$S=x^{2}+xy+y^{2}-3x-3y+15$
$S=x^{2}+xy-3x+y^{2}-3y+15$
$S=x^{2}+2x(\frac{1}{2}y-\frac{3}{2})+(\frac{1}{2}y-\frac{3}{2})^{2}-(\frac{1}{2}y-\frac{3}{2})^{2}+y^{2}-3y+15$
$S=[x+(\frac{1}{2}y+\frac{3}{2})]^{2}-\frac{1}{4}y^{2}+\frac{3}{2}y-\frac{9}{4}-3y+15$
$S=[x+(\frac{1}{2}y+\frac{3}{2})]^{2}+\frac{3}{4}y^{2}-\frac{3}{2}y+\frac{51}{4}$
$S=[x+(\frac{1}{2}y+\frac{3}{2})]^{2}+\frac{3}{4}y^{2}-\frac{3}{2}y+\frac{3}{4}+12$
$S=[x+(\frac{1}{2}y+\frac{3}{2})]^{2}+\frac{3}{4}(y^{2}-2y+1)+12$
$S=[x+(\frac{1}{2}y+\frac{3}{2})]^{2}+\frac{3}{4}(y-1)^{2}+12\geq12$
Dấu $"="$ xảy ra
`<=>`$\left \{ {{x+(\frac{1}{2}y+\frac{3}{2})=0} \atop {y-1=0}} \right.$
`<=>`$\left \{ {{x+(\frac{1}{2}.1+\frac{3}{2})=0} \atop {y=1}} \right.$
`<=>`$\left \{ {{x=1} \atop {y=1}} \right.$
`<=>`$x=y=1$
Vậy $MinS=12$ khi $x=y=1$
