\(1=\dfrac{1}{x}+\dfrac{4}{y}\ge2\sqrt{\dfrac{4}{xy}}=4\sqrt{\dfrac{1}{xy}}\left(Bđt.Cauchy\right)\)
\(\Leftrightarrow\sqrt{xy}\ge4\)
\(A=x+y\ge2\sqrt{xy}\left(Bđt.Cauchy\right)\)
\(\Leftrightarrow A=x+y\ge2.4=8\)
Vậy \(A_{min}=8\)
Dâu '=' xảy ra khi \(\dfrac{1}{x}=\dfrac{4}{y}\Leftrightarrow y=4x\)