Prove inequality: When $n > 2$, $n! < {left(frac{n+2}{sqrt{6}}right)}^n$

Prove: When $n > 2$,

$$n! < {\left(\frac{n+2}{\sqrt{6}}\right)}^n$$

PS: please do not use mathematical induction method.

EDIT: sorry, I forget another constraint, this problem should be solved by
algebraic mean inequality.

Thanks.

Answer

This used to be one of my favourite high-school problems. This is one approach: consider $y=\ln x$ and say that you want to integrate it between $1$ and $n$.

obviously the sum of the areas of trapezium $<\int_1^n\ln x\mathrm{d}x$. From this inequality, you get another inequality:
$$n!<\left(\frac{n^{n+\frac{1}{2}}}{e^{n-1}}\right)$$
Then just show the following inequality and you are done:
$$\left(\frac{n^{n+\frac{1}{2}}}{e^{n-1}}\right)<{\left(\frac{n+2}{\sqrt{6}}\right)}^n$$