<s>[INST] 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.
 [/INST] If you're allowed to use $\sum\limits_{k=1}^{n} k^2 =\frac{n(n+1)(2n+1)}{6} $,
$$1^2 + 2^2 + {…}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {…}+ {n}^2$$
because
$\dfrac{(n-1)n(n-\frac{1}{2})}{3} \lt \dfrac{n^3}{3}$
and
$\dfrac{n(n+1)(n+\frac{1}{2})}{3} \gt \dfrac{n^3}{3}$ 
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