Sử dụng bất đẳng thức Svac-xơ, ta có:
[math]\frac{1}{3x+y}\le \frac{1}{16}(\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{y})[/math][math]=>\frac{1}{3x+y}\le \frac{1}{16}(\frac{3}{x}+\frac{1}{y})[/math][math]=>\sum\frac{1}{3x+y}\le \frac{1}{16}\sum(\frac{3}{x}+\frac{1}{y})=\frac{1}{16}.4\sum\frac{1}{x}=\frac{1}{4}.3=\frac{3}{4} \textnormal{(đpcm)}[/math][math]\textnormal{Dấu bằng xảy ra}\Leftrightarrow x=y=z=1[/math]