Point in triangle 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? 
 A: 
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. 
