Q.Prove that $1^2 + 2^2 + \ldots + n^2 > \dfrac{n^3}{3}$, $n \in N$.

Let $P(n)$ be the statement $1^2+2^2+\ldots+n^2>\dfrac{n^3}{3}$.

Base case: For $n=1$,

Since $1>\frac13$, $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k\ge1$:

We must show $1^2+\ldots+k^2+(k+1)^2>\dfrac{(k+1)^3}{3}$.

Adding $(k+1)^2$ to both sides of the induction hypothesis:

It suffices to show $\dfrac{k^3}{3}+(k+1)^2\ge\dfrac{(k+1)^3}{3}$, i.e.

Compute $(k+1)^3-k^3=3k^2+3k+1$, so we need

which is true (in fact strictly $>0$) for every $k\ge1$. Since $3k+2>0$ strictly, …

Inequality induction. Add $(k+1)^2$ to the induction hypothesis, then separately prove the algebraic inequality $\frac{k^3}{3}+(k+1)^2>\frac{(k+1)^3}{3}$ by expanding the cube difference and sh …

• Treating the target inequality as an equality and trying to "balance" both sides instead of bounding the added term.
• Sign or expansion errors in $(k+1)^3-k^3=3k^2+3k+1$. …