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Algebra

Blazing goIITian

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15 Mar 2008 19:18:14 IST

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1384

Challenge 4

None

Let a,b,c be non negative real numbers such that

a² + b²+ c² + abc = 4

Prove that --

0 <= ab + bc + ca - abc <= 2

(there are more than one solutions)

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sandeep ramesh

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15 Mar 2008 19:20:48 IST

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LHS is done bcos abc < 4 and ofc by AM GM ab + bc + ca > = 3 (abc)^2/3 and 3 (abc)^2/3 > abc

sandeep ramesh

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15 Mar 2008 19:21:38 IST

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n u could have said positive nos bcos clearly the trivial case isnt valid :D

sandeep ramesh

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15 Mar 2008 19:28:11 IST

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2nd part:

a^2 + b^2 + c^2 > ab + bc +ca

implies ab+bc+ca - abc < = 4 - 2abc

But a^2+ b^2 + c^a > abc for abc < 4?

Hence ab+bc+ca<=2

Hari Shankar

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15 Mar 2008 19:29:00 IST

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AM-GM gives abc

Now a²+b²+c² = 4-abc

ab+bc+ca

Hence 4-2abc

ab+bc+ca-abc

4-2abc

Hence ab+bc+ca

beaten to the finish

sandeep ramesh

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15 Mar 2008 19:30:01 IST

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am i right then?

Hari Shankar

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15 Mar 2008 19:31:55 IST

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didnt want 2 point out, but looks highly suspect!

sandeep ramesh

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15 Mar 2008 19:32:28 IST

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what does that mean sirji?

Hari Shankar

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15 Mar 2008 19:34:56 IST

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your last two lines of the proof are making me dizzy. Cud u write it down more elaborately for dummies like yours truly

sandeep ramesh

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15 Mar 2008 19:37:26 IST

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well i kinda said in a genl way not wanting to find out the answer meself :P

i knew for sure that abc <4 and also a^2 + b^2 + c^2 > abc and so i didnt bother proving that but just stated it as a fact to prove later but u had already done that :D

And ofc the last line is the same as yours with using the eqn a^2 + b^2 + c^2 + abc =4

Hari Shankar

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15 Mar 2008 19:40:27 IST

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bitte langsam!

See a²+b²+c²>abc

Hence, a²+b²+c²+abc = 4>2abc

So ab+bc+ca = 4-2abc<0.

Was macht du Herr Rohit!

just realisedmy proof is nonsense

I get the feeling that the problem is flawed

sandeep ramesh

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15 Mar 2008 19:43:49 IST

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what are you trying to do there? and i dont understand hindi :D

no i said ab+bc+ca<=a^2+b^2+c^2<= 4 - abc

abc < 2 Hence the result follows

anchit saini

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15 Mar 2008 20:44:50 IST

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i'll post the soln (solutions!!)
tomorrow!!

till then keep posting!

Hari Shankar

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15 Mar 2008 21:08:38 IST

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I want to approach this like a police interrogator.

Does the name Zuming Feng mean anything to you?

sandeep ramesh

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15 Mar 2008 21:11:57 IST

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yes co author with Titu Andreescu in a combinatorics book

Hari Shankar

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15 Mar 2008 21:23:25 IST

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a trig book

sandeep ramesh

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15 Mar 2008 21:34:05 IST

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and also a combinatorics book sire i have that

Hari Shankar

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15 Mar 2008 21:39:13 IST

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the trig book is relevant here

sandeep ramesh

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15 Mar 2008 21:44:00 IST

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bcos this question was adapted from there?

or does a trig subsn too work?

Hari Shankar

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15 Mar 2008 21:59:39 IST

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both r true

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