Proving an inequality involving factorials: $(\frac{n}{2})^n \ge n! \geq (\frac{n}{3})^n$

Question

For $n \geq 6$, where $n$ is a natural number, prove that $(\frac{n}{2})^n \ge n! \geq (\frac{n}{3})^n$.

I tried using induction but could not do it.

Answer 1

Re-trying with induction...assume $(\frac{n}{2})^n \ge n!$ is valid for all natural number $k$ such that $n \ge k \ge 6$, then let's prove it's valid also for $n+1$. For the inductive hypothesis we have $$ \left(\frac{n}{2}\right)^n \ge n! $$ so $$ (n+1)\cdot \left(\frac{n}{2}\right)^n \ge (n+1)! $$ but since $\left(\frac{n+1}{2}\right)^{(n+1)} = \left(\frac{n+1}{2}\right)^{(n)}\cdot \left(\frac{n+1}{2}\right) \ge (n+1)\cdot \left(\frac{n}{2}\right)^n$ because $$\frac{1}{2}\left(\frac{n+1}{2}\right)^{(n)} \ge \left(\frac{n}{2}\right)^n$$ $$ \frac{(n+1)^n}{2} \ge n^n$$ $$ \left(\frac{n+1}{n}\right)^n \ge 2 $$ for all positive integers, $$ \left(\frac{n+1}{2}\right)^{(n+1)} \ge (n+1)\cdot \left(\frac{n}{2}\right)^n \ge (n+1)!$$ for all positive integers $\ge 6$ (it's trivial to prove $\left(\frac{6}{2}\right)^6 \ge 6!$).

As for the second part of the inequality you should re-try with induction...

Hint : assume $n! \geq (\frac{n}{3})^n$ true for all integers $k$ such that $n \ge k \ge 6$, then we have $(n+1)! \geq (n+1)\cdot \left(\frac{n}{3}\right)^n$ but, as before $(n+1)\cdot \left(\frac{n}{3}\right)^n \geq (\frac{n+1}{3})^{n+1}$, because $$n^n \geq \frac{(n+1)^n}{3}$$ $$ 3 \geq \left( \frac{n+1}{n}\right)^{n}$$ for all positive integers (since is trivial for the smallest positive numbers, and $lim_{n \to \infty} (\frac{n+1}{n})^n = e \lt 3$ )