Proof of $ n! \lt \Big(\frac{n+1}{2}\Big)^n$

Question

Use $AM $-$GM$ inequality to show that $$(\forall n \in \mathbb{N}) : n! \lt \Bigg(\frac{n+1}{2}\Bigg)^n$$ .

Answer 1

We have $$\sqrt[n]{n!}=\sqrt[n]{1\cdot 2\cdot …\cdot n}\le \frac{1+2+...+n}{n}=\frac{n(n+1)}{2n}$$ Can you finish?

Answer 2

Inequality problems are all about finding small patterns. Often you can start from some modest equations. Let $a,b$ are positive numbers.

Such that $a+b=m$

Then, $ab=a(m-a)=am-a^2=\frac{m^2}{4}-(a-m/2)^2$

This means $ab$ is maximum when $a=b=m/2$.

Take, $a+b=n+1$

Hence $$1\cdot2\cdots n=(1\cdot n)(2\cdot (n-1))\cdots \left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)\leq \left(\frac{n+1}{2}\right)^n$$

You can easily work out cases when $n$ is even or odd.

Answer 3

$$\sqrt[n]{x_0\cdot x_1\cdot x_2\cdot...\cdot x_n} \leq \frac {x_0+x_1+x_2+...+x_n}{n}$$

$$n! = n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot...\cdot 1$$

Using the AM-GM inequality, we can conclude:

$$\sqrt[n] {n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot...\cdot 1} \leq \frac{1+...+(n-3)+(n-2)+(n-1)+n}{n}$$

Simplify the right hand side.

$$\sqrt[n] {n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot...\cdot 1} \leq n\biggr(\frac{n+1}{2n}\biggr)$$

Simplify again and raise both sides to power $n$.

$$n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot...\cdot 1 \leq \biggr(\frac{n+1}{2}\biggr)^n$$

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

Proven.