Cho 3 số dương x,y,z thỏa mãn x+y+z=1.Cmr:$\sum \sqrt{\frac{xy}{xy+z}}\leq \frac{3}{2}$
Ta có: $\sqrt \frac {xy}{xy+z}=\sqrt \frac {xy}{xy+z(x+y+z)}= \sqrt \frac{xy}{(z+y)(z+x)} \leq (\frac{x}{x+z}+\frac{y}{y+z}).\frac{1}{2}$
Suy ra: $\sum \sqrt{\frac{xy}{xy+z}}\leq \sum (\frac{x}{x+z}+\frac{y}{y+z}).\frac{1}{2} \leq \frac{3}{2}$
Dấu "=" xảy ra khi $x=y=z= \frac{1}{3}$