Inequality with $\arctan$ I try to show that $x \cdot \arctan\left( \frac{1}{x^2} \right)$ is monotonically decreasing, but I can't solve this inequality with $\arctan$. Can somebody show me how to do this?
$$
x \in [1, \infty)
$$
$$
\arctan\left( \dfrac{1}{x^2} \right) \leq \dfrac{2x^2}{x^4+1}
$$

Putting $x=1/y$, it is enough to show that, for $0 < y \leq 1$
\begin{equation*}
\arctan(y^2) \leq \frac{2y^2}{1+y^4}
\end{equation*}
If we let 
\begin{equation*}
f(y) = \frac{2y^2}{1+y^4} - \arctan(y^2) 
\end{equation*}
then 
\begin{equation*}
f'(y) = \frac{2y(1-3y^4)}{(1+y^4)^2}
\end{equation*}
$f'(y)=0$ when $y = 0$ or $y = 3^{-1/4}$. At $y=0$ $f(y) = 0$ and at $y = 3^{-1/4}, f(y) > 0$. Also $f(1) = 1-\frac{\pi}{4} > 0$. Thus it follows that $f(y) > 0$ for $0 < y \leq 1$.
