<s>[INST] Prove $(1-\frac{8}{\pi^2})x^2\cot^2(x)+\frac{8}{\pi^2}x\cot(x)+\frac{4}{\pi^2}x^2-1\ge0$ for $x\in[0,\pi/2]$ 
How to prove the following inequality:
$$\left(1-\frac{8}{\pi^2}\right)x^2\cot^2(x)+\frac{8}{\pi^2}x\cot(x)+\frac{4}{\pi^2}x^2-1\ge0,\quad x\in[0,\pi/2]?$$

I evaluated the LHS of the above inequality numerically. I found that equality holds at $x=0$ and $x=\pi/2$, and strict inequality holds elsewhere. The LHS is not a monotonic function; as $x$ increases it first increases and then decreases. I derived the derivative of the LHS and the results are quite messy. Is there any easier ways of proving the inequality?
 [/INST] A partial answer - I am going to show that the given inequality holds for every $x\in(0,1]$.
Since the cotangent is the logarithmic derivative of the sine function we have
$$ \cot x = \frac{1}{x} + \sum_{n\geq 1}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right) \tag{1} $$
and
$$ \frac{1-\pi z\cot(\pi z)}{2} = \sum_{n\geq 1}\zeta(2n) z^{2n}.\tag{2} $$
This gives the Taylor series of $z^2\cot(z)^2$ too, and allow us to state that the given LHS is
$$ \left(-\frac{2}{3}+\frac{20}{3 \pi ^2}\right) x^2+\left(\frac{1}{15}-\frac{32}{45 \pi ^2}\right) x^4+\sum_{k\geq 3}c_k x^{2k},\qquad c_k\geq 0\tag{3} $$
for any $x\in\left(0,\frac{\pi}{2}\right)$, so the given inequality holds over the interval $\left(0,1\right]$ for sure.
That should be most of the job.

The given inequality is indeed equivalent to
$$\forall x\in\left(0,\frac{\pi}{2}\right),\quad x\cot(x)\geq \frac{-4+\sqrt{\left(\pi ^2-4\right)^2-4 \left(\pi ^2-8\right) x^2}}{\pi ^2-8} $$
where the RHS is a remarkably good approximation of the LHS.
The (generalized) Shafer-Fink inequality should completely settle the question, but I have to perform some numerical experiments. In a Shafer-Fink-suitable form, the previous line can be written as
$$ \forall x>0,\quad \frac{\arctan x}{x}\geq \frac{-4+\sqrt{\left(\pi^2-4\right)^2+4 \pi ^2 x^2}}{\pi^2-8+4 x^2}\tag{4} $$
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