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# 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 for any cyclic quadrilateral $ABCD$ 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 triangles $BAP$ and $BDC$ are similar, 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.

## Proof for All Dimensions?

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

Note that

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

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

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

$=(\bold a-\bold c)\cdot(\bold b-\bold d)$.

From the Triangle Inequality,

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

$\implies|\bold a-\bold b| |\bold c-\bold d|+|\bold a-\bold d| |\bold b-\bold c|\ge|\bold a-\bold c| |\bold b-\bold d|$

$\implies AB\cdot CD+AD\cdot BC\ge AC\cdot BD$.

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.