prove that $(p,mp]$ will always contain at least $y-1$ cubes 
If $n, m\in \mathbb{Z}_{>0}$ and $(n, mn]$ contains $y$ cubes, prove that $(p,mp]$ will always contain at least $y-1$ cubes (integers $a$ so that $a=k^3$ for some integer k) for all $p\in [n,\infty)$.

This works for small examples, such as for $n=7, m=4$ when the cubes are $8$ and $27$ as $(p, 4p]$ always contains at least one cube for $p\ge n$, but I'm not sure how to generalize the results. I know that between any two real numbers that are at least one apart there is an integer. Also, clearly, the ratio between $k^3$ and $(k+1)^3$ decreases as $k$ increases. For $n\ge 8$, there exists an integer $k$ so that $k \in (n, mn]$ is a perfect cube because $\sqrt[3]{mn} - \sqrt[3]{n} \ge \sqrt[3]{n} > 1$.

Lemma (without proof): Show that $ ( n, mn ] $ contains at least $y$ cubes iff
$$ \lfloor \sqrt[3]{mn} \rfloor  - \lfloor \sqrt[3] {n} \rfloor \geq y.$$
Corollary: The problem can be restated to:

For $ p \geq n \geq 1$, and $ m \geq 1$, show that
$$ \lfloor \sqrt[3]{mp} \rfloor  - \lfloor \sqrt[3] {p} \rfloor \geq \lfloor \sqrt[3]{mn} \rfloor  - \lfloor \sqrt[3] {n} \rfloor - 1.$$

Proof of inequality: This is  true because with $ \sqrt[3]{m} \geq 1$,
$$ \lfloor \sqrt[3]{mp} \rfloor  - \lfloor \sqrt[3]{mn} \rfloor  \geq \lfloor \sqrt[3]{mp} - \sqrt[3]{mn} \rfloor \geq \lfloor \sqrt[3]{m}  \times \lfloor \sqrt[3]{p} - \sqrt[3]{n} \rfloor  \rfloor \\ \geq  \lfloor \sqrt[3]{p} - \sqrt[3]{n} \rfloor \geq  \lfloor \sqrt[3]{p}  \rfloor -  \lfloor\sqrt[3]{n} \rfloor  - 1.$$

The absolute-function inequalities used at each stage of the inequality are derived from the well-known inequalities (for suitable values of $x, y$, and rearranged as needed):

*

*$ \lfloor x \rfloor + \lfloor y \rfloor  \leq \lfloor x+y \rfloor$

*$ \lfloor xy \rfloor \geq \lfloor x \lfloor y \rfloor \rfloor$.

*For $ x \geq 1$, $ \lfloor x \lfloor y \rfloor \rfloor \geq \lfloor y \rfloor$

*$ \lfloor x \rfloor + \lfloor y \rfloor  \geq \lfloor x+y \rfloor - 1$
