Prove that $\arctan(\operatorname{arctanh}(x))\geq \operatorname{arctanh}(\arctan(x))$ for $0\leq x <1$

Question

It's inspired by a general result proposed in a book (Dictionary of inequalities second edition by Peter Bullen p.101) related to composition of function :

In fact we have $x\in[0,1)$:

$$\arctan(\operatorname{arctanh}(x))\geq \operatorname{arctanh}(\arctan(x))\quad (1)$$

To prove it I have introduced a function :

$$f(x)=\arctan(\operatorname{arctanh}(x))-\operatorname{arctanh}(\arctan(x))$$

We differentiate to get:

$$f'(x)= -\frac{1}{((x^2-1)(\operatorname{arctanh}^2(x)+1))}+\frac{1}{((x^2+1)(\arctan^2(x)-1))}$$

And then I use :

Let $0\leq x <1$ then we have :

$$\operatorname{arctanh}(x)\geq x \geq \arctan(x)$$

Then I cannot conclude directly .

Question :

How to prove $(1)$ ?

Thanks in advance !

Answer 1

MacLaurin expansion $$\arctan\left(\operatorname{arctanh} x\right)=x+\frac{x^5}{15}+\frac{x^7}{45}+\frac{64 x^9}{2835}+\frac{71 x^{11}}{4725}+\frac{5209 x^{13}}{405405}+\frac{2203328 x^{15}}{212837625}+\ldots$$ All coefficients of the series are positive $$\operatorname{arctanh}\left(\arctan x\right)=x+\frac{x^5}{15}-\frac{x^7}{45}+\frac{64 x^9}{2835}-\frac{71 x^{11}}{4725}+\frac{5209 x^{13}}{405405}-\frac{2203328 x^{15}}{212837625}+\ldots$$ Coefficients have the same absolute value as the previous series, but alternating signs from $5$th degree on.

Thus for $x\in[0,1)$ $$\arctan\left(\operatorname{arctanh} x\right)\ge \operatorname{arctanh}\left(\arctan x\right)$$ while for $x\in(-1,0]$ the opposite happens.

Answer 2

Fact 1: $\frac{-3x^3+12x}{-x^3-x^2+12x+12}\sqrt{\frac{1+x}{1-x}} \ge \operatorname{arctanh}(x)$ for all $x \in [0, 1)$.
(Hint: Use $u^2 = \frac{1+x}{1-x}$ and then take derivative.)

Fact 2: $\frac{64x^5+735x^3+945x}{225x^4+1050x^2+945} \ge \arctan x$ for all $x \in [0, 1)$.
(Hint: Take derivative.)

Let $f(x) = \arctan(\operatorname{arctanh}(x)) - \operatorname{arctanh}(\arctan(x))$.

By Facts 1-2, we have, for all $x$ in $[0, 1)$, \begin{align} f'(x) &= \frac{1}{(1-x^2)(\operatorname{arctanh}^2(x)+1)} - \frac{1}{(x^2+1)(1 - \arctan^2(x))}\\ &\ge \frac{1}{(1-x^2)(\frac{(-3x^3+12x)^2}{(-x^3-x^2+12x+12)^2}\frac{1+x}{1-x}+1)} - \frac{1}{(x^2+1)(1 - (\frac{64x^5+735x^3+945x}{225x^4+1050x^2+945})^2)}\\ &\ge 0. \end{align} Also, $f(0) = 0$. Thus, $f(x) \ge 0$ for all $x$ in $[0, 1)$.

We are done.