Easy way to show $e^{-x^2}<\frac1{x^2}$ and $e^{-x^2}<\frac1{\sqrt{|x|}}$ I'm looking for an easy and basic way to show that
$$e^{-x^2}<\frac1{x^2}$$
and
$$e^{-x^2}<\frac1{\sqrt{|x|}}$$
for $x\neq 0$. I think there must be a simple trick which won't come to my mind right now.
 A: As regards the first one, for $x\not=0$,
$$e^{-x^2}<\frac1{x^2} \Leftrightarrow x^2<1+x^2\leq e^{x^2}=\sum_{k=0}^{\infty}\frac{x^{2k}}{k!}.$$
For the second one, note that 
$$\sqrt{|x|}\leq \frac{1+|x|}{2}<1+x^2\leq e^{x^2}.$$
A: Hint for the first. 
It is equivalent to prove that for $x>0$
$$-x <\ln(\frac {1}{x} )$$
or
$$\ln (x)<x .$$
is true if $0 <x \leq 1$ since $\ln x \leq 0$.
Now consider $f (x)=\ln (x)-x $ for $x>1$.
$$f'(x)=\frac {1-x}{x}<0$$
$f $ is decreasing and
$$f (x)<f (1)<0$$
A: For the first one, you want to prove that $\frac {1}{e^{x^2}} < \frac {1}{x^2}$ which is the same as proving that $e^{x^2}>x^2$.
Divide by $x^2$  $(x\neq0)$ and you get to prove that $\frac {e^{x^2}}{x^2}>1$.
Now you can look at that function, find its minimum and evaluate it, and get that it is bigger than 1.
A: Differentiate $x^2 e^{-x^2 }$ with respect to $x$, we get $2x e^{-x^2}-2x^3 e^{-x^2}=2x(e^{-x^2 }-x^2 e^{-x^2 }$. 
By analysis the zeros of this derivative, we found that $x^2 e^{-x^2 }$ attains its maximum at $|x|=1$, in which the relevant image is $\frac{1}{e}<1$.
As $x^2 > 0$ for all $x\neq 0$, we could get the proof of the first inequality. And by the similar method we could get the second one.
A: For positive $\alpha$, $$e^{-x^2}<|x|^{-2\alpha}=(x^2)^{-\alpha}$$ is equivalent to
$$x^2>\alpha\log x^2$$
or $$t>\alpha\log t.$$
By the study of the function $t/\log t$, which has a minimum at $t=e$, one sees that the inequality holds for all $t>0$ when $\alpha<e$. In particular, with $\alpha=1$ and $\alpha=1/4$.
A: If $y \ge 0$, then $y< e^y$, therefore $ye^{-y}<1$, put $y= x^2$ to get the first inequality.
For the second inequality, observe that $$e^{-kx^2}<1, \ \ \forall \ k>0 $$.
Thus multiplying $e^{-3x^2}$ to the inequality $x^2e^{-x^2}<1$ we have,
$$x^2e^{-4x^2}<1$$
Since all the quantities are positive, we have
$$0 <(x^2e^{-4x^2})^{1/4}<1$$
i.e 
$$\sqrt{|x|}e^{-x^2}<1 \implies e^{-x^2} < \frac{1}{\sqrt{|x|}}$$
