Question

Cho x,y>0.Tìm Min của \(A=\dfrac{xy}{x^2+y^2}+\dfrac{x^2+y^2}{xy}\)

Answer 1

\(\dfrac{x^2+y^2}{xy}=t;x,y>0\Rightarrow t\ge2\) khi x=y

\(A=t+\dfrac{1}{t}\ge2+\dfrac{1}{2}=\dfrac{5}{2}\)

\(A-\dfrac{5}{2}=\left(t-2\right)+\left(\dfrac{1}{t}-\dfrac{1}{2}\right)=\left(t-2\right)-\dfrac{\left(t-2\right)}{2t}=\dfrac{\left(2t-1\right)\left(t-2\right)}{2t}\)

\(t\ge2\Rightarrow\left\{{}\begin{matrix}2t-1>0\\t-2\ge0\\2t>0\end{matrix}\right.\)\(\Rightarrow\dfrac{\left(2t-1\right)\left(t-2\right)}{2t}\ge0\) đẳng thức khi t=2

\(\Rightarrow A-\dfrac{5}{2}\ge0\Rightarrow A\ge\dfrac{5}{2}\)

Vậy GTNN (A) =5/2 khi x=y