Geometric solution? Given coordinates of $A$, $B$, $C$, find $M$ on $y=x$ minimizing $AM+BM+CM$ I have the problem: 

Let be given three points $A(1,2)$, $B(3,4)$, $C(5,6)$. Find point $M$  on the line $y=x$   so that sum of distances $P=AM+BM+CM$ is smallest. 

I tried. We have
$$P=\sqrt{(x-1)^2 + (x-2)^2} + \sqrt{(x-3)^2 + (x-4)^2}  +\sqrt{(x-5)^2 + (x-6)^2}.$$
We know that
$$\sqrt{a^2 + b^2}+\sqrt{c^2 + d^2} \geqslant \sqrt{(a+c)^2 + (b+d)^2}.$$
The sign of equal occur when and only when $\dfrac{a}{c}=\dfrac{b}{d}$. 
We have
\begin{align*}
\sqrt{(x-1)^2 + (x-2)^2}  + \sqrt{(x-5)^2 + (x-6)^2} & = \sqrt{(x-1)^2 + (x-2)^2}  + \sqrt{(5-x)^2 + (6-x)^2} \\
& \geqslant \sqrt{(x-1 + 6-x)^2 + (x-2 + 5-x)^2}\\
& \geqslant \sqrt{34}.
\end{align*}
The sign of equal occur $$\dfrac{x-1}{6-x}=\dfrac{x-2}{5-x} \Leftrightarrow x=\dfrac{7}{2}.$$
Another way 
$$\sqrt{(x-3)^2 + (x-4)^2} =\sqrt{2x^2 - 14 x + 25} = \sqrt{2}\sqrt{\left (x-\dfrac{7}{2}\right)^2 + \dfrac{1}{4} } \geqslant \dfrac{1}{\sqrt{2}}.$$
The sign of equal occur $ x=\dfrac{7}{2}.$
Therefore, the least of the expression $P $ is $\dfrac{1}{\sqrt{2}}+\sqrt{34}$ at $x=\dfrac{7}{2}.$

How can I solve this problem geometrically?


Let's $a$ be a line $y=x$.
So, we claim that for the basis $M$ of the perpendicular dropped from $B$ to $a$ sum $AM + BM + CM$ is the smallest. It is easy to demonstrate this using additional point $A'$ which is symmetrical to the point $A$ with respect to the line $a$. One may see that $A'MC$ is the line segment (because of the symmetry of points $A$, $C$ with respect to point $B$). Thus, if $M' \neq M$ is arbitrary chosen point on $a$, then 
\begin{align}
AM' + BM' + CM' = & A'M' + M'C + BM' > \\  &A'M + MC + BM = AM + BM + CM,
\end{align}
so $M$ is our desired point. It's easy to calculate its coordinates, which are $(\frac{7}{2},\frac{7}{2})$ exactly as in your answer. 

