TITLE: Find the smallest value of the following expression $\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2}+ \sqrt{(y-3)^2 +9}$
QUESTION [10 upvotes]: Find the smallest value of the following expression:
  $$\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2} + \sqrt{(y-3)^2 +9}$$

I tried to derive the expression with respect to $x$ and $y$ and then equal the derivative to zero and find the critical points, but I could not do that. It is complicated. Is there a way?

REPLY [16 votes]: An elementary school solution. No fancy stuff! 
Let $A= (0,0), B=(9,-2), C= (-3,3)$ and let $D=(x,y)$, $E=(x,0)$, $F=(0,y)$. Then we are interested for which $D$ the expression $$f(D)= \color{red}{AD}+\color{green}{EB+FC}$$ takes a minimum. Since  $\color{red}{AD} =\color{green}{EF}$ we have $$f(D) = \color{green}{BE+EF+FC}$$

By the triangle inequality we have $$f(D)\geq BC = 13$$ with equality iff $F$ and $E$ are on a line $BC:\;5x+12y=51$ and that is iff $D=({21\over 5},{7\over 4})$.