How to prove the inequality $ \frac{h}{\sqrt{2 \pi t^3}} e^{- \frac{h^2}{2t}} \leq \frac{C}{h^2} e^{- \frac{h^2}{3t}}$?

Question

In a proof I try to understand appears (as far as I see) the statement that there exists a constant $C>0$, such that

$$ \frac{h}{\sqrt{2 \pi t^3}} e^{- \frac{h^2}{2t}} \leq \frac{C}{h^2} e^{- \frac{h^2}{3t}}$$

for all $t,h > 0$. But at first try I was not able to verify it by myself.

Answer 1

It suffices to find $C>0$ such that $$\frac{h^3}{\sqrt{t^3}}e^{-\frac{h^2}{6t}}\leq C.$$ Note that for $x>0$ we have $e^x>x+\frac{x^2}{2}\geq2\sqrt{\frac{x^3}{2}}=c\sqrt{x^3}$, so $$\frac{h^3}{\sqrt{t^3}}e^{-\frac{h^2}{6t}}<\frac{h^3}{\sqrt{t^3}}\frac{1}{c\sqrt{\left(\frac{h^2}{6t}\right)^3}}=C.$$

Answer 2

Finally I found that $ e^{-x} \leq \frac 1 x $ helps, where $x>0$. Thus

$$\frac h {\sqrt{2 \pi t^3 } } e^{- \frac{h^2}{2t}} = \frac 1 {\sqrt{2 \pi }} \frac h {t^{3/2}} e^{- \frac{h^2}{3t}} (e^{- \frac{h^2}{9t}})^{3/2} \leq C \frac h {t^{3/2}} \frac {t^{3/2}}{h^3} e^{-\frac{h^2}{3t}} = \frac C {h^2} e^{-\frac{h^2}{3t}}$$