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?
$$
x \in [1, \infty)
$$
$$
\arctan\left( \dfrac{1}{x^2} \right) \leq \dfrac{2x^2}{x^4+1}
$$
 A: 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 $0 < y \leq 1$.
A: It's a bit fiddly but here we go...
You need to prove
$$\arctan\Bigl(\frac1{x^2}\Bigr)\le\frac{2x^2}{1+x^4}\quad
  \hbox{for $x\ge1$}\ ,$$
that is,
$$\arctan\theta\le\frac{2\theta}{1+\theta^2}\quad\hbox{for $0<\theta\le1$}\ .$$
We have
$$\eqalign{
  \frac d{d\theta}(RHS)&=2\frac{1-\theta^2}{(1+\theta^2)^2}>0\quad
  \hbox{for $0<\theta<1$}\cr
  \frac d{d\theta}(LHS)&=\frac1{1+\theta^2}\cr
  LHS=RHS&=0\quad\hbox{when $\theta=0$}\ .\cr}$$
You can then check that


*

*for $0<\theta\le\frac1{\sqrt3}$ we have $\frac d{d\theta}(RHS)\ge\frac d{d\theta}(LHS)$, so $RHS>LHS$;

*for $\frac1{\sqrt3}<\theta\le1$ we have $RHS\ge\frac{\sqrt3}2>\frac\pi4\ge LHS$.

