Question

Cho x>0,y>0 và x+y<=1

tìm Min K =\(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

Answer 1

\(K=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

Ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\)

\(\Rightarrow\left(x+\frac{1}{x}\right)^2\ge4\)

\(\Rightarrow\left(y+\frac{1}{y}\right)^2\ge4\)

\(\Rightarrow M\ge8\)

Answer 2

\(K\ge\frac{1}{2}\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2=\frac{1}{2}\left(4x+\frac{1}{x}+4y+\frac{1}{y}-3\left(x+y\right)\right)^2\)

\(K\ge\frac{1}{2}\left(2\sqrt{\frac{4x}{x}}+2\sqrt{\frac{4y}{y}}-3.1\right)^2=\frac{25}{2}\)

\(\Rightarrow K_{min}=\frac{25}{2}\) khi \(x=y=\frac{1}{2}\)