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.

Hint. Consider the derivative of $$\arctan x-\frac{x}{1+x^2},$$ namely $$\frac{2x}{(1+x^2)^2},$$ which is positive for all $x>0.$ Thus, the difference is increasing as $x\to+\infty.$ Since the difference is $0$ when $x=0,$ it follows that the difference is positive for all $x>0,$ and the result immediately follows.
