$P=x^2+y^2+xy-2x-3y+1$
$ =x^2+xy-2x+y^2-3y+1$
$ =x^2+x(y-2)+\frac{1}{4}(y-2)^2-\frac{1}{4}(y-2)^2+y^2-3y+1$
$ =[x+\frac{1}{2}(y-2)]^2+\frac{3}{4}y^2-2y$
$ =[x+\frac{1}{2}(y-2)]^2+\frac{3}{4}(y^2-\frac{8}{3}y+\frac{16}{9})-\frac{4}{3}$
$ =[x+\frac{1}{2}(y-2)]^2+\frac{3}{4}(y-\frac{4}{3})^2-\frac{4}{3}$$\geq$ $\frac{-4}{3}$
Dấu "=" xảy ra khi :
$\left \{ {{x+\frac{1}{2}(y-2)=0} \atop {y-\frac{4}{3}=0}} \right.$ ⇔ $\left \{ {{x=\frac{1}{3}} \atop {y=\frac{4}{3}}} \right.$
Vậy GTNN của P là $\frac{-4}{3}$ khi $x=\frac{1}{3},y=\frac{4}{3}$
