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

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    $\begingroup$ Where is this problem originated ? $\endgroup$
    – Khosrotash
    Aug 24, 2017 at 8:57
  • $\begingroup$ You can check that for some values in desmos $\endgroup$
    – MAN-MADE
    Aug 24, 2017 at 8:58
  • $\begingroup$ 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 $1$ $\endgroup$
    – TStancek
    Aug 24, 2017 at 9:06
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    $\begingroup$ @JamesJ You should add what you tried or your thoughts. Many questions get closed due to the lack of showing effort. $\endgroup$
    – Scientifica
    Aug 24, 2017 at 9:13
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    $\begingroup$ @TStancek: $n!$ is asymptotic to $n^{1/2}(n/e)^n$, so the margin is not so large. ($12/5=2.4$) $\endgroup$
    – user65203
    Sep 5, 2017 at 14:10

1 Answer 1

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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!

$$1\cdot n+2(n-1)+...+n\cdot1=\sum_{k=1}^nk(n-k+1)=$$ $$=(n+1)\sum_{k=1}^nk-\sum_{k=1}^nk^2=(n+1)\cdot\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}=$$ $$=\frac{n(n+1)}{6}\cdot(3n+3-2n-1)=\frac{n(n+1)(n+2)}{6}.$$

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  • $\begingroup$ You also can make three times that obtain this :(4n+5)/9>=(n!)^(1/n) $\endgroup$
    – JamesJ
    Aug 24, 2017 at 9:53
  • $\begingroup$ @JamesJ It's weaker because $\frac{4}{9}>\frac{5}{12}$. $\endgroup$
    – Michael Rozenberg
    Aug 24, 2017 at 9:55
  • $\begingroup$ Yes it is:(n!)^(1/n)<=(5n+7)/12<=(4n+5)/9<=(n+1)/2 $\endgroup$
    – JamesJ
    Aug 24, 2017 at 9:57
  • $\begingroup$ I think if (5n+7)/12 is the best number? $\endgroup$
    – JamesJ
    Aug 24, 2017 at 10:00
  • $\begingroup$ @JamesJ We can get a best estimation by the Stirling approximation. $\endgroup$
    – Michael Rozenberg
    Aug 24, 2017 at 10:03

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