\(F\left(x\right)=\left(x^2+8x+16\right)+\left(y^2-6y+9\right)-35\)
\(=\left(x+4\right)^2+\left(y-3\right)^2-35\ge-35\forall x,y\)
Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+4\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)