Claim :
Let $0\leq x$ then prove (without using derivative) that :
$$\tanh^{-1}\left(\sqrt{\tanh(x)}\right)-\tan^{-1}\left(\sqrt{\tanh(x)}\right)\geq 0$$
Trick :
We have using integral :
$$\tanh^{-1}\left(\sqrt{\tanh(x)}\right)-\tan^{-1}\left(\sqrt{\tanh(x)}\right)=\int_{}^{}\sqrt{\tanh(x)}\,dx$$
So we have :
$$0\leq \int_{0}^{a}\sqrt{\tanh(x)}\,dx=\tanh^{-1}\left(\sqrt{\tanh(a)}\right)-\tan^{-1}\left(\sqrt{\tanh(a)}\right)$$
I'm really curious to see another proof :
Question :
Have you an alternative proof without derivative ?
Thanks in advance !