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Refined quadratic estimations of Shafer’s inequality

Journal of Inequalities and Applications20172017:40

https://doi.org/10.1186/s13660-017-1312-4

Received: 20 October 2016

Accepted: 30 January 2017

Published: 8 February 2017

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.

Keywords

Shafer’s inequalityan upper bound for arctangenta lower bound for arctangent

MSC

26D1542A10

1 Introduction

Shafer [13] 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 [48] and [912]. Especially, Zhu [12] 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 [8], 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 [1316] and [1719]. Especially, Zhu [19] 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. □

2.2 Proof of Theorem 2.2

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. □

2.3 Proof of Theorem 2.3

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. □

2.4 Proof of Theorem 2.4

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.

Declarations

Acknowledgements

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

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.

Authors’ Affiliations

(1)
General Education, Ube National College of Technology, Ube, Japan

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