# Is this true $n!\leq(\frac{5n+7}{12})^n,n∈N$?

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 $\frac{5n+7}{12}=\frac{\frac{n+1}2+\frac{n+2}3}2\geq \sqrt{\frac{(n+1)(n+2)}6}$ $=\sqrt{\frac{1}{n}\left(\frac{n(n+1)(n+2)}{6}\right)}$ $=\sqrt{\frac{1}{n}\sum_{k=1}^{n}k(n-k+1)}$ $\ge\sqrt{\sqrt[n]{(n!)^2}}=\sqrt[n]{n!}$.

• Where is this problem originated ? Aug 24, 2017 at 8:57
• You can check that for some values in desmos Aug 24, 2017 at 8:58
• Even though reduced to $5/12$ of the original value, it is still roughly $n^n$, which grows way faster than $n!$, no doubt there is $n_0$ such that for all $n\geq n_0$ this holds. The trick I guess is to prove $n_0$ is $1$ Aug 24, 2017 at 9:06
• @JamesJ You should add what you tried or your thoughts. Many questions get closed due to the lack of showing effort. Aug 24, 2017 at 9:13
• @TStancek: $n!$ is asymptotic to $n^{1/2}(n/e)^n$, so the margin is not so large. ($12/5=2.4$)
– user65203
Sep 5, 2017 at 14:10

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!
$$1\cdot n+2(n-1)+...+n\cdot1=\sum_{k=1}^nk(n-k+1)=$$ $$=(n+1)\sum_{k=1}^nk-\sum_{k=1}^nk^2=(n+1)\cdot\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}=$$ $$=\frac{n(n+1)}{6}\cdot(3n+3-2n-1)=\frac{n(n+1)(n+2)}{6}.$$
• @JamesJ It's weaker because $\frac{4}{9}>\frac{5}{12}$. Aug 24, 2017 at 9:55