Consider the four points $A=(2,4), B=(0,0), C=(6,0)$, and $D=(6,3)$ in the plane. For which point $P$ in the plane is the sum $P A+P B+P C+P D$ minimized?
The answer is $P=(4,2)$, which is the intersection of the line segments $A C$ and $B D$. To see this is optimal, note that by the triangle inequality we have

$$
P A+P C \geq A C \quad \text { and } \quad P B+P D \geq B D
$$

so $P A+P B+C D+P D \geq A C+B D$ with equality occurring for only the point $P$ we named earlier.