Bound on Stirling numbers of the first kind?

Let $n$ be divisible by $3$. Prove that

$$c(n, n/3) \geq \frac{n!}{3^{n/3}(n/3)!}$$

.

-
In order to make the best use of this website, it is important that you learn to explain your questions more effectively. Where did the problem arise? What have you tried? What difficulty did you encounter? And especially avoid copying-and-pasting the question verbatim: you're talking to humans, not a search engine. – Erick Wong Feb 4 '13 at 19:27

Recall that $c(n, n/3)$ counts the permutations of $\{1, 2, \dots, n\}$ into $n/3$ cycles, while $\frac{n!}{3^{n/3}(n/3)!}$ counts the permutations of $\{1, 2, \dots, n\}$ into $n/3$ 3-cycles.