193. An inequality with n! .

by Virgil Nicula, Dec 21, 2010, 8:58 PM

Proposed problem. Prove that for any $n\ge 3$ we have $\left(n!\right)^2 > n^n$ .

Proof 1. $(\forall )\ r \in \mathbb{Z}\ ,\ 0 \le r \le n-1$ we have $(n-r)(r+1) \ge n$ $\implies$ $\prod_{r=0}^{n-1} (n-r)(r+1) \ge \underbrace{n\cdot n \cdot n \cdots n}_{n \text{ times}}$ $\Longleftrightarrow (n!)^2 \ge n^n$ .

Proof 2. Denote $a_n=\frac {\left(n!\right)^2}{n^n}\ ,\ n\ge 2$ . Observe that $\frac {a_{n+1}}{a_n}=(n+1)\cdot\left(\frac {n}{n+1}\right)^n>\frac {n+1}{3}\ge 1$ $\implies$ $a_{n+1}>a_n$ $\implies$

$1=a_1=a_2<a_3<\ldots <a_{n-1}<a_n$ . Thus, for any $n\ge 3$ we have $a_n>1$ , i.e. $\left(n!\right)^2 > n^n$ . I used that $e_n=\left(\frac {n+1}{n}\right)^n\nearrow e\in (2,3)$ .

Indeed, apply $\mathrm{A.M.}\ge \mathrm{G.M.}$ to $a_k=1+\frac 1n$ for $k\in\overline{1,n}$ and $a_{n+1}=1$ . Obtain $\frac {n\cdot\left(1+\frac 1n\right)+1}{n+1}=\frac {n+2}{n+1}>\sqrt [n+1]{\left(1+\frac 1n\right)^n\cdot 1}$ $\implies$ $e_{n+1}=\boxed{\left(1+\frac {1}{n+1}\right)^{n+1}>\left(1+\frac 1n\right)^n}=e_n$ $\implies$ $e_{n+1}>e_n$ .
This post has been edited 14 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:33 PM

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