How to prove that $\frac{n^n}{3^n}\le n!$ for $\forall \ge6$ without using any calculus tools (without involving lim or Euler number). I was told that the proof is similar to proving the definition of e, but I still cannot figure out how to do it. Thank you for your help!
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Use induction. Just check it is true for $n=6$. Now suppose it is true for $n$, i.e. $$\frac{n^n}{n!}\leq 3^n.$$ Consider $$\frac{(n+1)^{n+1}}{(n+1)!}=\frac{(n+1)^n}{n!}\leq\frac{3n^n}{n!}=3\cdot\frac{n^n}{n!}\leq 3\cdot 3^n=3^{n+1}$$ where the first inequality follows from $$\frac{(n+1)^n}{n^n}=\left(1+\frac{1}{n}\right)^n\leq e\leq 3.$$
Paul
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Is $e$ the "Euler number" that the OP mentions? – robjohn Dec 02 '14 at 00:48
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In this answer, it is shown, using Bernoulli's Inequality that is proven there using induction, that $$ \left(1+\frac1n\right)^{n+1} $$ is a decreasing sequence. Therefore, for $n\ge5$, $$ \left(\frac{n+1}{n}\right)^{n+1}\le\left(\frac65\right)^6=2.985984 $$ Thus, for $n\ge5$, $$ \begin{align} \left.\frac{(n+1)^{n+1}}{(n+1)!}\middle/\frac{n^n}{n!}\right. &=\left(\frac{n+1}{n}\right)^n\\ &\lt\left(\frac{n+1}{n}\right)^{n+1}\\[9pt] &\lt3 \end{align} $$ Since $$ \frac{5^5}{5!}\lt3^5 $$ we have that for $n\ge5$, $$ \frac{n^n}{n!}\lt3^n\iff\frac{n^n}{3^n}\lt n! $$
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I think the proof you've been hinted towards is this one: applying the GM/HM inequality to $n+1$ copies of $\frac{n+1}{n}$ and one $1$ yields $$ \Bigl(\frac{n+1}{n}\Bigr)^{(n+1)/(n+2)} 1^{1/(n+2)} \ge \frac1{\frac{n+1}{n+2}\cdot\frac{n}{n+1} + \frac1{n+2}\cdot\frac11} $$ Simplify and you will find that you have shown $\bigl(1+\frac1n\bigr)^{n+1}$ to be decreasing; then proceed as in robjohn's answer.
(This proof is dual to a standard one that $\bigl(1+\frac1n\bigr)^n$ is increasing, which is often given to justify the definition of $e$ as the limit.)