Ta có:
\(P=\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}-3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+5\)
\(=\dfrac{x^2}{y^2}-3\dfrac{x}{y}+\dfrac{9}{4}+\dfrac{y^2}{x^2}-3\dfrac{y}{x}+\dfrac{9}{4}+\dfrac{1}{2}\)
\(=\left(\dfrac{x}{y}-\dfrac{3}{2}\right)^2+\left(\dfrac{y}{x}-\dfrac{3}{2}\right)^2+\dfrac{1}{2}\)
Với \(x;y\ne0\) ta có:
\(\left(\dfrac{x}{y}-\dfrac{3}{2}\right)^2\ge0;\left(\dfrac{y}{x}-\dfrac{3}{2}\right)^2\ge0\)
\(\Rightarrow\left(\dfrac{x}{y}-\dfrac{3}{2}\right)^2+\left(\dfrac{y}{x}-\dfrac{3}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)
Vậy Min P = \(\dfrac{1}{2}\)
Để \(P=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{3}{2}=0\\\dfrac{y}{x}-\dfrac{3}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{3}{2}\\\dfrac{y}{x}=\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow x=y=0\)
