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Algebra

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2 Mar 2008 20:10:49 IST

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Easy Inequality

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If a,b,c are +ve numbers such that a+b+c = 1, then prove that ab+bc+ca1/3

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Akhil

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2 Mar 2008 20:15:00 IST

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using AM>=GM
we get
GM<=(1/3)
HM<=GM
HM= 3abc/ab+bc+ca
max value of abc=GM^3= 1/27
so we have
3(1/27)/(ab+bc+ca) <= 1/3
or
ab+bc + ca<=1/3

Hari Shankar

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2 Mar 2008 20:18:53 IST

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check the last step

sreeraman nagasubramaniyan

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2 Mar 2008 20:37:27 IST

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a+b+c = 1

(a+b+c)² = 1 = a²+b²+c² + 2(ab+bc+ca)

(a²+b²+c²)/3 >= [(a+b+c)/3 ]²

(a²+b²+c²) >= 1/3

so 2(ab+bc+ca) <= 1-1/3

so ab+bc+ca <=1/3

Hari Shankar

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3 Mar 2008 08:08:21 IST

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Another way is to see that (a+b+c)²>=3(ab+bc+ca)

karthik vadlamani

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6 Mar 2008 19:34:17 IST

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use cauchy-schwarz inequality
cheers

Hari Shankar

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6 Mar 2008 19:59:41 IST

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could you elaborate that please

akki ~~ unlucky forever ~~

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13 Jan 2009 21:52:59 IST

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i am taking a,b,c are +ve reals

let a,b,c be the roots of a cubic equation,

f(x)=x³-(a+b+c)x²+(ab+bc+ac)x-abc

equation of slopes will be,

f^'(x) = 3x²-2(a+b+c)x+(ab+bc+ac)

now, as a,b,c r +ve reals, then equation of its slopes will also be reals

so, making $D\ge0$

$4(a+b+c)^2-12(ab+bc+ac)\ge0$

or, $(ab+bc+ac)\le\frac{(a+b+c)^2}{3}$

or, $(ab+bc+ac)\le\frac{1}{3}$

is this correct, or there is something wrong,???????????

is the conditions of real , missing in que??????????

Conjurer

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14 Jan 2009 08:49:46 IST

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If they are +ve then they are real too :S

Nice solution.

debmalya choudhuri

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14 Jan 2009 13:08:43 IST

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i think thhe method of akki is the best cheers dude proof done in a great way and i dont think there 's nething wrong in his assumptions as a,b, c are real and positive

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