Talk:Pedoe's inequality

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Field:  Geometry

Is the special case assertion true?

The article currently says without citation or explanation that the Hadwiger–Finsler inequality is a special case of Pedoe's inequality, which is

${\displaystyle A^{2}(b^{2}+c^{2}-a^{2})+B^{2}(a^{2}+c^{2}-b^{2})+C^{2}(a^{2}+b^{2}-c^{2})\geq 16Ff,}$

for two triangles with sides and area (a, b, c; f) and (A, B, C; F). For that to be true, there would have to be a class of (A, B, C; F) for which this collapses to the Hadwiger-Finsler inequality, which is

${\displaystyle a^{2}+b^{2}+c^{2}\geq (a-b)^{2}+(b-c)^{2}+(c-a)^{2}+4{\sqrt {3}}T.}$

I don't see how this special case assertion could be true. Anyone know? Loraof (talk) 13:43, 28 June 2015 (UTC)