[tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3[/tex]
Bunhia: [tex]A^2\leq 3(\sum \frac{1}{3x+y})= 3\left ( \sum \frac{1}{x+x+x+y} \right )\leq \frac{3}{16}\left ( \sum \frac{3}{x}+\frac{1}{y} \right )=\frac{3}{4}\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )=\frac{9}{4}[/tex]