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# Shafer-type inequalities for inverse trigonometric functions and Gauss lemniscate functions

Journal of Inequalities and Applications20162016:212

https://doi.org/10.1186/s13660-016-1157-2

• Received: 1 July 2016
• Accepted: 31 August 2016
• Published:

## Abstract

In this paper, we present Shafer-type inequalities for inverse trigonometric functions and Gauss lemniscate functions.

## Keywords

• inverse trigonometric functions
• lemniscate function
• inequalities

• 26D07

## 1 Introduction

Shafer  indicated several elementary quadratic approximations of selected functions without proof. Subsequently, Shafer  established these results as analytic inequalities. For example, Shafer  proved that for $$x>0$$,
$$\frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x.$$
(1.1)
The inequality (1.1) can also be found in . Also in , Shafer proved that for $$0< x<1$$,
$$\frac{8x}{3\sqrt{1-x^{2}}+\sqrt{25+\frac{5}{3}x^{2}}}< \arcsin x.$$
(1.2)
Zhu  proved that the function
$$F(x)=\frac{ (\frac{8x}{\arctan x}-3 )^{2}-25}{x^{2}}$$
is strictly decreasing for $$x>0$$, and
$$\lim_{x\to0^{+}}F(x)=\frac{80}{3}\quad \text{and} \quad \lim _{x\to \infty}F(x)=\frac{256}{\pi^{2}}.$$
From this one derives the following double inequality:
$$\frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x< \frac{8x}{3+\sqrt {25+\frac{256}{\pi^{2}}x^{2}}}, \quad x>0.$$
(1.3)
The constants $$80/3$$ and $$256/\pi^{2}$$ are the best possible. In , (1.3) is called Shafer-type inequality.
Using the Maple software, we find that
$$\arctan x \biggl(3+\sqrt{25+\frac{80}{3}x^{2}} \biggr)=8x+ \frac {32}{4\text{,}725}x^{7}-\frac{64}{4\text{,}725}x^{9}+ \frac{25\text{,}376}{1\text{,}299\text{,}375}x^{11}-\cdots.$$
This fact motivated us to present a new upper bound for arctanx, which is the first aim of the present paper.

### Theorem 1.1

For $$x>0$$,
$$\arctan x< \frac{8x+\frac{32}{4\text{,}725}x^{7}}{3+\sqrt{25+\frac{80}{3}x^{2}}}.$$
(1.4)

The second aim of the present paper is to develop (1.2) to produce a symmetric double inequality.

### Theorem 1.2

For $$0< x<1$$, we have
$$\frac{8x}{3\sqrt{1-x^{2}}+\sqrt{25+ax^{2}}}< \arcsin x< \frac{8x}{3\sqrt {1-x^{2}}+\sqrt{25+bx^{2}}}$$
(1.5)
with the best possible constants
$$a=\frac{5}{3}=1.666666\ldots \quad \textit{and}\quad b= \frac{256-25\pi ^{2}}{\pi^{2}}=0.938223\ldots.$$
(1.6)

Recently, some famous inequalities for trigonometric and inverse trigonometric functions have been improved (see, for example, ).

The lemniscate, also called the lemniscate of Bernoulli, is the locus of points $$(x, y)$$ in the plane satisfying the equation $$(x^{2} + y^{2})^{2} = x^{2} + y^{2}$$. In polar coordinates $$(r, \theta)$$, the equation becomes $$r^{2} = \cos(2\theta)$$ and its arc length is given by the function
$$\operatorname {arcsl}x= \int_{0}^{x}\frac{1}{\sqrt{1-t^{4}}}\,\mathrm{d}t,\quad |x|\leq1,$$
(1.7)
where arcslx is called the arc lemniscate sine function studied by Gauss in 1797-1798. Another lemniscate function investigated by Gauss is the hyperbolic arc lemniscate sine function, defined as
$$\operatorname {arcslh}x= \int_{0}^{x}\frac{1}{\sqrt{1+t^{4}}}\,\mathrm{d}t,\quad x \in \mathbb{R}.$$
(1.8)
Functions (1.7) and (1.8) can be found (see , Chapter 1, , p.259 and ).
Another pair of lemniscate functions, the arc lemniscate tangent arctl and the hyperbolic arc lemniscate tangent arctlh, have been introduced in , (3.1)-(3.2). Therein it has been proven that
$$\operatorname {arctl}x=\operatorname {arcsl}\biggl(\frac{x}{\sqrt{1+x^{4}}} \biggr), \quad x\in \mathbb{R}$$
(1.9)
and
$$\operatorname {arctlh}x=\operatorname {arcslh}\biggl(\frac{x}{\sqrt{1-x^{4}}} \biggr),\quad |x|< 1$$
(1.10)
(see , Proposition 3.1).

In analogy with (1.1), we here establish Shafer-type inequalities for the lemniscate functions, which is the last aim of the present paper.

### Theorem 1.3

For $$0< x<1$$,
$$\frac{10x}{5+\sqrt{25-10x^{4}}}< \operatorname {arcsl}x$$
(1.11)
and
$$\frac{10x}{5+\sqrt{25-15x^{4}}}< \operatorname {arctlh}x.$$
(1.12)

### Theorem 1.4

For $$x>0$$,
$$\frac{95x}{80+\sqrt{225+285x^{4}}}< \operatorname {arcslh}x.$$
(1.13)

We present the following conjecture.

### Conjecture 1.1

For $$x>0$$,
$$\operatorname {arcslh}x< \frac{95x+\frac{931}{2\text{,}925}x^{13}}{80+\sqrt{225+285x^{4}}}$$
(1.14)
and
$$\frac{1\text{,}210x}{940+9\sqrt{900+1\text{,}210x^{4}}}< \operatorname {arctl}x< \frac{1\text{,}210x+\frac {2\text{,}078\text{,}417}{280\text{,}800}x^{13}}{940+9\sqrt{900+1\text{,}210x^{4}}}.$$
(1.15)

## 2 Lemmas

The following lemmas have been proved in .

### Lemma 2.1

For $$|x|<1$$,
$$\operatorname {arcsl}x =\sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi }(4n+1)\cdot n!}x^{4n+1}=x+\frac{1}{10} x^{5}+ \frac{1}{24} x^{9}+\cdots.$$
(2.1)

### Lemma 2.2

For $$0< x<1$$,
$$\operatorname {arctlh}x=\sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{3}{4})}{\Gamma(\frac {3}{4})\cdot (4n+1)\cdot n!}x^{4n+1}=x+\frac{3}{20}x^{5}+ \frac{7}{96}x^{9}+\cdots.$$
(2.2)

## 3 Proofs of Theorems 1.1 to 1.4

### Proof of Theorem 1.1

The inequality (1.11) is obtained by considering the function $$f(x)$$ defined by
$$f(x)=\arctan x-\frac{8x+\frac{32}{4\text{,}725}x^{7}}{3+\sqrt{25+\frac {80}{3}x^{2}}}, \quad x>0.$$
Differentiation yields
$$f'(x)=-\frac{2g(x)}{(1+x^{2})(9+\sqrt{225+240x^{2}})^{2}\sqrt{225+240x^{2}}},$$
where
\begin{aligned} g(x) =&\bigl(-7\text{,}875-2\text{,}100x^{2}+112x^{6}+112x^{8} \bigr)\sqrt{225+240x^{2}} \\ &{} +118\text{,}125+94\text{,}500x^{2}+2\text{,}800x^{6}+5 \text{,}360x^{8}+2\text{,}560x^{10}. \end{aligned}
We now show that
$$g(x)>0, \quad x>0.$$
(3.1)
By an elementary change of variable
$$t=\sqrt{225+240x^{2}}, \quad t>15,$$
the inequality (3.1) is equivalent to
\begin{aligned}& \biggl[-7\text{,}875-2\text{,}100 \biggl(\frac{t^{2}-225}{240} \biggr)+112 \biggl( \frac {t^{2}-225}{240} \biggr)^{3}+112 \biggl(\frac{t^{2}-225}{240} \biggr)^{4} \biggr]t \\& \qquad {}+118\text{,}125+94\text{,}500 \biggl(\frac{t^{2}-225}{240} \biggr)+2 \text{,}800 \biggl(\frac{t^{2}-225}{240} \biggr)^{3}+5\text{,}360 \biggl( \frac{t^{2}-225}{240} \biggr)^{4} \\& \qquad {}+2\text{,}560 \biggl(\frac{t^{2}-225}{240} \biggr)^{5} \\& \quad =\frac {(2t^{6}+141t^{5}+4\text{,}515t^{4}+93\text{,}690t^{3}+1\text{,}562\text{,}400t^{2}+24\text{,}053\text{,}625t+362\text{,}626\text{,}875)(t-15)^{4}}{622\text{,}080\text{,}000} \\& \quad >0\quad \text{for } t>15, \end{aligned}
which is true. Hence, we have
$$g(x)>0\quad \text{and} \quad f'(x)< 0\quad \text{for } x>0.$$
So, $$f(x)$$ is strictly decreasing for $$x>0$$, and we have
$$f(x)< f(0)=0, \quad x>0.$$
The proof is complete. □

### Remark 3.1

Let $$x_{0}=1.4243\ldots$$ . Then we have
$$\frac{8x+\frac{32}{4\text{,}725}x^{7}}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \frac {8x}{3+\sqrt{25+\frac{256}{\pi^{2}}x^{2}}},\quad 0< x< x_{0}.$$
This shows that for $$0< x< x_{0}$$, the upper bound in (1.11) is better than the upper bound in (1.3). In fact, for $$x\to0$$, we have
\begin{aligned}& \arctan x-\frac{8x}{3+\sqrt{25+\frac{256}{\pi^{2}}x^{2}}}=O\bigl(x^{3}\bigr), \\& \arctan x-\frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}=O\bigl(x^{7}\bigr), \end{aligned}
and
$$\arctan x-\frac{8x+\frac{32}{4\text{,}725}x^{7}}{3+\sqrt{25+\frac{80}{3}x^{2}}}=O\bigl(x^{9}\bigr).$$

### Proof of Theorem 1.2

The double inequality (1.11) can be written for $$0< x<1$$ as
$$b< F(x)< a,$$
where
$$F(x)=\frac{ (\frac{8x}{\arcsin x}-3\sqrt{1-x^{2}} )^{2}-25}{x^{2}}, \quad 0< x< 1.$$
By an elementary change of variable,
$$x=\sin t,\quad 0< t< \frac{\pi}{2},$$
we have
$$G(t)=F(\sin t)=\frac{ (\frac{8\sin t}{t}-3\cos t )^{2}-25}{\sin^{2}t},\quad 0< t< \frac{\pi}{2}.$$
We now prove that $$F(x)$$ is strictly decreasing for $$0< x<1$$. It suffices to show that $$G(t)$$ is strictly decreasing for $$0< t<\pi/2$$. Differentiation yields
\begin{aligned} \begin{aligned}[b] -\frac{t^{3}\sin^{3}t}{16}G'(t)={}&8\sin^{3} t-3t^{2}\sin t-\bigl(2t^{3}+3t\bigr)\cos t+3t \cos^{3}t \\ ={}&8 \biggl(\frac{3\sin t-\sin(3t)}{4} \biggr)-3t^{2}\sin t- \bigl(2t^{3}+3t\bigr)\cos t \\ &{}+3t \biggl(\frac{\cos(3t)+3\cos t}{4} \biggr) \\ ={}&\bigl(6-3t^{2}\bigr)\sin t-2\sin(3t)+\frac{3}{4}t\cos(3t)- \biggl(\frac {3}{4}t+2t^{3} \biggr)\cos t \\ ={}&\frac{16}{945}t^{9}-\frac{16}{4\text{,}725}t^{11}+\sum _{n=6}^{\infty }(-1)^{n}u_{n}(t), \end{aligned} \end{aligned}
(3.2)
where
$$u_{n}(t)=\frac{21+2n+48n^{2}+64n^{3}+(6n-21)\cdot9^{n}}{4\cdot(2n+1)!}t^{2n+1}.$$
Elementary calculations reveal that for $$0< t<\pi/2$$ and $$n\geq6$$,
\begin{aligned} \frac{u_{n+1}(t)}{u_{n}(t)}&=\frac{t^{2} (135+290n+240n^{2}+64n^{3}+(54n-135)\cdot9^{n} )}{2(n+1)(2n+3) (21+2n+48n^{2}+64n^{3}+(6n-21)\cdot9^{n} )} \\ &< \frac{(\pi/2)^{2}}{n+1}\frac{135+290n+240n^{2}+64n^{3}+(54n-135)\cdot 9^{n}}{2(2n+3) (21+2n+48n^{2}+64n^{3}+(6n-21)\cdot9^{n} )} \\ &< \frac{135+290n+240n^{2}+64n^{3}+(54n-135)\cdot9^{n}}{2(2n+3) (21+2n+48n^{2}+64n^{3}+(6n-21)\cdot9^{n} )} \end{aligned}
and
\begin{aligned}& 2(2n+3) \bigl(21+2n+48n^{2}+64n^{3}+(6n-21) \cdot9^{n} \bigr) \\& \qquad {}- \bigl(135+290n+240n^{2}+64n^{3}+(54n-135) \cdot9^{n} \bigr) \\& \quad =\bigl(24n^{2}-102n+9\bigr)9^{n}+256n^{4}+512n^{3}+56n^{2}-194n-9>0. \end{aligned}
We then obtain, for $$0< t<\pi/2$$ and $$n\geq6$$,
$$\frac{u_{n+1}(t)}{u_{n}(t)}< 1.$$
Hence, for every $$t\in(0,\pi/2)$$, the sequence $$n\longmapsto u_{n}(t)$$ is strictly decreasing for $$n\geq6$$. We then obtain from (3.2)
$$-\frac{t^{3}\sin^{3}t}{16}G'(t)>t^{9} \biggl( \frac{16}{945}-\frac {16}{4\text{,}725}t^{2} \biggr)>0,\quad 0< t< \frac{\pi}{2},$$
which implies $$G'(t)<0$$ for $$0< t<\pi/2$$. Hence, $$G(t)$$ is strictly decreasing for $$0< t<\pi/2$$, and $$F(x)$$ is strictly decreasing for $$0< x<1$$. So, we have
$$\frac{256-25\pi^{2}}{\pi^{2}}=\lim_{t\to1}F(t)< F(x)=\frac{ (\frac{8x}{\arcsin x}-3\sqrt{1-x^{2}} )^{2}-25}{x^{2}}< \lim _{t\to 0}F(t)=\frac{5}{3}$$
for all $$x\in(0, 1)$$, with the constants $$5/3$$ and $$(256-25\pi^{2})/\pi^{2}$$ being best possible. The proof is complete. □

### Proof of Theorem 1.3

By (2.1), we find that for $$0< x<1$$,
\begin{aligned} \bigl(25-10x^{4}\bigr)- \biggl(\frac{10x}{\operatorname {arcsl}x}-5 \biggr)^{2}&>\bigl(25-10x^{4}\bigr)- \biggl( \frac{10x}{x+\frac{1}{10}x^{5}+\frac{1}{24}x^{9}}-5 \biggr)^{2} \\ &=\frac{10x^{8}(3\text{,}120-1\text{,}344x^{4}-120x^{8}-25x^{12})}{(120+12x^{4}+5x^{8})^{2}}. \end{aligned}
Noting that
$$3\text{,}120-1\text{,}344t-120t^{2}-25t^{3}>0\quad \text{for } 0< t< 1,$$
we obtain, for $$0< x<1$$,
$$\bigl(25-10x^{4}\bigr)- \biggl(\frac{10x}{\operatorname {arcsl}x}-5 \biggr)^{2}>0,$$
which implies (1.11).
By (2.2), we find that for $$0< x<1$$,
\begin{aligned} \bigl(25-15x^{4}\bigr)- \biggl(\frac{10x}{\operatorname {arctlh}x}-5 \biggr)^{2}&>\bigl(25-15x^{4}\bigr)- \biggl( \frac{10x}{x+\frac{1}{10}x^{5}+\frac{1}{24}x^{9}}-5 \biggr)^{2} \\ & =\frac{15x^{8}A(x)}{(24\text{,}960+3\text{,}744x^{4}+1\text{,}820x^{8}+1\text{,}155x^{12})^{2}}, \end{aligned}
(3.3)
where
\begin{aligned} A(x) =&\bigl(115\text{,}947\text{,}520-71\text{,}285\text{,}760x^{8}-4 \text{,}204\text{,}200x^{16}\bigr) \\ &{} +x^{4}\bigl(87\text{,}320\text{,}064-11\text{,}961 \text{,}040x^{8}-1\text{,}334\text{,}025x^{16}\bigr). \end{aligned}
Noting that for $$0< t<1$$,
$$115\text{,}947\text{,}520-71\text{,}285\text{,}760t-4\text{,}204 \text{,}200t^{2}>0$$
and
$$87\text{,}320\text{,}064-11 \text{,}961\text{,}040t-1\text{,}334\text{,}025t^{2}>0,$$
we obtain $$A(x)>0$$ for $$0< x<1$$. From (3.3), we obtain (1.12). The proof is complete. □

### Proof of Theorem 1.4

The inequality (1.13) is obtained by considering the function $$h(x)$$ defined by
$$h(x)=\operatorname {arcslh}x-\frac{95x}{80+\sqrt{225+285x^{4}}}, \quad x>0.$$
Differentiation yields
$$h'(x)=\frac{1}{\sqrt{1+x^{4}}}-\frac{475(16\sqrt {225+285x^{4}}+45-57x^{4})}{(80+\sqrt{225+285x^{4}})^{2}\sqrt{225+285x^{4}}}.$$
By an elementary change of variable
$$t=\sqrt{225+285x^{4}}, \quad x>0 \qquad \biggl(\text{or } x=\sqrt {\frac{t^{2}-225}{285}}, t>15 \biggr),$$
(3.4)
we have
\begin{aligned}& \frac{1}{\sqrt{1+x^{4}}}-\frac{475(16\sqrt {225+285x^{4}}+45-57x^{4})}{(80+\sqrt{225+285x^{4}})^{2}\sqrt{225+285x^{4}}} \\& \quad =\frac{285}{\sqrt{17\text{,}100+285t^{2}}}+\frac {95(t^{2}-80t-450)}{t(80+t)^{2}}=\frac{95I(t)}{t(80+t)^{2}}, \end{aligned}
where
$$I(t)=\frac{19\text{,}200t+480t^{2}+3t^{3}}{\sqrt {17\text{,}100+285t^{2}}}+t^{2}-80t-450, \quad t>15.$$
We now prove that
$$h'(x)>0, \quad x>0.$$
It suffices to show that
$$I(t)>0, \quad t>15.$$
Differentiation yields
\begin{aligned}& I'(t)=\frac{6(192\text{,}000+9\text{,}600t+90t^{2}+80t^{3}+t^{4})}{(60+t^{2})\sqrt {17\text{,}100+285t^{2}}}+2t-80, \\& I''(t)=\frac{6(576\text{,}000-565\text{,}200t-4\text{,}800t^{2}+150t^{3}+t^{5})}{(60+t^{2})^{2}\sqrt {17\text{,}100+285t^{2}}}+2, \end{aligned}
and
$$I'''(t)=\frac{10\text{,}800(-18\text{,}840-1\text{,}920t+1\text{,}271t^{2}+8t^{3})}{(60+t^{2})^{3}\sqrt {17\text{,}100+285t^{2}}}>0 \quad \text{for } t>15.$$
Thus, we have, for $$t>15$$,
$$I''(t)>I''(15)=0\quad \Longrightarrow\quad I'(t)>I'(15)=0\quad \Longrightarrow\quad I(t)>I(15)=0.$$
Hence, $$h'(x)>0$$ holds for $$x>0$$, and we have
$$h(x)>h(0)=0, \quad x>0.$$
The proof is complete. □

## Declarations

### Acknowledgements

The authors thank the referees for helpful comments. 