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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Another good one
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hsbhatt (5000)

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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}

Time wounds all heels
    
pardesi (541)

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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
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sandeepramesh (1247)

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EDITED!
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hsbhatt (5000)

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You chaps from AOPS are downright ruffians. good job fellas!

Time wounds all heels
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sboosy (3063)

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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 ...
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sandeepramesh (1247)

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why are we ruffians sir? Becos we attack all problems? Mr. Green Mr. Green..
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anchitsaini (4352)

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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

Impossible To be Impossible is Impossible
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hsbhatt (5000)

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marauders is more like it

Time wounds all heels
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