How can we show that for fixed $a>0$ and $\forall\zeta \in (0,a)$ it holds $$\int_0^\infty e^{-at^{2}}t^{2n}dt\leq \frac{1}{2}\frac{n!}{\zeta^{n}}\sqrt{\frac{\pi}{a-\zeta}}$$ where $n$ is a non negative integer?
Any hint or help is appreciated to kickstart my attempt.
Thanks in advance.
The LHS equals
$$ \frac{\Gamma\left(n+\tfrac{1}{2}\right)}{2 a^{n+1/2}}=\frac{\sqrt{\pi}\,n!\binom{2n}{n}}{2 a^{n+1/2}4^n} $$
so the inequality is equivalent to
$$ \frac{\binom{2n}{n}}{4^n}\leq \frac{a^{n+1/2}}{\zeta^n(a-\zeta)^{1/2}} $$
or by setting $\lambda=\frac{\zeta}{a}\in(0,1)$,
$$ \lambda^{n}(1-\lambda)^{1/2}\leq \frac{4^n}{\binom{2n}{n}}.$$
Since the maximum of the LHS is attained at $\lambda=\frac{2n}{2n+1}$, the problem boils down to showing
$$ \frac{(2n)^n}{(2n+1)^{n+1/2}}\leq \frac{4^n}{\binom{2n}{n}} $$
or
$$ \sqrt{2n+1}\left(1+\frac{1}{2n}\right)^n\geq \frac{1}{4^n}\binom{2n}{n} $$
which is fairly loose since the LHS behaves like $\sqrt{2en}$ while the RHS behaves like $\frac{1}{\sqrt{\pi n}}$.
Induction easily cracks it. Happy new year!
