Prove or disprove this inequality

Question

Let $p, q, a$, and $b$ be natural numbers such that $p<q$, $1<b<a$ and $b\nmid a$.
Is is true that $(bp+aq)^3> (a^3+b^3)q^3$?

This is what I tried: expanding the left-hand side, we conclude that for the inequality to be satisfied we need to have $p^3b+3pq+3q^2>q^3b$. But I can't go further. Can you please give me the idea to complete the argument?

Answer 1

To add to the counterexample provided in the other answer:

Dividing both sides by $q^3$, you can write the inequality as

$$\left(a+b\frac{p}{q}\right)^3>a^3+b^3.$$

From this it should be obvious that since you can make $q$ as large as you like relative to $p$, you can make $p/q\approx 0$, so that the left-hand side is approximately $a^3$, which cannot be larger than $a^3+b^3$.

(If the restriction were instead $q\leq p$, then the inequality would hold.)

Answer 2

No, e.g. you could have $a=3$, $b=2$, $p=1$. Then $$(bp+aq)^3 = (2+3q)^3 = 27 q^3 + 54 q^2+36 q+8< 35 q^3 = (a^3 + b^3) q^3$$ for sufficiently large $q$ ($q \ge 8$ turns out to be sufficient).