Q:

Prove for all natural numbers $n ≥ 2: n! <\big( \frac{n + 1}{2}\big)^n$

It asks to prove the above statement, and I was given a hint to use the AM$\ge$GM inequality, i.e. the geometric mean is always less than or equal to the arithmetic mean: $(a_1a_2· · · a_n)^\frac{1}{n} ≤
\frac{a_1 + a_2· · · + a_n}{n}$, with equality if and only if $a_1=a_2=...=a_n$.
I tried to use induction to prove this but got stuck on proving the inductive case, if $P(n)$ holds then $P(n+1)$ holds.
Could anyone help me in trying to prove this or at least set me on the right path?
Thanks in advance!

A:

By the AM/GM inequality applied to $1,2,\dots,n$ we have
$$\sqrt[n]{n!}<\frac{1+2+\dots+n}n$$
This can be rearranged (using the triangular sum $1+2+\dots+n=\frac{n(n+1)}2$) to get
$$n!<\left(\frac{n(n+1)}{2n}\right)^n$$
$$n!<\left(\frac{n+1}2\right)^n$$

