China Girls' Mathematical Olympiad

In view of $x^3+y^3>0$, it suffices to show that $(x^2+4y^2)(x-y)<x^3+y^3$. Expanding the left-hand side of the last inequality and canceling the like terms we obtain $4x{y}^2<x^{2}y+5y^3$. By the AM-GM inequality, we have $x^{2}y+5y^3\ge 2\sqrt{5x{y}^2}>4x{y}^2$.