TITLE: Is this true $n!\leq(\frac{5n+7}{12})^n,n∈N$?
QUESTION [5 upvotes]: Is the following inequality true? 
For all $n\in \Bbb N$ prove that:
$$n!\leq\left(\frac{5n+7}{12}\right)^n.$$
I know the answer,but I want to see other people how to prove the problem. 
In my proof I used $\frac{5n+7}{12}=\frac{\frac{n+1}2+\frac{n+2}3}2\geq \sqrt{\frac{(n+1)(n+2)}6}$
$=\sqrt{\frac{1}{n}\left(\frac{n(n+1)(n+2)}{6}\right)}$
$=\sqrt{\frac{1}{n}\sum_{k=1}^{n}k(n-k+1)}$
$\ge\sqrt{\sqrt[n]{(n!)^2}}=\sqrt[n]{n!}$.

REPLY [6 votes]: By AM-GM
$$\frac{1\cdot n+2(n-1)+...+n\cdot1}{n}\geq\sqrt[n]{(n!)^2}$$ or
$$\left(\sqrt{\frac{(n+1)(n+2)}{6}}\right)^n\geq n!.$$
Thus, it remains to prove that
$$\frac{5n+7}{12}\geq\sqrt{\frac{(n+1)(n+2)}{6}},$$
which is $$(n-1)^2\geq0.$$
Done!
$$1\cdot n+2(n-1)+...+n\cdot1=\sum_{k=1}^nk(n-k+1)=$$
$$=(n+1)\sum_{k=1}^nk-\sum_{k=1}^nk^2=(n+1)\cdot\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}=$$
$$=\frac{n(n+1)}{6}\cdot(3n+3-2n-1)=\frac{n(n+1)(n+2)}{6}.$$