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Algebra
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14 Mar 2008 19:25:22 IST
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(a2+b2)/2 >= ab
=> ab<=(a2+b2)/2
(a2-ab+b2) >=
(a2 -(a2+b2)/2 +b2) =
((a2+b2)/2)similarly
(b2-bc+c2) >=
((b2+c2)/2)using AM of mth power >= mth power of AM
(
((a2+b2)/2) +
((b2+c2)/2))/2 > = [((a2+b2)/2)+(b2+c2)/2) /2 ]1/2
((a2+b2)/2) +
((b2+c2)/2))/2 > = [((a2+b2)/2)+(b2+c2)/2) /2 ]1/2which is =
(a2+b2+b2+c2) ...now i think that rearrangement inequality can be employed to prove the final answer ...
(a2+b2+b2+c2) ...now i think that rearrangement inequality can be employed to prove the final answer ...














consider any point
bisect the angle and draw
now the given inequality reduces to the triangle inequality