Proof: As shown in Figure 1, for definiteness, let the height from vertex $A$ to the opposite face $BCD$ in tetrahedron $A-BCD$ be $AH = h$, and the distance between edges $AB$ and $CD$ be $d$.

In the plane $BCD$, draw a line $l \parallel CD$ through $B$, and draw $EF \perp CD$ at $F$, intersecting $l$ at $E$. Let $FG$ and $EK$ be the altitudes of $\triangle AEF$. It is easy to see that $l \perp$ plane $AEF$, so $FG \perp l$. Also, $FG \perp AE$, hence $FG$ is the distance from $E$ to plane $AEB$. Since $CD \parallel$ plane $AEB$, $FG$ is the distance between the skew lines $CD$ and $AB$. Similarly, $EK$ is equal to the height of tetrahedron $A-BCD$ from vertex $B$. Given that $EK \geqslant AH$, from $AH \cdot EF = EK \cdot AF$ we get
\begin{aligned} AF & \leqslant EF . \\ \text{ and } \frac{h}{d} & =\frac{AH}{FG}=\frac{AE}{EF}h$.