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Does Ptolemy's inequality work for any ordered quadruple of points (A,B,C,D) in the plane? If so, is the following proof of Pompeiu's theorem correct?

If we assume that the Ptolemy's theorem states that for every ordered quadruple (A,B,C,D) the following inequality holds: $$AB\cdot CD + BC\cdot DA \geq AC\cdot BD,$$

consider the equilateral triangle $ABC$ and a point $M$ in the plane. Ptolemy for $(A,B,C,M)$ gives $$AB\cdot CM + BC\cdot MA \geq AC\cdot BM \iff CM + MA \geq BM,$$ Ptolemy for $(A,B,M,C)$ gives $$AB\cdot CM + BM\cdot CA \geq AM\cdot BC \iff CM + BM \geq AM,$$ and Ptolemy for $(A,M,B,C)$ gives $$AM\cdot BC + MB\cdot AC \geq AB\cdot MC \iff AM + MB \geq MC,$$

and thus $MA,MB,MC$ can form the sides of an (eventually degenerate) triangle.

Is this how Ptolemy's inequality works? Is this proof for Pompeiu's theorem correct?

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  • $\begingroup$ the proof using Ptolemy's inequality is correct. There is another proof using a rotation of $\pi/3$ to produce the triangle with those sides. As a bonus, if we know the distances $MA$, $MB$, $MC$ we can find the side of the triangle $ABC$ (two solutions from a quadratic equation). $\endgroup$
    – orangeskid
    Apr 25, 2018 at 8:46

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