Inequality $(n!)^2\le \left[\frac{(n+1)(n+2)}{6}\right]^n$ Prove that
$$
(n!)^2\le \left[\frac{(n+1)(n+2)}{6}\right]^n
$$
holds for all $n\in\mathbb{Z^+}$.
I tried induction but there's no obvious way to go from $n$ to $n+1$. 
 A: Try to prove first that
$$n\cdot 1 + (n-1)\cdot 2 + \dots + 2\cdot (n-1) + 1\cdot n = \frac{n(n+1)(n+2)}6.$$
(I have posted a separate question about this sum.)
Then use the AM-GM inequality for the sum on the LHS.
A: Here is an idea I had (by no means a complete answer but a start nonetheless), where I assume you have verified the base case and all that jazz:
\begin{align}
[(k+1)!]^2 &= (k+1)^2(k!)^2\\[1em]
           &\leq (k+1)^2\left[\frac{(k+1)(k+2)}{6}\right]^k\\[1em]
           &\leq\frac{(k+2)(k+3)}{(k+1)(k+1)}\cdot\left\{(k+1)^2\left[\frac{(k+1)(k+2)}{6}\right]^k\right\}\\[1em]
           &= \frac{(k+2)^{k+1}(k+3)(k+1)^k}{6^k}\\[1em]
           &\leq\frac{(k+2)^{k+1}(k+3)^{k+1}}{6^{k+1}}\qquad\left[\text{since}\quad\frac{(k+3)(k+1)^k}{6^k}\leq\frac{(k+3)^{k+1}}{6^{k+1}}\right]\\
           &\vdots
\end{align}
where the inequality after the "since" is true for $k\geq 18$.

It's hardly optimal, but it is something at least. It introduces another inequality that may be approached via asymptotics or the like--the extra $(k+1)$ term really seems to get in the way of many lines of attack. 
Good luck!
