$\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$ ( cauchy schwarz)Vậy $min=36$ khi $x=\frac{1}{6};y=\frac{1}{3};z=\frac{1}{2}$
chứng minh : $x,y>0$
$(x+y)(\frac{a^2}{x}+\frac{b^2}{y})\geq (\sqrt{x}.\frac{a}{\sqrt{x}}+\sqrt{y}.\frac{b}{\sqrt{y}})^2=(a+b)^2$
$\frac{a^2}{x}+\frac{b^2}{y}\geq \frac{(a+b)^2}{x+y}$-----> đây là schwarz ( gọi chung BCS)