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Algebra

Hari Shankar's Avatar
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14 Mar 2008 14:06:34 IST
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\text {This is a lovely one} \\ \\

\text {Prove that}  \\ \\

\sqrt{a^2-ab+b^2} + \sqrt{b^2-bc+c^2} \geq \sqrt{a^2+ac+c^2}


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pardesi .svk's Avatar

Scorching goIITian

Joined: 7 Dec 2007
Posts: 219
14 Mar 2008 18:13:36 IST
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i am assuming that a,b,c are positive
consider any point P from there draw rays PQ and PR such that \angle QPR =\frac{2\pi}{3} and QP=a,PR=c
bisect the angle and draw SP=b.
now the given inequality reduces to the triangle inequality
QS + SR > QR
sandeep ramesh's Avatar

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Joined: 13 Mar 2008
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14 Mar 2008 18:15:09 IST
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EDITED!
Hari Shankar's Avatar

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14 Mar 2008 18:55:44 IST
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You chaps from AOPS are downright ruffians. good job fellas!
sreeraman nagasubramaniyan's Avatar

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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
which is =  (a2+b2+b2+c2) ...now i think that rearrangement inequality can be employed to prove the final answer ...
sandeep ramesh's Avatar

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14 Mar 2008 19:30:09 IST
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why are we ruffians sir? Becos we attack all problems? Mr. Green Mr. Green..
anchit saini's Avatar

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Joined: 1 Feb 2008
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14 Mar 2008 19:34:46 IST
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ruffian

noun

1. A rough, violent person who engages in destructive actions

just to add
those destructive actions are done on problems!!


Mr. Green Mr. Green
Hari Shankar's Avatar

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Joined: 28 Feb 2007
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14 Mar 2008 19:41:09 IST
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marauders is more like it



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