3. Given a triangle $ABC$ with sides $a=BC, b=CA, c=AB$. Inside the triangle, determine a point $P$ such that the value of the expression $a x^{2}+b y^{2}+c z^{2}$ is minimized, where $x, y, z$ are the distances from point $P$ to the lines $BC, CA, AB$ respectively.
Solution. Let $S$ denote the area of the given triangle. We need to determine when the expression $L=a x^{2}+b y^{2}+c z^{2}$ reaches its minimum, given that $a x+b y+c z=2 S$. We will prove that

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

with equality holding if and only if $x=y=z$. Indeed, inequality (1) is equivalent to the inequality

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

or to the inequality

$$
a b(x-y)^{2}+b c(y-z)^{2}+c a(z-x)^{2} \geq 0
$$

which is true, with equality holding if and only if $x=y=z$.

The proven inequality can be written as $L \geq \frac{4 S^{2}}{a+b+c}$. Therefore, the given expression $L$ has a value that is not less than $\frac{4 S^{2}}{a+b+c}$, and is equal to this number if and only if $x=y=z$, i.e., if and only if $P$ is the center of the inscribed circle of triangle $ABC$.