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December 22, 2014

December 22, 2014

Posted by **Anonymous** on Thursday, February 14, 2013 at 12:43pm.

Then prove that 1≤ [(a^2+b^2+c^2)/(ab + bc+ca)]

- maths -
**Steve**, Thursday, February 14, 2013 at 4:26pmso, 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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