Inequality for n factorial

Question

Can some one show me how to prove the following inequality

$$n!\geq 2(3^{n-2}) \quad \forall n\geq 2$$

Answer 1

Note that it is true for $n=2$. Let's assume the expression is true for $n-1$ (we'll use $n\geq 3$ here). Then:

$$n!=n\cdot(n-1)!\geq n\cdot 2(3^{n-3})\geq 3\cdot 2(3^{n-3})=2(3^{n-2})$$

and thus, by the principle of induction, the statement must be true for all $n$.

---

It's actually also pretty intuitive; On the left we have $$n\cdot (n-1)\cdot\cdots\cdot 3\cdot2$$ (total of $n-1$ terms, we're excluding the $1$) and on the right we have $$3\cdot3\cdot\cdots\cdot 3\cdot2$$ (also a total of $n-1$ terms). See why the inequality is true?

Answer 2

Let's proceed by induction on $n$. For $n=2$, we have:

$$ 2!=2\ge 2\cdot 3^{2-2}=2$$ and so the base case is true. Suppose now that the claim is true for some $k\ge 2$, that is:

$$ k!\ge 2\cdot 3^{k-2} $$

We'd like to show that this implies

$$(k+1)!\ge 2\cdot 3^{k-1} $$

Indeed, note that $ (k+1)!=(k+1)k!$ and by the induction hypothesis we have:

$$ (k+1)!=(k+1)k!\ge (k+1)\cdot \left(2\cdot 3^{k-2}\right)$$

and let's now use the fact that $k\ge 2$: $$ (k+1)!\ge (k+1)\cdot \left(2\cdot 3^{k-2}\right)\ge 3\cdot (2\cdot 3^{k-2})=2\cdot 3^{k-1}$$ and so our theorem is true.