Prove that $1^2 + 2^2 + {...}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {...}+ {n}^2 $ Prove that $1^2 + 2^2 + {...}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {...}+ {n}^2 $

I know I need to use induction for this proof, but it feels like a pretty complicated one.
Basis: For $n = 2$,
$$1^2 < \frac{8}3 < 1^2 + 2^2$$
Induction Hypothesis: Assume $P(n)$ holds for $n=k$, that is,
$$1^2 + 2^2 + {...}+ {(k - 1)}^2 < \frac{k^3}{3} < 1^2 + 2^2 + {...}+ {k}^2$$
We need to show that $P(n)$ also holds for $n=k+1$
Proof:
$$1^2+2^2+{...}+{(k)}^2=1^2+2^2+{...}+{(k-1)}^2+{k}^2$$
After this, I'm not sure how to use the assumed inequality to prove it because it's a less than inequality. If I could get a hint that'd be awesome.

HINT: Since we know that $1^2 + 2^2 + {...}+ {(k - 1)}^2 < \frac{k^3}{3} < 1^2 + 2^2 + {...}+ {k}^2$ by induction hypothesis, what we need to prove is that
$$k^2 \le \frac{3k^2+3k+1}{3} \le (k+1)^2$$
