Prove a lower bound for $\sum_{i=1}^n i^2$ 
Prove that $$\sum_{i=1}^n i^2 \geq \frac{n^3}{3}$$ for all $n \geq 1.$

What I know: I know the basic format of how to make a proof with the basis and inductive step but I am unsure of how to prove this particular statement and expand it. This is for a data structures class by the way.
My attempt so far has been $1^2 + 2^2 +\cdots+n^2$ is $i^2$ expanded. Can anyone provide some insight or link me to similar examples on how to go about structuring this? 
Thanks so much in advance I am really lost and haven't taken a proofs class before so it's all new to me.

Here's another way to look at it.
Suppose you guess that
$s(n)
=\sum_{i=1}^n i^2$
grows something like $n^3$.
You conjecture that
$s(n)
\ge cn^3$
for some positive $c$.
For $n=1$,
this becomes
$1 \ge c$,
so the induction base case
holds for any $c \le 1$.
Suppose this holds for $n$,
so that
$s(n) \ge c n^3$.
You now want to show that
this holds for
$n+1$,
so that $s(n+1) \ge c(n+1)^3$.
Since $s(n+1)
= s(n)+(n+1)^2$,
we need to 
find a value of $c$
such that
$s(n) \ge cn^3$
implies that
$s(n+1) \ge c(n+1)^3
$.
The basic identities needed
are that
$s(n+1)-s(n)
=(n+1)^2
$
and
$(n+1)^3-n^3
=3n^2+3n+1
$.
Since we want
$s(n+1)
\ge c(n+1)^3
$,
using the identities,
this is the same as
$s(n)+(n+1)^2
\ge c(n^3+3n^2+3n+1)
$.
Since we are assuming that
$s(n) \ge cn^3
$,
this is implied by
$cn^3+(n+1)^2
\ge c(n^3+3n^2+3n+1)
$,
or
$n^2+2n+1
\ge 
c(3n^2+3n+1)
$
or
$n^2(1-3c)+n(2-3c)+(1-c)
\ge 
0
$.
The easiest way to have this happen
is to have all those
coefficients $\ge 0$.
It turns out that
$c=\frac13$ is the largest value
that works for all three.
Therefore,
we have shown that
if 
$s(n) \ge \frac{n^3}{3}$,
then
$s(n+1) \ge \frac{(n+1)^3}{3}
$.
Finally,
since
$s(1) 
\ge \frac13
$,
$s(n)
\ge \frac{n^3}{3}
$ holds for all $n$
and we are done.
