Ptolemy's Inequality is a famous inequality attributed to the Greek mathematician Ptolemy.
The inequality states that in for four points in the plane,
with equality for any quadrilateral with diagonals and .
This also holds if are four points in space not in the same plane, but equality can't be achieved.
Proof for Coplanar Case
But since this is a spiral similarity, we also know that the triangles are also similar, which implies that
Now, by the triangle inequality, we have . Multiplying both sides of the inequality by and using and gives us
which is the desired inequality. Equality holds iff. , , and are collinear. But since the triangles and are similar, this would imply that the angles and are congruent, i.e., that is a cyclic quadrilateral.
Outline for 3-D Case
Construct a sphere passing through the points and intersecting segments and . We can now prove it through similar triangles, since the intersection of a sphere and a plane is always a circle.
Proof for All Dimensions?
Let any four points be denoted by the vectors .
From the Triangle Inequality,
Note about Higher Dimensions
Similar to the fact that that there is a line through any two points and a plane through any three points, there is a three-dimensional "solid" or 3-plane through any four points. Thus in an n-dimensional space, one can construct a 3-plane through the four points and the theorem is trivial, assuming the case has already been proven for three dimensions.