Inside a random triangle, consider a point P and its distances PM, PN and PQ to the 3 sides a, b and c. For which location of the point P the below sum gets minimal? $$\frac{a}{PM} + \frac{b}{PN} + \frac{c}{PQ}$$
I think it is the incenter but how do we prove it?
Note that \begin{align} a\cdot x+b\cdot y+c\cdot z&=2\,S=\mathrm{const}\quad\text{for fixed }a,b,c. \end{align}
\begin{align} \operatorname*{argmin}_{P}\left(\frac{a}x + \frac{b}y + \frac{c}z\right) &= \operatorname*{argmin}_{P}2\,S\cdot\left(\frac{a}x + \frac{b}y + \frac{c}z\right) \\ &= \operatorname*{argmin}_{P} \left( (\tfrac{x}y+\tfrac{y}x)\,ab+(\tfrac{y}z +\tfrac{z}y)bc+(\tfrac{z}x+\tfrac{x}z)\,ca +\underbrace{(a^2+b^2+c^2)}_{\mathrm{const}} \right) \\ &= \operatorname*{argmin}_{P} \left( (\tfrac{x}y+\tfrac{y}x)\,ab+(\tfrac{y}z +\tfrac{z}y)bc+(\tfrac{z}x+\tfrac{x}z)\,ca \right) ,\\ \end{align}
\begin{align} \left( (\tfrac{x}y+\tfrac{y}x)\,ab+(\tfrac{y}z +\tfrac{z}y)bc+(\tfrac{z}x+\tfrac{x}z)\,ca \right) &\overset{\text{AM-GM}}{\ge} \left( 3\, \sqrt[3]{ a^2\,b^2\,c^2\, (\tfrac{x}y+\tfrac{y}x) (\tfrac{y}z+\tfrac{z}y) (\tfrac{z}x+\tfrac{x}z) } \right) ,\\ \operatorname*{argmin}_{P} \left( 3\, \sqrt[3]{ a^2\,b^2\,c^2\, (\tfrac{x}y+\tfrac{y}x) (\tfrac{y}z+\tfrac{z}y) (\tfrac{z}x+\tfrac{x}z) } \right) &= \operatorname*{argmin}_{P} \left( (\tfrac{x}y+\tfrac{y}x) (\tfrac{y}z+\tfrac{z}y) (\tfrac{z}x+\tfrac{x}z) \right) ,\\ (\tfrac{x}y+\tfrac{y}x) (\tfrac{y}z+\tfrac{z}y) (\tfrac{z}x+\tfrac{x}z) &\overset{\text{AM-GM}}{\ge} 2\cdot2\cdot2=8 ,\\ \operatorname*{min}_{P} \left( (\tfrac{x}y+\tfrac{y}x) (\tfrac{y}z+\tfrac{z}y) (\tfrac{z}x+\tfrac{x}z) \right) &=8,\quad P:x=y=z . \end{align} That is, $x=y=z=r$ and $P$ coincides with the incenter of the triangle.