# Sharpness of Wilker and Huygens type inequalities

## Abstract

We present an elementary proof of Wilker's inequality involving trigonometric functions, and establish sharp Wilker and Huygens type inequalities.

Mathematics Subject Classification 2010: 26D05.

## 1. Introduction

Wilker in  proposed two open problems:

1. (a)

Prove that if 0 < x < π/2, then

$\begin{array}{c}{\left(\frac{\text{sin}x}{x}\right)}^{2}+\frac{\text{tan}x}{x}>2.\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}1\end{array}$
(1)
2. (b)

Find the largest constant c such that

${\left(\frac{\text{sin}x}{x}\right)}^{2}+\frac{\text{tan}x}{x}>2+c{x}^{3}\text{tan}x$

for 0 < x < π/2.

In , inequality (1) was proved, and the following inequality

$2+{\left(\frac{2}{\pi }\right)}^{4}{x}^{3}\text{tan}x<{\left(\frac{\text{sin}x}{x}\right)}^{2}+\frac{\text{tan}x}{x}<2+\frac{8}{45}{x}^{3}\text{tan}x\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{2.77695pt}{0ex}}\text{0}\phantom{\rule{2.77695pt}{0ex}}<\phantom{\rule{2.77695pt}{0ex}}x\phantom{\rule{2.77695pt}{0ex}}<\phantom{\rule{2.77695pt}{0ex}}\frac{\pi }{2},$
(2)

where the constants ${\left(\frac{2}{\pi }\right)}^{4}$ and $\frac{8}{45}$ are best possible, was also established.

Wilker type inequalities (1) and (2) have attracted much interest of many mathematicians and have motivated a large number of research papers involving different proofs and various generalizations and improvements (cf.  and the references cited therein). A brief survey of some old and new inequalities associated with trigonometric functions can be found in . These include (among other results) Wilker's inequality.

Another inequality which is of interest to us is Huygens  inequality, which asserts that

$2\left(\frac{\text{sin}x}{x}\right)+\frac{\text{tan}x}{x}>3\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{2.77695pt}{0ex}}\text{all}\phantom{\rule{1em}{0ex}}0<\left|x\right|<\frac{\pi }{2}.$
(3)

Neuman and Sándor  have pointed out that (3) implies (1). In , Zhu established some new inequalities of the Huygens type for trigonometric and hyperbolic functions. Baricz and Sándor  pointed out that inequalities (1) and (3) are simple consequences of the arithmetic-geometric mean inequality together with the well-known Lazarević-type inequality [, p. 238]

${\left(\text{cos}x\right)}^{1/3}<\frac{\text{sin}x}{x}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\text{for}\phantom{\rule{2.77695pt}{0ex}}\text{all}\phantom{\rule{1em}{0ex}}0<\left|x\right|<\frac{\pi }{2},$
(4)

or equivalently,

${\left(\frac{\text{sin}x}{x}\right)}^{2}\frac{\text{tan}x}{x}>1\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{2.77695pt}{0ex}}\text{all}\phantom{\rule{1em}{0ex}}0<\left|x\right|<\frac{\pi }{2}.$

Wu and Srivastava [, Lemma 3] established another inequality

${\left(\frac{x}{\text{sin}}\right)}^{2}+\frac{x}{\text{tan}x}>2\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.3em}{0ex}}\text{for}\phantom{\rule{2.77695pt}{0ex}}\text{all}\phantom{\rule{1em}{0ex}}0<\left|x\right|<\frac{\pi }{2}.$
(5)

In , Chen and Cheung showed that Wilker inequality (1), Huygens inequality (3), Lazarević-type inequality (4) and Wu-Srivastava inequality (5) can be grouped into the following inequality chain:

$\begin{array}{ll}\hfill \frac{{\left(\text{sin}x/x\right)}^{2}+\text{tan}x/x}{2}& >\frac{2\left(\text{sin}x/x\right)+\text{tan}x/x}{3}>\sqrt{{\left(\frac{\text{sin}x}{x}\right)}^{2}\frac{\text{tan}x}{x}}>1\phantom{\rule{2em}{0ex}}\\ >\frac{2}{1/{\left(\text{sin}x/x\right)}^{2}+1/\left(\text{tan}x/x\right)},\phantom{\rule{1em}{0ex}}0<\left|x\right|<\frac{\pi }{2},\phantom{\rule{2em}{0ex}}\end{array}$
(6)

in terms of the arithmetic, geometric and harmonic means.

In this article, we present an elementary proof of Wilker's inequality (2), and we establish sharp Wilker and Huygens type inequalities.

The following elementary power series expansions are useful in our investigation.

$\text{sin}x=\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{\left(2n+1\right)!},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left|x\right|<\infty ,$
(7)
$\text{cos}x=\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)!},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left|x\right|<\infty ,$
(8)
$\text{tan}x=\sum _{n=1}^{\infty }\frac{{2}^{2n}\left({2}^{2n}-1\right){\left(-1\right)}^{n-1}{B}_{2n}}{\left(2n\right)!}{x}^{2n-1},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left|x\right|<\frac{\pi }{2},$
(9)
$x\text{cot}x=\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{B}_{2n}\frac{{\left(2x\right)}^{2n}}{\left(2n\right)!},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}0<\left|x\right|<\pi .$
(10)

where B n (n = 0, 1,2,...) are Bernoulli numbers, defined by

$\frac{t}{{e}^{t}-1}=\sum _{n=0}^{\infty }{B}_{n}\frac{{t}^{n}}{n!}.$

The following lemma is also needed in the sequel.

Lemma 1.  Let a n and b n > 0, n = 0,1, 2,... be real numbers with ${\left\{\frac{{a}_{n}}{{b}_{n}}\right\}}_{n=1}^{\infty }$ being strictly increasing (respectively, decreasing). If the power series $A\left(x\right):={\sum }_{n=0}^{\infty }{a}_{n}{x}^{n}$ and $B\left(x\right):={\sum }_{n=0}^{\infty }{b}_{n}{x}^{n}$ are convergent for |x| < R, then the function A(x)/B(x) is strictly increasing (respectively, decreasing) on (0, R).

## 2. An elementary proof of Wilker's inequality (2)

Proof of (2). Consider the function

$f\left(x\right):=\frac{{\left(\frac{\text{sin}x}{x}\right)}^{2}+\frac{\text{tan}x}{x}-2}{{x}^{3}\text{tan}x},\phantom{\rule{1em}{0ex}}0

By using power series expansions (7) and (8), we obtain

$\begin{array}{ll}\hfill -{x}^{6}{\text{sin}}^{2}x{f}^{\prime }\left(x\right)& =-3{x}^{2}\text{sin}\left(2x\right)+5{\text{sin}}^{3}x\text{cos}x+5x-2{x}^{3}+2x{\text{cos}}^{4}x-7x{\text{cos}}^{2}x\phantom{\rule{2em}{0ex}}\\ =\left(-3{x}^{2}+\frac{5}{4}\right)\text{sin}\left(2x\right)-\frac{5}{8}\text{sin}\left(4x\right)-2{x}^{3}+\frac{9}{4}x+\frac{1}{4}x\text{cos}\left(4x\right)-\frac{5}{2}x\text{cos}\left(2x\right)\phantom{\rule{2em}{0ex}}\\ =\sum _{n=4}^{\infty }{\left(-1\right)}^{n}{u}_{n}\left(x\right)\phantom{\rule{2em}{0ex}}\\ =\frac{16}{945}{x}^{9}-\frac{16}{1575}{x}^{11}+\frac{16}{7425}{x}^{13}-\frac{11072}{42567525}{x}^{15}+\sum _{n=8}^{\infty }{\left(-1\right)}^{n}{u}_{n}\left(x\right),\phantom{\rule{2em}{0ex}}\end{array}$

where

${u}_{n}\left(x\right):=\frac{\left(\left(2n-9\right){4}^{n-1}+6{n}^{2}-2n\right){4}^{n}}{\left(2n+1\right)!}{x}^{2n+1}.$

Elementary calculations reveal that, for 0 < x < π/2 and n ≥ 8,

$\begin{array}{ll}\hfill \frac{{u}_{n+1}\left(x\right)}{{u}_{n}\left(x\right)}& =8{x}^{2}\frac{6{n}^{2}+10n+4+\left(2n-7\right){4}^{n}}{\left(n+1\right)\left(2n+3\right)\left(24{n}^{2}-8n+\left(2n-9\right){4}^{n}\right)}\phantom{\rule{2em}{0ex}}\\ <8{\left(\frac{\pi }{2}\right)}^{2}\frac{6{n}^{2}+10n+4+\left(2n-7\right){4}^{n}}{\left(n+1\right)\left(2n+3\right)\left(24{n}^{2}-8n+\left(2n-9\right){4}^{n}\right)}\phantom{\rule{2em}{0ex}}\\ =\frac{{\pi }^{2}}{2n+3}\frac{12{n}^{2}+20n+8+2\left(2n-7\right){4}^{n}}{\left(n+1\right)\left(24{n}^{2}-8n+\left(2n-9\right){4}^{n}\right)}.\phantom{\rule{2em}{0ex}}\end{array}$

Write

${\alpha }_{n}:=\frac{{\pi }^{2}}{2n+3}\frac{12{n}^{2}+20n+8+2\left(2n-7\right){4}^{n}}{\left(n+1\right)\left(24{n}^{2}-8n+\left(2n-9\right){4}^{n}\right)}.$

It is easy to see that for n ≥ 8,

$\frac{12{n}^{2}+20n+8+2\left(2n-7\right){4}^{n}}{\left(n+1\right)\left(24{n}^{2}-8n+\left(2n-9\right){4}^{n}\right)}<1.$

Hence for all 0 < x < π/2 and n ≥ 8,

$\frac{{u}_{n+1}\left(x\right)}{{u}_{n}\left(x\right)}<{\alpha }_{n}<\frac{{\pi }^{2}}{2n+3}<1.$

Therefore, for fixed x (0, π/2), the sequence n u n (x) is strictly decreasing with regard to n ≥ 8. Hence, for 0 < x < π/2,

$\begin{array}{ll}\hfill -{x}^{6}{\text{sin}}^{2}x{f}^{\prime }\left(x\right)& >\frac{16}{945}{x}^{9}-\frac{16}{1575}{x}^{11}+\frac{16}{7425}{x}^{13}-\frac{11072}{42567525}{x}^{15}\phantom{\rule{2em}{0ex}}\\ ={x}^{9}\left(\frac{16}{945}-\frac{16}{1575}{x}^{2}+\frac{16}{7425}{x}^{4}-\frac{11072}{42567525}{x}^{6}\right)\phantom{\rule{2em}{0ex}}\\ >0.\phantom{\rule{2em}{0ex}}\end{array}$

Hence f(x) is strictly decreasing on (0, π/2). Noting that $\underset{t\to {0}^{+}}{\text{lim}}f\left(t\right)=\frac{8}{45}$, we have

$\frac{16}{{\pi }^{4}}=\underset{t\to {\left(\pi /2\right)}^{-}}{\text{lim}}f\left(t\right)

for all $x\in \left(0,\frac{\pi }{2}\right)$, with the constants $\frac{16}{{\pi }^{4}}$ and $\frac{8}{45}$ being best possible. This completes the proof of (2).

## 3. Sharp Wilker's inequality

By using power series expansions (8) and (9), we have, for 0 < x < π/2,

$\begin{array}{ll}\hfill {\left(\frac{\text{sin}x}{x}\right)}^{2}+\frac{\text{tan}x}{x}& =\frac{1}{2{x}^{2}}=-\frac{1}{2{x}^{2}}\text{cos}\left(2x\right)+\frac{\text{tan}x}{x}\phantom{\rule{2em}{0ex}}\\ =2+\sum _{k=3}^{\infty }\frac{\left(2\left({2}^{2k}-1\right)\left|{B}_{2k}\right|-{\left(-1\right)}^{k}\right){2}^{2k-1}}{\left(2k\right)!}{x}^{2k-2}.\phantom{\rule{2em}{0ex}}\end{array}$

It is well known [, p. 805] that

$\frac{2\left(2k\right)!}{{\left(2\pi \right)}^{2k}}<\left|{B}_{2k}\right|<\frac{2\left(2k\right)!}{{\left(2\pi \right)}^{2k}\left(1-{2}^{1-2k}\right)},\phantom{\rule{1em}{0ex}}k\ge 1.$
(11)

By (11), we find that

$2\left({2}^{2k}-1\right)\left|{B}_{2k}\right|>2\left({2}^{2k}-1\right)\frac{2\left(2k\right)!}{{\left(2\pi \right)}^{2k}}>1,\phantom{\rule{1em}{0ex}}k\ge 3.$

Hence, we have

$\begin{array}{ll}\hfill {\left(\frac{\text{sin}x}{x}\right)}^{2}+\frac{\text{tan}x}{x}& >2+\sum _{k=3}^{n}\frac{\left(2\left({2}^{2k}-1\right)\left|{B}_{2k}\right|-{\left(-1\right)}^{k}\right){2}^{2k-1}}{\left(2k\right)!}{x}^{2k-2}\phantom{\rule{2em}{0ex}}\\ =2+\frac{8}{45}{x}^{4}+\frac{16}{315}{x}^{6}+\frac{104}{4725}{x}^{8}+\frac{592}{66825}{x}^{10}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\cdots +\frac{\left(2\left({2}^{2n}-1\right)\left|{B}_{2n}\right|-{\left(-1\right)}^{n}\right){2}^{2n-1}}{\left(2n\right)!}{x}^{2n-2}.\phantom{\rule{2em}{0ex}}\end{array}$
(12)

Motivated by (12), we are now in a position to establish our first main result.

Theorem 1. (i) For 0 < x < π/2, we have

$2+\frac{8}{45}{x}^{4}+\frac{16}{315}{x}^{5}\text{tan}x<{\left(\frac{\text{sin}x}{x}\right)}^{2}+\frac{\text{tan}x}{x}<2+\frac{8}{45}{x}^{4}+{\left(\frac{2}{\pi }\right)}^{6}{x}^{5}\text{tan}x.$
(13)

The constants $\frac{16}{315}$ and ${\left(\frac{2}{\pi }\right)}^{6}$ are best possible.

1. (ii)

For 0 < x < π/2, we have

$\begin{array}{ll}\hfill 2+\frac{8}{45}{x}^{4}+\frac{16}{315}{x}^{6}+\frac{104}{4725}{x}^{7}\text{tan}x& <{\left(\frac{\text{sin}x}{x}\right)}^{2}+\frac{\text{tan}x}{x}\phantom{\rule{2em}{0ex}}\\ <2+\frac{8}{45}{x}^{4}+\frac{16}{315}{x}^{6}+{\left(\frac{2}{\pi }\right)}^{8}{x}^{7}\text{tan}x.\phantom{\rule{2em}{0ex}}\end{array}$
(14)

The constants $\frac{104}{4725}$ and ${\left(\frac{2}{\pi }\right)}^{8}$ are best possible.

Proof. We only prove inequality (13). The proof of (14) is analogous.

Consider the function

$\begin{array}{ll}\hfill g\left(x\right):& =\frac{{\left(\frac{\text{sin}x}{x}\right)}^{2}+\frac{\text{tan}x}{x}-2-\frac{8}{45}{x}^{4}}{{x}^{5}\text{tan}x}\phantom{\rule{2em}{0ex}}\\ =\frac{\text{sin}\left(2x\right)}{2{x}^{7}}+\frac{1}{{x}^{6}}-\frac{2\text{cot}x}{{x}^{5}}-\frac{8\text{cot}x}{45x},\phantom{\rule{1em}{0ex}}0

By using power series expansions (7) and (10), we find that

$g\left(x\right)=\sum _{n=3}^{\infty }{\beta }_{n}{2}^{2n-1}{x}^{2n-6},$

where

${\beta }_{n}:=\frac{4\cdot \left|{B}_{2n}\right|}{\left(2n\right)!}+\frac{\left|{B}_{2\left(n-2\right)}\right|}{45\cdot \left(2n-4\right)!}+{\left(-1\right)}^{n}\frac{2}{\left(2n+1\right)!},\phantom{\rule{1em}{0ex}}n\ge 3.$

By (11), we obtain

$\begin{array}{ll}\hfill \frac{4\cdot \left|{B}_{2n}\right|}{\left(2n\right)!}+\frac{\left|{B}_{2\left(n-2\right)}\right|}{45\cdot \left(2n-4\right)!}& >\frac{4}{\left(2n\right)!}\frac{2\left(2n\right)!}{{\left(2\pi \right)}^{2n}}+\frac{1}{45\cdot \left(2n-4\right)!}\frac{2\left(2n-4\right)!}{{\left(2\pi \right)}^{2n-4}}\phantom{\rule{2em}{0ex}}\\ =\frac{180+2\cdot {\left(2\pi \right)}^{4}}{45\cdot {\left(2\pi \right)}^{2n}}.\phantom{\rule{2em}{0ex}}\end{array}$

By induction on n, it is easy to see that

$\frac{180+2\cdot {\left(2\pi \right)}^{4}}{45\cdot {\left(2\pi \right)}^{2n}}>\frac{2}{\left(2n+1\right)!}\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{2.77695pt}{0ex}}\text{all}\phantom{\rule{2.77695pt}{0ex}}n\ge 3.$

Hence β n > 0 for n ≥ 3, and we have

${g}^{\prime }\left(x\right)=\sum _{n=4}^{\infty }{b}_{n}{2}^{2n-1}\left(2n-6\right){x}^{2n-7}>0,\phantom{\rule{1em}{0ex}}0

Therefore, g(x) is strictly increasing on (0, π/2). Noting that $\underset{t\to {\left(\pi /2\right)}^{-}}{\text{lim}}g\left(t\right)={\left(\frac{2}{\pi }\right)}^{6}$, we have

$\frac{16}{315}=\underset{t\to {0}^{+}}{\text{lim}}g\left(t\right)

for all $x\in \left(0,\frac{\pi }{2}\right)$, with the constants $\frac{16}{315}$ and ${\left(\frac{2}{\pi }\right)}^{6}$ being best possible. This completes the proof of (13).

Remark 1. Inequality (14) is sharper than (13). On the other hand, there is no strict comparison between inequalities (2) and (13). There is no strict comparison between inequalities (2) and (14) either.

In view of inequalities (13) and (14), we propose the following conjecture.

Conjecture 1. For 0 < x < π/2 and n ≥ 3, we have

$\begin{array}{l}\phantom{\rule{1em}{0ex}}2+\sum _{k=3}^{n}\frac{\left(2\left({2}^{2k}-1\right)\left|{B}_{2k}\right|-{\left(-1\right)}^{k}\right){2}^{2k-1}}{\left(2k\right)!}{x}^{2k-2}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+\frac{\left(2\left({2}^{2n+2}-1\right)\left|{B}_{2n+2}\right|-{\left(-1\right)}^{n+1}\right){2}^{2n+1}}{\left(2n+2\right)!}{x}^{2n-1}\text{tan}x\phantom{\rule{2em}{0ex}}\\ <{\left(\frac{\text{sin}x}{x}\right)}^{2}+\frac{\text{tan}x}{x}\phantom{\rule{2em}{0ex}}\\ <2+\sum _{k=3}^{n}\frac{\left(2\left({2}^{2k}-1\right)\left|{B}_{2k}\right|-{\left(-1\right)}^{k}\right){2}^{2k-1}}{\left(2k\right)!}{x}^{2k-2}+{\left(\frac{2}{\pi }\right)}^{2n}{x}^{2n-1}\text{tan}x.\phantom{\rule{2em}{0ex}}\end{array}$

## 4. Sharp the Wu-Srivastava inequality

By using power series expansion (10), we obtain for 0 < x < π/2,

${\text{csc}}^{2}x=-{\left(\text{cot}x\right)}^{\prime }=\frac{1}{{x}^{2}}+\sum _{k=1}^{\infty }\frac{{2}^{2k}\left(2k-1\left|{B}_{2k}\right|\right)}{\left(2k\right)!}{x}^{2k-2}.$
(15)

Hence for 0 < x < π/2,

$\begin{array}{ll}\hfill {\left(\frac{x}{\text{sin}x}\right)}^{2}+\frac{x}{\text{tan}x}& ={x}^{2}{\text{csc}}^{2}x+x\text{cot}x\phantom{\rule{2em}{0ex}}\\ =2+\sum _{k=2}^{\infty }\frac{{2}^{2k+1}\left(k-1\right)\left|{B}_{2k}\right|}{\left(2k\right)!}{x}^{2k}\phantom{\rule{2em}{0ex}}\\ =2+\frac{2}{45}{x}^{4}+\frac{8}{945}{x}^{6}+\frac{2}{1575}{x}^{8}+\frac{16}{93555}{x}^{10}+\cdots \phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}\end{array}$
(16)

Motivated by (16), we establish our second main result:

Theorem 2. (i) For 0 < x < π/2, we have

${\left(\frac{x}{\text{sin}x}\right)}^{2}+\frac{x}{\text{tan}x}<2+\frac{2}{45}{x}^{3}\text{tan}x.$
(17)

The constant $\frac{2}{45}$ is best possible.

1. (ii)

For 0 < x < π/2, we have

${\left(\frac{x}{\text{sin}x}\right)}^{2}+\frac{x}{\text{tan}x}<2+\frac{2}{45}{x}^{4}+\frac{8}{945}{x}^{5}\text{tan}x.$
(18)

The constant $\frac{8}{945}$ is best possible.

Proof. We only prove inequality (18). The proof of (17) is analogous.

Consider the function

$G\left(x\right):=\frac{{\left(\frac{x}{\text{sin}x}\right)}^{2}+\frac{x}{\text{tan}x}-2-\frac{2}{45}{x}^{4}}{{x}^{5}\text{tan}x}=\frac{A\left(x\right)}{B\left(x\right)},$

where

$\begin{array}{ll}\hfill A\left(x\right):& ={\left(\frac{x}{\text{sin}x}\right)}^{2}+\frac{x}{\text{tan}x}-2-\frac{2}{45}{x}^{4}\phantom{\rule{2em}{0ex}}\\ =\sum _{n=1}^{\infty }\frac{{2}^{2n+5}\left(n+1\right)\left|{B}_{2\left(n+2\right)}\right|}{\left(2n+4\right)!}{x}^{2n+4}=\sum _{n=1}^{\infty }{a}_{n}{x}^{2n+4}\phantom{\rule{2em}{0ex}}\end{array}$

with

${a}_{n}:=\frac{{2}^{2n+5}\left(n+1\right)\left|{B}_{2\left(n+2\right)}\right|}{\left(2n+4\right)!},$

and

$B\left(x\right):={x}^{5}\text{tan}x=\sum _{n=1}^{\infty }\frac{{2}^{2n}\left({2}^{2n}-1\right)\left|{B}_{2n}\right|}{\left(2n\right)!}{x}^{2n+4}=\sum _{n=1}^{\infty }{b}_{n}{x}^{2n+4}$

with

${b}_{n}:=\frac{{2}^{2n}\left({2}^{2n}-1\right)\left|{B}_{2n}\right|}{\left(2n\right)!}.$

We claim that the function G(x) is strictly decreasing on (0, π/2). By Lemma 1, it suffices to show that

$\frac{{a}_{n}}{{b}_{n}}>\frac{{a}_{n+1}}{{b}_{n+1}},\phantom{\rule{1em}{0ex}}n\ge 1.$
(19)

It is known [, p. 96] that

$\frac{2\cdot \left(2n\right)!}{{\left(2\pi \right)}^{2n}}<\left|{B}_{2n}\right|<\frac{{\pi }^{2}\left(2n\right)!}{3{\left(2\pi \right)}^{2n}},\phantom{\rule{1em}{0ex}}n\ge 1.$
(20)

By using (20), we obtain

$\frac{{a}_{n}}{{b}_{n}}=\frac{{2}^{5}\left(n+1\right)\left|{B}_{2\left(n+2\right)}\right|}{\left(2n+4\right)!}\cdot \frac{\left(2n\right)!}{\left({2}^{2n}-1\right)\left|{B}_{2n}\right|}>\frac{192\left(n+1\right)}{{\pi }^{2}{\left(2\pi \right)}^{4}\left({4}^{n}-1\right)}$

and

$\frac{{a}_{n+1}}{{b}_{n+1}}=\frac{{2}^{5}\left(n+2\right)\left|{B}_{2\left(n+3\right)}\right|}{\left(2n+6\right)!}\cdot \frac{\left(2n+2\right)!}{\left({2}^{2n+2}-1\right)\left|{B}_{2\left(n+1\right)}\right|}<\frac{16{\pi }^{2}\left(n+2\right)}{3{\left(2\pi \right)}^{4}\left({4}^{n+1}-1\right)}.$

So (19) is a consequence of the elementary inequality

$\frac{192\left(n+1\right)}{{\pi }^{2}{\left(2\pi \right)}^{4}\left({4}^{n}-1\right)}>\frac{16{\pi }^{2}\left(n+2\right)}{3{\left(2\pi \right)}^{4}\left({4}^{n+1}-1\right)},$

which is equivalent to

$\frac{36\left(n+1\right)}{{4}^{n}-1}>\frac{{\pi }^{4}\left(n+2\right)}{{4}^{n+1}-1},\phantom{\rule{1em}{0ex}}n\ge 1.$
(21)

The proof of the inequality (21) is not difficult, and is left with the readers. This proves the claim.

Noting that $\underset{t\to {0}^{+}}{\text{lim}}G\left(t\right)=\frac{8}{945}$, we have

$G\left(x\right)<\underset{t\to {0}^{+}}{\text{lim}}G\left(t\right)=\frac{8}{945}\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{2.77695pt}{0ex}}\text{all}\phantom{\rule{2.77695pt}{0ex}}x\in \left(0,\frac{\pi }{2}\right)$

with the constant $\frac{8}{945}$ being best possible. This completes the proof of (18).

In view of inequalities (17) and (18), we propose the following conjecture.

Conjecture 2. For 0 < x < π/2 and n ≥ 1,

${\left(\frac{x}{\text{sin}x}\right)}^{2}+\frac{x}{\text{tan}x}<2+\sum _{k=2}^{n}\frac{\left(k-1\right)\cdot {2}^{2k+1}\left|{B}_{2k}\right|}{\left(2k\right)!}{x}^{2k}+\frac{n\cdot {2}^{2n+3}\left|{{B}_{2}}_{\left(n+1\right)}\right|}{\left(2n+2\right)!}{x}^{2n+1}\text{tan}x.$
(22)

## 5. Skarp Huygens inequality

By using power series expansions (7) and (9), for 0 < x < π/2, we have

$2\left(\frac{\text{sin}x}{x}\right)+\frac{\text{tan}x}{x}=3+\sum _{k=3}^{\infty }\left(\frac{{2}^{2k}\left({2}^{2k}-1\right)\left|{B}_{2k}\right|}{4k}-{\left(-1\right)}^{k}\right)\frac{2{x}^{2k-2}}{\left(2k-1\right)!}.$

By (11), we find that

$\frac{{2}^{2k}\left({2}^{2k}-1\right)\left|{B}_{2k}\right|}{4k}>\frac{{2}^{2k}\left({2}^{2k}-1\right)}{4k}\frac{2\left(2k\right)!}{{\left(2\pi \right)}^{2k}}=\frac{\left({2}^{2k}-1\right)\cdot \left(2k-1\right)!}{{\pi }^{2k}}>1,\phantom{\rule{1em}{0ex}}k\ge 3.$

Hence we have

$\begin{array}{ll}\hfill 2\left(\frac{\text{sin}x}{x}\right)+\frac{\text{tan}x}{x}& >3+\sum _{k=3}^{n}\left(\frac{{2}^{2k}\left({2}^{2k}-1\right)\left|{B}_{2k}\right|}{4k}-{\left(-1\right)}^{k}\right)\frac{2{x}^{2k-2}}{\left(2k-1\right)!}\phantom{\rule{2em}{0ex}}\\ =3+\frac{3}{20}{x}^{4}+\frac{3}{56}{x}^{6}+\frac{7}{320}{x}^{8}+\frac{3931}{443520}{x}^{10}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\cdots +\left(\frac{\left({2}^{2n}\left({2}^{2n}-1\right)\left|{B}_{2n}\right|\right)}{4n}-{\left(-1\right)}^{n}\right)\frac{2{x}^{2n-2}}{\left(2n-1\right)!}.\phantom{\rule{2em}{0ex}}\end{array}$
(23)

Motivated by (23), we establish our third main result:

Theorem 3. (i) For 0 < x < π/2, we have

$3+\frac{3}{20}{x}^{3}\text{tan}x<2\left(\frac{\text{sin}x}{x}\right)+\frac{\text{tan}x}{x}<3+{\left(\frac{2}{\pi }\right)}^{4}{x}^{3}\text{tan}x.$
(24)

The constants $\frac{3}{20}$ and ${\left(\frac{2}{\pi }\right)}^{4}$ are best possible.

1. (ii)

For 0 < x < π/2, we have

$3+\frac{3}{20}{x}^{4}+\frac{3}{56}{x}^{5}\text{tan}x<2\left(\frac{\text{sin}x}{x}\right)+\frac{\text{tan}x}{x}<3+\frac{3}{20}{x}^{4}+{\left(\frac{2}{\pi }\right)}^{6}{x}^{5}\text{tan}x.$
(25)

The constants $\frac{3}{56}$ and ${\left(\frac{2}{\pi }\right)}^{6}$ are best possible.

Proof. We only prove inequality (25). The proof of (24) is analogous.

Consider the function

$\begin{array}{ll}\hfill h\left(x\right):\phantom{\rule{0.3em}{0ex}}& =\frac{2\left(\frac{\text{sin}x}{x}\right)+\frac{\text{tan}x}{x}-3-\frac{3}{20}{x}^{4}}{{x}^{5}\text{tan}x}\phantom{\rule{2em}{0ex}}\\ =\frac{2\text{cos}x}{{x}^{6}}+\frac{1}{{x}^{6}}-\frac{3\text{cot}x}{{x}^{5}}-\frac{3\text{cot}x}{20x},\phantom{\rule{1em}{0ex}}0

By using power series expansions (8) and (10), we find that

$h\left(x\right)=\sum _{n=3}^{\infty }{c}_{n}{x}^{2n-6},$

where

${c}_{n}:=\frac{3\cdot {2}^{2n}\left|{B}_{2n}\right|}{\left(2n\right)!}+\frac{3\cdot {2}^{2n-4}\left|{B}_{2\left(n-2\right)}\right|}{20\cdot \left(2n-4\right)!}+{\left(-1\right)}^{n}\frac{2}{\left(2n\right)!},\phantom{\rule{1em}{0ex}}n\ge 3.$

By (11), we obtain

$\begin{array}{ll}\hfill \frac{3\cdot {2}^{2n}\left|{B}_{2n}\right|}{\left(2n\right)!}+\frac{3\cdot {2}^{2n-4}\left|{B}_{2\left(n-2\right)}\right|}{20\cdot \left(2n-4\right)!}& >\frac{3\cdot {2}^{2n}}{\left(2n\right)!}\frac{2\left(2n\right)!}{{\left(2\pi \right)}^{2n}}+\frac{3\cdot {2}^{2n-4}}{20\cdot \left(2n-4\right)!}\frac{2\left(2n-4\right)!}{{\left(2\pi \right)}^{2n-4}}\phantom{\rule{2em}{0ex}}\\ =\frac{60+3\cdot {\pi }^{4}}{10\cdot {\pi }^{2n}}.\phantom{\rule{2em}{0ex}}\end{array}$

By induction on n, it is easy to show that

$\frac{60+3\cdot {\pi }^{4}}{10\cdot {\pi }^{2n}}>\frac{2}{\left(2n\right)!}\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{2.77695pt}{0ex}}\text{all}\phantom{\rule{2.77695pt}{0ex}}n\ge 3.$

Hence c n > 0 for n ≥ 3, and we have

${h}^{\prime }\left(x\right)=\sum _{n=4}^{\infty }{c}_{n}\left(2n-6\right){x}^{2n-7}>0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{2.77695pt}{0ex}}\text{all}\phantom{\rule{2.77695pt}{0ex}}0

Therefore, h(x) is strictly increasing on (0, π/2). Noting that $\underset{t\to {0}^{+}}{\text{lim}}h\left(t\right)=\frac{3}{56}$ and $\underset{t\to \left(\pi /2\right)-}{\text{lim}}h\left(t\right)={\left(\frac{2}{\pi }\right)}^{6}$, we have

$\frac{3}{56}=\underset{t\to {0}^{+}}{\text{lim}}h\left(t\right)

for all $x\in \left(0,\frac{\pi }{2}\right)$ with the constants $\frac{3}{56}$ and ${\left(\frac{2}{\pi }\right)}^{6}$ being possible. This completes the proof of (25).

Remark 2. There is no strict comparison between inequalities (24) and (25).

In view of inequalities (24) and (25), we propose the following conjecture.

Conjecture 3. For 0 < x < π/2 and n ≥ 2, we have

$\begin{array}{l}\phantom{\rule{1em}{0ex}}3+\sum _{k=3}^{n}\left(\frac{{2}^{2k}\left({2}^{2k}-1\right)\left|{B}_{2k}\right|}{4k}-{\left(-1\right)}^{k}\right)\frac{2}{\left(2k-1\right)!}{x}^{2k-2}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+\left(\frac{{2}^{2n+2}\left({2}^{2n+2}-1\right)\left|{B}_{2n+2}\right|}{4\left(n+1\right)}-{\left(-1\right)}^{n+1}\right)\frac{2}{\left(2n+1\right)!}{x}^{2n-1}\text{tan}x\phantom{\rule{2em}{0ex}}\\ <2\left(\frac{\text{sin}x}{x}\right)+\frac{\text{tan}x}{x}\phantom{\rule{2em}{0ex}}\\ <3+\sum _{k=3}^{n}\left(\frac{{2}^{2k}\left({2}^{2k}-1\right)\left|{B}_{2k}\right|}{4k}-{\left(-1\right)}^{k}\right)\frac{2}{\left(2k-1\right)!}{x}^{2k-2}+{\left(\frac{2}{\pi }\right)}^{2n}{x}^{2n-1}\text{tan}x.\phantom{\rule{2em}{0ex}}\end{array}$
(26)

## References

1. Wilker JB: Problem E 3306. Am Math Mon 1989, 96: 55. 10.2307/2323260

2. Sumner JS, Jagers AA, Vowe M, Anglesio J: Inequalities involving trigonometric functions. Am Math Mon 1991, 98: 264–267. 10.2307/2325035

3. Guo BN, Qiao BM, Qi F, Li W: On new proofs of Wilker inequalities involving trigonometric functions. Math Inequal Appl 2003, 6: 19–22.

4. Mortitc C: The natural approach of Wilker-Cusa-Huygens inequalities. Math Inequal Appl 2011, 14: 535–541.

5. Neuman E: On Wilker and Huygens type inequalities. Math Inequal Appl, in press.

6. Pinelis I: L'Hospital rules of monotonicity and Wilker-Anglesio inequality. Am Math Mon 2004, 111: 905–909. 10.2307/4145099

7. Wu SH, Srivastava HM: A weighted and exponential generalization of Wilker's inequality and its applications. Integr Trans Spec Funct 2007, 18: 529–535. 10.1080/10652460701284164

8. Wu SH, Srivastava HM: A further refinement of Wilker's inequality. Integr Trans Spec Funct 2008, 19: 757–765. 10.1080/10652460802340931

9. Zhang L, Zhu L: A new elementary proof of Wilker's inequalities. Math Inequal Appl 2008, 11: 149–151.

10. Zhu L: A new simple proof of Wilker's inequality. Math Inequal Appl 2005, 8: 749–750.

11. Zhu L: On Wilker-type inequalities. Math Inequal Appl 2007, 10: 727–731.

12. Zhu L: Some new Wilker-type inequalities for circular and hyperbolic functions. Abstr Appl Anal 2009, 2009: 9. (Article ID 485842)

13. Zhu L: A source of inequalities for circular functions. Comput Math Appl 2009, 58: 1998–2004. 10.1016/j.camwa.2009.07.076

14. Srivastava R: Some families of integral, trigonometric and other related inequalities. Appl Math Inf Sci 2011, 5: 342–360.

15. Huygens C: Oeuvres Completes 1888–1940. Société Hollondaise des Science, Haga;

16. Neuman E, Sándor J: On some inequalities involving trigonometric and hyperbolic functions with emphasis on the Cusa-Huygens, Wilker, and Huygens inequalities. Math Inequal Appl 2010, 13: 715–723.

17. Zhu L: Some new inequalities of the Huygens type. Comput Math Appl 2009, 58: 1180–1182. 10.1016/j.camwa.2009.07.045

18. Baricz A, Sándor J: Extensions of generalized Wilker inequality to Bessel functions. J Math Inequal 2008, 2: 397–406.

19. Mitrinović DS: Analytic Inequalities. Springer-Verlag, Berlin; 1970.

20. Chen CP, Cheung WS: Wilker- and Huygens-type inequalities and solution to Oppenheim's problem. Integr Trans Spec Funct, in press.

21. Ponnusamy S, Vuorinen M: Asymptotic expansions and inequalities for hypergeometric functions. Mathematika 1997, 44: 278–301. 10.1112/S0025579300012602

22. Abramowitz M, Stegun IA: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. In Appl Math Ser National Bureau of Standards. Volume 55. 9th edition. Washington, D.C; 1972.

23. Kuang J-Ch: Applied Inequalities. 3rd edition. Shandong Science and Technology Press, Jinan City, Shandong Province, China (Chinese); 2004.

## Acknowledgements

The research is supported in part by the Research Grants Council of the Hong Kong SAR, Project No. HKU7016/07P

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Correspondence to Chao-Ping Chen.

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Chen, CP., Cheung, WS. Sharpness of Wilker and Huygens type inequalities. J Inequal Appl 2012, 72 (2012). https://doi.org/10.1186/1029-242X-2012-72 