Let f be an inversion through a circle with center D and radius r. For point X, let X' denote f(X).
Lemma:
Let P and Q be two arbitrary points such that P, Q, and D are not collinear. Thus,
|DP||DP'| = |DQ||DQ'| = r2
==> ΔDPQ ~ ΔDQ'P'
|P'Q'| |DP'| |DP||DP'| r2
==> -------- = ------- = ----------- = ----------
|PQ| |DQ| |DP||DQ| |DP||DQ|
r2|PQ|
==> |P'Q'| = ----------
|DP||DQ|
WOLOG let the labeling of ABCD be such that angle B is greater than or equal to the other three angles.
From the triangle inequality,
|A'B'| + |B'C'| ≥ |A'C'|
r2|AB| r2|BC| r2|AC|
==> ---------- + ---------- ≥ ----------
|DA||DB| |DB||DC| |DA||DC|
==> |AB||CD| + |AD||BC| ≥ |AC||BD|
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