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.
 A: 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.
A: The first thing to try with this sort of inequality is to put $y=\frac{x}{n}$ and try to prove the resulting equality for $0<y<1$.
This seems a good idea in this case because multiplying both sides by $x^2$ will give a power of $\frac{x}{n}$ on the right-hand side, and on the left-hand side $(x-n)$ is $n$ times $(\frac{x}{n}-1)$.
So:


*

*Multiply both sides by $x^2$.

*Substitute $x=ny$.

*Simplify.

A: I wanted to find another answer different from the elegant one given by Jack D’Aurizio. Here is what I got. 
$$\frac{1-e^{-x^2}}{x^2}e^{-(x-n)^2}<\frac{2}{n^2}\iff \frac{e^{x^2}-1}{x^2}\lt\frac{2e^{x^2+(x-n)^2}}{n^2}$$ 
Let $$f(x)= \frac{e^{x^2}-1}{x^2}\\g(x)= \frac{2e^{x^2+(x-n)^2}}{n^2}$$  Notice first that $f$ and $g$ are both convex functions and that $g(0)=g(n)=\frac {2e^{n^2}}{n^2}$.
Besides $$f(0)=\lim_{x\to 0}f(x)=1\lt \frac {2e^{n^2}}{n^2}=g(0)$$ furthermore $g$ takes its minimum at $x=\frac n2$ and $$f(\frac n2)=\frac{4(e^{\frac{n^2}{4}}-1)}{n^2}\lt \frac {2e^{\frac{n^2}{2}}}{n^2}=g(\frac n2)$$  (it is immediately verified that $2(e^{\frac x2}-1)\lt e^x$ for all x). 
It follows by convexity of both functions and by
$$f(0)\lt g(0) \text { and}\space f(\frac n2)\lt g(\frac n2)$$ that $$\color{red}{f(x)\lt g(x)\text { for}\space 0\lt x\le \frac n2}\qquad (*)$$ 
Besides $$f(n)= \frac{e^{n^2}-1}{n^2}\lt\frac {2e^{n^2}}{n^2}=g(n)$$ and similarly, by convexity, $f(\frac n2)\lt g(\frac n2)$ and $f(n)\lt g(n)$, we have
$$\color{red}{ f(x)\lt g(x)\text { for}\space \frac n2 \le x\lt  n}\qquad (**)$$
Thus, by $(*)$ and $(**)$, $$ f(x)\lt g(x)\text { for}\space  0  \lt x\lt  n$$
NOTE.-However that is not necessary in this case where the functions $f$ and $g$ are the same gender, it is better to add that $f(x)= g(x)\iff x=0$  to have fully valid argument used convexity.
