How to prove the inequality $ \frac{1-e^{-x^2}}{x^2}e^{-(x-n)^2}<\frac{2}{n^2}$ for $0Can somebody prove that this inequality is true for $0<x<n$?
$$ \frac{1-e^{-x^2}}{x^2}e^{-(x-n)^2}<\frac{2}{n^2}$$
I'm pretty much stuck.

We may notice that $\frac{1-e^{-x^2}}{x^2}\leq \frac{2}{2+x^2}$ for any $x\in\mathbb{R}$ and $e^{-(x-n)^2}\leq \frac{1}{1+(x-n)^2}$ for any $x\in[0,n]$, hence it is enough to prove that:
$$ \forall x\in(0,n),\qquad \frac{2}{2+x^2}\cdot\frac{1}{1+(n-x)^2}<\frac{2}{n^2} \tag{1}$$
or:
$$ \forall x\in(0,n),\qquad  (2+x^2)\cdot(1+(n-x)^2) > n^2\tag{2} $$
But due to the Cauchy-Schwarz inequality we have:

$$ \forall x\in(0,n),\qquad (\sqrt{x^2+1}^2+1)(1+(n-x)^2)\geq \left(\sqrt{x^2+1}+n-x\right)^2 \tag{3}$$

so $(2)$ is trivial, $(1)$ follows and the claim is proved.
