jmc

By the AM-HM inequality, $$\frac{x+y}{2}\ge \frac{2}{\frac{1}{x}+\frac{1}{y}}=\frac{2xy}{x+y},$$so $\frac{x+y}{xy}\ge \frac{4}{x+y}.$ Hence, $$\frac{x+y}{xyz}\ge \frac{4}{(x+y)z}.$$By the AM-GM inequality, $$\sqrt{(x+y)z}\le \frac{x+y+z}{2}=\frac{1}{2},$$so $(x+y)z\le \frac{1}{4}.$ Hence, $$\frac{4}{(x+y)z}\ge 16.$$Equality occurs when $x=y=\frac{1}{4}$ and $z=\frac{1}{2},$ so the minimum value is $\boxed{16}$.