Show that $n!\ge2^{\left(n-1\right)}\left(n-2\right)^2$ by using induction

Question

I was trying to do prove the following inequality but I am stuck.

$$\tag{n > 5} n!\ge2^{\left(n-1\right)}\left(n-2\right)^2$$

I've shown it for $n=6$:

$$6 !=720 \ge 2^5 \cdot4^2 = 2^9 = 512$$

But I am stuck at showing it for $n+1$:

$$\left(n+1\right)!\ge2^n\cdot\left(n-1\right)^2$$ $$n!\cdot\left(n+1\right)\ge2^n\cdot\left(n-1\right)^2$$ $$?$$

I would be very happy if someone could give me a hint on how to keep going from here.

Greetings, Finn

Answer 1

$$(n+1)!\geq (n+1)2^{n-1}(n-2)^2$$ and it remains to prove that $$(n+1)2^{n-1}(n-2)^2\geq2^{n}(n-1)^2$$ or $$n^3-5n^2+4n+2\geq0,$$ which is obvious for $n\geq6.$

Answer 2

The most reasonable path forward is to use what you know: the inductive hypothesis gives you $n! \geq 2^{n-1}(n-2)^2$, so $$ n!(n+1) \geq 2^{n-1}(n+1)(n-2)^2 = 2^n(n-1)^2 \cdot \dfrac{(n+1)(n-2)^2}{2(n-1)^2} $$

Our goal is now to show that the fraction $\dfrac{(n+1)(n-2)^2}{2(n-1)^2} \geq 1$. The piece of information we have that we haven't used yet is $n > 5$. See if you can proceed from here.