[tex]A=\frac{x^2}{y}+\frac{y^2}{x} \geq x+y[/tex]
Ta có: [tex]\frac{1}{2}(\sqrt{x}-\sqrt{y})^2+\frac{1}{2}(\sqrt{x}-1)^2+\frac{1}{2}(\sqrt{y}-1)^2\geq 0\\\rightarrow x+y-\sqrt{xy}-\sqrt{x}-\sqrt{y}+1\geq 0\\\rightarrow x+y\geq \sqrt{xy}+\sqrt{x}+\sqrt{y}-1=(\sqrt{x}+1)(\sqrt{y}+1) -2 \geq 4-2=2[/tex]
Dấu = khi x=y=1