[TEX]B=x^2-2x+y^2-4y+6=(x^2-2x+1)+(y^2-4y+4)+1=(x-1)^2+(y-2)^2+1[/TEX]
Do [TEX](x-1)^2 \geq 0, (y-2)^2 \geq 0 \Rightarrow B \geq 1[/TEX] với mọi [TEX]x,y[/TEX]
Dấu "=" xảy ra khi [tex]\left\{\begin{matrix} x-1=0 & \\y-2=0 & \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} x=1 & \\ y=2 & \end{matrix}\right.[/tex]