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.