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

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.
