\(\)\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\Rightarrow2\ge\dfrac{4}{x+y}\Leftrightarrow x+y\ge2\)(chắc bài cho x,y>0?
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\Rightarrow\sqrt{xy}\ge1\Leftrightarrow xy\ge1\)
\(D=\dfrac{1}{x+2y}+\dfrac{1}{2x+y}=\dfrac{1}{x+y+y}+\dfrac{1}{x+x+y}\le\dfrac{1}{y+2}+\dfrac{1}{x+2}\)
\(cm:\dfrac{1}{x+2}+\dfrac{1}{y+2}\le\dfrac{2}{3}\Leftrightarrow\dfrac{x+y+4}{\left(x+2\right)\left(y+2\right)}\le\dfrac{2}{3}\)
\(\Leftrightarrow2\left(x+2\right)\left(y+2\right)\ge3\left(x+y+4\right)\Leftrightarrow4x+4y+8+2xy\ge3x+3y+12\Leftrightarrow x+y+2xy\ge4\left(1\right)\)
\(x+y\ge2;xy\ge1\Rightarrow\left(1\right)đúng\Rightarrow D\le\dfrac{2}{3}\Rightarrow dấu"="xayra\Leftrightarrow x=y=1\)
