Ptolemy's Inequality

Ptolemy's Inequality is a famous inequality attributed to the Greek mathematician Ptolemy.

Theorem

The inequality states that in for four points $A, B, C, D$ in the plane,

$AB \cdot CD + BC \cdot DA \ge AC \cdot BD$ ,

with equality if and only if $ABCD$ is a cyclic quadrilateral with diagonals $AC$ and $BD$ .

This also holds if $A,B,C,D$ are four points in space not in the same plane, but equality can't be achieved.

Proof for Coplanar Case

We construct a point $P$ such that the triangles $APB, \; DCB$ are similar and have the same orientation. In particular, this means that

$BD = \frac{BA \cdot DC }{AP} \; (*)$ .

But since this is a spiral similarity, we also know that the triangles $ABD, \; PBC$ are also similar, which implies that

$BD = \frac{BC \cdot AD}{PC} \; (**)$ .

Now, by the triangle inequality, we have $AP + PC \ge AC$ . Multiplying both sides of the inequality by $BD$ and using $(*)$ and $(**)$ gives us

$BA \cdot DC + BC \cdot AD \ge AC \cdot BD$ ,

which is the desired inequality. Equality holds iff. $A$ , $P$ , and ${C}$ are collinear. But since the angles $BAP$ and $BDC$ are congruent, this would imply that the angles $BAC$ and $BDC$ are congruent, i.e., that $ABCD$ is a cyclic quadrilateral.

Outline for 3-D Case

Construct a sphere passing through the points $B,C,D$ and intersecting segments $AB,AC,AD$ and $E,F,G$ . We can now prove it through similar triangles, since the intersection of a sphere and a plane is always a circle.

General Proof

Let any four points be denoted by $\bold a,\bold b,\bold c,\bold d$ .

Note that

$(\bold a-\bold b)\cdot(\bold c-\bold d)+(\bold a-\bold d)\cdot(\bold b-\bold c)$

$=\bold a\cdot\bold c-\bold a\cdot\bold d-\bold b\cdot\bold c+\bold b\cdot\bold d+\bold a\cdot\bold b-\bold a\cdot\bold c-\bold d\cdot\bold b+\bold d\cdot\bold c$