Strange inequality in an NT problem
Solution 1:
Notice that $$a^2+ab+b^2\mid ab(a+b)\iff a^2+ab+b^2\mid a^3\iff a^2+ab+b^2\mid b^3$$ $$\iff a^2+ab+b^2\mid \gcd(a,b)^3$$
Thus, let $d:=\gcd(a,b)$, and let $x:=\frac{a}d, y:=\frac{b}d$. It follows that $$a^2+ab+b^2\mid \gcd(a,b)^3\iff x^2+xy+y^2\mid d\implies d>xy$$
You are left to show that $\lvert a-b\rvert>\sqrt[3]{ab}\iff \lvert dx-dy\rvert^3=\lvert a-b\rvert^3> ab = d^2\cdot xy\iff d\lvert x-y\rvert > xy$. But this is trivial, since $d>xy$ and $\lvert x-y\rvert\geqslant 1.$