jmc

By AM-GM, $$\begin{array}{rl}x+\frac{1}{y}& \ge 2\sqrt{\frac{x}{y}},\\ y+\frac{1}{z}& \ge 2\sqrt{\frac{y}{z}},\\ z+\frac{1}{x}& \ge 2\sqrt{\frac{z}{x}},\end{array}$$so $$\left(x+\frac{1}{y}\right)\left(y+\frac{1}{z}\right)\left(z+\frac{1}{x}\right)\ge 2\sqrt{\frac{x}{y}}\cdot 2\sqrt{\frac{y}{z}}\cdot 2\sqrt{\frac{z}{x}}=8.$$Equality occurs when $x=\frac{1}{y},$ $y=\frac{1}{z},$ and $z=\frac{1}{x},$ or $xy=1,$ $yz=1,$ and $xz=1.$ Multiplying all three equations, we get $x^2{y}^2{z}^2=1.$ Since $x,$ $y,$ and $z$ are all positive, $$xyz=1.$$Dividing $yz=1,$ we get $x=1.$ Similarly, $y=1$ and $z=1,$ so there is only $\boxed{1}$ triple $(x,y,z),$ namely $(1,1,1).$