Prove that $1^{2} + 2^{2} + \ldots + n^{2} > \frac{n^{3}}{3}$ for all $n \in N$.

Let $P(n)$ be the statement: $1^{2} + 2^{2} + \ldots + n^{2} > \frac{n^{3}}{3}$.
Step $1$: For $n = 1$,$1^{2} = 1$ and $\frac{1^{3}}{3} = \frac{1}{3}$. Since $1 > \frac{1}{3}$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$1^{2} + 2^{2} + \ldots + k^{2} > \frac{k^{3}}{3}$ $(1)$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$1^{2} + 2^{2} + \ldots + k^{2} + (k+1)^{2} > \frac{(k+1)^{3}}{3}$.
Starting from the left side:
$1^{2} + 2^{2} + \ldots + k^{2} + (k+1)^{2} > \frac{k^{3}}{3} + (k+1)^{2}$ (using $(1)$)
$= \frac{k^{3} + 3(k^{2} + 2k + 1)}{3} = \frac{k^{3} + 3k^{2} + 6k + 3}{3}$
$= \frac{(k^{3} + 3k^{2} + 3k + 1) + 3k + 2}{3} = \frac{(k+1)^{3} + 3k + 2}{3}$
Since $3k + 2 > 0$ for $k \in N$,we have $\frac{(k+1)^{3} + 3k + 2}{3} > \frac{(k+1)^{3}}{3}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.