\(\Leftrightarrow6\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+20=\dfrac{5\left(x+y\right)\left(xy+3\right)}{xy}\ge\dfrac{5\left(x+y\right)2\sqrt{3xy}}{xy}=10\sqrt{3}\left(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}\right)\)
Đặt \(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=t\ge2\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}=t^2-2\)
\(\Rightarrow6\left(t^2-2\right)+20\ge10\sqrt{3}t\)
\(\Rightarrow3t^2-5\sqrt{3}t+4\ge0\)
\(\Rightarrow\left(\sqrt{3}t-1\right)\left(\sqrt{3}t-4\right)\ge0\)
Do \(t\ge2\Rightarrow\sqrt{3}t-1>0\)
\(\Rightarrow\sqrt{3}t-4\ge0\Rightarrow t\ge\dfrac{4}{\sqrt{3}}\)
\(\Rightarrow t^2\ge\dfrac{16}{3}\Rightarrow t^2-2\ge\dfrac{10}{3}\)
\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}\ge\dfrac{10}{3}\) (do \(\dfrac{x}{y}+\dfrac{y}{x}=t^2-2\))
Vậy \(A_{min}=\dfrac{10}{3}\) khi \(\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)
