\frac{1}{2x+y+z}=\frac{1}{(x+y)+(x+z))}\leq \frac{1}{4}(\frac{1}{x+y}+\frac{1}{x+z})\leq \frac{1}{16}(\frac{2}{x}+\frac{1}{y}+\frac{1}{z})
\frac{1}{x+2y+z}=\frac{1}{(x+y)+(y+z)}\leq \frac{1}{16}(\frac{1}{x}+\frac{2}{y}+\frac{1}{z})
\frac{1}{x+y+2z}\leq...