5. Find all positive integer solutions $x, y, z$ that satisfy the equation:
$$
5(x y+y z+x z)=4 x y z
$$
5. The original equation is transformed into $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{4}{5}$.

Assume $x \leqslant y \leqslant z$, then
$$
\frac{3}{x} \geqslant \frac{1}{x}+\frac{1}{y}+\frac{1}{x}=\frac{4}{5} \text {. }
$$

Therefore, $x1$.
Thus, we get $1<x<4$.
We will discuss two cases:
Case 1. When $x=2$,
$$
\frac{2}{y} \geqslant \frac{1}{y}+\frac{1}{z}=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}
$$
$$
\frac{1}{y}<\frac{1}{y}+\frac{1}{z}=\frac{3}{10},
$$

From this, we get $3<y<7$.
If $y=4$, then $z=20$,
If $y=5$, then $z=10$,
If $y=6$, then $z$ is not an integer.
Case 2. When $x=3$,
$$
\frac{2}{y} \geqslant \frac{4}{5}-\frac{1}{3}=\frac{7}{15}, \text { and } \frac{1}{y}<\frac{7}{15} .
$$

Therefore, $2<y<5$.
When $y=3$ or $y=4$, $z$ does not have an integer solution.
Thus, the original equation has 12 solutions:
$$
\begin{array}{l}
(2,4,20)(2,20,4),(4,2,20),(20,2,4) \\
(4,20,2),(20,4,2),(2,5,10)(2,10,5) \\
(5,2,10),(10,2,5),(5,10,2),(10,5,2) \text {. }
\end{array}
$$
(Translated by Li Zhengyin from Qufu Normal University from Singapore
《Mathematical medley》1988-12)