\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow2x^2+2y^2\ge x^2+y^2+2xy\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\Rightarrow x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2=\frac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)