How to use mathematical induction with inequality?

Question

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) $$ \sum_{r=1}^{n} r^3 > \frac{n^4}{4} $$

Answer 1

Hint for the first question:

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

2. 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 $\displaystyle\sum\limits_{r=1}^{n-1}r^3+n^3<\frac{n^4}{4}+n^3$

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Hint for the second question:

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

2. 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 $\displaystyle\sum\limits_{r=1}^{n}r^3+(n+1)^3>\frac{n^4}{4}+(n+1)^3$