A Classic Substitution

We can rewrite the given equation as xyz-(x+y+z)=2.

The AM-GM inequality gives us $\frac{x+y+z}{3}=\sqrt[3]{xyz}$ .

Equality occurs only when x=y=z.

We can now solve for x in $x^3-3x-2=0$ which factors to give us $(x+1)^2(x-2)$ .

Now since x, y, and z are positive, $x=2$ is the only valid solution.

This gives us $0.5 \cdot 3 = \boxed{1.5}$ as the answer.

It can be shown that $(x, y, z) = (\frac{a+b}{c}, \frac{b+c}{a}, \frac{c+a}{b})$ , for some positive real numbers $a, b,$ and $c.$ Thus,

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}\geq\boxed{1.5}$