[tex](x-1)^{2}+(y-1)^{2}+(x-1)(y-1)+2008=(x-1)^{2}+2(x-1).\frac{1}{2}(y-1)+\frac{1}{4}(y-1)^{2}+\frac{3}{4}(y-1)^{2}+2008=(x-1+\frac{1}{2}y-\frac{1}{2})^{2}+\frac{3}{4}(y-1)^{2}+2008[/tex]
Hai cái bình phương đầu luôn [tex]\geq 0[/tex] nên BT[tex]\geq 2008[/tex]