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I've been stuck on this proof for a long time. I don't even really know where to start.

I've tried simplifying the inequality to:

$$\frac{n^n}{3^{n-1}} \leq n!.$$

But I couldn't find any leads to follow there. Could someone give me a hint on where to go from here. I'm just so lost. Any help would be greatly appreciated.

Paolo Leonetti
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    Induction on $n$ works like a charm, using the elementary upper bound $\left(1+\frac1n\right)^n\leqslant3$ for every $n\geqslant1$. – Did Mar 09 '18 at 07:40
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    Interestingly, rather than following the hint in the comment above or digesting the correct answer below, you quasi instantly accepted the flagrantly false "answer" you had just received. Did you at least check it? Already its first line is wrong. (Please note that regulars of the site are not surprised that it is, coming from this user.) – Did Mar 09 '18 at 07:52
  • Boy I'm really messing up everything tonight. I should have looked at that answer more carefully. That's my fault, I need to be better about properly understanding the answers. I won't let this happen again. –  Mar 09 '18 at 07:56
  • It is 9 am here, good morning :P – Paolo Leonetti Mar 09 '18 at 07:57
  • Good morning to you, thank you for your answer to my question. I really appreciate everyone's help. –  Mar 09 '18 at 07:58
  • @Did: 'quasi instantly' - that bombed! – Alex Mar 09 '18 at 08:00
  • Could this be thought of as similar to proving that $\left(\dfrac{n}{3}\right)^n$ is $\mathcal{O}\left(\dfrac{n!}{3}\right)$? Could the limits be used? i.e. if $\displaystyle\lim_{n\to\infty}\dfrac{\left(\dfrac{n}{3}\right)^n}{\dfrac{n!}{3}}$ is a constant, then the inequality holds. Just a thought. – an4s Mar 09 '18 at 08:35
  • @an4s The question is about finite $n$ cases, not limits when $n\to\infty$. – Did Mar 09 '18 at 09:11

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Fix $n=3k \ge 3$ (the other cases can be done similarly), then \begin{align} n!=(3k)!&\ge (3k)(3k-1) \cdots (k) \\ &= (2k)\prod_{i=1}^k (2k+i)(2k-i)\\ &\ge (2k)\prod_{i=1}^k (2k+k)(2k-k)\\ &=(2k)(3k^2)^k \\ &= (2\cdot 3^k) k^{3k}\\ & \ge (2\cdot 3) k^{3k} \\ &>3 \left(\frac{n}{3}\right)^n. \end{align}

Paolo Leonetti
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  • Some terms in the product in the first line could be non-integer. Is this certainly not a problem? – Alex Mar 09 '18 at 08:11
  • @Alex With "the other cases", I was referring to "non-multiples of $3$".. – Paolo Leonetti Mar 09 '18 at 08:12
  • I got it. So there's no single solution, i.e. for all $n$. – Alex Mar 09 '18 at 08:16
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    Not as it is. But minor modifications work with noninteger $k$: note that $$\frac{f(n+1)}{f(n)}\le \frac{f(n+2)}{f(n)}<2n$$ for all $n$, where $f(n):=(\frac{n}{3})^n$. Then, setting $k=\lfloor n/3\rfloor$, $$ n! \ge (3k!) \ge (2\cdot 3^k) k^{3k} (k-1)! \ge 9k\cdot k^{3k} \ge f(n). $$ – Paolo Leonetti Mar 09 '18 at 08:25
  • ah good one, thanks @Paolo – Alex Mar 09 '18 at 11:55