Posted by Anonymous on Thursday, February 14, 2013 at 12:43pm.
so, we want to show that
ab+bc+ca <= (a^2+b^2+c^2)
since all squares are positive,
0 <= (a-b)^2 + (a-c)^2 + (b-c)^2
0 <= a^2-2ab+b^2 + a^2-2ac+c^2 + b^2-2bc+c^2
0 <= 2(a^2+b^2+c^2) - 2(ab+ac+bc)
ab+ac+bc <= a^2+b^2+c^2
This is in fact true for any three real numbers, not just sides of a triangle
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