Question

Cho x,y>0 , x+y=1.  Chứng minh :  \(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{25}{2}\)

Answer 1

lớp 9 hở.... ặc ặc...............

Answer 2

nhầm,mấy bất đẳng thức tính cực trị từ cô si

Answer 3

Ta có \(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2=\frac{1}{2}\left(1+\frac{1}{xy}\right)^2\).

Theo bất đẳng Cô-Si ta có \(1=x+y\ge2\sqrt{xy}\to xy\le\frac{1}{4}\to\) 

\(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(1+\frac{1}{xy}\right)^2\ge\frac{1}{2}\left(1+4\right)^2=\frac{25}{2}.\)