1 Introduction
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 where arcsl 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 The functions (1.1) and (1.2) can be found (see [1], Chapter 1, [2], p. 259, and [3‐11]).
$$\begin{aligned} \operatorname {arcsl}x= \int_{0}^{x}\frac{1}{\sqrt{1-t^{4}}}\,\mathrm{d}t,\quad \vert x \vert \leq1, \end{aligned}$$
(1.1)
$$\begin{aligned} \operatorname {arcslh}x= \int_{0}^{x}\frac{1}{\sqrt{1+t^{4}}}\,\mathrm{d}t, \quad x\in \mathbb{R}. \end{aligned}$$
(1.2)
Another pair of lemniscate functions, the arc lemniscate tangent arctl and the hyperbolic arc lemniscate tangent arctlh, have been introduced in [4], (3.1)-(3.2). Therein it has been proven that and (see [4], Proposition 3.1).
$$\begin{aligned} \operatorname {arctl}x=\operatorname {arcsl}\biggl(\frac{x}{\sqrt[4]{1+x^{4}}} \biggr), \quad x\in \mathbb{R} \end{aligned}$$
(1.3)
$$\begin{aligned} \operatorname {arctlh}x=\operatorname {arcslh}\biggl(\frac{x}{\sqrt[4]{1-x^{4}}} \biggr),\quad |x|< 1 \end{aligned}$$
(1.4)
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Recently, numerous inequalities have been given for the lemniscate functions [6, 9‐11]. For example, Neuman [6] proved the following inequalities: and for \(0<|x|<1\).
$$ \biggl(\frac{5}{3+2(1-x^{4})^{1/2}} \biggr)^{1/2}< \frac{\operatorname {arcsl}x}{x}< \bigl(1-x^{4}\bigr)^{-1/10} $$
(1.5)
$$ \biggl(\frac{5}{3+2(1+x^{4})^{1/2}} \biggr)^{1/2}< \frac{\operatorname {arcslh}x}{x}< \bigl(1+x^{4}\bigr)^{-1/10} $$
(1.6)
Shafer [12] indicated several elementary quadratic approximations of selected functions without proof. Subsequently, Shafer [13] established these results as analytic inequalities. For example, Shafer [13] proved that, for \(x>0\), The inequality (1.7) can also be found in [14]. Zhu [15] developed (1.7) to produce a symmetric double inequality where the constants \(80/3\) and \(256/\pi^{2}\) are the best possible. In [15], (1.8) is called a Shafer-type inequality.
$$ \frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x. $$
(1.7)
$$ \frac{8x}{3+\sqrt{25+\frac{80}{3}x^{2}}}< \arctan x< \frac{8x}{3+\sqrt {25+\frac{256}{\pi^{2}}x^{2}}},\quad x>0, $$
(1.8)
Mortici and Srivastava [16] presented new bounds for arctanx. Some inequalities for trigonometric functions were refined in [17].
Very recently, Sun and Chen [18] established the following Shafer-type inequalities for the lemniscate functions:
and presented the following conjecture.
$$\begin{aligned} &\frac{10}{5+\sqrt{25-10x^{4}}}< \frac{\operatorname {arcsl}x}{x}, \quad 0< x< 1, \end{aligned}$$
(1.9)
$$\begin{aligned} &\frac{10}{5+\sqrt{25-15x^{4}}}< \frac{\operatorname {arctlh}x}{x},\quad 0< x< 1, \end{aligned}$$
(1.10)
$$\begin{aligned} &\frac{95}{80+\sqrt{225+285x^{4}}}< \frac{\operatorname {arcslh}x}{x},\quad x>0, \end{aligned}$$
(1.11)
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Conjecture 1.1
For
\(x>0\),
and
$$ \frac{\operatorname {arcslh}x}{x}< \frac{95+\frac{931}{2925}x^{12}}{80+\sqrt{225+285x^{4}}} $$
(1.12)
$$ \frac{1210}{940+9\sqrt{900+1210x^{4}}}< \frac{\operatorname {arctl}x}{x}< \frac {1210+\frac{2{,}078{,}417}{280{,}800}x^{12}}{940+9\sqrt{900+1210x^{4}}}. $$
(1.13)
Based on the Padé approximation method, in this paper we present new inequalities for Gauss lemniscate functions. We also prove Conjecture 1.1.
Some computations in this paper were performed using Maple software.
2 Padé approximant
For later use, we introduce the Padé approximant (see [19‐21]). Let f be a formal power series, The Padé approximation of order \((p, q)\) of the function f is the rational function, denoted by where \(p\geq0\) and \(q\geq1\) are any given integers, the coefficients \(a_{j}\) and \(b_{j}\) are given by (see [19, 21]) and we have Thus, the first \(p + q + 1\) coefficients of the series expansion of \([p/q]_{f}\) are identical to those of f. Moreover, we have (see [20]) with \(f_{n}(x) = c_{0}+ c_{1}x+ \cdots+ c_{n}x^{n}\), the nth partial sum of the series f (\(f_{n}\) is identically zero for \(n < 0\)).
$$\begin{aligned} f(t)=c_{0}+c_{1}t+c_{2}t^{2}+ \cdots. \end{aligned}$$
(2.1)
$$\begin{aligned}{} [p/q]_{f}(t)=\frac{\sum_{j=0}^{p}a_{j}t^{j}}{1+\sum_{j=1}^{q}b_{j}t^{j}}, \end{aligned}$$
(2.2)
$$\begin{aligned} \textstyle\begin{cases} a_{0}=c_{0},\\ a_{1}=c_{0}b_{1}+c_{1},\\ a_{2}=c_{0}b_{2}+c_{1}b_{1}+c_{2},\\ \vdots\\ a_{p} = c_{0}b_{p}+\cdots+ c_{p-1}b_{1} + c_{p,}\\ 0 = c_{p+1} + c_{p}b_{1} + \cdots+ c_{p-q+1}b_{q},\\ \vdots&\\ 0 = c_{p+q} + c_{p+q-1}b_{1} + \cdots+ c_{p}b_{q}, \end{cases}\displaystyle \end{aligned}$$
(2.3)
$$\begin{aligned}{} [p/q]_{f}(t)- f (t) = O\bigl(t^{p+q+1} \bigr). \end{aligned}$$
(2.4)
$$ \begin{aligned} &[p/q]_{f}(t)= \frac{\left \vert \begin{matrix} t^{q}f_{p-q}(t) & t^{q-1}f_{p-q+1}(t) &\cdots&f_{p}(t) \\ c_{p-q+1} & c_{p-q+2} &\cdots&c_{p+1} \\ \vdots&\vdots&\ddots&\vdots\\ c_{p} & c_{p+1} &\cdots&c_{p+q} \end{matrix} \right \vert } { \left \vert \begin{matrix} t^{q} & t^{q-1} &\cdots&1 \\ c_{p-q+1} & c_{p-q+2} &\cdots&c_{p+1} \\ \vdots&\vdots&\ddots&\vdots\\ c_{p} & c_{p+1} &\cdots&c_{p+q} \end{matrix} \right \vert } , \end{aligned} $$
(2.5)
Chen [9] presented the following power-series expansions (for \(\vert x\vert <1\)):
and
$$\begin{aligned} & \frac{\operatorname {arcsl}x}{x}=\sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi }(4n+1)\cdot n!}x^{4n}, \end{aligned}$$
(2.6)
$$\begin{aligned} & \frac{\operatorname {arcslh}x}{x}=\sum_{n=0}^{\infty}(-1)^{n} \frac{\Gamma (n+\frac{1}{2})}{\sqrt{\pi}(4n+1)\cdot n!}x^{4n}, \end{aligned}$$
(2.7)
$$\begin{aligned} &\frac{\operatorname {arctl}x}{x}=\sum_{n=0}^{\infty}(-1)^{n} \frac{\Gamma(n+\frac {3}{4})}{\Gamma(\frac{3}{4})\cdot (4n+1)\cdot n!}x^{4n} \end{aligned}$$
(2.8)
$$\begin{aligned} \frac{\operatorname {arctlh}x}{x}=\sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{3}{4})}{\Gamma (\frac{3}{4})\cdot (4n+1)\cdot n!}x^{4n}. \end{aligned}$$
(2.9)
We now consider the Padé approximant for the function \(\frac{\operatorname {arcsl}x}{x}\) at the point \(x=0\). Let with the coefficients \(c_{j}\) given by
$$\begin{aligned} f(t)=\sum_{j=0}^{\infty}c_{j}t^{j}=1+ \frac{1}{10} t+\frac{1}{24} t^{2}+\frac{5}{208}t^{3}+ \frac{35}{2176}t^{4}+\cdots, \end{aligned}$$
(2.10)
$$\begin{aligned} c_{j}=\frac{\Gamma(j+\frac{1}{2})}{\sqrt{\pi}(4j+1)\cdot j!}. \end{aligned}$$
(2.11)
Let us find the \((2, 2)\) Padé approximant for the function (2.10) at the point \(t=0\), Noting that holds, we have, by (2.3), that is, We thus obtain and we have, by (2.4), That is Replacing t by \(x^{4}\) in (2.15) yields
$$\begin{aligned}{} [2/2]_{f}(t)=\frac{\sum_{j=0}^{2}a_{j}t^{j}}{1+\sum_{j=1}^{2}b_{j}t^{j}}. \end{aligned}$$
$$\begin{aligned} c_{0}=1,\qquad c_{1}=\frac{1}{10},\qquad c_{2}=\frac{1}{24},\qquad c_{3}=\frac {5}{208},\qquad c_{4}=\frac{35}{2176}, \end{aligned}$$
(2.12)
$$\begin{aligned} \textstyle\begin{cases} a_{0}=1,\\ a_{1}=b_{1}+\frac{1}{10},\\ a_{2} =b_{2}+\frac{1}{10}b_{1}+ \frac{1}{24} , \\ 0 = \frac{5}{208}+\frac{1}{24}b_{1}+\frac{1}{10}b_{2},\\ 0 =\frac{35}{2176}+\frac{5}{508}b_{1}+\frac{1}{24}b_{2}, \end{cases}\displaystyle \end{aligned}$$
$$\begin{aligned} a_{0}=1,\qquad a_{1}=-\frac{55}{68},\qquad a_{2} = \frac{23{,}623}{265{,}200},\qquad b_{1}=\frac{309}{340},\qquad b_{2}= \frac{489}{3536}. \end{aligned}$$
$$ [2/2]_{f}(t)=\frac{1-\frac{55}{68}t+\frac{23{,}623}{265{,}200}t^{2}}{1-\frac {309}{340}t+\frac{489}{3536}t^{2}}, $$
(2.13)
$$ f(t)=[2/2]_{f}(t)+O \bigl(t^{5} \bigr). $$
(2.14)
$$ \sum_{j=0}^{\infty} \frac{\Gamma(j+\frac{1}{2})}{\sqrt{\pi }(4j+1)\cdot j!}t^{j}=\frac{1-\frac{55}{68}t+\frac{23{,}623}{265{,}200}t^{2}}{1-\frac {309}{340}t+\frac{489}{3536}t^{2}}+O \bigl(t^{5} \bigr). $$
(2.15)
$$\begin{aligned} \frac{\operatorname {arcsl}x}{x}&=\frac{1-\frac{55}{68}x^{4}+\frac {23{,}623}{265{,}200}x^{8}}{1-\frac{309}{340}x^{4}+\frac{489}{3536}x^{8}}+O \bigl(x^{20} \bigr) \\ &=\frac{265{,}200-214{,}500x^{4}+23{,}623x^{8}}{15(17{,}680-16{,}068x^{4}+2445x^{8})}+O \bigl(x^{20} \bigr). \end{aligned}$$
(2.16)
Remark 2.1
Using (2.5), we can also derive (2.13). Indeed, we have
$$\begin{aligned}{} [2/2]_{f}(t)&=\frac{\left \vert \begin{matrix} t^{2}f_{0}(t) & tf_{1}(t) &f_{2}(t) \\ c_{1} &c_{2} &c_{3} \\ c_{2} &c_{3} &c_{4} \end{matrix} \right \vert } { \left \vert \begin{matrix} t^{2} &t &1 \\ c_{1} &c_{2} &c_{3} \\ c_{2} &c_{3} &c_{4} \end{matrix} \right \vert } = \frac{\left \vert \begin{matrix} t^{2} &t (1+\frac{1}{10} t ) &1+\frac{1}{10} t+\frac{1}{24} t^{2}\\ \frac{1}{10} &\frac{1}{24} &\frac{5}{208} \\ \frac{1}{24} &\frac{5}{208} &\frac{35}{2176} \end{matrix} \right \vert } { \left \vert \begin{matrix} t^{2} &t &1 \\ \frac{1}{10} &\frac{1}{24} &\frac{5}{208} \\ \frac{1}{24} &\frac{5}{208} &\frac{35}{2176} \end{matrix} \right \vert } \\ &=\frac{1-\frac{55}{68}t+\frac{23{,}623}{265{,}200}t^{2}}{1-\frac {309}{340}t+\frac{489}{3536}t^{2}}. \end{aligned}$$
Following the same method as used in the derivation of the formula (2.16), we find
and
$$\begin{aligned} &\frac{\operatorname {arcslh}x}{x}=\frac{1+\frac{55}{68}x^{4}+\frac {23{,}623}{265{,}200}x^{8}}{1+\frac{309}{340}x^{4}+\frac {489}{3536}x^{8}}+O\bigl(x^{20} \bigr) \\ &\phantom{\frac{\operatorname {arcslh}x}{x}}=\frac{265{,}200+214{,}500x^{4}+23{,}623x^{8}}{15(17{,}680+16{,}068x^{4}+2445x^{8})}+O \bigl(x^{20} \bigr), \end{aligned}$$
(2.17)
$$\begin{aligned} & \frac{\operatorname {arctl}x}{x}=\frac{1+\frac{63}{130}x^{4}- \frac {139}{6240}x^{8}}{1+\frac{33}{52}x^{4}}+O\bigl(x^{16} \bigr), \end{aligned}$$
(2.18)
$$\begin{aligned} &\frac{\operatorname {arctl}x}{x}=\frac{1+\frac{79{,}047}{94{,}520}x^{4}+\frac {565{,}795}{5{,}898{,}048}x^{8}}{1+\frac{18{,}645}{18{,}904}x^{4}+\frac {336{,}105}{1{,}966{,}016}x^{8}}+O\bigl(x^{20} \bigr) \\ &\phantom{\frac{\operatorname {arctl}x}{x}}=\frac {29{,}490{,}240+24{,}662{,}664x^{4}+2{,}828{,}975x^{8}}{15(1{,}966{,}016+1{,}939{,}080x^{4}+336{,}105x^{8})}+O\bigl(x^{20}\bigr) \end{aligned}$$
(2.19)
$$\begin{aligned} \frac{\operatorname {arctlh}x}{x}&=\frac{1-\frac{79{,}047}{94{,}520}x^{4}+\frac {565{,}795}{5{,}898{,}048}x^{8}}{1-\frac{18{,}645}{18{,}904}x^{4}+\frac {336{,}105}{1{,}966{,}016}x^{8}}+O\bigl(x^{20} \bigr) \\ &=\frac {29{,}490{,}240-24{,}662{,}664x^{4}+2{,}828{,}975x^{8}}{15(1{,}966{,}016-1{,}939{,}080x^{4}+336{,}105x^{8})}+O\bigl(x^{20}\bigr). \end{aligned}$$
(2.20)
Conjecture 2.1
Let
and
Then the coefficients
\(a_{j}\)
and
\(\alpha_{j}\)
satisfy the following relation:
and the coefficients
\(b_{j}\)
and
\(\beta_{j}\)
satisfy the following relation:
$$\begin{aligned} \frac{\operatorname {arcsl}x}{x}&=\frac{1+\sum_{j=1}^{n}a_{j}x^{4j}}{1+\sum_{j=1}^{n}b_{j}x^{4j}}+O \bigl(x^{8n+4} \bigr) \end{aligned}$$
(2.21)
$$\begin{aligned} \frac{\operatorname {arcslh}x}{x}&=\frac{1+\sum_{j=1}^{n}\alpha_{j}x^{4j}}{1+\sum_{j=1}^{n}\beta_{j}x^{4j}}+O \bigl(x^{8n+4} \bigr). \end{aligned}$$
(2.22)
$$ a_{j}=(-1)^{j}\alpha_{j}, \quad j=1,2, \ldots, n, $$
(2.23)
$$ b_{j}=(-1)^{j}\beta_{j},\quad j=1,2, \ldots, n. $$
(2.24)
Conjecture 2.2
Let
and
Then the coefficients
\(p_{j}\)
and
\(r_{j}\)
satisfy the following relation:
and the coefficients
\(q_{j}\)
and
\(s_{j}\)
satisfy the following relation:
$$\begin{aligned} \frac{\operatorname {arctl}x}{x}&=\frac{1+\sum_{j=1}^{n}p_{j}x^{4j}}{1+\sum_{j=1}^{n}q_{j}x^{4j}}+O \bigl(x^{8n+4} \bigr) \end{aligned}$$
(2.25)
$$\begin{aligned} \frac{\operatorname {arctlh}x}{x}&=\frac{1+\sum_{j=1}^{n}r_{j}x^{4j}}{1+\sum_{j=1}^{n}s_{j}x^{4j}}+O \bigl(x^{8n+4} \bigr). \end{aligned}$$
(2.26)
$$ p_{j}=(-1)^{j}r_{j}, \quad j=1,2,\ldots, n, $$
(2.27)
$$ q_{j}=(-1)^{j}s_{j}, \quad j=1,2,\ldots, n. $$
(2.28)
3 Inequalities
Theorem 3.1
For
\(0< x<1\),
$$ \frac{265{,}200-214{,}500x^{4}+23{,}623x^{8}}{15(17{,}680-16{,}068x^{4}+2445x^{8})}< \frac {\operatorname {arcsl}x}{x}. $$
(3.1)
Proof
Consider the function Differentiation yields Elementary calculations reveal that where
$$ f(x)=\operatorname {arcsl}x-\frac {x(265{,}200-214{,}500x^{4}+23{,}623x^{8})}{15(17{,}680-16{,}068x^{4}+2445x^{8})},\quad 0< x< 1. $$
$$\begin{aligned} f'(x)={}&\frac{1}{\sqrt{1-x^{4}}} \\ &{} -\frac {312{,}582{,}400-411{,}873{,}280x^{4}+177{,}771{,} 984x^{8}-21{,}634{,}288x^{12}+3{,}850{,}549x^{16}}{(17{,}680- 16{,}068x^{4}+2445x^{8})^{2}}. \end{aligned}$$
$$\begin{aligned} & \biggl(\frac{1}{\sqrt{1-t}} \biggr)^{2}\\ &\qquad{}- \biggl(\frac {312{,}582{,}400-411{,}873{,}280t+177{,}771{,}984t^{2}-21{,}634{,}288t^{3}+3{,}850{,}549t^{4}}{(17{,}680-16{,}068t+2445t^{2})^{2}} \biggr)^{2} \\ &\quad =\frac{t^{5}g(t)}{(1-t)(17{,}680-16{,}068t+2445t^{2})^{4}}, \quad 0< t< 1, \end{aligned}$$
$$\begin{aligned} g(t)={}&1{,}744{,}280{,}123{,}040{,}000-2{,}406{,}774{,}938{,}256{,}000t \\ &{}+1{,}064{,}272{,}682{,}007{,}600t^{2} \\ &{} -145{,}697{,}716{,}749{,}000t^{3}+14{,}826{,}727{,}601{,}401t^{4}. \end{aligned}$$
We now prove that \(f'(x)>0\) for \(0< x<1\). It suffices to show that \(g(t)>0\) for \(0< t<1\). Differentiation yields and We then obtain, for \(0< t<1\), Hence, \(f'(x)>0\) for \(0< x<1\), and we have The proof is complete. □
$$\begin{aligned} g'(t)={}&-2{,}406{,}774{,}938{,}256{,}000+2{,}128{,}545{,}364{,}015{,}200t\\ &{}-437{,}093{,}150{,}247{,}000t^{2} \\ &{} +59{,}306{,}910{,}405{,}604t^{3} \end{aligned}$$
$$\begin{aligned} g''(t)={}&2{,}128{,}545{,}364{,}015{,}200-874{,}186{,}300{,}494{,}000t\\ &{}+177{,}920{,}731{,}216{,}812t^{2}>0,\quad 0< t< 1. \end{aligned}$$
$$\begin{aligned} &g'(t)< g'(1)=-656{,}015{,}814{,}082{,}196< 0\quad \Longrightarrow\\ & g(t)>g(1)=270{,}906{,}877{,}644{,}001>0. \end{aligned}$$
$$\begin{aligned} f(x)>f(0)=0, \quad 0< x< 1. \end{aligned}$$
Theorem 3.2
For
\(x>0\),
$$ \frac{\operatorname {arcslh}x}{x}< \frac {265{,}200+214{,}500x^{4}+23{,}623x^{8}}{15(17{,}680+16{,}068x^{4}+2445x^{8})}. $$
(3.2)
Proof
Consider the function Differentiation yields Elementary calculations reveal that where Hence, \(F'(x)<0\) for \(x>0\), and we have The proof is complete. □
$$ F(x)=\operatorname {arcslh}x-\frac {x(265{,}200+214{,}500x^{4}+23{,}623x^{8})}{15(17{,}680+16{,}068x^{4}+2445x^{8})},\quad x>0. $$
$$\begin{aligned} F'(x)={}&\frac{1}{\sqrt{1+x^{4}}} \\ &{} -\frac {312{,}582{,}400+411{,}873{,}280x^{4}+177{,}771{,}984x^{8}+21{,}634{,}288x^{12}+3{,}850{,}549x^{16}}{(17{,}680+16{,}068x^{4}+2445x^{8})^{2}}. \end{aligned}$$
$$\begin{aligned} & \biggl(\frac{1}{\sqrt{1+t}} \biggr)^{2}\\ &\qquad{}- \biggl(\frac {312{,}582{,}400+411{,}873{,}280t+177{,}771{,}984t^{2}+21{,}634{,}288t^{3}+3{,}850{,}549t^{4}}{(17{,}680+16{,}068t+2445t^{2})^{2}} \biggr)^{2} \\ &\quad =-\frac{t^{5}G(t)}{(1+t)(17{,}680+16{,}068t+2445t^{2})^{4}}, \end{aligned}$$
$$\begin{aligned} G(t)={}&1{,}744{,}280{,}123{,}040{,}000+2{,}406{,}774{,}938{,}256{,}000t+1{,}064{,}272{,}682{,}007{,}600t^{2} \\ &{} +145{,}697{,}716{,}749{,}000t^{3}+14{,}826{,}727{,}601{,}401t^{4}. \end{aligned}$$
$$\begin{aligned} F(x)< F(0)=0,\quad x>0. \end{aligned}$$
Remark 3.2
For
\(0< t<1\), we find
with
where
is a polynomial of the 16th degree, having all coefficients positive, and
for
\(0< t<1\). So, \(I(t)>0\)
for
\(0< t<1\). We then see that the inequality (3.2) is sharper than the right side of (1.6).
$$\begin{aligned} I(t):={}&\frac{1}{1+t}- \biggl(\frac {265{,}200+214{,}500t+23{,}623t^{2}}{15(17{,}680+16{,}068t+2445t^{2})} \biggr)^{10} \\ ={}&\frac{t^{2}P_{19}(t)}{576{,}650{,}390{,}625(1+t) (17{,}680+16{,}068t+2445t^{2} )^{10}} \end{aligned}$$
$$\begin{aligned} P_{19}(t)=P_{16}(t)+t^{17}P_{2}(t), \end{aligned}$$
$$\begin{aligned} P_{16}(t)={}&3{,}309{,}224{,}024{,}069{,}080{,}418{,}989{,}754{,}522{,}912{,}085{,}339{,}870{,}997{,}824{,}000t^{16} \\ &{} +\cdots\\ &{}+229{,}442{,}535{,}851{,}108{,}636{,}620{,}015{,}850{,}036{,}920{,}320{,}000{,}000{,}000{,}000{,}000{,}000 \end{aligned}$$
$$\begin{aligned} P_{2}(t)={}&77{,}541{,}624{,}086{,}159{,}498{,}428{,}020{,}328{,}992{,}837{,}339{,}887{,}447{,}064{,}000 \\ &{} -565{,}686{,}157{,}207{,}722{,}134{,}655{,}870{,}693{,}904{,}642{,}976{,}763{,}301{,}024t \\ &{} -54{,}119{,}091{,}759{,}561{,}776{,}058{,}592{,}767{,}712{,}571{,}305{,}215{,}681{,}649t^{2}>0 \end{aligned}$$
Theorem 3.3
For
\(x>0\),
$$\begin{aligned} &\frac{1+\frac{63}{130}x^{4}-\frac{139}{6240}x^{8}}{1+\frac {33}{52}x^{4}}< \frac{\operatorname {arctl}x}{x}< \frac {29{,}490{,}240+24{,}662{,}664x^{4}+2{,}828{,}975x^{8}}{15(1{,}966{,}016+1{,}939{,}080x^{4}+336{,}105x^{8})}. \end{aligned}$$
(3.3)
Proof
Consider the function Differentiation yields We then obtain Hence the first inequality in (3.3) holds for \(x>0\).
$$\begin{aligned} \lambda(x)=\operatorname {arctl}x-\frac{x(1+\frac{63}{130}x^{4}-\frac {139}{6240}x^{8})}{1+\frac{33}{52}x^{4}}. \end{aligned}$$
$$\begin{aligned} \lambda'(x)&=\frac{1}{(1+x^{4})^{3/4}}+\frac {x^{3}(48{,}672+14{,}456x^{4}+4587x^{8})}{30(52+33x^{4})^{2}}>0. \end{aligned}$$
$$\begin{aligned} \lambda(x)>\lambda(0)=0,\quad x>0. \end{aligned}$$
Consider the function Differentiation yields where
$$ T(x)=\operatorname {arctl}x-\frac {x(29{,}490{,}240+24{,}662{,}664x^{4}+2{,}828{,}975x^{8})}{15(1{,}966{,}016+1{,}939{,}080x^{4}+336{,}105x^{8})},\quad x>0. $$
$$\begin{aligned} T'(x)&=\frac{1}{(1+x^{4})^{3/4}}-\frac {P_{16}(x)}{(1{,}966{,}016+1{,}939{,}080x^{4}+336{,}105x^{8})^{2}}, \end{aligned}$$
$$\begin{aligned} P_{16}(x)={}&3{,}865{,}218{,}912{,}256+4{,}725{,}610{,}426{,}368x^{4}+1{,}899{,}763{,}315{,}008x^{8} \\ & {}+170{,}687{,}344{,}256x^{12}+63{,}388{,}842{,}825x^{16}. \end{aligned}$$
Elementary calculations reveal that where Hence, \(T'(x)<0\) for \(x>0\), and we have Hence, the second inequality in (3.3) holds for \(x>0\). The proof is complete. □
$$\begin{aligned} &\frac{1}{(1+x^{4})^{3}}- \biggl(\frac {P_{16}(x)}{(1{,}966{,}016+1{,}939{,}080x^{4}+336{,}105x^{8})^{2}} \biggr)^{4} \\ & \quad =-\frac{x^{20}P_{56}(x)}{(1+x^{4})^{3}(1{,}966{,}016+1{,}939{,}080x^{4}+336{,}105x^{8})^{8}}, \end{aligned}$$
$$\begin{aligned} P_{56}(x)={}&13{,}193{,}567{,}461{,}486{,}862{,}074{,}082{,}196{,}527{,}146{,}063{,}235{,}598{,}765{,}785{,}088 \\ &{}+75{,}159{,}817{,}580{,}420{,}162{,}914{,}879{,}309{,}165{,}363{,}929{,}497{,}102{,}849{,}146{,}880x^{4} \\ &{}+190{,}493{,}075{,}741{,}254{,}897{,}950{,}338{,}074{,}805{,}626{,}536{,}902{,}462{,}936{,}186{,}880x^{8} \\ &{}+283{,}233{,}781{,}637{,}227{,}052{,}608{,}425{,}496{,}608{,}925{,}420{,}321{,}925{,}549{,}260{,}800x^{12} \\ &{}+274{,}381{,}791{,}750{,}085{,}496{,}947{,}276{,}941{,}731{,}670{,}794{,}946{,}132{,}888{,}780{,}800x^{16} \\ &{}+182{,}271{,}787{,}590{,}701{,}615{,}339{,}301{,}540{,}193{,}194{,}594{,}557{,}235{,}390{,}578{,}688x^{20} \\ &+85{,}570{,}287{,}614{,}566{,}775{,}085{,}144{,}063{,}641{,}805{,}696{,}286{,}360{,}924{,}323{,}840x^{24} \\ &{}+29{,}163{,}131{,}006{,}055{,}200{,}534{,}183{,}374{,}987{,}447{,}946{,}919{,}333{,}657{,}968{,}640x^{28} \\ &{}+7{,}481{,}144{,}367{,}677{,}341{,}229{,}619{,}045{,}201{,}182{,}337{,}247{,}982{,}930{,}534{,}400x^{32} \\ &{}+1{,}502{,}545{,}339{,}351{,}309{,}468{,}552{,}186{,}115{,}563{,}901{,}082{,}330{,}882{,}201{,}600x^{36} \\ &{}+238{,}495{,}639{,}577{,}137{,}257{,}561{,}813{,}891{,}822{,}862{,}592{,}696{,}929{,}928{,}896x^{40} \\ &{}+29{,}999{,}531{,}147{,}567{,}967{,}753{,}948{,}099{,}263{,}441{,}234{,}315{,}939{,}560{,}800x^{44} \\ &{}+3{,}208{,}050{,}013{,}558{,}968{,}652{,}633{,}219{,}730{,}412{,}159{,}840{,}683{,}611{,}875x^{48} \\ &{}+222{,}336{,}558{,}000{,}169{,}152{,}230{,}844{,}556{,}985{,}454{,}831{,}178{,}171{,}875x^{52} \\ &{}+16{,}145{,}492{,}412{,}888{,}980{,}411{,}169{,}048{,}998{,}579{,}532{,}875{,}390{,}625x^{56}. \end{aligned}$$
$$\begin{aligned} T(x)< T(0)=0,\quad x>0. \end{aligned}$$
Theorem 3.4
For
\(0< x<1\),
$$\begin{aligned} &\frac {29{,}490{,}240-24{,}662{,}664x^{4}+2{,}828{,}975x^{8}}{15(1{,}966{,}016-1{,}939{,}080x^{4}+336{,}105x^{8})}< \frac {\operatorname {arctlh}x}{x}. \end{aligned}$$
(3.4)
Proof
Consider the function Differentiation yields where
$$ H(x)=\operatorname {arctlh}x-\frac {x(29{,}490{,}240-24{,}662{,}664x^{4}+2{,}828{,}975x^{8})}{15(1{,}966{,}016-1{,}939{,}080x^{4}+336{,}105x^{8})},\quad 0< x< 1. $$
$$\begin{aligned} H'(x)&=\frac{1}{(1-x^{4})^{3/4}}-\frac {Q(x^{4})}{(1{,}966{,}016-1{,}939{,}080x^{4}+336{,}105x^{8})^{2}}, \end{aligned}$$
$$\begin{aligned} Q(t)={}&3{,}865{,}218{,}912{,}256-4{,}725{,}610{,}426{,}368t+1{,}899{,}763{,}315{,}008t^{2} \\ &{} -170{,}687{,}344{,}256t^{3}+63{,}388{,}842{,}825t^{4},\quad 0< t< 1. \end{aligned}$$
Elementary calculations reveal that where Since \(R(t)>0\) for \(0< t<1\), we have \(H'(x)>0\) for \(0< x<1\). We then obtain The proof is complete. □
$$\begin{aligned} &\frac{1}{(1-t)^{3}}- \biggl(\frac {Q(t)}{(1{,}966{,}016-1{,}939{,}080t+336{,}105t^{2})^{2}} \biggr)^{4} \\ &\quad =\frac{t^{5}R(t)}{(1-t)^{3}(1{,}966{,}016-1{,}939{,}080t+336{,}105t^{2})^{8}}, \end{aligned}$$
$$\begin{aligned} R(t)={}&301{,}748{,}693{,}573{,}399{,}407{,}094{,}173{,}717{,}482{,}883{,}533{,}487{,}653{,}121 \\ &{}+9{,}932{,}615{,}535{,}811{,}554{,}413{,}076{,}061{,}553{,}884{,}665{,}646{,}471{,}005{,}365(1-t) \\ &{}+151{,}766{,}449{,}766{,}787{,}034{,}704{,}161{,}483{,}173{,}419{,}604{,}277{,}111{,}004{,}855(1-t)^{2} \\ &{}+680{,}433{,}563{,}535{,}649{,}162{,}659{,}902{,}808{,}405{,}182{,}093{,}255{,}793{,}778{,}810(1-t)^{3} \\ &{}+1856{,}390{,}570{,}444{,}186{,}005{,}047{,}799{,}006{,}039{,}729{,}435{,}155{,}242{,}856{,}245(1-t)^{4} \\ &{}+3{,}124{,}371{,}679{,}128{,}783{,}209{,}196{,}299{,}976{,}215{,}123{,}075{,}149{,}799{,}026{,}971(1-t)^{5} \\ &{}+3{,}542{,}098{,}875{,}374{,}455{,}780{,}446{,}672{,}503{,}755{,}643{,}630{,}859{,}189{,}032{,}595(1-t)^{6} \\ &{}+2{,}471{,}319{,}403{,}553{,}122{,}128{,}066{,}406{,}333{,}619{,}269{,}039{,}662{,}418{,}255{,}020(1-t)^{7} \\ &{}+1{,}090{,}939{,}975{,}419{,}395{,}315{,}836{,}646{,}390{,}524{,}713{,}606{,}873{,}688{,}710{,}195(1-t)^{8} \\ &{}+188{,}439{,}516{,}872{,}719{,}220{,}883{,}777{,}738{,}283{,}311{,}857{,}263{,}725{,}003{,}515(1-t)^{9} \\ &{}+72{,}805{,}480{,}166{,}035{,}035{,}195{,}735{,}976{,}881{,}949{,}595{,}398{,}022{,}003{,}221(1-t)^{10} \\ &{}+2{,}968{,}223{,}270{,}581{,}948{,}926{,}689{,}804{,}107{,}877{,}843{,}092{,}991{,}437{,}050(1-t)^{11} \\ &{}+ \bigl(1{,}786{,}914{,}569{,}129{,}666{,}891{,}048{,}623{,}948{,}471{,}984{,}527{,}027{,}924{,}375 \\ &{} -3{,}700{,}335{,}780{,}276{,}573{,}525{,}522{,}128{,}994{,}658{,}629{,}077{,}296{,}875(1-t) \bigr) (1-t)^{12} \\ &{}+16{,}145{,}492{,}412{,}888{,}980{,}411{,}169{,}048{,}998{,}579{,}532{,}875{,}390{,}625(1-t)^{14}. \end{aligned}$$
$$\begin{aligned} H(x)>H(0)=0, \quad 0< x< 1. \end{aligned}$$
4 Proof of Conjecture 1.1
Proof of (1.12)
It suffices to show by (3.2) that
i.e.,
$$\begin{aligned} \frac{265{,}200+214{,}500x^{4}+23{,}623x^{8}}{15(17{,}680+16{,}068x^{4}+2445x^{8})}< \frac {95+\frac{931}{2925}x^{12}}{80+\sqrt{225+285x^{4}}},\quad x>0, \end{aligned}$$
$$\begin{aligned} \frac{95+\frac{931}{2925}x^{12}}{\frac {265{,}200+214{,}500x^{4}+23{,}623x^{8}}{15(17{,}680+16{,}068x^{4}+2445x^{8})}}-80>\sqrt {225+285x^{4}},\quad x>0. \end{aligned}$$
(4.1)
Elementary calculations show that where We see from \(P_{24}(x)>0\) that (4.1) holds. The proof is complete. □
$$\begin{aligned} & \biggl(\frac{95+\frac{931}{2925}x^{12}}{\frac {265{,}200+214{,}500x^{4}+23{,}623x^{8}}{15(17{,}680+16{,}068x^{4}+2445x^{8})}}-80 \biggr)^{2}- \bigl( \sqrt{225+285x^{4}} \bigr)^{2} \\ & \quad =\frac{x^{16}P_{24}(x)}{(265{,}200+214{,}500x^{4}+23{,}623x^{8})^{2}}, \end{aligned}$$
$$\begin{aligned} P_{24}(x)={}&2{,}214{,}994{,}396{,}680{,}000+2{,}167{,}794{,}625{,} 751{,}625x^{4}\\ &{}+771{,}850{,}648{,}332{,}400x^{8} \\ & {}+100{,}410{,}388{,}038{,}870x^{12}+15{,}721{,}941{,}655{,}056x^{16} +3{,}584{,}399{,}789{,}880x^{20} \\ &{} +272{,}711{,}522{,}475x^{24}. \end{aligned}$$
Proof of (1.13)
First of all, we prove the second inequality in (1.13). It suffices to show by the right-hand side of (3.3) that
i.e.,
$$ \frac {29{,}490{,}240+24{,}662{,}664x^{4}+2{,}828{,}975x^{8}}{15(1{,}966{,}016+1{,}939{,}080x^{4}+336{,}105x^{8})}< \frac {1210+\frac{2{,}078{,}417}{280{,}800}x^{12}}{940+9\sqrt{900+1210x^{4}}},\quad x>0, $$
$$ \frac{1210+\frac{2{,}078{,}417}{280{,}800}x^{12}}{\frac {29{,}490{,}240+24{,}662{,}664x^{4}+2{,}828{,}975x^{8}}{15(1{,}966{,}016+1{,}939{,}080x^{4}+336{,}105x^{8})}}-940>9\sqrt {900+1210x^{4}},\quad x>0. $$
(4.2)
Elementary calculations show that where We see from \(P_{40}(x)>0\) that (4.2) holds. Hence, the second inequality in (1.13) holds.
$$\begin{aligned} & \biggl(\frac{1210+\frac{2{,}078{,}417}{280{,}800}x^{12}}{\frac {29{,}490{,}240+24{,}662{,}664x^{4}+2{,}828{,}975x^{8}}{15(1{,}966{,}016+1{,}939{,}080x^{4}+336{,}105x^{8})}}-940 \biggr)^{2}- \bigl(9 \sqrt{900+1210x^{4}} \bigr)^{2} \\ &\quad =\frac{P_{40}(x)}{350{,}438{,}400(29{,}490{,}240+24{,}662{,}664x^{4}+2{,}828{,}975x^{8})^{2}}, \end{aligned}$$
$$\begin{aligned} P_{40}(x)={}&423{,}992{,}204{,}507{,}234{,}653{,}175{,}808{,}000{,}000\\ &{}+813{,}161{,}362{,}018{,}201{,}812{,}231{,}782{,}400{,}000x^{4} \\ &{}+516{,}869{,}957{,}972{,}853{,}387{,}975{,}720{,}960{,}000x^{8}\\ &{}+125{,}747{,}707{,}439{,}033{,}797{,}403{,}639{,}808{,}000x^{12} \\ &{}+18{,}812{,}522{,}309{,}950{,}882{,}139{,}627{,}520{,}000x^{16}\\ &{}+6{,}902{,}175{,}873{,}182{,}801{,}970{,}021{,}120{,}000x^{20} \\ &{}+1{,}857{,}663{,}393{,}190{,}652{,}946{,}224{,}195{,}584x^{24}\\ &{}+192{,}485{,}925{,}752{,}231{,}924{,}989{,}587{,}840x^{28} \\ &{}+21{,}951{,}612{,}856{,}626{,}590{,}316{,}104{,}640x^{32}\\ &{}+5{,}630{,}747{,}696{,}194{,}377{,}041{,}485{,}200x^{36} \\ &{}+487{,}994{,}939{,}463{,}408{,}187{,}266{,}225x^{40}. \end{aligned}$$
Second, we prove the first inequality in (1.13). We consider two cases.
Case 1. \(0< x<1\).
It suffices to show by the left-hand side of (3.3) that
i.e.,
$$\begin{aligned} &\frac{1210}{940+9\sqrt{900+1210x^{4}}}< \frac{1+\frac {63}{130}x^{4}-\frac{139}{6240}x^{8}}{1+\frac{33}{52}x^{4}},\quad 0< x< 1, \end{aligned}$$
$$\begin{aligned} 9\sqrt{900+1210x^{4}}>\frac{1210}{\frac{1+\frac{63}{130}x^{4}-\frac {139}{6240}x^{8}}{1+\frac{33}{52}x^{4}}}-940,\quad 0< x< 1. \end{aligned}$$
(4.3)
Elementary calculations show that which shows that (4.3) holds.
$$\begin{aligned} & \bigl(9\sqrt{900+1210x^{4}} \bigr)^{2}- \biggl( \frac{1210}{\frac {1+\frac{63}{130}x^{4}-\frac{139}{6240}x^{8}}{1+\frac {33}{52}x^{4}}}-940 \biggr)^{2} \\ & \quad =\frac {1210x^{12}(128{,}621{,}376-81{,}039{,}502x^{4}+1{,}565{,}001x^{8})}{(6240+3024x^{4}-139x^{8})^{2}}>0,\quad 0< x< 1, \end{aligned}$$
Case 2. \(x\geq1\).
Consider the function \(U(x)\) defined by Differentiation yields Noting that holds, we obtain
$$ U(x)=\operatorname {arctl}x-\frac{1210x}{940+9\sqrt{900+1210x^{4}}}. $$
$$\begin{aligned} U'(x)&=\frac{1}{(1+x^{4})^{3/4}}+\frac{12{,}100(1089x^{4}-94\sqrt {900+1210x^{4}}-810)}{(940+9\sqrt{900+1210x^{4}})^{2}\sqrt{900+1210x^{4}}}. \end{aligned}$$
(4.4)
$$\begin{aligned} 1089x^{4}-94\sqrt{900+1210x^{4}}-810>0,\quad x\geq2, \end{aligned}$$
$$\begin{aligned} U'(x)>0,\quad x\geq2. \end{aligned}$$
We now show that \(U'(x)>0\) is also valid for \(1\leq x<2\). It suffices to show that where with and
$$\begin{aligned} y(x)>0, \quad 1\leq x< 2, \end{aligned}$$
$$\begin{aligned} y(x)=y_{1}(x)+y_{2}(x), \end{aligned}$$
$$\begin{aligned} y_{1}(x)=\frac{(940+9\sqrt{900+1210x^{4}})^{2}\sqrt {900+1210x^{4}}}{12{,}100(1+x^{4})^{3/4}}+1089x^{4}-810 \end{aligned}$$
$$\begin{aligned} y_{2}(x)=-94\sqrt{900+1210x^{4}}. \end{aligned}$$
Differentiation yields where Hence, we have \(y'_{1}(x)>0\) for \(1\leq x<2\).
$$\begin{aligned} y'_{1}(x)=\frac{x^{3}(940+9\sqrt{900+1210x^{4}})y_{3}(x)}{1210 (1+x^{4} )^{7/4}\sqrt{900+1210x^{4}}}+32{,}596x^{3}, \end{aligned}$$
$$\begin{aligned} y_{3}(x)={}&4104\sqrt{900+1210x^{4}}+3267x^{4}\sqrt {900+1210x^{4}}-113{,}740x^{4}-26{,}320 \\ >{}&4104\sqrt{1210x^{4}}+3267x^{4}\sqrt{1210x^{4}}-113{,}740x^{4}-26{,}320 \\ ={}&81{,}081\sqrt{10}-140{,}060+(305{,}910\sqrt{10}-454{,}960) (x-1) \\ &{} +(584{,}199\sqrt{10}-682{,}440) (x-1)^{2}+(718{,}740\sqrt {10}-454{,}960) (x-1)^{3} \\ &{} +(539{,}055\sqrt{10}-113{,}740) (x-1)^{4}+215{,}622\sqrt {10}(x-1)^{5}+35{,}937 \sqrt{10}(x-1)^{6} \\ >{}&0 \quad\text{for } 1\leq x< 2. \end{aligned}$$
Let \(1\leq r \leq x \leq s \leq2\). Since \(y_{1}(x)\) is increasing and \(y_{2}(x)\) is decreasing for \(1\leq x\leq2\), we obtain We divide the interval \([1, 2]\) into 100 subintervals: By direct computation we get Hence, This proves \(U'(x)>0\) for \(1\leq x<2\).
$$\begin{aligned} y(x)\geq y_{1}(r)+y_{2}(s)=:\sigma_{1}(r,s). \end{aligned}$$
$$\begin{aligned}{} [1, 2]=\bigcup_{k=0}^{99} \biggl[1+ \frac{k}{100}, 1+\frac{k+1}{100} \biggr]>0 \quad\text{for } k = 0, 1, 2, \ldots, 99. \end{aligned}$$
$$\begin{aligned} \sigma_{1} \biggl(1+\frac{k}{100}, 1+\frac{k+1}{100} \biggr)>0 \quad\text{for } k = 0, 1, 2, \ldots, 99. \end{aligned}$$
$$\begin{aligned} y(x)>0\quad \text{for } x\in \biggl[1+\frac{k}{100}, 1+\frac{k+1}{100} \biggr] \text{ and } k = 0, 1, 2, \ldots, 99. \end{aligned}$$
Acknowledgements
The authors thank the referees for helpful comments.
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.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors read and approved the final manuscript.