Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có:
\(P=\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{4}{x^2+y^2}=\frac{4}{20}=\frac{1}{5}\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix}x,y>0\\x=y\\x^2+y^2=20\end{matrix}\right.\)\(\Rightarrow x=y=\sqrt{10}\)
Vậy \(Min_P=\frac{1}{5}\) khi \(x=y=\sqrt{10}\)