# Inequality for function of $\arctan(x)$

I want to show that $$f(x) = \frac{1}{\arctan(x)} - \frac{1}{x}$$ is increasing on $$(0, \infty)$$. I can see this clearly by plotting it, but I'm struggling to write it out rigorously. It obviously suffices to show its derivative is always positive in this range (which is also clear from plotting it). We have $$f'(x) = \frac{(1+x^2)\arctan^2(x) -x^2}{x^2(1+x^2)\arctan^2(x)}$$ so again it suffices to show that $$g(x) \equiv (1+x^2)\arctan^2(x) -x^2 \ge 0 \quad \forall x >0$$ (and, yet again, this is clear from plotting it). I've jumped down the rabbit hole of taking the derivative of $$g$$ as well (since it is $$0$$ at $$x = 0$$ so it would again suffice to show that $$g' \ge 0$$) and it doesn't yield anything immediately useful for me. Please help if you can

$${1\over 1+x^2}\ge {1-x^2\over (1+x^2)^2}\quad \forall x >0$$ 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.