Proof As shown in Figure 26-1, establish a complex plane, and let the complex numbers corresponding to points $A, B, C$ be $a, b, c$, and the complex number corresponding to point $P$ be $z$. Noting the properties of the 3rd roots of unity, we have $|a-z|+|b-z|+|c-z|$
$$\begin{array}{l} =|a-z|+|(b-z) \omega|+\left|(c-z) \omega^{2}\right| \\ \geqslant\left|\left(a+b \omega+c \omega^{2}\right)-z\left(1+\omega+\omega^{2}\right)\right| \\ =\left|a+b \omega+c \omega^{2}\right| . \end{array}$$

Since the right side of inequality $(*)$ is a constant, from the proof process of $(*)$, the condition for equality in inequality $(*)$ is that the three complex numbers $a-z, (b-z) \omega$, and $(c-z) \omega^{2}$ correspond to vectors in the same direction. By the geometric meaning of complex number multiplication, this means that the vectors $\overrightarrow{P A}, \overrightarrow{P B}, \overrightarrow{P C}$ corresponding to $a-z, b-z, c-z$ have pairwise angles of $120^{\circ}$, i.e., $P$ coincides with point $M$. Therefore, $P A+P B+P C \geqslant M A+M B+M C$.