$x+y+xy \ge 7\\ \Leftrightarrow x(y+1) \ge 7 -y\\ \Leftrightarrow x \ge \dfrac{7 -y}{y+1}(y>0 \Rightarrow y+1 >0)\\ S=x+2y\\ =\dfrac{7 -y}{y+1}+2y\\ =\dfrac{-(1 +y)+8}{y+1}+2y\\ =-1+\dfrac{8}{y+1}+2y+2-2\\ =\dfrac{8}{y+1}+2(y+1)-3\\ \ge 2\sqrt{\dfrac{8}{y+1}.2(y+1)}-3(Cauchy)\\ =5$
Dấu $"="$ xảy ra
$\Leftrightarrow \left\{\begin{array}{l} \dfrac{8}{y+1}=2(y+1)\\ x = \dfrac{7 -y}{y+1}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} (y+1)^2=16\\ x = \dfrac{7 -y}{y+1}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} y=3\\ x = 1\end{array} \right.$
Vậy $min_S=5$ xảy ra khi $x=1;y=3$
