I'm working on proving the following inequality:$$((k/2)!)^2k^{-k}\ge(2e)^{-k}$$ for any positive even integer $k$. I can use Stirling formula to prove it for large $k$'s, but I want a proof that works for all positive even $k$. Any help is greatly appreciated.
PS: It seems that we can prove it by induction on $k$. So I'm more interested in a "non-induction" proof.
First let's rearrange:
$$((k/2)!)^2k^{-k} \ge (2e)^{-k} \iff ((k/2)!)^2 \ge \left(\dfrac{k}{2e}\right)^k \iff (k/2)! \ge \left(\dfrac{k}{2e}\right)^{k/2}$$
After substituting $n=\dfrac{k}{2}$ you'll get the equivalent inequality $$n! \ge \left(\dfrac{n}{e}\right)^{n} \iff \dfrac{n!}{n^n} \ge \dfrac{1}{e^n}$$
Since $e^x \ge \dfrac{x^n}{n!}$ we know
$$e^n \ge \dfrac{n^n}{n!} \iff \dfrac{n!}{n^n} \ge \dfrac{1}{e^n}$$