maths
posted by Anonymous .
If a, b and c are sides of a triangle
Then prove that 1≤ [(a^2+b^2+c^2)/(ab + bc+ca)]

so, we want to show that
ab+bc+ca <= (a^2+b^2+c^2)
since all squares are positive,
0 <= (ab)^2 + (ac)^2 + (bc)^2
0 <= a^22ab+b^2 + a^22ac+c^2 + b^22bc+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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