\(A=x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2=\frac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)
\(\left\{{}\begin{matrix}x;y\ge0\\x+y=1\end{matrix}\right.\) \(\Rightarrow0\le x;y\le1\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\end{matrix}\right.\) \(\Rightarrow A=x^2+y^2\le x+y=1\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)