Talk:Hadwiger–Finsler inequality

Weizenbock

Most results in mathematics are named after mathematicians. "Weizenbock's inequality" is an exception to the rule. It's named after a type of beer (dark wheat beer) -- presumably because it's easy to prove (as in even when you've been drinking, or perhaps right after breakfast, since some German families drink wheat beer in the morning).

Anyway, I think this is sort of funny, but I can't put in the article because I can't cite a source. Can anyone find a citation for the origin of "Weizenbock's inequality"? DavidCBryant 12:19, 18 August 2007 (UTC)

@DavidCBryant: Weizenbock is a mispelling of the correct German spelling Weitzenböck, which is after all the last name of the mathematician Roland Weitzenböck rather than a beer.--Kmhkmh (talk) 02:47, 3 December 2015 (UTC)

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

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

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

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

I don't see how this special case assertion could be true. Anyone know? Loraof (talk) 13:40, 28 June 2015 (UTC)

At first glance without checking the details. You get the Weitzenböck inequality as a special case of Pedoe if one of the triangles is equilateral and Weitzenböck is equivalent to the Hadwinger-Finsler (despite being a less tight upper bound).--Kmhkmh (talk) 03:06, 3 December 2015 (UTC)