As shown in Figure 2, let the angle bisectors of $\angle A O B$, $\angle B O C$, and $\angle C O A$ be $O E$, $O F$, and $O G$ respectively, and the perpendiculars from point $O$ to $A B$, $B C$, and $C A$ be $H$, $I$, and $J$ respectively.
Given that $\angle A O B = \angle B O C = \angle C O A = 120^{\circ}$,
$S_{\triangle A O B} = S_{\triangle A O E} + S_{\triangle B O E}$,

we have
$\frac{1}{2} O A \cdot O B \sin \angle A O B$
$= \frac{1}{2} O A \cdot O E \sin 60^{\circ} + \frac{1}{2} O B \cdot O E \sin 60^{\circ}$
$\Rightarrow O A + O B = \frac{O A \cdot O B}{O E}$.
In $\triangle A O B$, we have
$S_{\triangle A O B} = \frac{1}{2} O A \cdot O B \sin \angle A O B = \frac{1}{2} A B \cdot O H$
$\Rightarrow O A \cdot O B = \frac{2 \sqrt{3}}{3} A B \cdot O H$
$\Rightarrow O A + O B = \frac{2 \sqrt{3} A B}{3} \cdot \frac{O H}{O E}$.
Since $O H \leqslant O E$, it follows that $O A + O B \leqslant \frac{2 \sqrt{3}}{3} A B$, with equality if and only if $O H = O E$.
Similarly, $O B + O C \leqslant \frac{2 \sqrt{3}}{3} B C$, with equality if and only if $O I = O F$;
$O C + O A \leqslant \frac{2 \sqrt{3}}{3} C A$, with equality if and only if $O J = O G$.
Therefore, $O A + O B + O C \geqslant \frac{\sqrt{3}}{3}(A B + B C + C A)$,

with equality if and only if $\triangle A B C$ is an equilateral triangle.
Thus, the maximum value of $\frac{O A + O B + O C}{A B + B C + C A}$ is $\frac{\sqrt{3}}{3}$.
One, (40 points) Given that $A$, $B$, and $C$ are any three non-collinear points on a plane, and point $O$ is inside $\triangle ABC$, satisfying
$$
\begin{array}{l}
\angle A O B=\angle B O C=\angle C O A=120^{\circ} . \\
\text { Find the maximum value of } \frac{O A+O B+O C}{A B+B C+C A} \text { . }
\end{array}
$$