Ta có: $A=\frac{x^2+y^2}{1-(x+y)+xy}\geq \frac{(x+y)^2}{2(1-(x+y)+\frac{(x+y)^2}{4})}=\frac{(x+y)^2}{2(1-\frac{x+y}{2})^2}$$\Rightarrow A\geq \frac{2S^2}{(2-S)^2} (S=x+y>0)$
Ta có: $1=x^3+y^3=(x+y)(x^2+y^2-xy)\geq \frac{1}{4}(x+y)^3\Rightarrow S\leq \sqrt[3]{4}$
Khảo sát thôi bạn nhé.