South African Mathematics Olympiad

Suppose the numbers are $x$, $y$, $z$, with $0<x<y<z$ and $x+y+z=7$. If $x\ge 2$, then $y\ge 3$ and $z\ge 4$, so $x+y+z\ge 9$, which is too large. Therefore $x=1$, which leaves $y+z=6$. Since $2\le y<z$, the only possibility is $y=2$ and $z=4$. The product $xyz=1\times 2\times 4=8$.