Inequality for function of $\arctan(x)$

$${1\over 1+x^2}\ge {1-x^2\over (1+x^2)^2}\quad \forall x >0$$ which is derivative of $${\arctan(x)}\ge {x\over 1+x^2}\quad \forall x >0$$ $${2\arctan(x)\over 1+x^2}\ge {2x\over (1+x^2)^2}\quad \forall x >0$$ which is derivative of $$\arctan^2(x) \ge {x^2\over 1+x^2}\quad \forall x >0$$

$$(1+x^2)\arctan^2(x) -x^2 \ge 0 \quad \forall x >0$$

Consider instead $ \displaystyle g(x) = \arctan{x} - \frac{x^2}{1 + x^2}$. Note that $g(0) = 0$, so it suffices to show that $g'(x) = 0$ for $x \ge 0$.

Now, $\displaystyle g'(x) = \frac{2[(1 + x^2)\arctan{x} - x]}{(1 + x^2)^2}$. It thus suffices to consider $$h(x) = \arctan{x} - \frac{x}{(1 + x^2)},$$ and show that $h(x) \ge 0$ for $x \ge 0$. But $h(0) = 0$, and $$h'(x) = \frac{2x^2}{(1 + x^2)^2} \ge 0$$ for all $x$. This completes the proof.