Let $p$ be a positive integer and $p \ge 2$. I want to show that $$ 2(p/2)^{p/2} e^{-p/2} \le \sqrt{p} \Gamma(p/2), $$ where $\Gamma(x) = \int_0^{\infty} t^{x-1}e^{-t} ~ dt$ is known as Gamma function. I checked it on wolramalpha.com and its appeared to correct, see this plot.
What I've tried: I tried to prove it by induction:
For $p=2$ its clear.
For $p$ we have $2(p/2)^{p/2} e^{-p/2} \le \sqrt{p} \Gamma(p/2)$ and want to prove for $p+1$: $2(p/2 + 1/2)^{p/2 + 1/2} e^{-p/2 - 1/2} \le \sqrt{p + 1} \Gamma(p/2 + 1/2)$.
So, for using the known result for $p$ I do the following:
$$ 2(p/2 + 1/2)^{p/2 + 1/2} e^{-p/2 - 1/2} \le 2\sqrt{e \cdot \frac{p+1} 2} (p/2)^{p/2} e^{-p/2 - 1/2} = \sqrt{\frac{p+1} 2} 2(p/2)^{p/2} e^{-p/2}, $$ where in the first step I used $(1 + p^{-1})^{p/2} \le \sqrt{e}$. Then, using the induction assumption we have that we need to show $$ \sqrt{\frac{p+1} 2} \sqrt{p} \Gamma(p/2) \le \sqrt{p+1} \Gamma\bigg(\frac{p+1} 2\bigg), $$ or, equivalentnly, $$ \sqrt{\frac p 2} \Gamma\bigg(\frac p 2\bigg) \le \Gamma\bigg(\frac{p+1} 2 \bigg), $$ which seems to be wrong, see the plot from here.
Is the step $(1 + p^{-1})^{p/2} \le \sqrt{e}$ very rough? Are there any other possible approaches for proving the initial inequality?