\(P=x^2+xy+y^2-3\left(x+y\right)+3\)
\(\Leftrightarrow2P=2x^2+2xy+2y^2-6\left(x+y\right)+6\)
\(=\left(x^2+2xy+y^2\right)-4\left(x+y\right)+4+\left(x^2-2x+1\right)+\left(y^2-2y+1\right)\)
\(=\left(x+y-2\right)^2+\left(x-1\right)^2+\left(y-1\right)^2\ge0\)
Dấu = xảy ra khi \(x=y=1\)