Ay! Turns out, the answer was 16, occurring when x=y=1/4 and z=1/2.
We use a combination of the AM-HM and the AM-GM in that order.
Starting, we have by the AM-HM \(\frac{x + y}{2} \ge \frac{2}{\frac{1}{x} + \frac{1}{y}} = \frac{2xy}{x + y}\)
So we have \(\frac{x + y}{xy} \ge \frac{4}{x + y}\).
Multiplying by 1/z, we get \(\frac{x + y}{xyz} \ge \frac{4}{(x + y)z}\).
By the AM-GM inequality,
\(\sqrt{(x + y)z} \le \frac{x + y + z}{2} = \frac{1}{2}\),
meaning \((x + y)z \le \frac{1}{4}\).
Hence,
\(\frac{4}{(x + y)z} \ge 16\).
Using what we had before, we can write
\(\frac{x + y}{xy} \ge \frac{4}{x + y} \ge 16\).
Equality occurs when x=y, and the minimum value is 16.
If you ask me, this solution was quite interesting. I thought of using the QM-AM-GM-HM relationship, as well as the Cauchy-Schwarz Inequality, but not in this way. Thanks for your help!
+876 
