Ta có: A=\(\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)\)
Ap1 dụng bdt cô si ta có:
\(x+\dfrac{1}{x}\ge2\sqrt{x.\dfrac{1}{x}}=2\\ y+\dfrac{1}{y}\ge2\sqrt{y.\dfrac{1}{y}}=2\)
=> A\(\ge2.2=4\) Dấu bằng xảy ra khi \(x=1;y=1\)
Có \(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\Leftrightarrow\dfrac{15}{16xy}\ge\dfrac{15}{4}\)
\(A=xy+\dfrac{1}{xy}+\dfrac{x}{y}+\dfrac{y}{x}=xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\ge2\sqrt{xy.\dfrac{1}{16xy}}+2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}+\dfrac{15}{4}\)
\(\Leftrightarrow A\ge\dfrac{25}{4}\)
Dấu "=" xảy ra khi\(x=y=\dfrac{1}{2}\)
