How to use mathematical induction with inequality? I am stuck with this question.
Given that $n$ is a positive integer where $n≥2$, prove by the method of mathematical induction that
(a) $$ \sum_{r=1}^{n-1} r^3 < \frac{n^4}{4} $$ 
(b) $$ \sum_{r=1}^{n} r^3 > \frac{n^4}{4} $$
 A: Hint for the first question:


*

*Assume $\displaystyle\sum\limits_{r=1}^{n-1}r^3<\frac{n^4}{4}$

*Prove $\displaystyle\sum\limits_{r=1}^{n}r^3<\frac{(n+1)^4}{4}$
$\displaystyle\sum\limits_{r=1}^{n}r^3=$
$\displaystyle\sum\limits_{r=1}^{n-1}r^3+n^3<$
$\displaystyle\frac{n^4}{4}+n^3=$
$\displaystyle\frac{n^4+4n^3}{4}<$
$\displaystyle\frac{n^4+4n^3+6n^2+4n+1}{4}=$
$\displaystyle\frac{(n+1)^4}{4}$
Note that the assumption is used in order to infer $\displaystyle\sum\limits_{r=1}^{n-1}r^3+n^3<\frac{n^4}{4}+n^3$

Hint for the second question:


*

*Assume $\displaystyle\sum\limits_{r=1}^{n}r^3>\frac{n^4}{4}$

*Prove $\displaystyle\sum\limits_{r=1}^{n+1}r^3>\frac{(n+1)^4}{4}$
$\displaystyle\sum\limits_{r=1}^{n+1}r^3=$
$\displaystyle\sum\limits_{r=1}^{n}r^3+(n+1)^3>$
$\displaystyle\frac{n^4}{4}+(n+1)^3=$
$\displaystyle\frac{n^4+4n^3+12n^2+12n+4}{4}>$
$\displaystyle\frac{n^4+4n^3+6n^2+4n+1}{4}=$
$\displaystyle\frac{(n+1)^4}{4}$
Note that the assumption is used in order to infer $\displaystyle\sum\limits_{r=1}^{n}r^3+(n+1)^3>\frac{n^4}{4}+(n+1)^3$
