Inequality proof regarding factorial

Question

Prove that $n! ≤ n^n/2^n$ for all natural numbers $n ≥ 6$.

I know the proof using induction. I was looking for a more number theoretic approach to it.

We know, $n! ≤ n^n$ and $n! ≥ 2^n$, can we somehow use that?

Answer 1

By AM-GM $$\sqrt[n-1]{n!}\leq\frac{2+3+...+n}{n-1}=\frac{\frac{n(n+1)}{2}-1}{n-1}=\frac{n+2}{2}$$ Thus, it's enough to prove that $$\left(\frac{n+2}{2}\right)^{n-1}\leq\left(\frac{n}{2}\right)^n$$ or $$\left(1+\frac{2}{n}\right)^n\leq1+\frac{n}{2}$$ and since $\left(1+\frac{2}{n}\right)^n<e^2,$ the inequality is proven for $n>2e^2-2.$

Easy to check that the inequality is true for $6\leq n<2e^2-2$ and we are done!