Prove : $\forall x>0$, $\arctan(x)>\frac{x}{1+x^2}$ I think the question is itself wrong as, $\frac{x}{1+x^2}$ is the derivative of $\arctan x$. If not, I don't know how to proceed.

Letting $x=\tan\theta$ for some $\theta\in\left(-\pi/2,\pi/2\right)$ we are left with
$$ 2\theta > \sin(2\theta) $$
which holds for any $\theta\in(0,\pi/2)$, by convexity.
