Given a convex quadrilateral, prove the point on the plane with minimal total distance to the four vertices (that is, summing each of the four distances) is the intersection of the diagonals.
This problem was posed by Hadamard. My proof is below. It seems to me simpler than the existing proofs, and so I request verification and critique.
Given points $A,C$, for all points $P$, $PA + PC \leq AC$. Equality is reached for any point on segment $AC$.
Consider quadrilateral $ABCD$, with diagonals $AC$ and $BD$. From the above, if the two diagonals intersect at point $I$, no point on the plane has lower $IA + IC$, and no point on the plane has lower $IB + ID$, so clearly no point on the plane has lower $IA + IB + IC + ID$.
Update Reworded as per advice of Calvin Lin:
Given points $A,C$, for all points $P$, $PA + PC \geq AC$, with equality iff $P$ is on segment $AC$.
Consider quadrilateral $ABCD$, with diagonals $AC$ and $BD$ intersecting at $I$. For any point $P$: $$ PA + PC \geq AC = IA + IC \\ PB + PD \geq BD = IB + ID $$ with $I$ being the unique point in $AC \cap BD$ and therefore having equality in both equations. Therefore, for $P \neq I$, $$ PA + PB + PC + PD > I. $$