Example 3. Given: A regular tetrahedron $ABCD$ with each edge of length 1. Try to find the point $S$ whose sum of distances to the four vertices is minimized.
Solution: Obviously, the sum of distances from any point outside the regular tetrahedron to the vertices of the regular tetrahedron $ABCD$ cannot reach the minimum. Therefore, we only need to consider any point $P$ inside the regular tetrahedron.

As shown in the figure, let $E, F$ be the midpoints of $AB, CD$ respectively. It is easy to know that $EF=\frac{\sqrt{2}}{2}$. Let the projection of $P$ on the plane $ECD$ be $P_{1}$. Then $P_{1}C \leqslant PC, P_{1}D \leqslant PD$.

Since $PP_{1} \perp$ plane $ECD, AB \perp$ plane $ECD$, hence $PP_{1} \parallel AB$.

Also, the plane $ECD$ perpendicularly bisects the segment $AB$, so $P_{1}A = P_{1}B$.

Therefore, $P_{1}A + P_{1}B \leqslant PA + PB$ (in triangles with the same base and height, the perimeter is minimized when it is an isosceles triangle). Thus,
$$
\begin{array}{l}
P_{1}A + P_{1}B + P_{1}C + P_{1}D \\
\leqslant PA + PB + PC + PD.
\end{array}
$$

Now, let the projection of $P_{1}$ on the plane $ABF$ be $P_{2}$. Similarly,
$$
\begin{array}{l}
P_{2}A + P_{2}B + P_{2}C + P_{2}D \\
\leqslant P_{1}A + P_{1}B + P_{1}C + P_{1}D.
\end{array}
$$

Since the plane $ECD \perp$ plane $ABF$, $P_{2}$ lies on the intersection line $EF$, i.e., after two adjustments, the point we are looking for should fall on $EF$.

Let $P_{2}F = x$, then $P_{2}E = \frac{\sqrt{2}}{2} - x$. By the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
P_{2}A + P_{2}D + P_{2}C + PD \\
= 2(P_{2}A + P_{2}D) \geqslant \sqrt{6}.
\end{array}
$$

Equality holds if and only if $x = \frac{\sqrt{2}}{4}$. Therefore, when $P_{2}$ is the midpoint of $EF$, i.e., the centroid $P_{s}$ of the tetrahedron $ABCD$, the sum of distances to the vertices is minimized, with the minimum value being $\sqrt{6}$. This shows that any point $P$ inside the regular tetrahedron, after three adjustments, will be adjusted to the centroid $S$ of the tetrahedron, where the sum of distances to the four vertices is minimized.