# 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?

• This clearly holds asymptotically; by convexity, if it holds for some $n$ it holds for all larger $n$ (for $n \ge 1$ at least) and the inequality holds for $n = 2$ so we are done. – Brevan Ellefsen Aug 10 '19 at 5:02

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

• Oh that is much simpler than what I was trying to write, thank you so much! – mathnerd Aug 10 '19 at 4:49
• @Alex Now, please accept my answer. Click the green tick beside it. – Parcly Taxel Aug 10 '19 at 4:50

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

• Thank you so much!! – mathnerd Aug 10 '19 at 4:50

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 $$0\le(\sqrt a-\sqrt b)^2\implies \sqrt{ab}\le\frac{a+b}2$$.