# Refined quadratic estimations of Shafer’s inequality

## Abstract

We establish an inequality by quadratic estimations; the double inequality

$$\frac{\pi^{2} x}{4 +\sqrt{(\pi^{2} -4)^{2} + (2\pi x)^{2}}} < \arctan{x} < \frac{\pi^{2} x}{4 +\sqrt{32+ (2\pi x)^{2}}}$$

holds for $$x>0$$, where the constants $$(\pi^{2} -4)^{2}$$ and 32 are the best possible.

## 1 Introduction

Shafer  showed that the inequality

$$\arctan{x} > \frac{8 x}{3 +\sqrt{25+ \frac{80}{3} x^{2}}}$$
(1.1)

holds for $$x>0$$. Various Shafer-type inequalities are known, and they have been applied, extended and refined, see  and . Especially, Zhu  showed an upper bound for inequality (1.1) and proved that the following double inequality

$$\frac{8 x}{3 +\sqrt{25 + \frac{80}{3}x^{2}}} < \arctan{x} < \frac{8 x}{3 +\sqrt{25+ \frac{256}{\pi^{2}} x^{2}}}$$
(1.2)

holds for $$x>0$$, where the constants $$80/3$$ and $$256/\pi^{2}$$ are the best possible. Recently, in , Sun and Chen proved that the following inequality

$$\arctan{x}< \frac{8 x+\frac{32}{4725}x^{7}}{3 +\sqrt{25+ \frac{80}{3} x^{2}}}$$
(1.3)

holds for $$x>0$$; moreover, they showed that the inequality

$$\frac{8 x+\frac{32}{4725}x^{7}}{3 +\sqrt{25+ \frac{80}{3} x^{2}}} < \frac{8 x}{3 +\sqrt{25+ \frac{256}{\pi^{2}} x^{2}}}$$
(1.4)

holds for $$0< x< x_{0} \cong 1.4243$$. In this paper, we shall establish the refinements of inequalities (1.2) and (1.3).

## 2 Results and discussion

Motivated by (1.2), (1.3) and (1.4), in this paper, we give inequalities involving arctangent. The following are our main results.

### Theorem 2.1

For $$x>0$$, we have

$$\frac{\pi^{2} x}{4 +\sqrt{(\pi^{2} -4)^{2} + (2\pi x)^{2}}} < \arctan{x} < \frac{\pi^{2} x}{4 +\sqrt{32+ (2\pi x)^{2}}} ,$$
(2.1)

where the constants $$(\pi^{2} -4)^{2}$$ and 32 are the best possible.

### Theorem 2.2

For $$x> \alpha$$, we have

$$\frac{\pi^{2} x}{4 +\sqrt{(\pi^{2} -4)^{2} + (2\pi x)^{2}}} >\frac{8 x}{3 +\sqrt{25 + \frac{80}{3} x^{2}}} ,$$
(2.2)

where the constant $$\alpha = \sqrt{\frac{9600 -1860 \pi^{2} +90 \pi^{4}}{2304 -480 \pi^{2} +25 \pi^{4}}} \cong 2.26883$$ is the best possible.

### Theorem 2.3

For $$x> \beta$$, we have

$$\frac{8 x}{3 +\sqrt{25 + \frac{256}{\pi^{2}} x^{2}}} >\frac{\pi^{2} x}{4 +\sqrt{32 + (2\pi x)^{2}}} ,$$
(2.3)

where the constant $$\beta =\sqrt{ \frac{4096 +1536 \pi^{2} -528 \pi^{4} +24 \pi^{6} +\pi^{8}}{4096 \pi^{2} -768 \pi^{4} +36 \pi^{6}}} \cong 1.30697$$ is the best possible.

### Theorem 2.4

For $$x> \gamma$$, we have

$$\frac{8 x+\frac{32}{4725}x^{7}}{3 +\sqrt{25+ \frac{80}{3} x^{2}}} >\frac{\pi^{2} x}{4 +\sqrt{32 + (2\pi x)^{2}}} ,$$
(2.4)

where the constant $$\gamma \cong 1.38918$$ is the best possible and satisfies the equation

\begin{aligned} &151200 -14175 \pi^{2} + 128 \gamma^{6} - 1575 \sqrt{15} \pi^{2} \sqrt{15 + 16 \gamma^{2}} \\ &\quad {}+ 75600 \sqrt{8 + \pi^{2} \gamma^{2}} + 64 \gamma^{6} \sqrt{8 + \pi^{2} \gamma^{2}}=0. \end{aligned}

From Theorems 2.1, 2.2, 2.3 and 2.4, we can get the following proposition, immediately.

### Proposition 2.5

The double inequality (2.1) is sharper than (1.2) for $$x > \alpha$$. Moreover, the right-hand side of (2.1) is sharper than (1.3) for $$x>\gamma$$.

### 2.1 Proof of Theorem 2.1

Becker-Stark’s inequality is known as the inequality

$$\frac{8}{\pi^{2} -4x^{2}} < \frac{\tan{x}}{x} < \frac{\pi^{2}}{\pi^{2} -4x^{2}}$$
(2.5)

which holds for $$0< x < \pi/2$$. Also, Becker-Stark’s inequality (2.5) has various applications, extensions and refinements, see  and . Especially, Zhu  gave the following refinement of (2.5): The inequality

\begin{aligned} \frac{8}{\pi^{2} -4 x^{2}} + \frac{2}{\pi^{2}} -\lambda \bigl( \pi^{2} -4 x^{2}\bigr) < \frac{\tan{x}}{x} < \frac{8}{\pi^{2} -4 x^{2}} +\frac{2}{\pi^{2}} -\mu \bigl(\pi^{2} -4 x^{2}\bigr) \end{aligned}
(2.6)

holds for $$0< x<\pi/2$$, where the constants $$\lambda = (\pi^{2} -9)/(6\pi^{4})$$ and $$\mu = (10 -\pi^{2} )/\pi^{4}$$ are the best possible. In this paper, the result of Zhu (2.6) plays an important role in the proof of Theorem 2.1.

### Proof of Theorem 2.1

The equation

\begin{aligned} \arctan{x} = \frac{\pi^{2} x}{4 +\sqrt{c+ (2 \pi x)^{2}}} \end{aligned}

is equivalent to

\begin{aligned} c= \frac{\pi^{4} x^{2} -8 \pi^{2} x \arctan{x} +16 \arctan^{2} {x} -4 \pi^{2} x^{2} \arctan^{2} {x}}{\arctan^{2} {x}} . \end{aligned}

We set $$t= \arctan{x}$$, then

\begin{aligned} c &= \frac{\pi^{4} \tan^{2}{t}}{t^{2}} -\frac{8 \pi^{2} \tan{t}}{t} +16 -4 \pi^{2} \tan^{2}{t} \\ &= 16 +F_{1}(t) . \end{aligned}

First, we assume that $$0< t\leq 1/2$$. Here, the derivative of $$F_{1}(t)$$ is

\begin{aligned} F_{1}'(t) &= -\frac{8 \pi^{2} \sec^{2}{t}}{t} +\frac{8 \pi^{2} \tan{t}}{t^{2}} -8 \pi^{2} \sec^{2}{t} \tan{t} +\frac{2 \pi^{4} \sec^{2}{t} \tan{t}}{t^{2}} - \frac{2 \pi^{4} \tan^{2}{t}}{t^{3}} \\ &= \frac{\sin{t}}{\cos^{2}{t}} \biggl( -\frac{8 \pi^{2} }{t \sin{t}} +\frac{8 \pi^{2} \cos{t}}{t^{2}} - \frac{8 \pi^{2}}{\cos{t}} +\frac{2 \pi^{4}}{t^{2} \cos{t}} -\frac{2 \pi^{4} \sin{t}}{t^{3}} \biggr) \\ &= \frac{\sin{t}}{\cos^{2}{t}} F_{2}(t) . \end{aligned}

Since we have

\begin{aligned} t-\frac{t^{3}}{6} < \sin{t} < t-\frac{t^{3}}{6}+\frac{t^{5}}{120} \end{aligned}

and

\begin{aligned} 1 -\frac{t^{2}}{2} +\frac{t^{4}}{24} -\frac{t^{6}}{720} < \cos{t} < 1 - \frac{t^{2}}{2} +\frac{t^{4}}{24} \end{aligned}

for $$0 < t <\pi/2$$, the following inequality holds:

\begin{aligned} F_{2}(t) &< -\frac{8 \pi^{2}}{t (t-\frac{t^{3}}{6} +\frac{t^{5}}{120} )} +\frac{8 \pi^{2} ( 1 -\frac{t^{2}}{2} +\frac{t^{4}}{24} )}{t^{2}} \\ & \quad {}-\frac{8 \pi^{2}}{ (1-\frac{t^{2}}{2}+\frac{t^{4}}{24} )} +\frac{2 \pi^{4}}{t^{2} ( 1 -\frac{t^{2}}{2} +\frac{t^{4}}{24} -\frac{t^{6}}{720} )} -\frac{2 \pi^{4} (t-\frac{t^{3}}{6} )}{t^{3}} \\ &= \frac{\pi^{2} F_{3}(t)}{3 (120-20 t^{2}+t^{4} ) (24-12 t^{2}+t^{4} ) (-720+360 t^{2}-30 t^{4}+t^{6} )} , \end{aligned}

where $$F_{3}(t) = 82944000-8294400 \pi^{2} -72990720 t^{2} +7084800 \pi^{2} t^{2} +24883200 t^{4} -2246400 \pi^{2} t^{4} -4832640 t^{6} +371520 \pi^{2} t^{6} +596736 t^{8} -35904 \pi^{2} t^{8} -48192 t^{10} +2076 \pi^{2} t^{10} +2472 t^{12} -68 \pi^{2} t^{12} -74 t^{14} +\pi^{2} t^{14} +t^{16}$$. We set $$s=t^{2}$$, then

\begin{aligned} F_{3}(t) &> 82944000-8294400 \biggl( \frac{315}{100} \biggr)^{2} -72990720 s+7084800 \biggl( \frac{314}{100} \biggr)^{2} s \\ & \quad {}+24883200 s^{2}-2246400 \biggl( \frac{315}{100} \biggr)^{2} s^{2}-4832640 s^{3}+371520 \biggl( \frac{314}{100} \biggr)^{2} s^{3} \\ &\quad {} +596736 s^{4}-35904 \biggl( \frac{315}{100} \biggr)^{2} s^{4}-48192 s^{5}+2076 \biggl( \frac{314}{100} \biggr)^{2} s^{5} \\ &\quad {} +2472 s^{6}-68 \biggl( \frac{315}{100} \biggr)^{2} s^{6}-74 s^{7}+ \biggl( \frac{314}{100} \biggr)^{2} s^{7}+s^{8} \\ & = 642816-\frac{78435648 s}{25}+2593296 s^{2} -\frac{146200176 s^{3}}{125} + \frac{6011964 s^{4}}{25} \\ &\quad {} -\frac{17327169 s^{5}}{625} +\frac{179727 s^{6}}{100}-\frac{160351 s^{7}}{2500}+s^{8} \\ & = \frac{1}{2500} \bigl(1607040000 -7843564800 s +6483240000 s^{2} -2924003520 s^{3} \\ &\quad {} +601196400 s^{4} -69308676 s^{5} +4493175 s^{6} -160351 s^{7} +2500 s^{8}\bigr) \\ & = \frac{1}{2500} \biggl( 1607040000 -7843564800 s + \biggl( \frac{7}{8} \biggr) 6483240000 s^{2} \\ &\quad {} + s^{2} \biggl( \biggl( \frac{1}{8} \biggr) 6483240000 -2924003520 s +601196400 s^{2} -69308676 s^{3} \biggr) \\ &\quad {} +s^{6}\bigl(4493175 -160351 s +2500 s^{2}\bigr) \biggr) \\ & = \frac{1}{2500} \bigl( F_{4}(s) + s^{2} F_{5}(s) +s^{6} F_{6}(s) \bigr) . \end{aligned}

We shall show that the functions $$F_{4}(s)>0$$, $$F_{5}(s)>0$$ and $$F_{6}(s)>0$$. Here,

\begin{aligned} F_{4}(s) &= 5400 \bigl(297600-1452512 s+1050525 s^{2} \bigr) \\ & =5400 F_{7}(t) . \end{aligned}

The derivative of $$F_{7}(t)$$ is

\begin{aligned} F'_{7}(s) &= 2( -726256 +1050525s) \\ &\leq 2 \biggl( -726256 +1050525 \biggl( \frac{1}{4} \biggr) \biggr) \\ &= -\frac{1854499}{2} . \end{aligned}

Since $$F_{7}(s)$$ is strictly decreasing for $$0< s< 1/4$$ and $$F_{7}(1/4) = 2077/16$$, we have $$F_{4}(s) >0$$.

\begin{aligned} F_{5}(s) &= 36 \bigl(22511250-81222320 s+16699900 s^{2}-1925241 s^{3}\bigr) \\ &> 36 \bigl(22511250-81222320 s -1925241 s^{3}\bigr) \\ &\geq 36 \biggl( 22511250-81222320 \biggl( \frac{1}{4} \biggr) -1925241 \biggl( \frac{1}{4} \biggr)^{3} \biggr) \\ &= \frac{1854335151}{16} \end{aligned}

and

\begin{aligned} F_{6}(s) &> 4493175-160351 \biggl( \frac{1}{4} \biggr) \\ &= \frac{17812349}{4} . \end{aligned}

Therefore, we can get $$F_{3}(t)>0$$. By $$120-20 t^{2}+t^{4}>0$$, $$24-12 t^{2}+t^{4}>0$$ and $$-720+360 t^{2}-30 t^{4}+t^{6}<0$$, thus $$F_{2}(t) <0$$ and $$F_{1}(t)$$ is strictly decreasing for $$0 < t <1/2$$. From $$F_{1}(0+) = (\pi ^{2} -4 )^{2}-16$$, we can get

\begin{aligned} F_{1} \biggl( \frac{1}{2} \biggr) \leq F_{1}(t) < \bigl(\pi ^{2} -4 \bigr)^{2} -16 \end{aligned}

for $$0< t \leq 1/2$$. Next, we assume that $$1/2 < t < \pi/2$$. From inequality (2.6), we have

\begin{aligned} &-8 \pi^{2} \biggl\{ \frac{2}{\pi^{2}} +\frac{8}{\pi^{2} -4 t^{2}} - \frac{ (10 -\pi^{2} ) (\pi^{2} -4 t^{2} )}{\pi^{4}} \biggr\} \\ &\qquad {} +\pi^{2} (\pi -2 t) (\pi +2 t) \biggl\{ \frac{2}{\pi^{2}} + \frac{8}{\pi^{2} -4 t^{2}} -\frac{ (-9 +\pi^{2} ) (\pi^{2} -4 t^{2} )}{6 \pi^{4}} \biggr\} ^{2} \\ &\quad < F_{1}(t) \\ &\quad < -8 \pi^{2} \biggl\{ \frac{2}{\pi^{2}} +\frac{8}{\pi^{2} -4 t^{2}} - \frac{ (-9 +\pi^{2} ) (\pi^{2}-4 t^{2} )}{6 \pi^{4}} \biggr\} \\ &\qquad {}+\pi^{2} (\pi -2 t) (\pi +2 t) \biggl\{ \frac{2}{\pi^{2}} + \frac{8}{\pi^{2} -4 t^{2}} -\frac{ (10 -\pi^{2} ) (\pi^{2} -4 t^{2} )}{\pi^{4}} \biggr\} ^{2} \end{aligned}

and

\begin{aligned} \frac{G_{1}(t)}{36\pi^{6}} < F_{1}(t) < \frac{G_{2}(t)}{3 \pi^{6}} , \end{aligned}

where

\begin{aligned} G_{1}(t)& = 4761\pi^{6} -426 \pi^{8} + \pi^{10}-18252\pi^{4} t^{2} +1944\pi^{6} t^{2} -12 \pi^{8} t^{2} \\ &\quad {} +7344\pi^{2} t^{4} -1248\pi^{4} t^{4} +48\pi^{6} t^{4} -5184 t^{6} +1152\pi^{2} t^{6} -64\pi^{4} t^{6} \end{aligned}

and

\begin{aligned} G_{2}(t) &= -276 \pi^{6} +4 \pi^{8} +3 \pi^{10} -624 \pi^{4} t^{2} +416 \pi^{6} t^{2} -36 \pi^{8} t^{2} \\ &\quad {} +12480 \pi^{2} t^{4} -2688 \pi^{4} t^{4} +144 \pi^{6} t^{4}-19200 t^{6}+3840 \pi^{2} t^{6}-192 \pi^{4} t^{6} . \end{aligned}

We set $$s=t^{2}$$, then

\begin{aligned} G_{1}(t) &= 4761 \pi^{6} -426 \pi^{8} + \pi^{10} -12 \pi^{4} \bigl(1521 -162 \pi^{2} + \pi^{4} \bigr) s \\ &\quad {} +48 (-3+\pi ) \pi^{2} (3+\pi ) \bigl(-17 +\pi^{2} \bigr) s^{2} -64 (-3+\pi )^{2} (3 +\pi)^{2} s^{3} \\ &= G_{3}(s) \end{aligned}

and

\begin{aligned} G_{2}(t) &= -276 \pi^{6} +4 \pi^{8} +3 \pi^{10}-4 \pi^{4} \bigl(156 -104 \pi^{2} +9 \pi^{4} \bigr)s \\ &\quad {} +48 \pi^{2} \bigl(\pi^{2} -10 \bigr) \bigl(-26+3 \pi^{2} \bigr) s^{2} -192 \bigl(\pi^{2} -10 \bigr)^{2} s^{3} \\ &= G_{4}(s) . \end{aligned}

The derivatives of $$G_{3}(s)$$ are

\begin{aligned} G'_{3}(s) &= 12\bigl(-1521 \pi^{4} +162 \pi^{6} -\pi^{8} +1224 \pi^{2} s -208 \pi^{4} s \\ &\quad {} +8 \pi^{6} s -1296 s^{2} +288 \pi^{2} s^{2} -16 \pi^{4} s^{2}\bigr) \end{aligned}

and

\begin{aligned} G''_{3}(t) = 96 (-3+\pi ) (3+\pi ) \bigl(-17 \pi^{2} +\pi^{4} +36 s -4 \pi^{2} s \bigr) . \end{aligned}

From the inequality

\begin{aligned} -17 \pi ^{2}+\pi ^{4} +\bigl(36 -4 \pi ^{2}\bigr) s & < -17 \pi ^{2}+\pi ^{4} +\bigl(36 -4 \pi ^{2} \bigr) \biggl( \frac{1}{4} \biggr) \\ & = 9-18 \pi ^{2}+\pi ^{4} \\ & \cong -71.2438 , \end{aligned}

$$G''_{3}(s)<0$$ and $$G'_{3}(s)$$ is strictly decreasing for $$1/4 < s < \pi^{2}/4$$. Since $$G'_{3}(1/4) = 12(-81 +324 \pi^{2}-1574 \pi^{4} +164 \pi^{6} -\pi^{8}) \cong -24310.3$$, $$G'_{3}(s)<0$$ and $$G_{3}(s)$$ is strictly decreasing for $$1/4 < s < \pi^{2}/4$$. Therefore, we have $$G_{1}(t) > G_{3}(\pi^{2}/4) =576 \pi^{6}$$ for $$1/2< t< \pi/2$$. Next, the derivatives of $$G_{4}(s)$$ are

\begin{aligned} G'_{4}(s)& =4\bigl(-156 \pi^{4} +104 \pi^{6}-9 \pi^{8}+6240 \pi^{2} s-1344 \pi^{4} s+72 \pi^{6} s \\ &\quad {} -14400 s^{2} +2880 \pi^{2} s^{2} -144 \pi^{4} s^{2}\bigr) \end{aligned}

and

\begin{aligned} G''_{4}(s) = 96 \bigl( 10 - \pi^{2} \bigr) \bigl( 26 \pi^{2} -3 \pi^{4} -120 s +12 \pi^{2} s \bigr) . \end{aligned}

From the inequality

\begin{aligned} 26 \pi^{2} -3 \pi^{4} -120 s +12 \pi^{2} s & < 26 \pi^{2} -3 \pi^{4} +\bigl(-120 +12 \pi^{2}\bigr) \biggl( \frac{1}{4} \biggr) \\ & = -30+29 \pi ^{2}-3 \pi ^{4} \\ & \cong -36.0087 , \end{aligned}

$$G''_{4}(s)<0$$ and $$G'_{4}(s)$$ is strictly decreasing for $$1/4 < s < \pi^{2}/4$$. Since $$G'_{4}(1/4) = 4(-900+1740 \pi ^{2}-501 \pi ^{4}+122 \pi ^{6}-9 \pi ^{8}) \cong -2544.56$$, $$G'_{4}(s)<0$$ and $$G_{4}(s)$$ is strictly decreasing for $$1/4 < s < \pi^{2}/4$$. Therefore, we have $$G_{2}(t) > G_{4}(\pi^{2}/4) =48 \pi^{6}$$ for $$1/2< t< \pi/2$$. By the squeeze theorem, $$F_{1}(t) >16$$ for $$1/2 < t< \pi/2$$. Also, we have

\begin{aligned} F_{1}(t) < \frac{G_{2}(\frac{1}{2})}{3\pi^{6}} \end{aligned}

for $$1/2 < t <\pi/2$$ and

\begin{aligned} F_{1}(0+)-\frac{G_{2}(\frac{1}{2})}{3\pi^{6}} &=\bigl(\pi^{2} -4 \bigr)^{2} -16 -\frac{G_{2}(\frac{1}{2})}{3\pi^{6}} \\ & = \bigl(\pi^{2} -4 \bigr)^{2} -16 -\frac{-300 +840 \pi^{2}-327 \pi^{4} -163 \pi^{6}-5 \pi^{8} +3 \pi^{10}}{3\pi^{6}} \\ & =\frac{300-840 \pi^{2} +327 \pi^{4} +163 \pi^{6} -19 \pi^{8}}{3\pi^{6}} . \end{aligned}

By $$300-840 \pi^{2} +327 \pi^{4} +163 \pi^{6} -19 \pi^{8} \cong 286.654$$, we have

\begin{aligned} F_{1}(0+)>\frac{G_{2}(\frac{1}{2})}{3\pi^{6}} . \end{aligned}

Thus, we can get $$16 < F_{1}(t) < F_{1}(0+)$$ for $$0< t<\pi/2$$. The proof of Theorem 2.1 is complete. □

### Proof of Theorem 2.2

We have

\begin{aligned} F_{1}(x)&= \frac{\pi^{2} x}{4 +\sqrt{(\pi^{2} -4)^{2} + (2\pi x)^{2}}} -\frac{8 x}{3 +\sqrt{25 + \frac{80}{3} x^{2}}} \\ & = \frac{x ( -96 +9 \pi^{2} +\sqrt{15} \pi^{2} \sqrt{15 +16 x^{2}} -24 \sqrt{16-8 \pi^{2} +\pi^{4} +4 \pi^{2} x^{2}} )}{ (9 +\sqrt{15} \sqrt{15+16 x^{2}} ) (4 +\sqrt{16-8 \pi^{2}+\pi^{4}+4 \pi^{2} x^{2}} )} \\ & = \frac{x F_{2}(x)}{ (9 +\sqrt{15} \sqrt{15+16 x^{2}} ) (4 +\sqrt{16-8 \pi^{2}+\pi^{4}+4 \pi^{2} x^{2}} )} . \end{aligned}

The derivative of $$F_{2}(x)$$ is

\begin{aligned} F_{2}'(x) &= \frac{16 \pi^{2} x (-6 \sqrt{15+16 x^{2}}+\sqrt{15} \sqrt{16-8 \pi^{2} +\pi^{4} +4 \pi^{2} x^{2}} )}{\sqrt{15+16 x^{2}} \sqrt{16-8 \pi^{2} +\pi^{4} +4 \pi^{2} x^{2}}} \\ &= \frac{16 \pi^{2} x F_{3}(x)}{\sqrt{15+16 x^{2}} \sqrt{16-8 \pi ^{2}+\pi ^{4}+4 \pi ^{2} x^{2}}} . \end{aligned}

Here, we have $$15 (16-8 \pi ^{2}+\pi ^{4}+4 \pi ^{2} x^{2} ) - 36(15+16 x^{2} ) = 3 (-100-40 \pi ^{2}+5 \pi ^{4}-192 x^{2}+20 \pi ^{2} x^{2} )$$. Since $$-192 +20 \pi^{2} >0$$ and $$-100-40 \pi ^{2}+5 \pi ^{4}-192 x^{2}+20 \pi ^{2} x^{2}=0$$ for $$x=\sqrt{\frac{100+40 \pi ^{2}-5 \pi ^{4}}{20 \pi ^{2}-192}} \cong 1.198$$, we have $$F_{3}(x)<0$$ for $$0< x<\sqrt{\frac{100+40 \pi ^{2}-5 \pi ^{4}}{20 \pi ^{2}-192}}$$ and $$F_{3}(x)>0$$ for $$x>\sqrt{\frac{100+40 \pi ^{2}-5 \pi ^{4}}{20 \pi ^{2}-192}}$$. Therefore, $$F_{2}(x)$$ is strictly decreasing for $$0 < x < \sqrt{\frac{100+40 \pi ^{2}-5 \pi ^{4}}{20 \pi ^{2}-192}}$$ and strictly increasing for $$x > \sqrt{\frac{100+40 \pi ^{2}-5 \pi ^{4}}{20 \pi ^{2}-192}}$$. From $$F_{2}(0+)=0$$ and

\begin{aligned} F_{2}(\alpha) &= -96+9 \pi^{2} +\sqrt{15} \pi^{2} \sqrt{15+16 \alpha^{2} } -24 \sqrt{16 -8 \pi^{2} + \pi^{4} +4 \pi^{2} \alpha^{2}} \\ &= -96+9 \pi^{2} +\sqrt{15} \pi^{2} \biggl( \frac{\sqrt{15} (112-11 \pi ^{2} )}{-48+5 \pi ^{2}} \biggr) -24 \biggl( \frac{192+32 \pi ^{2}-5 \pi ^{4}}{-48+5 \pi ^{2}} \biggr) \\ &= 0 , \end{aligned}

we can get $$F_{2}(x)>0$$ for $$x> \alpha$$ and α is the best possible. The proof of Theorem 2.2 is complete. □

### Proof of Theorem 2.3

We have

\begin{aligned} F_{1}(x) &= \frac{8 x}{3 +\sqrt{25 + \frac{256}{\pi^{2}} x^{2}}} -\frac{\pi^{2} x}{4 +\sqrt{32 + (2\pi x)^{2}}} \\ &= \frac{\pi x (32-3 \pi^{2} -\pi \sqrt{25 \pi^{2} +256 x^{2}}+16 \sqrt{8 +\pi^{2} x^{2}} )}{ 2 (3 \pi +\sqrt{25 \pi^{2} +256 x^{2}} ) (2+\sqrt{8+\pi^{2} x^{2}} )} \\ &= \frac{\pi x F_{2}(x)}{ 2 (3 \pi +\sqrt{25 \pi^{2} +256 x^{2}} ) (2+\sqrt{8+\pi^{2} x^{2}} )} . \end{aligned}

The derivative of $$F_{2}(x)$$ is

\begin{aligned} F_{2}'(x) &= \frac{16 \pi x (\pi \sqrt{25 \pi^{2} +256 x^{2}} -16 \sqrt{8+\pi^{2} x^{2}} )}{\sqrt{25 \pi^{2} +256 x^{2}} \sqrt{8+\pi^{2} x^{2}}} \\ &= \frac{16 \pi x F_{3}(x)}{\sqrt{25 \pi^{2} +256 x^{2}} \sqrt{8+\pi^{2} x^{2}}}. \end{aligned}

Since $$\pi^{2} (25 \pi^{2} +256 x^{2}) - {16}^{2} (8+\pi^{2} x^{2}) = -2048 +25 \pi^{4} \cong 387.227$$, we can get $$\pi^{2} (25 \pi ^{2}+256 x^{2}) > {16}^{2} (8+\pi ^{2} x^{2})$$ for $$x>0$$. Therefore, $$F_{3}(x)>0$$ and $$F_{2}'(x)>0$$ for $$x>0$$. Since $$F_{2}(x)$$ is strictly increasing for $$x>0$$ and

\begin{aligned} F_{2}(\beta) &= 32-3 \pi^{2} -\pi \sqrt{25 \pi^{2} +256 \beta^{2}} +16 \sqrt{8 +\pi^{2} \beta^{2}} \\ &= 32-3 \pi^{2} -\pi \biggl( \frac{512 +96 \pi^{2} -17 \pi^{4}}{\pi ( -32 +3 \pi^{2} )} \biggr) +16 \biggl( \frac{192-12 \pi^{2} -\pi^{4}}{2 (-32+3 \pi ^{2} )} \biggr) \\ &= 0 , \end{aligned}

we can get $$F_{2}(x)>0$$ for $$x> \beta$$ and β is the best possible. The proof of Theorem 2.3 is complete. □

### Lemma 2.6

For $$x>0$$, we have

\begin{aligned} \frac{75600 \pi ^{2} x}{\sqrt{8+\pi ^{2} x^{2}}}+\frac{64 \pi ^{2} x^{7}}{\sqrt{8+\pi ^{2} x^{2}}} >\frac{25200 \sqrt{15} \pi ^{2} x}{\sqrt{15+16 x^{2}}} . \end{aligned}

### Proof

We have

\begin{aligned} \biggl( \frac{75600 \pi ^{2} x}{\sqrt{8+\pi ^{2} x^{2}}}+\frac{64 \pi ^{2} x^{7}}{\sqrt{8+\pi ^{2} x^{2}}} \biggr)^{2} - \biggl( \frac{25200 \sqrt{15} \pi ^{2} x}{\sqrt{15+16 x^{2}}} \biggr)^{2} &= \frac{256 \pi ^{4} x^{2} F_{1}(x)}{ (15+16 x^{2} ) (8+\pi ^{2} x^{2} )} , \end{aligned}

where $$F_{1}(x)= 37209375+357210000 x^{2}-37209375 \pi ^{2} x^{2}+567000 x^{6}+604800 x^{8}+240 x^{12}+256 x^{14}$$. Here, we have

\begin{aligned} F_{1}(x) & > 37209375+357210000 x^{2}-37209375 \pi ^{2} x^{2}+567000 x^{6} \\ & = 70875 \bigl(525+5040 x^{2}-525 \pi ^{2} x^{2}+8 x^{6}\bigr) . \end{aligned}

We set $$t=x^{2}$$ and $$F_{2}(t) = 525+5040 t -525 \pi^{2} t +8 t^{3}$$, then the derivative of $$F_{2}(t)$$ is $$F'_{2}(t) = 5040-525 \pi^{2} +24 t^{2}$$. Since $$F'_{2}(t)=0$$ for $$t= \frac{1}{2} \sqrt{\frac{1}{2} (-1680+175 \pi ^{2} )} \cong 2.4285$$, we have $$F'_{2}(t)<0$$ for $$0< t< \frac{1}{2} \sqrt{\frac{1}{2} (-1680+175 \pi ^{2} )}$$ and $$F'_{2}(t)>0$$ for $$t>\frac{1}{2} \sqrt{\frac{1}{2} (-1680+175 \pi ^{2} )}$$. Hence,

\begin{aligned} F_{2}(t) & \geq F_{2} \biggl(\frac{1}{2} \sqrt{ \frac{1}{2} \bigl(-1680+175 \pi ^{2} \bigr)} \biggr) \\ & = \frac{35}{2} \bigl(30+48 \sqrt{70 \bigl(-48+5 \pi ^{2} \bigr)}-5 \pi ^{2} \sqrt{70 \bigl(-48+5 \pi ^{2} \bigr)} \bigr) \\ & \cong 295.843 \end{aligned}

for $$t>0$$. Therefore, $$F_{1}(x) >0$$ and the proof of Lemma 2.6 is complete. □

### Proof of Theorem 2.4

We have

\begin{aligned} F_{1}(x) &= \frac{8x +\frac{32 x^{7}}{4725}}{3+\sqrt{25 +\frac{80}{3}x^{2}}}-\frac{\pi ^{2} x}{4 +\sqrt{32 +(2 \pi x)^{2}}} \\ &= \frac{x F_{2}(x)}{3150 (9 +\sqrt{15} \sqrt{15+16 x^{2}}) (2 +\sqrt{8 + \pi^{2} x^{2}})} , \end{aligned}

where $$F_{2}(x) =151200-14175 \pi^{2} +128 x^{6} -1575 \sqrt{15} \pi^{2} \sqrt{15+16 x^{2}} +75600 \sqrt{8 +\pi^{2} x^{2}} +64 x^{6} \sqrt{8 +\pi^{2} x^{2}}$$. The derivative of $$F_{2}(x)$$ is

\begin{aligned} F_{2}'(x)& = 768 x^{5}-\frac{25200 \sqrt{15} \pi^{2} x}{\sqrt{15+16 x^{2}}}+ \frac{75600 \pi^{2} x}{\sqrt{8 +\pi^{2} x^{2}}}+\frac{64 \pi^{2} x^{7}}{\sqrt{8 +\pi^{2} x^{2}}}+384 x^{5} \sqrt{\pi^{2} x^{2}+8} \\ &> -\frac{25200 \sqrt{15} \pi^{2} x}{\sqrt{15 +16 x^{2}}}+\frac{75600 \pi^{2} x}{\sqrt{8 +\pi^{2} x^{2}}}+\frac{64 \pi^{2} x^{7}}{\sqrt{8 +\pi^{2} x^{2}}} . \end{aligned}

By Lemma 2.6, we have $$F'_{2}(x)>0$$ and $$F_{2}(x)$$ is strictly increasing for $$x>0$$. From $$F_{2}(0+)=37800 (4 +4 \sqrt{2} -\pi^{2} ) \cong -8041.96$$, $$F_{2}(\gamma)=0$$ and $$F(\infty)=\infty$$, we can get $$F_{2}(x)>0$$ for $$x>\gamma$$. The proof of Theorem 2.4 is complete. □

## 3 Conclusions

In this paper, we established some inequalities involving arctangent. The double inequality in Theorem 2.1 provides sharper quadratic estimations than (1.2) and (1.3) for a location away from zero. By Theorems 2.2, 2.3 and 2.4, we obtained Proposition 2.5 immediately.

## References

1. Shafer, RE: On quadratic approximation. SIAM J. Numer. Anal. 11(2), 447-460 (1974)

2. Shafer, RE: Analytic inequalities obtained by quadratic approximation. Publ. Elektroteh. Fak. Univ. Beogr., Ser. Mat. Fiz. 577-598, 96-97 (1977)

3. Shafer, RE: On quadratic approximation, II. Publ. Elektroteh. Fak. Univ. Beogr., Ser. Mat. Fiz. 602-633, 163-170 (1978)

4. Guo, B-N, Luo, Q-M, Qi, F: Sharpening and generalizations of Shafer-Fink’s double inequality for the arc sine function. Filomat 27(2), 261-265 (2013)

5. Malešević, BJ: Application of λ-method on Shafer-Fink’s inequality. Publ. Elektroteh. Fak. Univ. Beogr., Mat. 8, 103-105 (1997)

6. Malešević, BJ: An application of λ-method on inequalities of Shafer-Fink’s type. Math. Inequal. Appl. 10(3), 529-534 (2007)

7. Nishizawa, Y: Sharpening of Jordan’s type and Shafer-Fink’s type inequalities with exponential approximations. Appl. Math. Comput. 269, 146-154 (2015)

8. Sun, J-L, Chen, C-P: Shafer-type inequalities for inverse trigonometric functions and Gauss lemniscate functions. J. Inequal. Appl. 2016, 212 (2016)

9. Zhu, L: On Shafer-Fink inequalities. Math. Inequal. Appl. 8(4), 571-574 (2005)

10. Zhu, L: On Shafer-Fink-type inequality. J. Inequal. Appl. 2007, 67430 (2007)

11. Zhu, L: New inequalities of Shafer-Fink type for arc hyperbolic sine. J. Inequal. Appl. 2008, 368275 (2008)

12. Zhu, L: On a quadratic estimate of Shafer. J. Math. Inequal. 2(4), 571-574 (2008)

13. Chen, C-P, Cheung, W-S: Sharp Cusa and Becker-Stark inequalities. J. Inequal. Appl. 2011, 136 (2011)

14. Chen, C-P, Sándor, J: Sharp inequalities for trigonometric and hyperbolic functions. J. Math. Inequal. 9(1), 203-217 (2015)

15. Debnath, L, Mortici, C, Zhu, L: Refinements of Jordan-Steckin and Becker-Stark inequalities. Results Math. 67(1-2), 207-215 (2015)

16. Sun, Z-J, Zhu, L: Simple proofs of the Cusa-Huygens-type and Becker-Stark-type inequalities. J. Math. Inequal. 7(4), 563-567 (2013)

17. Zhu, L: Sharp Becker-Stark-type inequalities for Bessel functions. J. Inequal. Appl. 2010, 838740 (2010)

18. Zhu, L, Hua, J: Sharpening the Becker-Stark inequalities. J. Inequal. Appl. 2010, 931275 (2010)

19. Zhu, L: A refinement of the Becker-Stark inequalities. Math. Notes 93(3-4), 421-425 (2013)

## Acknowledgements

I would like to thank referees for their careful reading of the manuscript and for their remarks and suggestions.

## Author information

Authors

### Corresponding author

Correspondence to Yusuke Nishizawa. 