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

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 $0< x < \pi/2$.

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 [1–8]. 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

3.1 Proof of Theorem 2.1

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. □

3.2 Proof of Theorem 2.2

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.