In 2010, Masjed-Jamei [1] obtained the following inequality:
$$ ( \arctan x ) ^{2}\leq \frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}},\quad \vert x \vert < 1. $$
(1.1)
[1] also reminded us that the above inequality is established in a larger interval \((-\infty ,\infty )\) because it was detected by Maple software. Inequality (1.1) gives the upper bound for the square of the inverse tangent function arctanx by the inverse hyperbolic sine function \(\sinh ^{-1}x=\ln (x+\sqrt{1+x^{2}})\).
In this paper, we first affirm Masjed-Jamei’s quest, conclude that the scope of the inequality is indeed the large interval \((-\infty , \infty )\), and give a simple proof of this result. Second, we get the strengthening of the inequality that we have just given. Then, we obtain some natural generalizations of this inequality. At the same time, we show the analogue for inverse hyperbolic tangent function \(\operatorname{arctanh}x= ( 1/2 ) \ln ((1+x)/(1-x))\) and inverse sine function arcsinx. Finally, we propose a conjecture on this topic.
Theorem 1.1
The inequality
$$ ( \arctan x ) ^{2}\leq \frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}} $$
(1.2)
holds for all
\(x\in (-\infty ,\infty )\), and the power number 2 is the best in (1.2).
Theorem 1.2
Let
\(0< r<\infty \), \(\lambda = 1\), and
\(\mu =r\ln ( r+\sqrt{r^{2}+1} ) / ( \sqrt{r^{2}+1} ( \arctan r ) ^{2} ) \). Then the double inequality
$$ \lambda ( \arctan x ) ^{2}\leq \frac{x\ln (x+\sqrt{1+x ^{2}})}{\sqrt{1+x^{2}}}\leq \mu ( \arctan x ) ^{2} $$
(1.3)
holds for all
\(x\in (-r,r)\), where
λ
and
μ
are the best constants in (1.3).
Theorem 1.3
Let
\(-\infty <\)
\(x<\infty \). Then we have
$$\begin{aligned}& -\frac{1}{45}x^{6} \leq ( \arctan x ) ^{2}-\frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}}\leq -\frac{1}{45}x^{6}+ \frac{4}{105}x^{8}, \end{aligned}$$
(1.4)
$$\begin{aligned}& -\frac{1}{45}x^{6}+\frac{4}{105}x^{8}- \frac{11}{225}x^{10} \leq ( \arctan x ) ^{2}- \frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}} \\& \hphantom{-\frac{1}{45}x^{6}+\frac{4}{105}x^{8}-\frac{11}{225}x^{10}}\leq -\frac{1}{45}x^{6}+ \frac{4}{105}x^{8}-\frac{11}{225}x^{10}+ \frac{586}{10 {,}395}x^{12}. \end{aligned}$$
(1.5)
Theorem 1.4
The inequality
$$ ( \operatorname{arctanh}x ) ^{2}\leq \frac{x\arcsin x}{\sqrt{1-x ^{2}}} $$
(1.6)
holds for all
\(x\in (-1,1)\), and the power number 2 is the best in (1.6).
Theorem 1.5
Let
\(0< r<1\), \(\alpha = 1\), and
\(\beta =r ( \arcsin r ) / ( \sqrt{1-r^{2}} ( \operatorname{arctanh}r ) ^{2} ) \). Then the double inequality
$$ \alpha ( \operatorname{arctanh}x ) ^{2}\leq \frac{x\arcsin x}{\sqrt{1-x ^{2}}}\leq \beta ( \operatorname{arctanh}x ) ^{2} $$
(1.7)
holds for all
\(x\in (-r,r)\), where
α
and
β
are the best constants in (1.7).
Theorem 1.6
Let
n, N
be two integers, \(n,N\geq 3\), and
$$ v_{n}=\frac{1}{n} \biggl( \frac{n!2^{n-1}}{ ( 2n-1 ) !!}- \biggl( 1+\frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) \biggr) . $$
(1.8)
Then the inequality
$$ \frac{x\arcsin x}{\sqrt{1-x^{2}}}- ( \operatorname{arctanh}x ) ^{2}\geq \sum_{n=3}^{N}v_{n}x^{2n} $$
(1.9)
holds for all
\(x\in (-1,1)\).