Ptolemy's inequality by inversion

The inversion. Assume $B,C,D\ne A$. The inversion with centre $A$ and radius $1$ sends a point $X\ne A$ to $X'$ on ray $AX$ with $AX\cdot AX'=1$.

Lemma. For $X,Y\ne A$: $X'Y'=\dfrac{XY}{AX\cdot AY}$. (SAS similarity of $AX'Y'$ and $AYX$.)

The inequality. The triangle inequality in the images gives $B'D'\le B'C'+C'D'$. By the lemma,
$$\frac{BD}{AB\cdot AD}\le\frac{BC}{AB\cdot AC}+\frac{CD}{AC\cdot AD}.$$
Multiply by $AB\cdot AC\cdot AD$:
$$BD\cdot AC\le BC\cdot AD+CD\cdot AB.$$
$\blacksquare$

Remark (equality). Equality holds exactly when $A,B,C,D$ lie on a circle in this order (or on a line): Ptolemy's theorem.

Lesson. Inversion turns a product of distances into a sum of distances; the triangle inequality in the image gives the inequality in the original configuration.