prove that $A(n) : \left(\frac n3\right)^n\lt n!\lt \left(\frac{n}{2}\right)^n$ for all $n\ge 6$

Question

prove that $A(n) : \left(\frac n3\right)^n\lt n!\lt \left(\frac{n}{2}\right)^n$ for all $n\ge 6$

first check $n=6$ : $2^6<6!<3^6$ ok

then $n\gt 6$ assume $A(m)$ is true, then show $A(m+1)$ is true

I can not prove the next step...help me

Answer 1

Continuing from my comment.

$$\frac{(n/2)^n} {((n-1)/2)^{n-1}} = \frac{(n/2)^n((n-1)/2)} {((n-1)/2)^{n}} = \frac{n^n((n-1)/2)} {(n-1)^{n}} = \frac{(n-1)/2}{(1-\frac{1}{n})^n} = (1-\frac{1}{n})^{-n}(n-1)/2 > e(n-1)/2 > n$$

If you replace $2$ with $3$ above, you get $$\frac{(n/3)^n} {((n-1)/3)^{n-1}} = (1-\frac{1}{n})^{-n}(n-1)/3 \approx \frac{e(n-1)}{3} < n$$

so you can finish your induction.