Question

cho x,y>0; x+y=1. Tìm GTNN của: \(\frac{1}{xy}+\frac{1}{x^2+y^2}\)

Answer 1

\(P=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{x^2+y^2+2xy}+\frac{1}{2.\frac{1}{4}\left(x+y\right)^2}=6\)

\(P_{min}=6\) khi \(x=y=1\)

Answer 2

\(x=y=\frac{1}{2}\) chứ :>>>