Dấu “=” xảy ra $\Leftrightarrow \left\{ \begin{align}& x+y-2=0 \\ & x-1=0 \\ & y-1=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x=1 \\ & y=1 \\
\end{align} \right.$

$C={{x}^{2}}+xy+{{y}^{2}}-3x-3y$

$C=\frac{1}{2}\left( 2{{x}^{2}}+2xy+2{{y}^{2}}-6x-6y \right)$

$C=\frac{1}{2}\left[ \left( {{x}^{2}}+{{y}^{2}}+4+2xy-4x-4y \right)+\left( {{x}^{2}}-2x+1 \right)+\left( {{y}^{2}}-2y+1 \right)-6 \right]$

$C=\frac{1}{2}\left[ {{\left( x+y-2 \right)}^{2}}+{{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}} \right]-3$

Ta có: ${{\left( x+y-2 \right)}^{2}}\ge 0;\,\,{{\left( x-1 \right)}^{2}}\ge 0;\,\,\left( y-1 \right)\ge 0$ với mọi $x,y$

Suy ra $\frac{1}{2}\left[ {{\left( x+y-2 \right)}^{2}}+{{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}} \right]-3\ge -3$

Dấu “=” xảy ra $\Leftrightarrow \left\{ \begin{align}& x+y-2=0 \\ & x-1=0 \\ & y-1=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x=1 \\ & y=1 \\
\end{align} \right.$

Vậy GTNN của $C=-3$ khi $x=y=1$.