Proof that $n! > 3n$ for $n\ge4 $ using the Principle of Mathematical Induction

Question

Use induction to prove that $n! > 3n$ for $n\ge4 $.

I have done the base case and got both sides being equal to $24>12$ for $n=4$. However, when doing the inductive step I can't seem to find the right form to match the expression on the right hand side.

So far I have:

Need to show: $(n+1)!>3(n+1)$.

When doing the inductive step:

$(n+1)! = (n+1)n!$

we know that $n!$ is larger than $3n$, then

$(n+1)n! >(n+1)3n$.

Here is where I don't know what to do next, could anyone shed some insight on how to continue after this part? Thanks.

Answer 1

In your proof you could do the following: $$n!\geq 3n$$ $$(n+1)!\geq 3n(n+1)$$

Now note that $n\geq 1$, therefore $3n(n+1)\geq 3(n+1)$.....

Answer 2

Assume $n!>3n$, then $n!\ge3$ as $3n>3$ for all $n\in\mathbb{N_{\geq4}}$.

This follows

\begin{align} &n!>3\\&(n+1)n!>3(n+1)\\&(n+1)!>3(n+1) \end{align}