Triangle inequality: 9R^2 >= a^2+b^2+c^2+(a-b)^2+(b-c)^2+(c-a)^2

[Royal]

Prove that in any triangle, the following inequality holds:

\(\displaystyle 9\, R^2\, \geq \, a^2\, +\, b^2\, +\, c^2\, +\, (a\, -\, b)^2\, +\, (b\, -\, c)^2\, +\, (c\, -\, a)^2\)

 

[Deleted member 4993]

Prove that in any triangle, the following inequality holds:
$$9R^2 \geq a^2+b^2+c^2+(a-b)^2+(b-c)^2+(c-a)^2$$

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[HallsofIvy]

Such a statement makes no sense until you have said what "a", "b", "c", and "R" mean. I can guess that a, b, and c are the lengths of the sides of the triangle but what is "R"? The radius of the circum-circle?