# Factorial lower bound: $n! \ge {\left(\frac n2\right)}^{\frac n2}$

A professor in class gave the following lower bound for the factorial $$n! \ge {\left(\frac n2\right)}^{\frac n2}$$ but I don't know how he came up with this formula. The upper bound of $n^n$ was quite easy to understand. It makes sense. Can anyone explain why the formula above is the lower bound?

Any help is appreciated.

• Can you prove it by induction? – Ian Coley Nov 21 '13 at 0:24

Suppose first that $n$ is even, say $n=2m$. Then

$$n!=\underbrace{(2m)(2m-1)\ldots(m+1)}_{m\text{ factors}}m!\ge(2m)(2m-1)\ldots(m+1)>m^m=\left(\frac{n}2\right)^{n/2}\;.$$

Now suppose that $n=2m+1$. Then

$$n!=\underbrace{(2m+1)(2m)\ldots(m+1)}_{m+1\text{ factors}}m!\ge(m+1)^{m+1}>\left(\frac{n}2\right)^{n/2}\;.$$

Hint: For a positive integers $a$, we have $a(n-a)\ge \frac{n}{2}$. This is because for $0\le x\le n$, the function $x(n-x)$ is increasing up to $x=\frac{n}{2}$, and then decreasing.

Pair the numbers $a$ and $n-a$ in the factorial.

• Can you please prove the inequality in the hint? I don't see how $x(n-x)$ having a maxima at $x=n/2$ explain the inequality. – Shiv Tavker May 7 '20 at 22:28
• @ShivTavker You can assume x is a real number and diferentiate the equation and equal it to 0: -x^2 +xn = 0 -> -2x + n = 0 -> x = n / 2. – Bitcoin Cash - ADA enthusiast 2 days ago