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# Sharp Becker-Stark’s type inequalities with power exponential functions

Journal of Inequalities and Applications20152015:402

https://doi.org/10.1186/s13660-015-0932-9

• Accepted: 6 December 2015
• Published:

## Abstract

In this paper, we give some inequalities with power exponential functions derived from the left hand side of Becker-Stark’s inequality:
$$\frac{8}{\pi^{2} -4x^{2}} < \frac{\tan{x}}{x} < \frac{\pi^{2}}{\pi^{2} -4x^{2}}$$
for $$0< x < \pi/2$$.

## Keywords

• Becker-Stark’s inequality
• Bernoulli’s inequality
• monotonically decreasing function
• monotonically increasing function

• 26D05

## 1 Introduction

Becker-Stark’s inequality is well known:
$$\frac{8}{\pi^{2} -4x^{2}} < \frac{\tan{x}}{x} < \frac{\pi^{2}}{\pi^{2} -4x^{2}}$$
(1.1)
for $$0< x < \pi/2$$. The research of Becker-Stark’s inequality is one of the active areas in mathematical analysis [18]. Recently, Zhu [6] gave the following refinement of Becker-Stark’s inequality: For $$0< x<\pi/2$$, the inequalities
\begin{aligned} \frac{8}{\pi^{2} -4 x^{2}} + \frac{2}{\pi^{2}} -\frac{\pi^{2} -9}{6 \pi^{4}} \bigl(\pi ^{2} -4 x^{2}\bigr) < \frac{\tan{x}}{x} \end{aligned}
(1.2)
and
\begin{aligned} \frac{\tan{x}}{x} < \frac{8}{\pi^{2} -4 x^{2}} +\frac{2}{\pi^{2}} - \frac{10 -\pi^{2}}{\pi^{4}} \bigl(\pi^{2} -4 x^{2}\bigr) \end{aligned}
(1.3)
hold, where the constants $$-(\pi^{2} -9)/(6\pi^{4})$$ and $$-(10 -\pi^{2} )/\pi ^{4}$$ are the best possible. Moreover, from the right hand side of the inequality (1.1), Chen and Cheung [2] gave the following inequality: For $$0< x < \pi/2$$, the inequality
$$\biggl( \frac{\pi^{2}}{\pi^{2} -4x^{2}} \biggr)^{\theta}< \frac{\tan{x}}{x} < \biggl( \frac{\pi^{2}}{\pi^{2} -4x^{2}} \biggr)^{\vartheta}$$
(1.4)
holds, where the constants $$\theta= \pi^{2}/12$$ and $$\vartheta=1$$ are the best possible. In [5], Sun and Zhu gave a simple proof of the results. The above inequality (1.4) is created based on the right hand side of Becker-Stark’s inequality (1.1). However, in this paper we establish some inequalities created based on the left hand side of the inequality (1.1).

## 2 Results and discussion

Motivated by (1.4), in this paper, we give some inequalities with power exponential functions derived from the left hand side of Becker-Stark’s inequality (1.1). Since we note that $$8/(\pi^{2} -4x^{2})<1$$ for $$0< x< (\sqrt{\pi^{2} -8})/2$$ and $$8/(\pi^{2} -4x^{2})>1$$ for $$(\sqrt{\pi^{2} -8})/2 < x <\pi/2$$, we obtain the two inequalities as follows.

### Theorem 2.1

For $$0< x < (\sqrt{\pi^{2}-8})/2$$, we have
$$\biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\theta} < \frac{\tan{x}}{x} < \biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\vartheta(x)}$$
with the best possible constant $$\theta= 0$$ and the function
$$\vartheta(x) = \frac{2}{2x - \sqrt{\pi^{2}-8}}+\frac{2}{\sqrt{\pi ^{2}-8}} .$$

### Theorem 2.2

For $$(\sqrt{\pi^{2}-8})/2 < x < \pi/2$$, we have
$$\biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\theta} < \frac{\tan{x}}{x} < \biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\vartheta(x)}$$
with the best possible constant $$\theta= 1$$ and the function
$$\vartheta(x) = \frac{2}{2x - \sqrt{\pi^{2}-8}}-\frac{\sqrt{\pi ^{2}-8}-\pi+2}{\pi-\sqrt{\pi^{2}-8}}.$$
From Theorems 2.1 and 2.2, we have the best possible constant θ such that
$$\biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\theta} < \frac{\tan{x}}{x} .$$
If $$0< x< (\sqrt{\pi^{2} -8})/2$$, the constant θ must be $$\theta< 0$$ in order to satisfy $$1 \leq\tan{x}/x <(8/(\pi^{2} -4x^{2}))^{\theta}$$. On the other hands, if $$(\sqrt{\pi^{2} -8})/2 < x <\pi/2$$, the constant θ must be $$1 < \theta$$ in order to satisfy $$8/(\pi^{2} -4x^{2}) \leq\tan{x}/x < (8/(\pi^{2} -4x^{2}))^{\theta}$$. Here, we obtain the two inequalities as follows.

### Theorem 2.3

For $$1/2 < x < (\sqrt{\pi^{2}-8})/2$$, we have
$$\biggl( \frac{8}{\pi^{2} -4 x^{2}} \biggr)^{\frac{\vartheta(x)}{8}} < \frac {\tan{x}}{x} ,$$
where the function $$\vartheta(x)$$ is in Theorem  2.1.

### Corollary 2.4

For $$0 < x < \pi/2$$, we do not have the best possible constant ϑ such that
$$\frac{\tan{x}}{x} < \biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\vartheta} .$$

## 3 Proofs of main theorems

### Proof of Theorem 2.1

We set
$$f(x) = \biggl( \frac{8}{\pi^{2}-4 x^{2}} \biggr)^{\frac{2}{2x - \sqrt{\pi^{2} -8}} +\frac{2}{\sqrt{\pi^{2} -8}}} -\frac{\tan{x}}{x} .$$
From
$$\frac{2}{2x - \sqrt{\pi^{2}-8}}+\frac{2}{\sqrt{\pi^{2}-8}} < 0$$
for $$0< x < (\sqrt{\pi^{2}-8})/2$$, by Bernoulli’s inequality, we have
$$\biggl( \frac{8}{\pi^{2}-4 x^{2}} \biggr)^{\frac{2}{2x - \sqrt{\pi^{2} -8}} +\frac{2}{\sqrt{\pi^{2} -8}}} > 1 + \biggl( \frac{8}{\pi^{2}-4 x^{2}} -1 \biggr) \biggl( \frac{2}{2x -\sqrt{\pi^{2} -8} } +\frac{2}{\sqrt{\pi^{2} -8}} \biggr) .$$
By the right hand side of the inequality (1.1), for $$0< x < (\sqrt {\pi^{2}-8})/2$$,
\begin{aligned} f(x) &> 1 + \biggl( \frac{8}{\pi^{2}-4 x^{2}} -1 \biggr) \biggl( \frac{2}{2x -\sqrt{\pi^{2} -8} } + \frac{2}{\sqrt{\pi^{2} -8}} \biggr) -\frac{\pi^{2}}{\pi ^{2} -4 x^{2}} \\ &= \frac{4 x ( 2 \sqrt{\pi^{2} -8} x^{2}-4 x^{2} -\pi^{2} x +8 x +\pi^{2} -8 )}{\sqrt{\pi^{2} -8} (\pi-2 x) ( \sqrt{\pi^{2} -8} -2 x ) (2x +\pi)} \\ &= \frac{4 x g(x)}{\sqrt{\pi^{2} -8} (\pi-2 x) ( \sqrt{\pi^{2} -8} -2 x ) (2x +\pi)} , \end{aligned}
where
$$g(x) = 2 \sqrt{\pi^{2}-8} x^{2}-4 x^{2}- \pi^{2} x+8 x+\pi^{2}-8 .$$
From $$\sqrt{\pi^{2} -8} -2 x >0$$ for $$0< x<(\sqrt{\pi^{2} -8})/2$$, it suffices to show that
$$g(x) >0 .$$
Here, the derivative of $$g(x)$$ is
$$g'(x) = 8 -\pi^{2} + 4 \bigl(\sqrt{\pi^{2} -8} -2 \bigr) x .$$
By $$8 -\pi^{2}<0$$ and $$\sqrt{\pi^{2} -8} -2 <0$$, we have $$g'(x) < 0$$ for any $$0< x<(\sqrt{\pi^{2} -8})/2$$. Since $$g(x)$$ is strictly decreasing for $$0< x<(\sqrt{\pi^{2} -8})/2$$, we have
$$g(x) > g \biggl( \frac{\sqrt{\pi^{2}-8}}{2} \biggr) =0 .$$
Therefore, we can get
$$\frac{\tan{x}}{x} < \biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\vartheta (x)} ,$$
where
$$\vartheta(x) = \frac{2}{2x - \sqrt{\pi^{2}-8}}+\frac{2}{\sqrt{\pi^{2}-8}} .$$
Since $$\tan{x}/x$$ is strictly increasing for $$0< x<\pi/2$$, we have
$$\frac{8}{\pi^{2} -4x^{2}} < 1 < \frac{\tan{x}}{x}$$
for any $$0< x<( \sqrt{\pi^{2} -8})/2$$. Hence, for $$0< x<( \sqrt{\pi^{2} -8} )/2$$, we obtain
$$\biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\theta} < \frac{\tan{x}}{x} < \biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\vartheta(x)} ,$$
where the constant $$\theta=0$$. Since $$\vartheta(x)$$ is strictly decreasing for $$0< x<(\sqrt{\pi^{2} -8})/2$$ and
$$\vartheta(x) < \vartheta(0) = 0 ,$$
the constant $$\theta=0$$ is the best possible. Therefore, the proof of Theorem 2.1 is complete. □

### Proof of Theorem 2.2

We set
$$f(x) = \biggl( \frac{8}{\pi^{2}-4 x^{2}} \biggr)^{\frac{2}{2x - \sqrt{\pi ^{2}-8}}-\frac{\sqrt{\pi^{2}-8}-\pi+2}{\pi-\sqrt{\pi^{2}-8}}} -\frac{\tan {x}}{x} .$$
From
$$\frac{2}{2x - \sqrt{\pi^{2}-8}}-\frac{\sqrt{\pi^{2}-8}-\pi+2}{\pi-\sqrt {\pi^{2}-8}}>1$$
for $$(\sqrt{\pi^{2}-8})/2 < x < \pi/2$$, by Bernoulli’s inequality, we have
\begin{aligned} & \biggl( \frac{8}{\pi^{2}-4 x^{2}} \biggr)^{\frac{2}{2x - \sqrt{\pi ^{2}-8}}-\frac{\sqrt{\pi^{2}-8}-\pi+2}{\pi-\sqrt{\pi^{2}-8}}} \\ &\quad > 1 + \biggl( \frac{8}{\pi^{2}-4 x^{2}} -1 \biggr) \biggl( \frac{2}{2x - \sqrt{\pi^{2}-8}}- \frac{\sqrt{\pi^{2}-8}-\pi+2}{\pi-\sqrt{\pi^{2}-8}} \biggr) . \end{aligned}
By the inequality (1.3), for $$(\sqrt{\pi^{2}-8})/2 < x <\pi/2$$,
\begin{aligned}[b] f(x) >{}& 1 + \biggl(\frac{8}{\pi^{2}-4 x^{2}}-1 \biggr) \biggl(\frac{2}{2x - \sqrt{\pi^{2}-8}}- \frac{\sqrt{\pi^{2}-8}-\pi+2}{\pi-\sqrt{\pi ^{2}-8}} \biggr) \\ &{} - \biggl( \frac{8}{\pi^{2} -4 x^{2}} +\frac{2}{\pi^{2}} -\frac{10 -\pi ^{2}}{\pi^{4}} \bigl( \pi^{2} -4 x^{2}\bigr) \biggr) \\ ={}& \frac{g(x)}{\pi^{4} (\sqrt{\pi^{2}-8}-\pi ) (\sqrt {\pi^{2}-8}-2 x ) (2 x+\pi)} , \end{aligned}
where
\begin{aligned} g(x) ={}& 16 \pi^{3} x^{4} -16 \pi^{2} \sqrt{ \pi^{2} -8} x^{4} +160 \sqrt{\pi^{2} -8} x^{4}-160 \pi x^{4} \\ &{} +16 \pi^{4} x^{3}-224 \pi^{2} x^{3} -16 \pi^{3} \sqrt{\pi^{2} -8} x^{3} +160 \pi \sqrt{ \pi^{2} -8} x^{3} +640 x^{3} \\ &{} +8 \pi^{4} x^{2} -40 \pi^{3} x^{2} +8 \pi^{2} \sqrt{\pi^{2} -8} x^{2} +320 \pi x^{2} \\ &{} -4 \pi^{6} x +48 \pi^{4} x-128 \pi^{2} x +4 \pi^{5} \sqrt{\pi^{2} -8} x \\ &{} -32 \pi^{3} \sqrt{\pi^{2} -8} x-\pi^{7}-2 \pi^{6} +16 \pi^{5} +16 \pi^{4} \\ &{} -64 \pi^{3} +\pi^{6} \sqrt{\pi^{2} -8} -8 \pi^{4} \sqrt{\pi^{2} -8} . \end{aligned}
From $$(\sqrt{\pi^{2}-8}-\pi ) (\sqrt{\pi^{2}-8}-2 x ) >0$$ for $$(\sqrt{\pi^{2} -8})/2 < x < \pi/2$$, it suffices to show that
$$g(x) >0 .$$
We have the derivatives
\begin{aligned} g'(x) ={}& 4\bigl(16 \pi^{3} x^{3} -16 \pi^{2} \sqrt{\pi^{2} -8} x^{3} +160 \sqrt{ \pi^{2} -8} x^{3} -160 \pi x^{3} \\ &{} +12 \pi^{4} x^{2} -168 \pi^{2} x^{2} -12 \pi^{3} \sqrt{\pi^{2} -8} x^{2} +120 \pi \sqrt{ \pi^{2} -8} x^{2} +480 x^{2} \\ &{} +4 \pi^{4} x -20 \pi^{3} x +4 \pi^{2} \sqrt{ \pi^{2} -8} x +160 \pi x \\ &{} -\pi^{6} +12 \pi^{4} -32 \pi^{2} + \pi^{5} \sqrt{\pi^{2} -8}-8 \pi^{3} \sqrt{\pi ^{2} -8}\bigr) \\ ={}& 4h(x) \end{aligned}
and
\begin{aligned} h'(x) ={}& 4\bigl(12 \pi^{3} x^{2} -12 \pi^{2} \sqrt{\pi^{2} -8} x^{2} +120 \sqrt{ \pi^{2} -8} x^{2} -120 \pi x^{2} \\ &{} +6 \pi^{4} x -84 \pi^{2} x -6 \pi^{3} \sqrt{ \pi^{2} -8} x +60 \pi\sqrt{\pi ^{2} -8} x +240 x \\ &{} +\pi^{4} -5 \pi^{3} +\pi^{2} \sqrt{ \pi^{2} -8} +40 \pi\bigr) \\ ={}& 4 k(x) . \end{aligned}
From
$$-12 \bigl(\pi^{2}-10 \bigr) \bigl(\sqrt{\pi^{2}-8}-\pi \bigr) \cong -2.77627 < 0$$
and
$$-6 \bigl(\pi^{2}-10 \bigr) \bigl(\pi \sqrt{\pi^{2}-8}- \pi^{2}+4 \bigr) \cong-1.23145 < 0 ,$$
we have
\begin{aligned} k(x) ={}& {-}12 \bigl(\pi^{2} -10 \bigr) \bigl(\sqrt{\pi^{2} -8}-\pi \bigr) x^{2} -6 \bigl(\pi^{2} -10 \bigr) \bigl(\pi \sqrt{\pi^{2} -8} -\pi^{2} +4 \bigr) x \\ &{} +\pi^{2} \sqrt{\pi^{2} -8}+\pi^{4} -5 \pi^{3} +40 \pi \\ >{}& {-}12 \bigl(\pi^{2} -10 \bigr) \bigl(\sqrt{\pi^{2} -8}-\pi \bigr) \biggl( \frac{\pi}{2} \biggr)^{2} -6 \bigl( \pi^{2}-10 \bigr) \bigl(\pi \sqrt{\pi ^{2}-8}- \pi^{2}+4 \bigr) \biggl( \frac{\pi}{2} \biggr) \\ &{} +\pi^{2} \sqrt{\pi^{2} -8}+\pi^{4} -5 \pi^{3} +40 \pi \\ \cong{}& 72.7519 . \end{aligned}
Since $$h(x)$$ is strictly increasing for $$(\sqrt{\pi^{2} -8})/2 < x < \pi /2$$, we have
$$h(x) > h \biggl( \frac{\sqrt{\pi^{2}-8}}{2} \biggr) \cong191.598 .$$
Thus, $$g(x)$$ is strictly increasing for $$(\sqrt{\pi^{2} -8})/2 < x < \pi /2$$ and we have
$$g(x) > g \biggl( \frac{\sqrt{\pi^{2}-8}}{2} \biggr) = 0 .$$
Therefore, we can get
$$\frac{\tan{x}}{x} < \biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\vartheta (x)} ,$$
where
$$\vartheta(x) = \frac{2}{2x - \sqrt{\pi^{2}-8}}-\frac{\sqrt{\pi ^{2}-8}-\pi+2}{\pi-\sqrt{\pi^{2}-8}} .$$
Since we have
$$1 < \frac{8}{\pi^{2} -4x^{2}} < \frac{\tan{x}}{x}$$
for any $$(\sqrt{\pi^{2} -8})/2 < x<\pi/2$$, we obtain
$$\biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\theta} < \frac{\tan{x}}{x} < \biggl( \frac{8}{\pi^{2} -4x^{2}} \biggr)^{\vartheta(x)} ,$$
where the constant $$\theta=1$$. Since $$\vartheta(x)$$ is strictly decreasing for $$(\sqrt{\pi^{2} -8})/2 < x< \pi/2$$ and
$$\vartheta(x) > \vartheta \biggl( \frac{\pi}{2} \biggr) = 1 ,$$
the constant $$\theta=1$$ is the best possible. Hence, the proof of Theorem 2.2 is complete. □

### 3.3 Proof of Theorem 2.3 and Corollary 2.4

We need two lemmas to prove Theorem 2.3.

### Lemma 3.1

For $$-1/5 < t < 0$$, we have
$$\ln (t +1) > \frac{9}{8}t .$$

### Proof

We set
$$f(x) = \ln (t+1)-\frac{9}{8} t,$$
then
$$f'(t) = -\frac{9 t+1}{8 (t+1)} .$$
From $$f'(t) >0$$ for $$-1/5 < t < -1/9$$ and $$f'(t) <0$$ for $$-1/9 < t <0$$, $$f(t)$$ is strictly increasing for $$-1/5 < t < -1/9$$ and $$f(t)$$ is strictly decreasing for $$-1/9 < t < 0$$. Since
\begin{aligned} f \biggl( -\frac{1}{5} \biggr) & = \frac{9}{40} -\ln \biggl( \frac {5}{4} \biggr) \cong0.00185645 \end{aligned}
and
$$f(0) =0 ,$$
we can get $$f(t)>0$$ for $$-1/5 < t < 0$$. □

### Lemma 3.2

For $$0 < s < 1/5$$, we have
$$\ln (s +1) > \frac{8}{9}s .$$

### Proof

We set
$$f(s) = \ln (s +1) - \frac{8}{9}s,$$
then
$$f'(s) = -\frac{8 s-1}{9 (s+1)} .$$
From $$f'(s) >0$$ for $$0 < s < 1/8$$ and $$f'(s) <0$$ for $$1/8 < s <1/5$$, $$f(s)$$ is strictly increasing for $$0 < s < 1/8$$ and $$f(s)$$ is strictly decreasing for $$1/8 < s <1/5$$. Since
\begin{aligned} f \biggl( \frac{1}{5} \biggr) & = \ln \biggl(\frac{6}{5} \biggr) - \frac{8}{45} \cong0.00454378 \end{aligned}
and
$$f(0) =0 ,$$
we can get $$f(s)>0$$ for $$0 < s < 1/5$$. □

### Proof of Theorem 2.3

We set
\begin{aligned} f(x) & = \ln \frac{\tan{x}}{x} - \biggl( \frac{\vartheta(x)}{8} \biggr) \ln \frac{8}{\pi^{2} -4 x^{2}} \\ & = \ln \frac{\tan{x}}{x} - \biggl( \frac{1}{8x -4 \sqrt{\pi^{2} -8}} +\frac{1}{4 \sqrt{\pi^{2} -8}} \biggr) \ln \frac{8}{\pi^{2} -4 x^{2}} . \end{aligned}
If
$$t= -1 +\frac{8}{\pi^{2} -4 x^{2}} ,$$
then $$-11/100 < t < 0$$ for $$1/2 < x < (\sqrt{\pi^{2}-8})/2$$, by Lemma 3.1, we can get
\begin{aligned} \ln \frac{8}{\pi^{2} -4 x^{2}} > \frac{9}{8} \biggl(-1 + \frac{8}{\pi ^{2} -4 x^{2}} \biggr) . \end{aligned}
If
$$s= -1+\frac{8}{\pi^{2} -4 x^{2}} + \frac{2}{\pi^{2}} -\frac{\pi^{2} -9}{6 \pi ^{4}} \bigl( \pi^{2} -4 x^{2}\bigr) ,$$
then $$0 < s < 1/5$$ for $$1/2 < x < (\sqrt{\pi^{2} -8})/2$$, by Lemma 3.2 and the inequality (1.2), we can get
\begin{aligned} \ln \frac{\tan{x}}{x} & > \ln \biggl( \frac{8}{\pi^{2} -4 x^{2}} + \frac{2}{\pi^{2}} - \frac{\pi^{2} -9}{6 \pi^{4}} \bigl(\pi^{2} -4 x^{2}\bigr) \biggr) \\ & >\frac{8}{9} \biggl(-1+\frac{8}{\pi^{2} -4 x^{2}} + \frac{2}{\pi^{2}} - \frac {\pi^{2} -9}{6 \pi^{4}} \bigl(\pi^{2} -4 x^{2}\bigr) \biggr) . \end{aligned}
Since
$$\frac{1}{8x -4 \sqrt{\pi^{2} -8}} +\frac{1}{4 \sqrt{\pi^{2} -8}} < 0$$
and
$$\frac{9}{8} \biggl(-1 +\frac{8}{\pi^{2} -4 x^{2}} \biggr) < \ln \frac {8}{\pi^{2} -4 x^{2}} < 0$$
for $$1/2 < x < (\sqrt{\pi^{2} -8})/2$$, we obtain
\begin{aligned} f(x) >{}& \frac{8}{9} \biggl(-1+\frac{8}{\pi^{2} -4 x^{2}}+\frac{2}{\pi^{2}} - \frac{\pi^{2} -9}{6 \pi^{4}} \bigl(\pi^{2} -4 x^{2}\bigr) \biggr) \\ &{} - \biggl( \frac{1}{8x -4 \sqrt{\pi^{2} -8}} +\frac{1}{4 \sqrt{\pi^{2} -8}} \biggr) \times \frac{9}{8} \biggl(-1 +\frac{8}{\pi^{2} -4 x^{2}} \biggr) \\ ={}& \frac{g(x)}{432 \pi^{4} \sqrt{\pi^{2} -8} (\pi-2 x) (\sqrt{\pi ^{2} -8}-2 x ) (\pi+2 x)} , \end{aligned}
where
\begin{aligned}[b] g(x) ={}& {-}18{,}432 \sqrt{\pi^{2} -8} x^{5} +2{,}048 \pi^{2} \sqrt{\pi^{2} -8} x^{5} \\ &{} -73{,}728 x^{4} +17{,}408 \pi^{2} x^{4} -1{,}024 \pi^{4} x^{4} \\ &{} +972 \pi^{4} x^{3} +15{,}360 \pi^{2} \sqrt{ \pi^{2} -8} x^{3} -4{,}096 \pi^{4} \sqrt{\pi ^{2} -8} x^{3} \\ &{} +61{,}440 \pi^{2} x^{2} -24{,}064 \pi^{4} x^{2} +2{,}048 \pi^{6} x^{2} \\ &{} +1{,}944 \pi^{4} x -243 \pi^{6} x -8{,}832 \pi^{4} \sqrt{\pi^{2} -8} x +896 \pi^{6} \sqrt{\pi^{2} -8} x \\ &{} -35{,}328 \pi^{4} +8{,}000 \pi^{6} -448 \pi^{8} . \end{aligned}
It suffices to show that $$g(x) >0$$ for $$1/2 < x < (\sqrt{\pi^{2} -8})/2$$. We have derivatives
\begin{aligned}& \begin{aligned}[b] g'(x) ={}& {-}92{,}160 \sqrt{\pi^{2} -8} x^{4} +10{,}240 \pi^{2} \sqrt{\pi^{2} -8} x^{4} \\ &{} -294{,}912 x^{3} +69{,}632 \pi^{2} x^{3} -4{,}096 \pi^{4} x^{3} \\ &{} +2{,}916 \pi^{4} x^{2} +46{,}080 \pi^{2} \sqrt{ \pi^{2} -8} x^{2} -12{,}288 \pi^{4} \sqrt { \pi^{2} -8} x^{2} \\ & +122{,}880 \pi^{2} x -48{,}128 \pi^{4} x +4{,}096 \pi^{6} x \\ &{} +1{,}944 \pi^{4} -243 \pi^{6} -8{,}832 \pi^{4} \sqrt{ \pi^{2} -8} +896 \pi^{6} \sqrt {\pi^{2} -8} , \end{aligned}\\& \begin{aligned}[b] g''(x) ={}& 8 \bigl( -46{,}080 \sqrt{\pi^{2} -8} x^{3} +5{,}120 \pi^{2} \sqrt{\pi^{2} -8} x^{3} \\ &{} -110{,}592 x^{2} +26{,}112 \pi^{2} x^{2} -1{,}536 \pi^{4} x^{2} \\ &{} +729 \pi^{4} x +11{,}520 \pi^{2} \sqrt{\pi^{2} -8} x -3{,}072 \pi^{4} \sqrt{\pi^{2} -8} x \\ &{} +15{,}360 \pi^{2} -6{,}016 \pi^{4} +512 \pi^{6} \bigr) \\ ={}& 8 h(x), \end{aligned} \end{aligned}
and
\begin{aligned} \frac{h'(x)}{3} ={}& {-}46{,}080 \sqrt{\pi^{2} -8} x^{2} +5{,}120 \pi^{2} \sqrt{\pi^{2} -8} x^{2} \\ &{} -73{,}728 x +17{,}408 \pi^{2} x -1{,}024 \pi^{4} x \\ &{} +243 \pi^{4} +3{,}840 \pi^{2} \sqrt{\pi^{2} -8} -1{,}024 \pi^{4} \sqrt{\pi^{2} -8} \\ < {}& {-}46{,}080 \sqrt{\pi^{2} -8} \biggl( \frac{1}{2} \biggr)^{2}+5{,}120 \pi^{2} \sqrt{\pi^{2} -8} \biggl( \frac{1}{2} \sqrt{\pi^{2} -8} \biggr)^{2} \\ &{} -73{,}728 \biggl( \frac{1}{2} \biggr) +17{,}408 \pi^{2} \biggl( \frac{1}{2} \sqrt{\pi^{2} -8} \biggr) -1{,}024 \pi^{4} \biggl( \frac{1}{2} \biggr) \\ &{} +243 \pi^{4}+3{,}840 \pi^{2} \sqrt{\pi^{2} -8}-1{,}024 \pi^{4} \sqrt{\pi^{2} -8} \\ ={}& {-}147{,}456+22{,}528 \pi^{2} -768 \pi^{6} +\pi^{4} \biggl( 5{,}632 +\frac{729}{2} \sqrt{\pi^{2} -8} \biggr) \\ \cong{}& {-}13{,}629.3 . \end{aligned}
Thus, $$h(x)$$ is strictly decreasing for $$1/2 < x < (\sqrt{\pi^{2} -8})/2$$. From
$$h \biggl( \frac{1}{2} \biggr) \cong-33{,}392 ,$$
we have $$g''(x) < 0$$ for $$1/2 < x < (\sqrt{\pi^{2} -8})/2$$. Therefore, $$g'(x)$$ is strictly decreasing for $$x_{1} < x < (\sqrt{\pi^{2} -8})/2$$. From
$$g' \biggl( \frac{1}{2} \biggr) \cong5{,}734.6$$
and
$$g' \biggl( \frac{\sqrt{\pi^{2} -8}}{2} \biggr) \cong-67{,}578 ,$$
there exists uniquely a real number $$x_{1}$$ with $$1/2 < x_{1} < (\sqrt{\pi ^{2} -8})/2$$ such that $$g'(x_{1}) = 0$$. Hence, $$g(x)$$ is strictly increasing for $$1/2 < x < x_{1}$$ and $$g(x)$$ is strictly decreasing for $$x_{1} < x < (\sqrt{\pi^{2} -8})/2$$. From
$$g \biggl( \frac{1}{2} \biggr) \cong4{,}939$$
and
$$g \biggl( \frac{\sqrt{\pi^{2} -8}}{2} \biggr) =0 ,$$
we can get $$g(x) >0$$ for $$1/2 < x < (\sqrt{\pi^{2} -8})/2$$. Hence, the proof of Theorem 2.3 is complete. □

### Proof of Corollary 2.4

By Theorem 2.3, for $$1/2 < x < (\sqrt{\pi^{2} -8})/2$$, we have the following:
\begin{aligned} \frac{\ln \frac{\tan{x}}{x}}{\ln \frac{8}{\pi^{2} -4 x^{2}} } & < \frac{1}{8x -4 \sqrt{\pi^{2} -8}} +\frac{1}{4 \sqrt{\pi^{2} -8}} \\ & = \biggl( -\frac{1}{4} \biggr) \biggl( \frac{x}{\sqrt{\pi^{2} -8}} \biggr) \biggl( \frac{1}{\frac{\sqrt{\pi^{2} -8}}{2} -x} \biggr) . \end{aligned}
Therefore
$$\lim_{x \to(\sqrt{\pi^{2} -8})/2 -0} \frac{\ln \frac{\tan {x}}{x}}{\ln \frac{8}{\pi^{2} -4 x^{2}} }= -\infty .$$
The proof of Corollary 2.4 is complete. □

## 4 Conclusions

In this paper, we gave four inequalities derived from the left hand side of Becker-Stark’s inequality (1.1), which are natural generalizations of the inequality (1.1). Since the value of $$8/(\pi^{2} -4x^{2})$$ is less than 1 for $$0< x< (\sqrt {\pi^{2} -8})/2$$ and the value of $$8/(\pi^{2} -4x^{2})$$ is larger than 1 for $$(\sqrt{\pi^{2} -8})/2 < x <\pi/2$$, we established the inequalities in Theorems 2.1 and 2.2. By Theorem 2.3, we obtained Corollary 2.4 immediately.

## Declarations

### Acknowledgements

The author would like to thank Professor Mitsuhiro Miyagi and the referees for their helpful suggestions and good advice.

## Authors’ Affiliations

(1)
General Education, Ube National College of Technology, Tokiwadai 2-14-1, Ube Yamaguchi, 755-8555, Japan

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