Q:

Proving that if $x>0$ then $\frac{x-(x^2+1)\arctan(x)}{x^2(x^2+1)}$ is less than $0$

How do I show that for $x>0$:
$$\frac{x-(x^2+1)\arctan(x)}{x^2(x^2+1)} < 0$$
I tried to do it somehow using the fact that $$\frac{\arctan(x)}{x} < 1$$ but still didn't figure it out...

A:

We have $$\begin{align}\frac{x-(x^2+1)\arctan x}{x^2(x^2+1)} < 0&\iff x<(x^2+1)\arctan x\\&\iff (1+x^2)\arctan x-x>0\end{align}$$ This is true iff $$\frac d{dx}\left[(1+x^2)\arctan x-x\right]=2x\arctan x+\frac{1+x^2}{1+x^2}-1=2x\arctan x>0$$ which is true since both $x$ and $\arctan x$ are greater than $0$ when $x>0$. Hence result.

