Ta có:\(x+\dfrac{1}{y}\le\dfrac{1}{2}\)\(\Rightarrow\)\(2xy+2\le y\)\(\le\dfrac{y^2+16}{8}\)(Cauchy)
\(\Rightarrow\)\(16xy+16\le y^2+16\Rightarrow16x\le y\)
Ta có:\(\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x}{y}+\dfrac{y}{256x}+\dfrac{255y}{256x}\)\(\ge2\sqrt{\dfrac{x.y}{y.256x}}+\dfrac{255.16x}{256x}=\dfrac{1}{8}+\dfrac{255}{16}=\dfrac{257}{16}\)
Dấu ''='' xảy ra \(\Leftrightarrow\) y=4;x=\(\dfrac{1}{4}\)