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?

Let consider the reverse $\frac{n^3}{3} > 3n-3$ then we can prove the induction step as follows
$$\frac{(n+1)^3}{3} =\frac{n^3}{3}+\frac{3n^2+3n+1}{3}> 3n-3+\frac{3n^2+3n+1}{3}>3n-3+3=3(n+1)-3$$
we have used that for $n\ge 2$
$$\frac{3n^2+3n+1}{3}>3 \iff3n^2+3n-8>0$$
