\(2\sqrt{xy}\le x+y\le1\Rightarrow\sqrt{xy}\le\frac{1}{2}\Rightarrow xy\le\frac{1}{4}\Rightarrow\frac{1}{xy}\ge4\)
\(A=xy+\frac{1}{xy}=xy+\frac{1}{16xy}+\frac{15}{16xy}\ge2\sqrt{\frac{xy}{16xy}}+\frac{15}{16}.4=\frac{17}{4}\)
\(\Rightarrow A_{min}=\frac{17}{4}\) khi \(x=y=\frac{1}{2}\)
b/ \(2y=xy-x=x\left(y-1\right)\Rightarrow x=\frac{2y}{y-1}=2+\frac{2}{y-1}\)
Đồng thời \(x;y>0\Rightarrow2y=x\left(y-1\right)>0\Rightarrow y-1>0\)
\(\Rightarrow S=2+\frac{2}{y-1}+2y=4+\frac{2}{y-1}+2\left(y-1\right)\ge4+2\sqrt{\frac{4\left(y-1\right)}{y-1}}=8\)
\(\Rightarrow S_{min}=8\) khi \(\frac{2}{y-1}=2\left(y-1\right)\Rightarrow y=2\Rightarrow x=4\)
c/ \(x+y+xy\ge7\Leftrightarrow x\left(y+1\right)\ge7-y\Leftrightarrow x\ge\frac{7-y}{y+1}=\frac{8}{y+1}-1\)
\(\Rightarrow S=x+2y\ge2y+\frac{8}{y+1}-1=2\left(y+1\right)+\frac{8}{y+1}-3\)
\(\Rightarrow S\ge2\sqrt{\frac{16\left(y+1\right)}{y+1}}-3=5\)
\(\Rightarrow S_{min}=5\) khi \(\left\{{}\begin{matrix}y=1\\x=5\end{matrix}\right.\)
