Let $\triangle A B C$ have sides $B C=a, C A=b, A B=c$, as shown in Figure 1. When $\angle A, \angle B, \angle C$ are all less than $120^{\circ}$, $F$ is the Fermat point of $\triangle A B C$, $F A=u, F B=v, F C=w$. Then
$$
u+v+w=\frac{\sqrt{2}}{2} \sqrt{a^{2}+b^{2}+c^{2}+4 \sqrt{3} S} .
$$
Proof: As shown in Figure 2, construct a regular $\triangle MNL$, and let $P$ be a point inside it such that $PM = a$, $PN = b$, and $PL = c$.
Rotate $\triangle PML$ around $M$ by $60^\circ$ clockwise to get $\triangle P'MN$.
Then $\triangle PMP'$ is an equilateral triangle with side length $a$, and $\triangle PP'N$ is a triangle with sides $PP' = a$, $PN = b$, and $P'N = c$.
Thus, $\triangle PP'N \cong \triangle CBA$.
$$
\begin{array}{l}
\therefore \angle NPM = \angle NPP' + \angle MPP' \\
= \angle ACB + 60^\circ = \angle C + 60^\circ.
\end{array}
$$

In $\triangle MPN$, by the Law of Cosines, we have
$$
\begin{array}{l}
MN^2 = PM^2 + PN^2 - 2 \cdot PM \cdot PN \cdot \cos \angle MPN \\
= a^2 + b^2 - 2ab \cos (60^\circ + C) \\
= a^2 + b^2 - 2ab \left(\frac{1}{2} \cos C - \frac{\sqrt{3}}{2} \sin C\right) \\
= a^2 + b^2 - ab \cos C + \sqrt{3} ab \sin C \\
= a^2 + b^2 - \frac{1}{2} \left(a^2 + b^2 - c^2 + 2\sqrt{3} S\right) \\
= \frac{1}{2} \left(a^2 + b^2 + c^2\right) + 2\sqrt{3} S. \\
\therefore MN = \frac{\sqrt{2}}{2} \sqrt{a^2 + b^2 + c^2 + 4\sqrt{3} S}. \\
\because MN = NL = ML = u + v + w^{[3]}, \\
\therefore u + v + w
\end{array}
$$