\(A=xx+yy=x^2+y^2\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(1\cdot x+1\cdot y\right)^2=4\)
\(\Rightarrow2\left(x^2+y^2\right)\ge4\)
\(\Rightarrow A\ge2\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix}x+y=2\\x=y\end{matrix}\right.\)\(\Rightarrow x=y=1\)