1 Introduction

In 2010, Masjed-Jamei [1] studied the relation of inverse tangent function \(\arctan x\) and inverse hyperbolic sine function \(\sinh ^{-1}(x)\) and proved an inequality as follows.

$$\begin{aligned} \left( \arctan x\right) ^2\leqslant \frac{x\sinh ^{-1}(x)}{\sqrt{1+x^2}},\ \ x\in (-1,1). \end{aligned}$$
(1.1)

The study related to (1.1) attracted much attention in last decade. At first, Zhu and Males̆ević [3] proved that (1.1) holds for any \(x\in (-\infty , +\infty )\). They also obtained some refinements of (1.1).

Proposition 1.1

[3, Theorem 1.3] For any \(x\in (-\infty ,+\infty )\), we have

$$\begin{aligned}{} & {} -\frac{1}{45}x^6\leqslant \left( \arctan x\right) ^2-\frac{x\sinh ^{-1} x}{\sqrt{1+x^2}}\leqslant -\frac{1}{45}x^6+\frac{4}{105}x^8, \end{aligned}$$
(1.2)
$$\begin{aligned}{} & {} -\frac{1}{45}x^6+\frac{4}{105}x^8-\frac{11}{225}x^{10}\leqslant \left( \arctan x\right) ^2-\frac{x\sinh ^{-1} x}{\sqrt{1+x^2}}\nonumber \\{} & {} \qquad \leqslant -\frac{1}{45}x^6+\frac{4}{105}x^8-\frac{11}{225}x^{10}+\frac{586}{10395}x^{12}. \end{aligned}$$
(1.3)

Define

$$\begin{aligned} v_n=\frac{1}{n}\left[ \frac{n!2^{n-1}}{(2n-1)!!}-\left( 1+\frac{1}{3}+\cdots +\frac{1}{2n-1}\right) \right] , n\geqslant 3. \end{aligned}$$
(1.4)

By using flexible analysis tools, Zhu and Males̆ević [4] extended (1.2) and (1.3) to general form as follows.

Proposition 1.2

[4, Theorem 1.1] For any \(x\in (-\infty ,+\infty )\), we have

$$\begin{aligned} \sum _{n=3}^{2m+1}(-1)^nv_nx^{2n}\leqslant \left( \arctan x\right) ^2-\frac{x\sinh ^{-1} x}{\sqrt{1+x^2}}\leqslant \sum _{n=3}^{2m+2}(-1)^nv_nx^{2n}. \end{aligned}$$
(1.5)

Proposition 1.3

[5, Theorem 2.1] The double inequality

$$\begin{aligned} \frac{x\sinh ^{-1}(x)}{\sqrt{1+x^2+\frac{1}{45}x^2}}<\left( \arctan x\right) ^2<\frac{x\sinh ^{-1}(x)}{\sqrt{1+x^2}} \end{aligned}$$
(1.6)

holds for any \(x\in (0,+\infty )\) with best constants 0 and 1/45.

Please see [6, 7] for more generalizations.

Motivated by (1.1)-(1.6), Zhu and Males̆ević [3] also studied the relation of inverse hyperbolic tangent function \(\tanh ^{-1}(x)\) and inverse sine function \(\arcsin x\) as follows.

Proposition 1.4

[3, Theorem 1.4] The inequality

$$\begin{aligned} \left[ \tanh ^{-1}(x)\right] ^2<\frac{x\arcsin x}{\sqrt{1-x^2}} \end{aligned}$$
(1.7)

holds for any \(x\in (0,1)\) with the the best power number 2.

Proposition 1.5

[3, Theorem 1.6] The inequality

$$\begin{aligned} \frac{x\arcsin x}{\sqrt{1-x^2}}-\left[ \tanh ^{-1}(x)\right] ^2<\sum _{n=3}^{N}v_nx_{2n} \end{aligned}$$
(1.8)

holds for any \(x\in (0,1)\).

Moreover, by investigating the power series of the following function

$$\begin{aligned}\frac{\left[ \tanh ^{-1}(x)\right] ^2}{\dfrac{\arcsin x}{\sqrt{1-x^2}}}=x-\frac{1}{45}x^5-\frac{22}{945}x^7-\frac{61}{2835}x^9+O(x^{10}), \end{aligned}$$

L. Zhu [2] obtained the following interesting double inequality of Masjed-Jamei type.

Proposition 1.6

[2, Theorem 1] The double inequality

$$\begin{aligned} \frac{\left( x-x^5\right) \arcsin x}{\sqrt{1-x^2}}<\left[ \tanh ^{-1}(x)\right] ^2<\frac{\left( x-\dfrac{1}{45}x^5\right) \arcsin x}{\sqrt{1-x^2}} \end{aligned}$$
(1.9)

holds for any \(x\in (0,1)\) with best constants \(-1\) and \(-\frac{1}{45}\).

We gave a now proof of (1.9) in [8] and provided a refinement in [9].

Proposition 1.7

[9, Theorem 1] The double inequality

$$\begin{aligned} \frac{\left( x-\dfrac{1}{45}x^5-\dfrac{44}{45}x^7\right) \arcsin x}{\sqrt{1-x^2}}<\left[ \tanh ^{-1}(x)\right] ^2<\frac{\left( x-\dfrac{1}{45}x^5-\dfrac{22}{945}x^7\right) \arcsin x}{\sqrt{1-x^2}}\nonumber \\ \end{aligned}$$
(1.10)

holds for any \(x\in (0,1)\) with best constants \(-\frac{44}{45}\) and \(-\frac{22}{945}\).

The goal of this paper is to prove a new lower bound of \(\left[ \tanh ^{-1}(x)\right] ^2\), which is a conjecture of Chen-Males̆ević [5, Conjecture 2.1].

Theorem 1.8

If \(x\in (0,1)\), then

$$\begin{aligned} \frac{x\arcsin x}{\left( 1-\frac{41}{45}x^2\right) ^{45/82}}<\left[ \tanh ^{-1}(x)\right] ^2. \end{aligned}$$
(1.11)

Remark 1.9

  1. (1)

    Numerical experiments show that inequality (1.11) is stronger than the left-hand side inequality of (1.10).

  2. (2)

    L. Zhu [10] claimed that he have proved Chen-Males̆ević’s conjecture. Unfortunately, his proof in [10] is false.

2 Proof of theorem 1.8

Lemma 2.1

Let

$$\begin{aligned}f(x)=\frac{\left[ \tanh ^{-1}(x)\right] ^2\left( 1-\frac{41}{45}x^2\right) ^{45/82}}{x}-\arcsin x,\end{aligned}$$

then f(x) is strictly increasing on (0, 1).

Proof

Step 1: Let \(t=\tanh ^{-1}(x)\in (0,+\infty )\), then \(x=\tanh (t)\). Define

$$\begin{aligned}F(t):=f(\tanh t)=\frac{t^2\left( 1-\frac{41}{45}\tanh ^2 t\right) ^{45/82}}{\tanh t}-\arcsin (\tanh t).\end{aligned}$$

In order to prove that f(x) is strictly increasing on (0, 1), we only need to prove F(t) is strictly increasing on \((0,+\infty )\).

Step 2: By direct computation, we have

$$\begin{aligned}&F^{\prime }(t)\cdot \tanh ^2t=\left[ 2t\left( 1-\frac{41}{45}\tanh ^2 t\right) ^{45/82}+\frac{45}{82}t^2\left( 1-\frac{41}{45}\tanh ^2 t\right) ^{-37/82}\right. \\&\qquad \left. \cdot \left( -\frac{82}{45}\tanh t\cdot \frac{1}{\cosh ^2t}\right) \right] \tanh t-t^2\left( 1-\frac{41}{45}\tanh ^2 t\right) ^{45/82}\\&\qquad \cdot \frac{1}{\cosh ^2t}-\frac{1}{\cosh t}\cdot \tanh ^2 t\\&\quad =2t\left( 1-\frac{41}{45}\tanh ^2 t\right) ^{45/82}\tanh t-t^2\left( 1-\frac{41}{45}\tanh ^2 t\right) ^{-37/82}\cdot \tanh ^2 t\cdot \frac{1}{\cosh ^2t}\\&\qquad -t^2\left( 1-\frac{41}{45}\tanh ^2 t\right) ^{45/82} \cdot \frac{1}{\cosh ^2t}-\frac{1}{\cosh t}\cdot \tanh ^2 t. \end{aligned}$$

Then

$$\begin{aligned} \varphi (t):&=F^{\prime }(t)\cdot \tanh ^2t\cdot \left( 1-\frac{41}{45}\tanh ^2 t\right) ^{37/82}\cdot \cosh ^4t\\&=2t\left( \sinh t\cosh ^3t-\frac{41}{45}\sinh ^3 t\cosh t\right) -t^2\left( \cosh ^2t+\frac{4}{45}\sinh ^2 t\right) \\&\quad -\sinh ^2t\cosh t\left( 1-\frac{41}{45}\tanh ^2 t\right) ^{37/82}. \end{aligned}$$

In order to prove F(t) is strictly increasing on \((0,+\infty )\), it is suffice to prove \(F^{\prime }(t)>0\) for \(t\in (0,+\infty )\), which is equivalent to \(\varphi (t)>0\) for \(t\in (0,+\infty )\).

Step 3: Denote

$$\begin{aligned} \varphi _1(t)&=2t\left( \sinh t\cosh ^3t-\frac{41}{45}\sinh ^3t\cosh t\right) -t^2\left( \cosh ^2t+\frac{4}{45}\sinh ^2 t\right) ,\\ \varphi _2(t)&=\sinh ^2t\cosh t\left( 1-\frac{41}{45}\tanh ^2 t\right) ^{37/82}, \end{aligned}$$

then

$$\begin{aligned}\varphi (t)=\varphi _1(t)-\varphi _2(t).\end{aligned}$$

Obviously, \(\varphi _2(t)>0\) on \((0,+\infty )\). And for any \(t\in (0,+\infty )\),

$$\begin{aligned} \varphi _1(t)&=\frac{1}{45}t\left( \sinh (4t)+43\sinh (2t)\right) -\frac{1}{90}t^2\left( 49\cosh (2t)+41\right) \\&=\frac{1}{90}\left[ 2t\left( \sum _{n=0}^{\infty }\frac{(4t)^{2n+1}}{(2n+1)!}+43\sum _{n=0}^{\infty }\frac{(2t)^{2n+1}}{(2n+1)!}\right) -t^2\left( 49\sum _{n=0}^{\infty }\frac{(2t)^{2n}}{(2n)!}+41\right) \right] \\&=t^2+\frac{1}{90}\sum _{n=1}^{\infty }\left( \frac{2\cdot 4^{2n+1}+86\cdot 2^{2n+1}}{(2n+1)!}-\frac{49\cdot 2^{2n}}{(2n)!}\right) t^{2n+2}\\&>0, \end{aligned}$$

since for any \(n\geqslant 1\), we have

$$\begin{aligned}\frac{2\cdot 4^{2n+1}+86\cdot 2^{2n+1}}{(2n+1)!}-\frac{49\cdot 2^{2n}}{(2n)!}>0.\end{aligned}$$

Step 4: Define

$$\begin{aligned}\psi _1(t)=\ln \varphi _1(t), \ \ \psi _2(t)=\ln \varphi _2(t)\end{aligned}$$

and

$$\begin{aligned}\psi (t)=\psi _1(t)-\psi _2(t),\end{aligned}$$

then \(\varphi (t)>0\) on \((0,+\infty )\) is equivalent to \(\psi (t)>0\) on \((0,+\infty )\). Since

$$\begin{aligned}\lim _{t\rightarrow 0^+}\psi (t)=\lim _{t\rightarrow 0^+}\ln \frac{\varphi _1(t)}{\varphi _2(t)}=0,\end{aligned}$$

it is suffice to prove that \(\psi ^{\prime }(t)>0\) on \((0,+\infty )\).

Step 5: From

$$\begin{aligned} \psi _1(t)&=\ln \varphi _1(t)\\&=\ln \left[ 2t\left( \sinh t\cosh ^3t-\frac{41}{45}\sinh ^3 t\cosh t\right) -t^2\left( \cosh ^2t+\frac{4}{45}\sinh ^2 t\right) \right] \\&=\ln \left[ \frac{1}{45}t\left( \sinh (4t)+43\sinh (2t)\right) -\frac{1}{90}t^2\left( 49\cosh (2t)+41\right) \right] \end{aligned}$$

and

$$\begin{aligned}\psi _2(t)=\ln \varphi _2(t)=2\ln \sinh t+\ln \cosh t+\frac{37}{82}\ln \left( 1-\frac{41}{45}\tanh ^2 t\right) ,\end{aligned}$$

we get

$$\begin{aligned} \psi _1^{\prime }(t)&=\frac{\frac{1}{45}\left( \sinh (4t)+43\sinh (2t)\right) +\frac{1}{45}t\left( 4\cosh (4t)+86\cosh (2t)\right) -\frac{1}{45}t\left( 49\cosh (2t)+41\right) -\frac{4}{45}t^2\sinh (2t)}{\frac{1}{45}t\left( \sinh (4t)+43\sinh (2t)\right) -\frac{1}{90}t^2\left( 49\cosh (2t)+41\right) }\\&=\frac{\left( \sinh (4t)+43\sinh (2t)\right) +t\left( 4\cosh (4t)+37\cosh (2t)-41\right) -49t^2\sinh (2t)}{t\left( \sinh (4t)+43\sinh (2t)\right) -\frac{1}{2}t^2\left( 49\cosh (2t)+41\right) } \end{aligned}$$

and

$$\begin{aligned} \psi _2^{\prime }(t)&=\frac{2\cosh t}{\sinh t}+\frac{\sinh t}{\cosh t}+\frac{37}{82}\cdot \frac{-\frac{82}{45}\tanh t\cdot \frac{1}{\cosh ^2t}}{\left( 1-\frac{41}{45}\tanh ^2 t\right) }\\&=\frac{2\cosh t}{\sinh t}+\frac{\sinh t}{\cosh t}-\frac{37\tanh t}{4\cosh ^2t+41}\\ {}&=\frac{\left( 2\cosh ^2 t+\sinh ^2t\right) \left( 4\cosh ^2t+41\right) -37\sinh ^2t}{\sinh t\cosh t\left( 4\cosh ^2t+41\right) }\\ {}&=\frac{\left( \cosh (2t)+1 +\frac{\cosh (2t)-1}{2}\right) \left( 2\cosh (2t)+43\right) -37\frac{\cosh (2t)-1}{2}}{\sinh t\cosh t\left( 2\cosh (2t)+43\right) }\\&=\frac{\left( 3\cosh (2t)+1\right) \left( 2\cosh (2t)+43\right) -37\cosh (2t)+37}{\sinh (4t)+43\sinh (2t)}\\&=\frac{3\cosh (4t)+94\cosh (2t)+83}{\sinh (4t)+43\sinh (2t)}. \end{aligned}$$

Then

$$\begin{aligned} \psi ^{\prime }(t)&=\psi _1^{\prime }(t)-\psi _2^{\prime }(t)\\&=\frac{A_2t^2+A_1t+A_0}{45\varphi _1(t)\left[ \sinh (4t)+43\sinh (2t)\right] }, \end{aligned}$$

where

$$\begin{aligned} A_0&=\left( \sinh (4t)+43\sinh (2t)\right) ^2\\&=\sinh ^2(4t)+86\sinh (4t)\sinh (2t)+1849\sinh ^2(2t)\\&=\frac{\cosh (8t)-1}{2}+43\left( \cosh (6t)-\cosh (2t)\right) +1849\cdot \frac{\cosh (4t)-1}{2}\\&=\frac{\cosh (8t)}{2}+43\cosh (6t)+\frac{1849}{2}\cosh (4t)-43\cosh (2t)-925\\&=\frac{1}{2}\sum _{n=0}^{\infty }\frac{(8t)^{2n}}{(2n!)}+43\sum _{n=0}^{\infty }\frac{(6t)^{2n}}{(2n!)}++\frac{1849}{2}\sum _{n=0}^{\infty }\frac{(4t)^{2n}}{(2n!)} -43\sum _{n=0}^{\infty }\frac{(2t)^{2n}}{(2n!)}-925, \end{aligned}$$
$$\begin{aligned} A_1&=\left( \sinh (4t)+43\sinh (2t)\right) [\left( 4\cosh (4t)+37\cosh (2t)-41\right) \\&\quad -\left( 3\cosh (4t)+94\cosh (2t)+83\right) ]\\&=\left( \sinh (4t)+43\sinh (2t)\right) \left( \cosh (4t)-57\cosh (2t)-124\right) \\&=\frac{\sinh (8t)}{2}-\frac{2451}{2}\sinh (4t)-\frac{57}{2}(\sinh (6t)+\sinh (2t))\\&\quad +\frac{43}{2}(\sinh (6t)-\sinh (2t))-124\left( \sinh (4t)+43\sinh (2t)\right) \\&=\frac{\sinh (8t)}{2}-7\sinh (6t)-\frac{2699}{2}\sinh (4t)-5382\sinh (2t)\\&=\frac{1}{2}\sum _{n=0}^{\infty }\frac{(8t)^{2n+1}}{(2n+1)!}-7\sum _{n=0}^{\infty }\frac{(6t)^{2n+1}}{(2n+1)!} -\frac{2699}{2}\sum _{n=0}^{\infty }\frac{(4t)^{2n+1}}{(2n+1)!}-5382\sum _{n=0}^{\infty }\frac{(2t)^{2n+1}}{(2n+1)!}, \end{aligned}$$

and

$$\begin{aligned} A_2&=\frac{1}{2}\left( 49\cosh (2t)+41\right) \left( 3\cosh (4t)+94\cosh (2t)+83\right) \\&\quad -49\sinh (2t)\left( \sinh (4t)+43\sinh (2t)\right) \\&=\frac{49}{4}\cosh (6t)+\frac{319}{2}\cosh (4t)+\frac{16087}{4}\cosh (2t)+\frac{7813}{2}\\&=\frac{49}{4}\sum _{n=0}^{\infty }\frac{(6t)^{2n}}{(2n!)}+\frac{319}{2}\sum _{n=0}^{\infty }\frac{(4t)^{2n}}{(2n!)}+\frac{16087}{4}\sum _{n=0}^{\infty }\frac{(2t)^{2n}}{(2n!)}+\frac{7813}{2}.\\ \end{aligned}$$

Step 6: Let

$$\begin{aligned}\theta (t):=A_2t^2+A_1t+A_0=\sum _{n=1}^{\infty }a_{2n+2}t^{2n+2},\end{aligned}$$

where

$$\begin{aligned} a_{2n+2}&=\frac{49}{4}\cdot \frac{6^{2n}}{(2n)!}+\frac{319}{2}\cdot \frac{4^{2n}}{(2n!)}+\frac{16087}{4}\cdot \frac{2^{2n}}{(2n!)}\\&\quad +\frac{1}{2}\cdot \frac{8^{2n+1}}{(2n+1)!}-7\cdot \frac{6^{2n+1}}{(2n+1)!}-\frac{2699}{2}\cdot \frac{4^{2n+1}}{(2n+1)!}-5382\cdot \frac{2^{2n+1}}{(2n+1)!}\\&\quad +\frac{1}{2}\frac{8^{2n+2}}{(2n+2)!}+43\frac{6^{2n+2}}{(2n+2)!}+\frac{1849}{2}\frac{4^{2n+2}}{(2n+2)!}-43\cdot \frac{2^{2n+2}}{(2n+2)!}. \end{aligned}$$

It is easy to check that

$$\begin{aligned}a_{2n+2}>0,\ \ 1\leqslant n\leqslant 5\end{aligned}$$

and

$$\begin{aligned} a_{2n+2}>&\frac{6^{2n}}{(2n)!}\cdot \left( \frac{49}{4}-\frac{42}{2n+1}\right) \\&+\frac{1}{2}\cdot \frac{4^{2n+1}}{(2n+1)!}\cdot \left( 2^{2n+1}-2699\right) \\&+\frac{2^{2n}}{(2n)!}\cdot \left( \frac{16087}{4}-\frac{10764}{2n+1}\right) \\ >&0 \end{aligned}$$

for any \(n\geqslant 6\). Therefore, \(\theta (t)>0\) for any \(t\in (0,+\infty )\), which implies

$$\begin{aligned}\psi ^{\prime }(t)>0\ \ \text {for any}\ \ t\in (0,+\infty ).\end{aligned}$$

The proof of Lemma 2.1 is completed. \(\square \)

Proof of theorem 1.8

By Lemma 2.1 and

$$\begin{aligned}\lim _{x\rightarrow 0^{+}}f(x)=0,\end{aligned}$$

we get \(f(x)>0\) for any \(x\in (0,1)\), which implies

$$\begin{aligned}\frac{x\arcsin x}{\left( 1-\frac{41}{45}x^2\right) ^{45/82}}<\left[ \tanh ^{-1}(x)\right] ^2.\end{aligned}$$

The proof is completed. \(\square \)