One, (40 points) Find all positive rational triples $(x, y, z)(x \leqslant y \leqslant z)$ such that
$$
x+y+z 、 \frac{1}{x}+\frac{1}{y}+\frac{1}{z} 、 x y z
$$

are all integers.
Consider the polynomial with roots $x$, $y$, $z$:
$$
\begin{array}{l}
f(t)=(t-x)(t-y)(t-z) \\
=t^{3}-(x+y+z) t^{2}+(x y+y z+z x) t-x y z .
\end{array}
$$

Notice that, $x y+y z+z x=x y z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is an integer. Therefore, $f(t)$ is a monic polynomial with integer coefficients. Since its roots are rational, the denominators of the roots must be divisors of the leading coefficient, hence the roots must be integers, i.e., $x, y, z$ are all integers.
Let $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=k$.
Since $x \leqslant y \leqslant z$, we have
$$
\frac{3}{x} \geqslant \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=k \geqslant 1 \Rightarrow x \leqslant 3 \text {. }
$$
(1) When $x=1$, $\frac{1}{y}+\frac{1}{z}$ is an integer.

If $y=1$, then $z=1$, yielding $(1,1,1)$;
If $y=2$, by $x+y+z=3+z$ being an integer, we know $z$ is an integer, hence $z=2$, yielding $(1,2,2)$;
If $y \geqslant 3$, then $\frac{1}{y}+\frac{1}{z} \leqslant \frac{1}{3}+\frac{1}{3}$, which is a contradiction.
(2) When $x=2$,
$$
\begin{array}{l}
\frac{2}{y} \geqslant \frac{1}{y}+\frac{1}{z}=k-\frac{1}{2} \geqslant 1-\frac{1}{2}=\frac{1}{2} \\
\Rightarrow y \leqslant 4 .
\end{array}
$$

If $y=2$, then $z=1$, which contradicts $z \geqslant y=2$.
If $y=3$, then $z=6$, yielding $(2,3,6)$;
If $y=4$, then $z=4$, yielding $(2,4,4)$.
(3) When $x=3$,
$$
\begin{array}{l}
\frac{2}{y} \geqslant \frac{1}{y}+\frac{1}{z}=k-\frac{1}{3} \geqslant 1-\frac{1}{3}=\frac{2}{3} \\
\Rightarrow y \leqslant 3 .
\end{array}
$$

Since $y \geqslant x=3$, then $y=3, z=3$. Hence $(3,3,3)$.
In summary,
$$
\begin{aligned}
(x, y, z)= & (1,1,1),(1,2,2),(2,3,6), \\
& (2,4,4),(3,3,3) .
\end{aligned}
$$