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

Question

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!

Answer 1

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 $$n!<\left(\frac{n(n+1)}{2n}\right)^n$$ $$n!<\left(\frac{n+1}2\right)^n$$

Answer 2

We can simply apply the $AM -GM$ inequality for first n natural numbers as they are all positive

$GM ≤ AM$ Now $GM \to \sqrt[n]{\prod^n_{i=1} i} = \sqrt[n]{n!}$

$AM \to \frac{\sum_{r=1}^n r}{n} = \frac{\frac{n(n+1)}{2}}{n}= \frac{(n+1)}{2} $

Now $AM-GM$ gives $\sqrt[n]{n!} ≤ \frac{(n+1)}{2}$ But as $n≥ 2$ and as you point out in the hint of the question, the equality will never hold. So it is safe to write $$\sqrt[n]{n!} < \frac{(n+1)}{2}$$ or $$n! < ( \frac{(n+1)}{2})^n $$

Answer 3

You can also break this down to two-number inequalities, combining factors symmetrically from each end of the sequence $1,2,...,n-1,n$. Take the factorial twice to avoid to have to split in the middle, to get $$ (n!)^2=\prod_{k=1}^nk(n+1-k)\le\prod_{k=1}^n\left(\frac{n+1}2\right)^2, $$ the last step using $0\le(\sqrt a-\sqrt b)^2\implies \sqrt{ab}\le\frac{a+b}2$.