Let $D$ be the closed unit disk in the plane, and let $z_1,z_2,\ldots,z_n$ be fixed points in $D$. Prove that there exists a point $z$ in $D$ such that the sum of the distances from $z$ to each of the $n$ points is greater or equal than $n$.
Let us take any domain D (it does not have to be a disk) containing n points labelled A1,A2 ...An . Then we take two new points B and C in this domain and separated by a distance of 2 ie; BC=2;
Then consider the triangle BA1C, we can apply the triangular inequality BA1+CA1 larger or equal to 2;
We can then sum all these inequalities over the n points and we get:
(BA1+BA2+...BAN) +(CA1+CA2+...CAn) larger or equal to 2n.
One of the sums in bracket is larger or equal to n say WLOG the B sum, then B is the required point.
In the case where D is a unit disk, then BC is a diameter and B is on the edge of the disk.
Let us take any domain D (it does not have to be a disk) containing n points labelled A1,A2 ...An . Then we take two new points B and C in this domain and separated by a distance of 2 ie; BC=2; Then consider the triangle BA1C, we can apply the triangular inequality BA1+CA1 larger or equal to 2; We can then sum all these inequalities over the n points and we get: (BA1+BA2+...BAN) +(CA1+CA2+...CAn) larger or equal to 2n. One of the sums in bracket is larger or equal to n say WLOG the B sum, then B is the required point. In the case where D is a unit disk, then BC is a diameter and B is on the edge of the disk.