parmenides51 31.08.2020 10:01 Three circles that touch each other externally have all their centers on one fourth circle with radius $R$. Show that the total area of the three circle disks is smaller than $4\pi R^2$.
WolfusA 31.08.2020 10:38 By Heron's formula$$R=\frac{(x+y)(y+z)(z+x)}{4\sqrt{(x+y+z)xyz}}.$$We must prove $$x^2+y^2+z^2< \frac{(x+y)^2(y+z)^2(z+x)^2}{4{(x+y+z)xyz}}.$$Equivalent with$$\sum x^4 (y-z)^2 + 2 \sum x^3 y^3 + 2 x^3 y^2 z + 2 x^3 y z^2 + 2 x^2 y^3 z + 10 x^2 y^2 z^2 + 2 x^2 y z^3 + 2 x y^3 z^2 + 2 x y^2 z^3 >0.\square$$#1720