$\text{Đề bài cho x,y dương có tổng bằng 1 tìm Min của A}\\
\text{x+y=1}\rightarrow xy\leq \frac{1}{4}\\
\text{Ta có A= }(\dfrac{1}{x}+1)(\dfrac{1}{y}+1)\\
\quad\quad\quad=\dfrac{(x+1)(y+1)}{xy}\\
\quad\quad\quad=\dfrac{(x+\frac{1}{2}+\frac{1}{2})(y+\frac{1}{2}+\frac{1}{2})}{xy}\\
\quad\quad\quad\geq \dfrac{9\sqrt[3]{\frac{xy}{16}}}{xy}\\
\quad\quad\quad=\dfrac{9}{\sqrt[3]{16}.\sqrt[3]{(xy)^{2}}}\\
\quad\quad\quad\geq 9\\
\text{Dấu = xảy ra x=y=$\frac{1}{2}$}$
