How to go about proving that $(\frac{n}{3})^n$ $\leq$ $\frac{n!}{3}$ for $n \in \mathbb{N}_{>0}$

Question

I've been stuck on this proof for a long time. I don't even really know where to start.

I've tried simplifying the inequality to:

$$\frac{n^n}{3^{n-1}} \leq n!.$$

But I couldn't find any leads to follow there. Could someone give me a hint on where to go from here. I'm just so lost. Any help would be greatly appreciated.

Answer 1

Fix $n=3k \ge 3$ (the other cases can be done similarly), then \begin{align} n!=(3k)!&\ge (3k)(3k-1) \cdots (k) \\ &= (2k)\prod_{i=1}^k (2k+i)(2k-i)\\ &\ge (2k)\prod_{i=1}^k (2k+k)(2k-k)\\ &=(2k)(3k^2)^k \\ &= (2\cdot 3^k) k^{3k}\\ & \ge (2\cdot 3) k^{3k} \\ &>3 \left(\frac{n}{3}\right)^n. \end{align}