# Proof of Ptolemy's inequality?

Can anyone prove the Ptolemy inequality, which states that for any convex quadrulateral $ABCD$, the following holds:$$\overline{AB}\cdot \overline{CD}+\overline{BC}\cdot \overline{DA} \ge \overline{AC}\cdot \overline{BD}$$

I know this is a generalization of Ptolemy’s theorem, whose proof I know. But I have no idea on this one, can anyone help?

#### Solutions Collecting From Web of "Proof of Ptolemy's inequality?"

Represent the four vertices as the complex numbers $a, b, c, d$. Notice that

$$(a-b)(c-d) + (a-d)(b-c) = (a-c)(b-d),$$

which you can simply multiply out. Then we must have that

\begin{align} |(a-b)(c-d) + (a-d)(b-c)| &= |(a-c)(b-d)| = |a-c||b-d| \\ |(a-b)(c-d)| + |(a-d)(b-c)| &\geq \\ |a-b||c-d| + |a-d||b-c| & \geq \end{align}

which is equivalent to your question. To pass from the first line to the second, we used the triangle inequality. To pass from the second to the third (and to get the second equality on the first line), we used that absolute values are multiplicative.

In general, many elementary geometry facts are bared easily with complex numbers, or vectors in general for higher dimensions (but being able to just multiply is pretty nice).