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From a point $M$ inside a given triangle $ABC$, perpendiculars $M A_{1}, M B_{1}, M C_{1}$ are dropped to the lines $BC, CA, AB$. For which points $M$ inside the given triangle $ABC$ does the value $\frac{a}{M A_{1}}+\frac{b}{M B_{1}}+\frac{c}{M C_{1}}$ attain its minimum value?
Let $x=M A_{1}, y=M B_{1}$ and $z=M C_{1}$. Then $a x+b y+c z=2 S_{B M C}+2 S_{A M C}+2 S_{A M B}=2 S_{A B C}$. Therefore, $\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right) \cdot 2 S_{A B C}=\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)(a x+b y+c z)=$ $=a^{2}+b^{2}+c^{2}+a b\left(\frac{x}{y}+\frac{y}{x}\right)+b c\left(\frac{y}{z}+\frac{z}{y}\right)+a c\left(\frac{z}{x}+\frac{x}{z}\right) \geq a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c$,

with equality achieved only when $x=y=z$, that is, when $M$ is the incenter of triangle $A B C$.

## Answer

$M$ is the incenter of triangle $A B C$.