Prove that the inequality $\frac{n^3}{3} < 3n-3$ applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold I have been given the following task:
Prove that the following inequality applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold:
$$\frac{n^3}{3} < 3n-3$$
My idea was to prove the statement by using induction.
For $n=1$ it follows:
$\frac{1^3}{3} = \frac{1}{3} \nless 0 = 3 * 1-3$
For $n=2$ it follows:
$\frac{2^3}{3} = \frac{8}{3} < 3 = 3 * 2-3$
For $n=3$ it follows:
$\frac{3^3}{3} = \frac{27}{3} = 9 \nless 6 = 3 * 3-3$
I now assume that the inequality does not hold for $n \in \mathbb{N} \backslash \{2\}.$
n $\rightarrow$ n+1
$\frac{(n+1)^3}{3} = \frac{n^3+3n^2+3n+1}{3} = \frac{n^3}{3}+ \frac{3n^2+3n+1}{3} < 3n-3 + \frac{3n^2+3n+1}{3}=\frac{3(3n-3)+3n^2+3n+1}{3}=\frac{3n^2+12n-8}{3}$
But how do I continue from this step or is this even the wrong approach?

Show that $n^3/3 \ge 3n-3$ for $n\ge 3$, or equivalently
$n(n^2-9)\ge -9$, or
$n(n+3)(n-3) \ge -9$.
For $n \ge 3:$ 
$n(n+3)(n-3) \ge 0$ and we are done.
