Inequality $\frac{x^3+y^3}{x-y}>4$ Let $x>y>0$ and $xy\geq 1$. Prove that $$\frac{x^3+y^3}{x-y}>4.$$
Of course we can factor $(x^3+y^3)=(x+y)(x^2-xy+y^2)$, but it is not very useful. For fixed $x-y$, we can try to find the minimum value of $x,y$ such that $xy\geq 1$, and show that $\frac{x^3+y^3}{x-y}>4$ for those values. But that amounts to solving the quadratic $x(x-k)= 1$, and the answer would be quite ugly to substitute into the inequality.

$$x^3+y^3 - 4(x-y) \ge 2(x^2 - xy +y^2) - 4(x-y) $$
$$= 2(x^2 - 2xy +y^2 - 2(x-y) + xy) \ge 2(x-y-1)^2$$
Where, we used $x+y \ge 2\sqrt{xy} \ge 2$ and $xy \ge 1$
