A= \(x^3+y^3+2xy=\left(x+y\right)\left(x^2+y^2-xy\right)+2xy\)
= \(2\left(x^2+y^2-xy\right)+2xy=2x^2+2y^2-2xy+2xy\)
= \(\left(1^2+1^2\right)\left(x^2+y^2\right)\) \(\ge\left(x+y\right)^2\) = 22 = 4(Ap dụng BĐT bunhiacopsky), dấu "=" xảy ra khi \(\dfrac{1}{x}=\dfrac{1}{y}\Leftrightarrow x=y=1\)
Vậy MinA = 4 <=> x = y =1.