \(n\geq1\).
$$ \frac{2^{2n-1}-1}{2^{2n-1}}\frac{(2\pi)^{2}}{2n(2n-1)}< \frac {|B_{2n-2}|}{|B_{2n}|}< \frac{2^{2n-3}}{2^{2n-3}-1} \frac{(2\pi)^{2}}{2n(2n-1)} $$
Lemma 2.5
Let
\(n\in\mathbb{N}\), \(B_{n}\)
be the Bernoulli numbers, and
\(a_{n}\)
and
\(b_{n}\)
be respectively defined by
Then the sequence
\(\{b_{n}/a_{n}\}\)
is strictly increasing for
\(n\geq3\).
$$\begin{aligned}& a_{n}=2^{2n}-2n^{2}-3n-2, \end{aligned}$$
(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)
Proof
Let \(n\geq3\) and Then from (2.1)-(2.3) and Remark 2.1 we get Let Then we clearly see that
for \(n\geq3\).
$$ u_{n}=\frac{b_{n+1}}{a_{n+1}}-\frac{b_{n}}{a_{n}}. $$
(2.3)
$$\begin{aligned} u_{n}={}&\frac{(2n-1)2^{2n+2}+2n^{2}+5n+7}{2^{2n+2}-2n^{2}-7n-7}-\frac {(n+1)(2n+1) (2^{2n-1}-1 )}{ 2^{2n+2}-2n^{2}-7n-7}\frac{|B_{2n}|}{|B_{2n+2}|} \\ &{}-\frac{(2n-3)2^{2n}+2n^{2}+n+4}{2^{2n}-2n^{2}-3n-2}+\frac{n(2n-1) (2^{2n-3}-1 )}{ 2^{2n}-2n^{2}-3n-2}\frac{|B_{2n-2}|}{|B_{2n}|} \\ >{}&\frac{2}{a_{n}a_{n+1}} \bigl[4\times 2^{4n}-\bigl(6n^{3}+7n^{2}+5n+11 \bigr)2^{2n}-\bigl(2n^{2}-2n-7\bigr) \bigr] \\ &{}+\frac{\pi^{2}}{2^{2n+2}}\frac {(6n^{2}+5n-39)2^{4n}+(20n^{2}+70n+134)2^{2n}-(32n^{2}+112n+112)}{a_{n}a_{n+1}}. \end{aligned}$$
(2.4)
$$ u_{n}^{\ast}=4\times 2^{4n}-\bigl(6n^{3}+7n^{2}+5n+11 \bigr)2^{2n}-\bigl(2n^{2}-2n-7\bigr). $$
(2.5)
$$\begin{aligned}& u_{3}^{\ast}=315>0, \end{aligned}$$
(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)
It is not difficult to verify that and for all \(n\geq3\).
$$ a_{n}>0 $$
(2.9)
$$ \bigl(6n^{2}+5n-39\bigr)2^{4n}+\bigl(20n^{2}+70n+134 \bigr)2^{2n}-\bigl(32n^{2}+112n+112\bigr)>0 $$
(2.10)
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
Then
\(v_{n}>u_{n}\)
for all
\(n\geq3\).
$$\begin{aligned}& c_{n}=2n\bigl(2^{2n}-1\bigr)-2n(2n-1) \bigl(2^{2n-3}-1 \bigr)\frac{|B_{2n-2}|}{|B_{2n}|}, \end{aligned}$$
(2.11)
$$\begin{aligned}& v_{n}=\frac{c_{n+1}}{a_{n+1}}-\frac{c_{n}}{a_{n}}. \end{aligned}$$
(2.12)
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
Note that and for all \(n\geq8\).
$$\begin{aligned}& v_{3}-u_{3}=\frac{8}{105},\qquad v_{4}-u_{4}= \frac{104}{3{,}045}, \qquad v_{5}-u_{5}=\frac{15{,}496}{1{,}102{,}145}, \end{aligned}$$
(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)
$$ a_{n}>0,\quad \bigl(180n^{2}+598n+1{,}166\bigr)2^{2n}-9 \bigl(32n^{2}+112n+112\bigr)>0, $$
(2.17)
$$ 6n^{2}+5n-335\geq6\times8^{2}+5\times8-335=89 $$
(2.18)
Lemma 2.7
Let
\(n\in\mathbb{N}\), and
\(w_{n}\)
be defined by
Then
\(w_{n}>0\)
for all
\(n\geq5\).
$$\begin{aligned} w_{n}={}&32\times 2^{6n}- \bigl(48n^{3}+206n^{2}+165n+2{,}183 \bigr)2^{4n}\\ &{}+ \bigl(3{,}284n^{2}+5{,}526n+4{,}716 \bigr)2^{2n}- \bigl(1{,}320n^{2}+1{,}980n+1{,}320 \bigr). \end{aligned}$$
Proof
Let Then we clearly see that
for all \(n\geq5\).
$$ w_{n}^{\ast}=32\times4^{n}- \bigl(48n^{3}+206n^{2}+165n+2{,}183 \bigr). $$
$$\begin{aligned}& w_{n}=2^{4n}w_{n}^{\ast}+ \bigl(3{,}284n^{2}+5{,}526n+4{,}716 \bigr)2^{2n}- \bigl(1{,}320n^{2}+1{,}980n+1{,}320 \bigr), \end{aligned}$$
(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)
Inequalities (2.20) and (2.21) lead to the conclusion that for all \(n\geq5\).
$$ w_{n}^{\ast}>0 $$
(2.22)
Note that for all \(n\geq5\).
$$ \bigl(3{,}284n^{2}+5{,}526n+4{,}716 \bigr)2^{2n}- \bigl(1{,}320n^{2}+1{,}980n+1{,}320 \bigr)>0 $$
(2.23)
Lemma 2.8
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
where \(w_{n}\) is given in Lemma 2.7.
$$\begin{aligned}& 35v_{3}-37u_{3}=0,\qquad 35v_{4}-37u_{4}=- \frac{288}{145}, \end{aligned}$$
(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)
Let
Then from the Wilker inequality and Lemma 2.3 we clearly see that for all \(x\in(0, \pi/2)\) and
where \(a_{n}\), \(b_{n}\), and \(c_{n}\) are respectively given by (2.1), (2.2), and (2.11).
$$\begin{aligned}& A(x)=(x-\sin x\cos x) (\sin x-x\cos x)^{2}\cos x, \end{aligned}$$
(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)
$$ A(x)>0,\qquad B(x)>0, \qquad C(x)>0 $$
$$\begin{aligned}& \frac{A(x)}{\sin^{3}x\cos^{2}x}=\sum_{n=3}^{\infty} \frac {2^{2n}|B_{2n}|a_{n}}{(2n)!}x^{2n},\qquad \frac{B(x)}{\sin^{3}x\cos^{2}x}=\sum _{n=3}^{\infty}\frac {2^{2n}|B_{2n}|b_{n}}{(2n)!}x^{2n}, \end{aligned}$$
(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)
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
Then the following statements are true:
$$ f(x)=\frac{q B(x)+C(x)}{A(x)}. $$
(2.32)
(1)
if
\(q=-1\), then
\(f(x)\)
is strictly increasing from
\((0, \pi/2)\)
onto
\((2q+12/5, 3-\pi^{2}/4)\);
(2)
if
\(q>-1\), then
\(f(x)\)
is strictly increasing from
\((0, \pi/2)\)
onto
\((2q+12/5, \infty)\);
(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
for all \(n\geq3\).
$$\begin{aligned}& f(x)=\frac{\sum_{n=3}^{\infty}(qb_{n}+c_{n})x^{2n}}{\sum_{n=3}^{\infty }a_{n}x^{2n}}, \end{aligned}$$
(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)
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 for \(n\geq3\).
$$ \frac{qb_{n+1}+c_{n+1}}{a_{n+1}}-\frac{qb_{n}+c_{n}}{a_{n}}\geq v_{n}-u_{n}>0 $$
(2.40)
Case 2 \(q\leq-37/35\). Then (2.34) and (2.35), together with Lemma 2.8, lead to
for \(n\geq4\).
$$\begin{aligned}& \frac{qb_{4}+c_{4}}{a_{4}}-\frac{qb_{3}+c_{3}}{a_{3}}\leq v_{3}-\frac{37}{35}u_{3}=0, \end{aligned}$$
(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)
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
and Then we clearly see that
$$\begin{aligned}& S_{p}(x)=\frac{1- (\frac{\sin x}{x} )^{p}}{p}\quad (p\neq 0),\qquad S_{0}(x)=\lim _{p\rightarrow0}S_{p}(x)=\log\frac{x}{\sin x}, \end{aligned}$$
(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)
$$ W_{p, q}(x)=\frac{S_{p}(x)}{T_{q}(x)}. $$
$$\begin{aligned} &S_{p}\bigl(0^{+}\bigr)=T_{q}\bigl(0^{+} \bigr)=0, \\ &W_{p,q}(x)=\frac{S_{p}(x)}{T_{q}(x)}=\frac {S_{p}(x)-S_{p}(0^{+})}{T_{q}(x)-T_{q}(0^{+})}= \textstyle\begin{cases} \frac{q}{p}\frac{1- (\frac{\sin x}{x} )^{p}}{ (\frac{\tan x}{x} )^{q}-1},& pq\neq0, \\ \frac{1}{p}\frac{1- (\frac{\sin x}{x} )^{p}}{\log\frac{\tan x}{x}},& p\neq0, q=0,\\ q \frac{\log\frac{x}{\sin x}}{ (\frac{\tan x}{x} )^{q}-1},& p=0, q\neq0,\\ \frac{\log (\frac{x}{\sin x} )}{\log (\frac{\tan x}{x} )},& p=q=0, \end{cases}\displaystyle \end{aligned}$$
(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)
\(W_{p, q}(x)\)
is strictly decreasing on
\((0, \pi/2)\)
if
\(q\geq -1\)
and
\(p+2q+12/5\geq0\);
(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)
\(W_{p, q}(x)\)
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 where \(A(x)\) and \(f(x)\) are respectively given by (2.27) and (2.32).
$$\begin{aligned} \biggl[\frac{S_{p}^{\prime}(x)}{T_{q}^{\prime}(x)} \biggr]^{\prime} &= \biggl[\frac{\sin x-x\cos x}{x-\sin x\cos x} \biggl(\frac{\sin x}{x} \biggr)^{p-q}\cos^{q+1}x \biggr]^{\prime} \\ &=-\frac{x^{q-p-1}\sin^{p-q-1}x\cos^{q}x}{(x-\sin x\cos x)^{2}}A(x)\bigl[f(x)+p\bigr], \end{aligned}$$
(2.48)
(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 for \(x\in(0, \pi/2)\).
$$ \biggl[\frac{S_{p}^{\prime}(x)}{T_{q}^{\prime}(x)} \biggr]^{\prime}< -\frac {x^{q-p-1}\sin^{p-q-1}x\cos^{q}x}{(x-\sin x\cos x)^{2}}A(x) \biggl(p+2q+\frac{12}{5} \biggr)\leq0 $$
(2.49)
(2) If \(-37/35< q\leq-1\) and \(p\leq\pi^{2}/4-3\), then (2.48) and Lemma 2.9(1) lead to for \(x\in(0, \pi/2)\).
$$\begin{aligned} \biggl[\frac{S_{p}^{\prime}(x)}{T_{q}^{\prime}(x)} \biggr]^{\prime} &\geq -\frac{x^{q-p-1}\sin^{p-q-1}x\cos^{q}x}{(x-\sin x\cos x)^{2}}A(x) \biggl[p+\frac{C(x)-B(x)}{A(x)} \biggr] \\ &>-\frac{x^{q-p-1}\sin^{p-q-1}x\cos^{q}x}{(x-\sin x\cos x)^{2}}A(x) \biggl(p+3-\frac{\pi^{2}}{4} \biggr)\geq0 \end{aligned}$$
(2.50)
(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 for \(x\in(0, \pi/2)\).
$$ \biggl[\frac{S_{p}^{\prime}(x)}{T_{q}^{\prime}(x)} \biggr]^{\prime}>-\frac {x^{q-p-1}\sin^{p-q-1}x\cos^{q}x}{(x-\sin x\cos x)^{2}}A(x) \biggl(p+2q+\frac{12}{5} \biggr)\geq0 $$
(2.51)
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
holds for all
\(x\in(0, \pi/2)\)
if
\((p, q)\in G_{1}\cup G_{2}\), and inequality (3.16) is reversed if
\((p, q)\in G_{3}\cup G_{4}\).
$$ \frac{2q}{p+2q} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \frac{p}{p+2q} \biggl(\frac{\tan x}{x} \biggr)^{q}>1 $$
(3.16)
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 for \(x\in(0, \pi/2)\).
$$ w_{p, q}(x)=\frac{q}{p}\frac{1-(\frac{\sin x}{x})^{p}}{ (\frac{\tan x}{x} )^{q}-1}< \frac{1}{2} $$
(3.19)
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 and Therefore, inequality (3.16) follows from (3.20) and (3.21). □
$$ w_{p, q}(x)=\frac{q}{p}\frac{1-(\frac{\sin x}{x})^{p}}{ (\frac{\tan x}{x} )^{q}-1}>\frac{1}{2} $$
(3.20)
$$ pq(p+2q)< 0. $$
(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
holds for all
\(x\in(0, \pi/2)\)
if
\((p, q)\in G_{5}\cup G_{6}\), and inequality (3.22) is reversed if
\((p, q)\in G_{7}\cup G_{8}\).
$$ \biggl(\frac{\pi}{2} \biggr)^{p} \biggl(\frac{\sin x}{x} \biggr)^{p}+ \biggl[1- \biggl(\frac{\pi}{2} \biggr)^{p} \biggr] \biggl(\frac{\tan x}{x} \biggr)^{q}< 1 $$
(3.22)
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 and Therefore, inequality (3.22) follows easily from (3.23) and (3.24).
$$ w_{p, q}(x)=\frac{q}{p}\frac{1-(\frac{\sin x}{x})^{p}}{ (\frac{\tan x}{x} )^{q}-1}>\frac{q}{p} \biggl[ \biggl(\frac{2}{\pi} \biggr)^{p}-1 \biggr] $$
(3.23)
$$ p>0. $$
(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 and
$$ w_{p, q}(x)=\frac{q}{p}\frac{1-(\frac{\sin x}{x})^{p}}{ (\frac{\tan x}{x} )^{q}-1}< \frac{q}{p} \biggl[ \biggl(\frac{2}{\pi} \biggr)^{p}-1 \biggr] $$
(3.25)
$$ p< 0. $$
(3.26)
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.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.