Question

cho x,y >0 và x+y=1

cm [ x+ 1/x ]² + [ y + 1/y ]² ≥ 25/2

Answer 1

Lời giải: 

Áp dụng BĐT Bunhiacopxky:

$[(x+\frac{1}{x})^2+(y+\frac{1}{y})^2](1+1)\geq (x+\frac{1}{x}+y+\frac{1}{y})^2$

$\Leftrightarrow (x+\frac{1}{x})^2+(y+\frac{1}{y})^2\geq \frac{1}{2}(x+y+\frac{1}{x}+\frac{1}{y})^2=\frac{1}{2}(1+\frac{1}{xy})^2$

Mà: 
$xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}$ theo BĐT Cô-si

$\Rightarrow (x+\frac{1}{x})^2+(y+\frac{1}{y})^2\geq \frac{1}{2}(1+\frac{1}{\frac{1}{4}})^2=\frac{25}{2}$ (đpcm)

Dấu "=" xảy ra khi $x=y=\frac{1}{2}$