Let $P$ be an interior point of triangle $ABC$, and let the distances from $P$ to the sides $a, b, c$ be $x, y, z$ respectively. Determine the points $P$ for which the sum $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$ is minimal.
Using the fact that for any point $P$, $a x+b y+c z=2 t$ (where $t$ is the area of triangle $ABC$).

Write the inequality between the weighted arithmetic and harmonic means for the numbers $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ with weights $a, b, c$:

$$
\frac{a \cdot \frac{1}{x}+b \cdot \frac{1}{y}+c \cdot \frac{1}{z}}{a+b+c} \geq \frac{a+b+c}{a x+b y+c z}
$$

which is

$$
\frac{a}{x}+\frac{b}{y}+\frac{c}{z} \geq \frac{(a+b+c)^{2}}{2 t}
$$

and equality holds if and only if $\frac{1}{x}=\frac{1}{y}=\frac{1}{z}$, i.e., $x=y=z$.

Thus, the sum $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$ is minimized precisely when $x=y=z$, which means $P$ is the incenter of the inscribed circle.