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

Question

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...

Answer 1

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.

Answer 2

HINT: definie $$f(x)=\arctan(x)-\frac{x}{x^2+1}$$ and Show that $$f'(x)=\frac{2 x^2}{\left(x^2+1\right)^2}>0$$ for $x>0$

Answer 3

The question is equivalent to proving $\arctan(x)>\frac{x}{x^2+1}$ for any $x>0$, but this is trivial since $$ \arctan(x)=\int_{0}^{x}\frac{dt}{t^2+1}> \int_{0}^{x}\frac{dt}{x^2+1}=\frac{x}{x^2+1}.$$