Talk:Pedoe's inequality

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) 13:43, 28 June 2015 (UTC)