# Sharp Shafer-Fink type inequalities for Gauss lemniscate functions

## Abstract

In this paper, we establish sharp Shafer-Fink type inequalities for Gauss lemniscate functions.

MSC:26D07.

## 1 Introduction and definitions

In geometry, the lemniscate of Bernoulli is a plane curve defined by two given points ${F}_{1}$ and ${F}_{2}$, known as foci, at distance 2a from each other as the locus of points P so that $P{F}_{1}\cdot P{F}_{2}={a}^{2}$. This gives the equation ${\left({x}^{2}+{y}^{2}\right)}^{2}=2{a}^{2}\left({x}^{2}-{y}^{2}\right)$. In polar coordinates $\left(r,\theta \right)$, the equation becomes ${r}^{2}=2acos\left(2\theta \right)$. The arc length from the origin to a point on the Bernoulli lemniscate ${r}^{2}=cos\left(2\theta \right)$ is given by the function

$arcslx={\int }_{0}^{x}\frac{\mathrm{d}t}{\sqrt{1-{t}^{4}}},\phantom{\rule{1em}{0ex}}|x|\le 1,$
(1.1)

where arcslx is called the arc lemniscate sine function studied by CF Gauss in 1797-1798. Another lemniscate function investigated by Gauss is the hyperbolic arc lemniscate sine function, defined as

$arcslhx={\int }_{0}^{x}\frac{\mathrm{d}t}{\sqrt{1+{t}^{4}}},\phantom{\rule{1em}{0ex}}x\in \mathbb{R}.$
(1.2)

The functions (1.1) and (1.2) can be found in [[1], p.259], [[2], (2.5)-(2.6)], [37] and [[8], Ch. 1].

Following Neuman [3], Gauss’ arc lemniscate tangent and the hyperbolic arc lemniscate tangent functions are defined by

$arctlx=arcsl\left(\frac{x}{\sqrt[4]{1+{x}^{4}}}\right),\phantom{\rule{1em}{0ex}}x\in \mathbb{R}$
(1.3)

and

$arctlhx=arcslh\left(\frac{x}{\sqrt[4]{1-{x}^{4}}}\right),\phantom{\rule{1em}{0ex}}|x|<1,$
(1.4)

respectively.

For $0\le x\le 1$, it is known in the literature that

$\frac{3x}{2+\sqrt{1-{x}^{2}}}\le \frac{6\left(\sqrt{1+x}-\sqrt{1-x}\right)}{4+\sqrt{1+x}+\sqrt{1-x}}\le arcsinx\le \frac{\pi x}{2+\sqrt{1-{x}^{2}}}.$
(1.5)

The first and second inequalities in equation (1.5) were established by Shafer (see, e.g., [[9], p.247]), while the third inequality was proved by Fink [10]. In recent years, Shafer-Fink’s inequalities have attracted much attention of the mathematical community. By using the λ-method of Mitrinović and Vasić [9], Malešević [11] improved the upper bound for arcsinx and established the following inequality: For $0\le x\le 1$,

$\frac{3x}{2+\sqrt{1-{x}^{2}}}\le arcsinx\le \frac{\left(\pi /\left(\pi -2\right)\right)x}{\left(2/\left(\pi -2\right)\right)+\sqrt{1-{x}^{2}}}\le \frac{\pi x}{2+\sqrt{1-{x}^{2}}}.$
(1.6)

In [1215], other upper bounds for arcsinx were established: For $0\le x\le 1$,

$\begin{array}{rl}\frac{3x}{2+\sqrt{1-{x}^{2}}}& \le \frac{6\left(\sqrt{1+x}-\sqrt{1-x}\right)}{4+\sqrt{1+x}+\sqrt{1-x}}\le arcsinx\\ \le \frac{\left(\pi \left(2-\sqrt{2}\right)/\left(\pi -2\sqrt{2}\right)\right)\left(\sqrt{1+x}-\sqrt{1-x}\right)}{\left(\sqrt{2}\left(4-\pi \right)/\left(\pi -2\sqrt{2}\right)\right)+\sqrt{1+x}+\sqrt{1-x}}\\ \le \frac{\pi \left(\sqrt{2}+1/2\right)\left(\sqrt{1+x}-\sqrt{1-x}\right)}{4+\sqrt{1+x}+\sqrt{1-x}}\le \frac{\pi x}{2+\sqrt{1-{x}^{2}}}.\end{array}$
(1.7)

In [16], Pan and Zhu gave some further generalizations of these results and obtained two new Shafer-Fink type double inequalities. In [17], Zhu provided a solution to an open problem posed by Oppenheim in [18]. At the same time, some Shafer-Fink inequalities were deduced from the solution of Oppenheim’s problem. Chen and Cheung [19] provided a laconic proof to Oppenheim’s problem. Recently, Qi and Guo [20, 21] presented a sharpening and generalizations of Shafer-Fink’s inequality.

Related to the inverse sine inequality, the inverse tangent inequality is also of much interest. In the literature, we have

(1.8)

The first inequality in equation (1.8) was presented without proof by Shafer [22]. Three proofs of it were later given in [23]. The second inequality in equation (1.8) can be found in, e.g., [[24], p.288]. Shafer’s inequality (1.8) was recently sharpened and generalized by Qi et al. in [25]. For each $\theta >0$, Chen et al. [26] determined the largest number ${\theta }_{1}$ and the smallest number ${\theta }_{2}$ such that the inequalities

$\frac{{\theta }_{1}x}{1+\theta \sqrt{1+{x}^{2}}}\le arctanx\le \frac{{\theta }_{2}x}{1+\theta \sqrt{1+{x}^{2}}}$
(1.9)

are valid for all $x\ge 0$. Zhu [[27], Theorems 1.9 and 1.10] established Shafer-Fink type inequalities for the inverse hyperbolic sine function.

Recently, numerous inequalities have been given for the lemniscate functions. For example, Neuman [5] proved the following inequalities:

${\left(\frac{5}{3+2{\left(1-{x}^{4}\right)}^{1/2}}\right)}^{1/2}<\frac{arcslx}{x}<{\left(1-{x}^{4}\right)}^{-1/10}$
(1.10)

and

${\left(\frac{5}{3+2{\left(1+{x}^{4}\right)}^{1/2}}\right)}^{1/2}<\frac{arcslhx}{x}<{\left(1+{x}^{4}\right)}^{-1/10}$
(1.11)

for $0<|x|<1$.

Chen [28, 29] established Wilker and Huygens type inequalities for Gauss lemniscate functions. For example, Chen [28] proved that for $0<|x|<1$,

$2+\frac{1}{20}{x}^{3}arctlx<{\left(\frac{arcslx}{x}\right)}^{2}+\frac{arctlx}{x}$
(1.12)

with the best possible constant $\frac{1}{20}$. Chen [29] proved that for $0<|x|<1$,

$\frac{2\left(arcslx/x\right)+arctlx/x}{3}>\frac{3\left(arcslx/x\right)+2\left(arctlx/x\right)}{5}>1.$
(1.13)

In this paper, we establish sharp Shafer-Fink type inequalities for Gauss lemniscate functions.

The following lemma is required in our present investigation.

Lemma 1.1 ([3032])

Let $-\mathrm{\infty }, and let $f,g:\left[a,b\right]\to \mathbb{R}$ be continuous on $\left[a,b\right]$, differentiable on $\left(a,b\right)$. Let ${g}^{\prime }\left(x\right)\ne 0$ on $\left(a,b\right)$. If ${f}^{\prime }\left(x\right)/{g}^{\prime }\left(x\right)$ is increasing (decreasing) on $\left(a,b\right)$, then so are

$\left[f\left(x\right)-f\left(a\right)\right]/\left[g\left(x\right)-g\left(a\right)\right]\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\left[f\left(x\right)-f\left(b\right)\right]/\left[g\left(x\right)-g\left(b\right)\right].$

If ${f}^{\prime }\left(x\right)/{g}^{\prime }\left(x\right)$ is strictly monotone, then the monotonicity in the conclusion is also strict.

Remark 1.1 A generalization of the familiar trigonometric and hyperbolic functions was described by Lindqvist [33]. The generalized p-trigonometric functions occur as an eigenfunction of the Dirichlet problem for the one-dimensional p-Laplacian. Recently, the p-trigonometric functions have been studied extensively, see for example [3437] and their references. Very recently, Takeuchi [38] (see also [39]) introduced the $\left(p,q\right)$-trigonometric functions that coincide with the p-trigonometric functions for $p=q$ and are connected with the Dirichlet problem for the $p,q$-Laplacian. These $\left(p,q\right)$-trigonometric functions have been the subject of intense investigations (see, for example, [35, 3942]). For $p,q>1$ the function ${arcsin}_{p,q}$ is defined in [38, 39] by

${arcsin}_{p,q}\left(x\right)={\int }_{0}^{x}\frac{\mathrm{d}t}{{\left(1-{t}^{q}\right)}^{1/p}},\phantom{\rule{1em}{0ex}}|x|\le 1.$
(1.14)

Similarly, for $p,q>1$ the function ${arcsinh}_{p,q}$ is defined by [40]

${arcsinh}_{p,q}\left(x\right)={\int }_{0}^{x}\frac{\mathrm{d}t}{{\left(1+{t}^{q}\right)}^{1/p}},\phantom{\rule{1em}{0ex}}x\ge 0.$
(1.15)

Clearly,

$arcslx={arcsin}_{2,4}\left(x\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}arcslhx={arcsinh}_{2,4}\left(x\right).$

## 2 Main results

Theorem 2.1 For $0,

$\frac{{\alpha }_{1}\left(\sqrt{1+x}-\sqrt{1-x}\right)}{4-\sqrt{1+x}-\sqrt{1-x}}
(2.1)

with the best possible constants

${\alpha }_{1}=2\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{\beta }_{1}=\frac{2\sqrt{2}-1}{4}B\left(\frac{1}{4},\frac{1}{2}\right)=2.39712057\dots .$
(2.2)

Here

$B\left(x,y\right)={\int }_{0}^{1}{t}^{x-1}{\left(1-t\right)}^{y-1}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t$

is the beta function.

Proof For $0, let

$f\left(x\right)=\frac{{f}_{1}\left(x\right)}{{f}_{2}\left(x\right)},$

where

${f}_{1}\left(x\right)=arcslx\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{f}_{2}\left(x\right)=\frac{\sqrt{1+x}-\sqrt{1-x}}{4-\sqrt{1+x}-\sqrt{1-x}}.$

Then,

$\frac{2{f}_{1}^{\prime }\left(x\right)}{{f}_{2}^{\prime }\left(x\right)}=\frac{{\left(4-\sqrt{1+x}-\sqrt{1-x}\right)}^{2}\sqrt{1+x}\sqrt{1-x}}{\sqrt{1-{x}^{4}}\left(-1+\sqrt{1-x}+\sqrt{1+x}\right)}=:{f}_{3}\left(x\right).$

Differentiation yields

${f}_{3}^{\prime }\left(x\right)=\frac{4-\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1-{x}^{2}}\sqrt{1-{x}^{4}}\left({x}^{2}+1\right){\left(-1+\sqrt{1-x}+\sqrt{1+x}\right)}^{2}}{f}_{4}\left(x\right),$

where

$\begin{array}{rl}{f}_{4}\left(x\right)=& \left(6{x}^{3}+{x}^{2}-4x+1\right)\sqrt{1-x}+\left(6{x}^{3}-{x}^{2}-4x-1\right)\sqrt{1+x}\\ -\left({x}^{3}-3x\right)\sqrt{1-{x}^{2}}-6{x}^{3}+6x.\end{array}$

Motivated by the investigations in [12], we are in a position to prove ${f}_{4}\left(x\right)>0$ for $x\in \left(0,1\right)$. Let

$A\left(x\right)=\left\{\begin{array}{cc}\mu ,\hfill & x=0,\hfill \\ \frac{{f}_{4}\left(x\right)}{{x}^{3}},\hfill & 0

where μ is constant determined with limit:

$\mu =\underset{x\to 0}{lim}\frac{{f}_{4}\left(x\right)}{{x}^{3}}=\frac{27}{8}.$

Using Maple we determine Taylor approximation for the function $A\left(x\right)$ by the polynomial of the fourth order:

${P}_{1}\left(x\right)=\frac{27}{8}-\frac{159}{128}{x}^{2}-\frac{465}{1\text{,}024}{x}^{4},$

which has a bound of absolute error

${\epsilon }_{1}=\frac{1\text{,}719}{1\text{,}024}$

for values $x\in \left[0,1\right]$. It is true that

$A\left(x\right)-\left({P}_{1}\left(x\right)-{\epsilon }_{1}\right)\ge 0,\phantom{\rule{1em}{0ex}}0\le x\le 1$

and

${P}_{1}\left(x\right)-{\epsilon }_{1}=\frac{1\text{,}737}{1\text{,}024}-\frac{159}{128}{x}^{2}-\frac{465}{1\text{,}024}{x}^{4}>0,\phantom{\rule{1em}{0ex}}0

Hence, for $x\in \left(0,1\right)$ it is true that $A\left(x\right)>0$ and therefore ${f}_{4}\left(x\right)>0$ and ${f}_{3}^{\prime }\left(x\right)>0$ for $x\in \left(0,1\right)$. Therefore, the function $\frac{{f}_{1}^{\prime }\left(x\right)}{{f}_{2}^{\prime }\left(x\right)}$ is strictly increasing on $\left(0,1\right)$. By Lemma 1.1, the function

$f\left(x\right)=\frac{{f}_{1}\left(x\right)}{{f}_{2}\left(x\right)}=\frac{{f}_{1}\left(x\right)-{f}_{1}\left(0\right)}{{f}_{2}\left(x\right)-{f}_{2}\left(0\right)}$

is strictly increasing on $\left(0,1\right)$. And hence,

$2=\underset{x\to 0}{lim}f\left(x\right)

for $0. By rearranging terms in the last expression, Theorem 2.1 follows. □

Theorem 2.2 For $0,

$\frac{{\alpha }_{2}\left(\sqrt{1+x}-\sqrt{1-x}\right)}{4-\sqrt{1+x}-\sqrt{1-x}}
(2.3)

with the best possible constants

${\alpha }_{2}=2\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{\beta }_{2}=\frac{2\sqrt{2}-1}{4}B\left(\frac{1}{4},\frac{1}{4}\right)=3.39004\dots .$
(2.4)

Here $B\left(x,y\right)$ denotes the beta function.

Proof For $0, let

$F\left(x\right)=\frac{{F}_{1}\left(x\right)}{{F}_{2}\left(x\right)},$

where

${F}_{1}\left(x\right)=arctlhx\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{F}_{2}\left(x\right)=\frac{\sqrt{1+x}-\sqrt{1-x}}{4-\sqrt{1+x}-\sqrt{1-x}}.$

Then,

$\frac{2{F}_{1}^{\prime }\left(x\right)}{{F}_{2}^{\prime }\left(x\right)}=\frac{{\left(4-\sqrt{1+x}-\sqrt{1-x}\right)}^{2}\sqrt{1+x}\sqrt{1-x}}{{\left(1-{x}^{4}\right)}^{3/4}\left(-1+\sqrt{1+x}+\sqrt{1-x}\right)}=:{F}_{3}\left(x\right).$

Differentiation yields

${F}_{3}^{\prime }\left(x\right)=\frac{4-\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1-x}\sqrt{1+x}{\left(-1+\sqrt{1-x}+\sqrt{1+x}\right)}^{2}{\left(1-{x}^{4}\right)}^{3/4}\left({x}^{2}+1\right)}{F}_{4}\left(x\right),$

where

$\begin{array}{rl}{F}_{4}\left(x\right)=& \left(11{x}^{3}-{x}^{2}-4x-1\right)\sqrt{1+x}+\left(11{x}^{3}+{x}^{2}-4x+1\right)\sqrt{1-x}\\ -3x\left({x}^{2}-1\right)\sqrt{1-{x}^{2}}-12{x}^{3}+6x.\end{array}$

Motivated by the investigations in [12], we are in a position to prove ${F}_{4}\left(x\right)>0$ for $x\in \left(0,1\right)$. Let

$B\left(x\right)=\left\{\begin{array}{cc}\lambda ,\hfill & x=0,\hfill \\ \frac{{F}_{4}\left(x\right)}{{x}^{3}},\hfill & 0

where λ is a constant determined by the limit

$\lambda =\underset{x\to 0}{lim}\frac{{F}_{4}\left(x\right)}{{x}^{3}}=\frac{43}{8}.$

Using Maple we determine a Taylor approximation for the function $B\left(x\right)$ by the polynomial of fourth order:

${P}_{2}\left(x\right)=\frac{43}{8}-\frac{191}{128}{x}^{2}-\frac{609}{1\text{,}024}{x}^{4},$

which has a bound of the absolute error of

${\epsilon }_{2}=\frac{9\text{,}511}{1\text{,}024}-5\sqrt{2}$

for values $x\in \left[0,1\right]$. It is true that

$B\left(x\right)-\left({P}_{2}\left(x\right)-{\epsilon }_{2}\right)\ge 0,\phantom{\rule{1em}{0ex}}0\le x\le 1$

and

${P}_{2}\left(x\right)-{\epsilon }_{2}=-\frac{4\text{,}007}{1\text{,}024}+5\sqrt{2}-\frac{191}{128}{x}^{2}-\frac{609}{1\text{,}024}{x}^{4}>0,\phantom{\rule{1em}{0ex}}0\le x\le 1.$

Hence, for $x\in \left(0,1\right)$ it is true that $B\left(x\right)>0$ and therefore ${F}_{4}\left(x\right)>0$ and ${F}_{3}^{\prime }\left(x\right)>0$ for $x\in \left(0,1\right)$. Therefore, the function $\frac{{F}_{1}^{\prime }\left(x\right)}{{F}_{2}^{\prime }\left(x\right)}$ is strictly increasing on $\left(0,1\right)$. By Lemma 1.1, the function

$F\left(x\right)=\frac{{F}_{1}\left(x\right)}{{F}_{2}\left(x\right)}=\frac{{F}_{1}\left(x\right)-{F}_{1}\left(0\right)}{{F}_{2}\left(x\right)-{F}_{2}\left(0\right)}$

is strictly increasing on $\left(0,1\right)$. And hence,

$2=\underset{x\to 0}{lim}F\left(x\right)

for $0. By rearranging terms in the last expression, Theorem 2.2 follows. □

Theorem 2.3 For $0<|x|<1$,

$\frac{{a}_{1}}{4+\sqrt{1-{x}^{4}}}<\frac{arcslx}{x}<\frac{{b}_{1}}{4+\sqrt{1-{x}^{4}}}$
(2.5)

and

$\frac{{a}_{2}}{\frac{7}{3}+{\left(1-{x}^{4}\right)}^{1/3}}<\frac{arcslx}{x}<\frac{{b}_{2}}{\frac{7}{3}+{\left(1-{x}^{4}\right)}^{1/3}}$
(2.6)

with the best possible constants

${a}_{1}=5,\phantom{\rule{2em}{0ex}}{b}_{1}=B\left(\frac{1}{4},\frac{1}{2}\right)=5.2441151\dots$
(2.7)

and

${a}_{2}=\frac{7}{12}B\left(\frac{1}{4},\frac{1}{2}\right)=3.0590671\dots ,\phantom{\rule{2em}{0ex}}{b}_{2}=\frac{10}{3}=3.3333333\dots .$
(2.8)

Here $B\left(x,y\right)$ denotes the beta function.

Proof For $0, let

${L}_{1}\left(x\right)=\frac{4+\sqrt{1-{x}^{4}}}{x}arcslx.$

Differentiation yields

${L}_{1}^{\prime }\left(x\right)=\frac{{x}^{4}+4\sqrt{1-{x}^{4}}+1}{{x}^{2}\sqrt{1-{x}^{4}}}{L}_{2}\left(x\right),$

where

${L}_{2}\left(x\right)=-arcslx+\frac{\left(4+\sqrt{1-{x}^{4}}\right)x}{{x}^{4}+4\sqrt{1-{x}^{4}}+1}.$

Elementary calculation shows that

${L}_{2}^{\prime }\left(x\right)=\frac{24{x}^{4}\left(1-\sqrt{1-{x}^{4}}\right)}{{\left({x}^{4}+4\sqrt{1-{x}^{4}}+1\right)}^{2}\sqrt{1-{x}^{4}}}>0,\phantom{\rule{1em}{0ex}}0

Hence, ${L}_{2}\left(x\right)>{L}_{2}\left(0\right)=0$ and ${L}_{1}^{\prime }\left(x\right)>0$ for $0. Therefore, the function ${L}_{1}\left(x\right)$ is strictly increasing on $\left(0,1\right)$. And hence,

$5=\underset{x\to 0}{lim}{L}_{1}\left(x\right)<{L}_{1}\left(x\right)=\frac{4+\sqrt{1-{x}^{4}}}{x}arcslx<\underset{x\to 1}{lim}{L}_{1}\left(x\right)=B\left(\frac{1}{4},\frac{1}{2}\right)$

for $0. Hence, inequality (2.5) holds with the best possible constants given in equation (2.7).

For $0, let

${M}_{1}\left(x\right)=\frac{\frac{7}{3}+{\left(1-{x}^{4}\right)}^{1/3}}{x}arcslx.$

Differentiation yields

${M}_{1}^{\prime }\left(x\right)=\frac{{x}^{4}+7{\left(1-{x}^{4}\right)}^{2/3}+3}{3{x}^{2}{\left(1-{x}^{4}\right)}^{2/3}}{M}_{2}\left(x\right),$

where

${M}_{2}\left(x\right)=-arcslx+\frac{x{\left(1-{x}^{4}\right)}^{1/6}\left(7+3{\left(1-{x}^{4}\right)}^{1/3}\right)}{{x}^{4}+7{\left(1-{x}^{4}\right)}^{2/3}+3}.$

Elementary calculation shows that

${M}_{2}^{\prime }\left(x\right)=-\frac{2{x}^{4}}{3{\left(1-{x}^{4}\right)}^{3/2}{\left({x}^{4}+7{\left(1-{x}^{4}\right)}^{2/3}+3\right)}^{2}}{M}_{3}\left(x\right),$

where

${M}_{3}\left(x\right)=3{x}^{8}-66{x}^{4}+63+\left(84-56{x}^{4}\right){\left(1-{x}^{4}\right)}^{2/3}-147{\left(1-{x}^{4}\right)}^{4/3}.$

We claim that ${M}_{3}\left(x\right)>0$ for $0. By an elementary change of variable

$x={\left(1-{t}^{3}\right)}^{1/4},\phantom{\rule{1em}{0ex}}0
(2.9)

we find that

where

${M}_{4}\left(t\right)=28{t}^{2}+60{t}^{3}+56{t}^{5}+3{t}^{6}-147{t}^{4}={t}^{2}\left(1-t\right)\left(-3{t}^{3}-59{t}^{2}+88t+28\right).$

Obviously, ${M}_{4}\left(t\right)>0$ for $0. This proves the claim.

Hence, ${M}_{2}^{\prime }\left(x\right)<0$ for $0. This implies that ${M}_{2}\left(x\right)<{M}_{2}\left(0\right)=0$ and ${M}_{1}^{\prime }\left(x\right)<0$ for $0. Therefore, the function ${M}_{1}\left(x\right)$ is strictly decreasing on $\left(0,1\right)$. And hence,

$\frac{7}{12}B\left(\frac{1}{4},\frac{1}{2}\right)=\underset{x\to 1}{lim}{M}_{1}\left(x\right)<{M}_{1}\left(x\right)=\frac{\frac{7}{3}+{\left(1-{x}^{4}\right)}^{1/3}}{x}arcslx<\underset{x\to 0}{lim}{M}_{1}\left(x\right)=\frac{10}{3}$

for $0. Hence, inequality (2.6) holds with the best possible constants given in equation (2.8). □

Remark 2.1 (i) There is no strict comparison between the two lower bounds in equations (2.5) and (2.6). Also, there is no strict comparison between the two upper bounds in equations (2.5) and (2.6).

1. (ii)

The lower bound in equation (1.10) is sharper than the one in equation (2.5), since

$\frac{5}{3+2{\left(1-{x}^{4}\right)}^{1/2}}-{\left(\frac{5}{4+\sqrt{1-{x}^{4}}}\right)}^{2}=\frac{5{\left(1-\sqrt{1-{x}^{4}}\right)}^{2}}{{\left(4+\sqrt{1-{x}^{4}}\right)}^{2}\left(3+2\sqrt{1-{x}^{4}}\right)}>0,\phantom{\rule{1em}{0ex}}0<|x|<1.$

There is no strict comparison between the two upper bounds in equations (1.10) and (2.5).

1. (iii)

By two elementary changes of variable,

$t=1-{x}^{4},\phantom{\rule{1em}{0ex}}0

we find that

$\begin{array}{c}\frac{\frac{10}{3}}{\frac{7}{3}+{\left(1-{x}^{4}\right)}^{1/3}}-\frac{1}{{\left(1-{x}^{4}\right)}^{1/10}}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{10{t}^{1/10}-7-3{t}^{1/3}}{\left(7+3{t}^{1/3}\right){t}^{1/10}}=\frac{10{u}^{3}-7-3{u}^{10}}{\left(7+3{u}^{10}\right){u}^{3}}\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{\left(3{u}^{8}+6{u}^{7}+9{u}^{6}+12{u}^{5}+15{u}^{4}+18{u}^{3}+21{u}^{2}+14u+7\right){\left(1-u\right)}^{2}}{\left(7+3{u}^{10}\right){u}^{3}}<0.\hfill \end{array}$

Hence, the upper bound in equation (2.6) is sharper than the one in equation (1.10). There is no strict comparison between the two lower bounds in equations (1.10) and (2.6).

Theorem 2.4 For $x\ne 0$,

$\frac{a}{\frac{17}{3}+{\left(1+{x}^{4}\right)}^{1/4}}<\frac{arcslhx}{x}<\frac{b}{\frac{17}{3}+{\left(1+{x}^{4}\right)}^{1/4}}$
(2.10)

with the best possible constants

$a=\frac{1}{4}B\left(\frac{1}{4},\frac{1}{4}\right)=1.8540746\dots \phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}b=\frac{20}{3}=6.6666666\dots .$
(2.11)

Here $B\left(x,y\right)$ denotes the beta function.

Proof The inequality (2.10) is obtained by considering the function $p\left(x\right)$ defined by

$p\left(x\right)=\frac{\left(\frac{17}{3}+{\left(1+{x}^{4}\right)}^{1/4}\right)}{x}arcslhx,\phantom{\rule{1em}{0ex}}x>0.$

Differentiation yields

${p}^{\prime }\left(x\right)=-\frac{17{\left(1+{x}^{4}\right)}^{3/4}+3}{3{x}^{2}{\left(1+{x}^{4}\right)}^{3/4}}arcslhx+\frac{\left(\frac{17}{3}+{\left(1+{x}^{4}\right)}^{1/4}\right)}{x}\frac{1}{\sqrt{1+{x}^{4}}}=\frac{17{\left(1+{x}^{4}\right)}^{3/4}+3}{3{x}^{2}{\left(1+{x}^{4}\right)}^{3/4}}q\left(x\right),$

where

$q\left(x\right)=-arcslhx+\frac{x{\left(1+{x}^{4}\right)}^{1/4}\left(17+3{\left(1+{x}^{4}\right)}^{1/4}\right)}{17{\left(1+{x}^{4}\right)}^{3/4}+3}.$

Elementary calculation shows that

${q}^{\prime }\left(x\right)=-\frac{{x}^{4}}{{\left(1+{x}^{4}\right)}^{3/2}{\left(17{\left(1+{x}^{4}\right)}^{3/4}+3\right)}^{2}}r\left(x\right),$

where

$r\left(x\right)=-27-27{x}^{4}-51{\left(1+{x}^{4}\right)}^{3/4}+578{\left(1+{x}^{4}\right)}^{3/2}.$

We claim that $r\left(x\right)>0$ for $x>0$. By an elementary change of variable

$t={\left(1+{x}^{4}\right)}^{1/4},\phantom{\rule{1em}{0ex}}x>0,$
(2.12)

we find that

where

$s\left(t\right)={t}^{3}\left(-51-27t+578{t}^{3}\right),\phantom{\rule{1em}{0ex}}t>1.$

Obviously, $s\left(t\right)>0$ for $t>1$. This proves the claim.

Hence, ${q}^{\prime }\left(x\right)<0$ and $q\left(x\right) for $x>0$. Therefore, ${p}^{\prime }\left(x\right)<0$ for $x>0$, and we have

$\frac{1}{4}B\left(\frac{1}{4},\frac{1}{4}\right)=\underset{x\to \mathrm{\infty }}{lim}p\left(x\right)

Hence, the inequality (2.10) holds with the best possible constants given in equation (2.11). □

Remark 2.2 Inequality (1.11) is sharper than inequality (2.10).

Theorem 2.5 For $x\ne 0$,

$\frac{\alpha }{\frac{2}{3}+{\left(1+{x}^{4}\right)}^{1/4}}<\frac{arctlx}{x}<\frac{\beta }{\frac{2}{3}+{\left(1+{x}^{4}\right)}^{1/4}}$
(2.13)

with the best possible constants

$\alpha =\frac{1}{4}B\left(\frac{1}{4},\frac{1}{2}\right)=1.3110287\dots \phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\beta =\frac{5}{3}=1.6666666\dots .$
(2.14)

Here $B\left(x,y\right)$ denotes the beta function.

Proof The inequality (2.13) is obtained by considering the function $P\left(x\right)$ defined by

$P\left(x\right)=\frac{\left(\frac{2}{3}+{\left(1+{x}^{4}\right)}^{1/4}\right)}{x}arctlx,\phantom{\rule{1em}{0ex}}x>0.$

Differentiation yields

${P}^{\prime }\left(x\right)=-\frac{2{\left(1+{x}^{4}\right)}^{3/4}+3}{3{x}^{2}{\left(1+{x}^{4}\right)}^{3/4}}arctlx+\frac{\left(\frac{2}{3}+{\left(1+{x}^{4}\right)}^{1/4}\right)}{x}\frac{1}{{\left(1+{x}^{4}\right)}^{3/4}}=\frac{2{\left(1+{x}^{4}\right)}^{3/4}+3}{3{x}^{2}{\left(1+{x}^{4}\right)}^{3/4}}Q\left(x\right),$

where

$Q\left(x\right)=-arctlx+\frac{\left(2+3{\left(1+{x}^{4}\right)}^{1/4}x}{2{\left(1+{x}^{4}\right)}^{3/4}+3}.$

Elementary calculation shows that

${Q}^{\prime }\left(x\right)=-\frac{6{x}^{4}\left(2\sqrt{1+{x}^{4}}+{\left(1+{x}^{4}\right)}^{3/4}-3\right)}{{\left(1+{x}^{4}\right)}^{3/4}{\left(2{\left(1+{x}^{4}\right)}^{3/4}+3\right)}^{2}}<0,\phantom{\rule{1em}{0ex}}x>0.$

Hence, $Q\left(x\right) for $x>0$. Therefore, ${P}^{\prime }\left(x\right)<0$ for $x>0$, and we have

$\frac{1}{4}B\left(\frac{1}{4},\frac{1}{2}\right)=\underset{x\to \mathrm{\infty }}{lim}P\left(x\right)

Hence, the inequality (2.13) holds with the best possible constants given in equation (2.14). □

Theorem 2.6 For $0<|x|<1$,

$\frac{{a}_{3}}{\frac{7}{3}+\sqrt{1-{x}^{4}}}<\frac{arctlhx}{x}<\frac{{b}_{3}}{\frac{7}{3}+\sqrt{1-{x}^{4}}}$
(2.15)

and

$\frac{{a}_{4}}{\frac{2}{3}+{\left(1-{x}^{4}\right)}^{1/4}}<\frac{arctlhx}{x}<\frac{{b}_{4}}{\frac{2}{3}+{\left(1-{x}^{4}\right)}^{1/4}}$
(2.16)

with the best possible constants

${a}_{3}=\frac{10}{3}=3.3333333\dots ,\phantom{\rule{2em}{0ex}}{b}_{3}=\frac{7}{12}B\left(\frac{1}{4},\frac{1}{4}\right)=4.3261742\dots$
(2.17)

and

${a}_{4}=\frac{1}{6}B\left(\frac{1}{4},\frac{1}{4}\right)=1.236049\dots ,\phantom{\rule{2em}{0ex}}{b}_{4}=\frac{5}{3}=1.666666\dots .$
(2.18)

Proof For $0, let

${J}_{1}\left(x\right)=\frac{\frac{7}{3}+\sqrt{1-{x}^{4}}}{x}arctlhx.$

Differentiation yields

${J}_{1}^{\prime }\left(x\right)=\frac{3{x}^{4}+7\sqrt{1-{x}^{4}}+3}{3{x}^{2}\sqrt{1-{x}^{4}}}{J}_{2}\left(x\right),$

where

${J}_{2}\left(x\right)=-arctlhx+\frac{\left(7+3\sqrt{1-{x}^{4}}\right)x}{{\left(1-{x}^{4}\right)}^{1/4}\left(3{x}^{4}+7\sqrt{1-{x}^{4}}+3\right)}.$

Elementary calculation shows that

${J}_{2}^{\prime }\left(x\right)=\frac{3{x}^{4}}{{\left(1-{x}^{4}\right)}^{7/4}{\left(3{x}^{4}+7\sqrt{1-{x}^{4}}+3\right)}^{2}}{J}_{3}\left(x\right),$

where

${J}_{3}\left(x\right)=-25{x}^{4}-3{x}^{8}+28+\left(42{x}^{4}-28\right)\sqrt{1-{x}^{4}}.$

We claim that ${J}_{3}\left(x\right)>0$ for $0. By an elementary change of variable

$x={\left(1-{t}^{2}\right)}^{1/4},\phantom{\rule{1em}{0ex}}0
(2.19)

we find that

where

${J}_{4}\left(t\right)=14t+31{t}^{2}-42{t}^{3}-3{t}^{4}=t\left(1-t\right)\left(3{t}^{2}+45t+14\right).$

Obviously, ${J}_{4}\left(t\right)>0$ for $0. This proves the claim.

Hence, ${J}_{2}^{\prime }\left(x\right)>0$ for $0. This implies that ${J}_{2}\left(x\right)>{J}_{2}\left(0\right)=0$ and ${J}_{1}^{\prime }\left(x\right)>0$ for $0. Therefore, the function ${J}_{1}\left(x\right)$ is strictly increasing on $\left(0,1\right)$. And hence,

$\frac{10}{3}=\underset{x\to 0}{lim}{J}_{1}\left(x\right)<{J}_{1}\left(x\right)=\frac{\frac{7}{3}+\sqrt{1-{x}^{4}}}{x}arctlhx<\underset{x\to 1}{lim}{J}_{1}\left(x\right)=\frac{7}{12}B\left(\frac{1}{4},\frac{1}{4}\right)$

for $0. Hence, inequality (2.15) holds with the best possible constants given in equation (2.17).

For $0, let

${T}_{1}\left(x\right)=\frac{\frac{2}{3}+{\left(1-{x}^{4}\right)}^{1/4}}{x}arctlhx.$

Differentiation yields

${T}_{1}^{\prime }\left(x\right)=\frac{2{\left(1-{x}^{4}\right)}^{3/4}+3}{3{x}^{2}{\left(1-{x}^{4}\right)}^{3/4}}{T}_{2}\left(x\right),$

where

${T}_{2}\left(x\right)=-arctlhx+\frac{x\left(2+3{\left(1-{x}^{4}\right)}^{1/4}\right)}{2{\left(1-{x}^{4}\right)}^{3/4}+3}.$

Elementary calculation shows that

${T}_{2}^{\prime }\left(x\right)=-\frac{6{x}^{4}\left(3-2\sqrt{1-{x}^{4}}-{\left(1-{x}^{4}\right)}^{3/4}\right)}{{\left(1-{x}^{4}\right)}^{3/4}{\left(2{\left(1-{x}^{4}\right)}^{3/4}+3\right)}^{2}}<0,\phantom{\rule{1em}{0ex}}0

which implies that ${T}_{2}\left(x\right)<{T}_{2}\left(0\right)=0$ and ${T}_{1}^{\prime }\left(x\right)<0$ for $0. Therefore, the function ${T}_{1}\left(x\right)$ is strictly decreasing on $\left(0,1\right)$. And hence,

$\frac{1}{6}B\left(\frac{1}{4},\frac{1}{4}\right)=\underset{x\to 1}{lim}{T}_{1}\left(x\right)<{T}_{1}\left(x\right)=\frac{\frac{2}{3}+{\left(1-{x}^{4}\right)}^{1/4}}{x}arctlhx<\underset{x\to 0}{lim}{T}_{1}\left(x\right)=\frac{5}{3}$

for $0. Hence, inequality (2.16) holds with the best possible constants given in equation (2.18). □

Remark 2.3 There is no strict comparison between the two lower bounds in equations (2.15) and (2.16). Also, there is no strict comparison between the two upper bounds in equations (2.15) and (2.16).

## References

1. Borwein JM, Borwein PB: Pi and the AGM: A Study in the Analytic Number Theory and Computational Complexity. Wiley, New York; 1987.

2. Carlson BC: Algorithms involving arithmetic and geometric means. Am. Math. Mon. 1971, 78: 496-505. 10.2307/2317754

3. Neuman E: On Gauss lemniscate functions and lemniscatic mean. Math. Pannon. 2007, 18: 77-94.

4. Neuman E: Two-sided inequalities for the lemniscate functions. J. Inequal. Spec. Funct. 2010, 1: 1-7.

5. Neuman E: On Gauss lemniscate functions and lemniscatic mean II. Math. Pannon. 2012, 23: 65-73.

6. Neuman E: Inequalities for Jacobian elliptic functions and Gauss lemniscate functions. Appl. Math. Comput. 2012, 218: 7774-7782. 10.1016/j.amc.2012.01.041

7. Neuman E: On lemniscate functions. Integral Transforms Spec. Funct. 2013, 24: 164-171. 10.1080/10652469.2012.684054

8. Siegel CL 1. In Topics in Complex Function Theory. Wiley, New York; 1969.

9. Mitrinović DS, Vasić PM: Analytic Inequalities. Springer, New York; 1970.

10. Fink AM: Two inequalities. Publ. Elektroteh. Fak. Univ. Beogr., Mat. 1995, 6: 48-49.

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

12. Malešević BJ: One method for proving inequalities by computer. J. Inequal. Appl. 2007., 2007: Article ID 78691 10.1155/2007/78691

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

14. Zhu L: On Shafer-Fink inequalities. Math. Inequal. Appl. 2005, 8: 571-574.

15. Zhu L: On Shafer-Fink-type inequality. J. Inequal. Appl. 2007., 2007: Article ID 67430 10.1155/2007/67430

16. Pan W, Zhu L: Generalizations of Shafer-Fink-type inequalities for the arc sine function. J. Inequal. Appl. 2009., 2009: Article ID 705317 10.1155/2009/705317

17. Zhu L: A solution of a problem of Oppenheim. Math. Inequal. Appl. 2007, 10: 57-61.

18. Oppenheim A: E1277. Am. Math. Mon. 1957, 64: 504. 10.2307/2308467

19. Chen CP, Cheung WS: Wilker- and Huygens-type inequalities and solution to Oppenheim’s problem. Integral Transforms Spec. Funct. 2012, 23: 325-336. 10.1080/10652469.2011.586637

20. Qi F, Guo BN: Sharpening and generalizations of Shafer’s inequality for the arc sine function. Integral Transforms Spec. Funct. 2012, 23: 129-134. 10.1080/10652469.2011.564578

21. Qi F, Guo BN: Sharpening and generalizations of Shafer-Fink’s double inequality for the arc sine function. Filomat 2013, 27: 261-265. 10.2298/FIL1302261G

22. Shafer RE: Problem E1867. Am. Math. Mon. 1966, 73: 309. 10.2307/2315358

23. Shafer RE: Problems E1867 (solution). Am. Math. Mon. 1967, 74: 726-727.

24. Kuang JC: Applied Inequalities. 3rd edition. Shangdong Science and Technology Press, Jinan; 2004.

25. Qi F, Zhang SQ, Guo BN: Sharpening and generalizations of Shafer’s inequality for the arc tangent function. J. Inequal. Appl. 2009., 2009: Article ID 930294 10.1155/2009/930294

26. Chen CP, Cheung WS, Wang W: On Shafer and Carlson inequalities. J. Inequal. Appl. 2011., 2011: Article ID 840206 10.1155/2011/840206

27. Zhu L: New inequalities of Shafer-Fink type for arc hyperbolic sine. J. Inequal. Appl. 2008., 2008: Article ID 368275 10.1155/2008/368275

28. Chen CP: Wilker and Huygens type inequalities for the lemniscate functions. J. Math. Inequal. 2012, 6: 673-684.

29. Chen CP: Wilker and Huygens type inequalities for the lemniscate functions, II. Math. Inequal. Appl. 2013, 16: 577-586.

30. Anderson GD, Qiu SL, Vamanamurthy MK, Vuorinen M: Generalized elliptic integral and modular equations. Pac. J. Math. 2000, 192: 1-37.

31. Anderson GD, Vamanamurthy MK, Vuorinen M: Conformal Invariants, Inequalities, and Quasiconformal Maps. Wiley, New York; 1997.

32. Anderson GD, Vamanamurthy MK, Vuorinen M: Monotonicity rules in calculus. Am. Math. Mon. 2006, 113: 805-816. 10.2307/27642062

33. Lindqvist P: Some remarkable sine and cosine functions. Ric. Mat. 1995, 44: 269-290.

34. Biezuner RJ, Ercole G, Martins EM: Computing the first eigenvalue of the p -Laplacian via the inverse power method. J. Funct. Anal. 2009, 257: 243-270. 10.1016/j.jfa.2009.01.023

35. Bushell PJ, Edmunds DE: Remarks on generalised trigonometric functions. Rocky Mt. J. Math. 2012, 42: 25-57. 10.1216/RMJ-2012-42-1-25

36. Drábek P, Manásevich R: On the closed solution to some p -Laplacian nonhomogeneous eigenvalue problems. Differ. Integral Equ. 1999, 12: 773-788.

37. Lang J, Edmunds DE Lecture Notes in Mathematics 2016. In Eigenvalues, Embeddings and Generalised Trigonometric Functions. Springer, Berlin; 2011.

38. Takeuchi S: Generalized Jacobian elliptic functions and their application to bifurcation problems associated with p -Laplacian. J. Math. Anal. Appl. 2012, 385: 24-35. 10.1016/j.jmaa.2011.06.063

39. Edmunds DE, Gurka P, Lang J: Properties of generalized trigonometric functions. J. Approx. Theory 2012, 164: 47-56. 10.1016/j.jat.2011.09.004

40. Baricz A, Bhayo BA, Klén R: Convexity properties of generalized trigonometric and hyperbolic functions. Aequ. Math. 2013. 10.1007/s00010-013-0222-x

41. Bhayo BA, Vuorinen M: On generalized trigonometric functions with two parameters. J. Approx. Theory 2012, 164: 1415-1426. 10.1016/j.jat.2012.06.003

42. Klén R, Vuorinen M, Zhang X: Inequalities for the generalized trigonometric and hyperbolic functions. J. Math. Anal. Appl. 2014, 409: 521-529. 10.1016/j.jmaa.2013.07.021

## Acknowledgements

The authors would like to thank the referees for their careful reading of the manuscript and insightful comments.

## Author information

Authors

### Corresponding author

Correspondence to Ji-En Deng.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors read and approved the final manuscript.

## Rights and permissions

Reprints and Permissions

Deng, JE., Chen, CP. Sharp Shafer-Fink type inequalities for Gauss lemniscate functions. J Inequal Appl 2014, 35 (2014). https://doi.org/10.1186/1029-242X-2014-35