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Algebra

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Joined:	28 Feb 2007
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21 Mar 2008 14:13:38 IST

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504

Easy Ineq

None

If 1/a + 1/b + 1/c = 1, then P.T. (a-1) (b-1)(c-1)8

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pardesi .svk

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Joined: 7 Dec 2007

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21 Mar 2008 14:21:45 IST

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THIS FORUM NEEDS TO BE LaTeX Enabled...why isn't it there The image ?http://www.mathlinks.ro/images/smiles/huh.gif? cannot be displayed, because it contains errors.

The image ?http://www.mathlinks.ro/images/smiles/huh.gif? cannot be displayed, because it contains errors.

ok coming to the question set 1/x=a 1/y=b,1/z=c
so we have x+y+z=1
we have by am-gm
(x+y)(y+z)(z+x)>= 8xyz
so (1-z)(1-y)(1-x)>=8xyz
dividing by xyz we get
(a-1) (b-1)(c-1)

Abhijith

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Joined: 23 Jan 2007

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21 Mar 2008 14:32:08 IST

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The question asks us to prove

i.e

Using

Hence proved.

sreeraman nagasubramaniyan

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21 Mar 2008 14:36:14 IST

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$[tex] \\ \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1 \\ \\ \mbox{Thus} \ abc>=27 \\ \\ 1 - \frac{1}{a}+ 1-\frac{1}{b} + 1- \frac{1}{c} = 2 \\ \\ \frac{(a-1)(b-1)(c-1)}{abc} >= \frac{8}{27} \\ \\ \mbox{But since } \ abc>=27 \ \mbox{We get} \\ \\ (a-1)(b-1)(c-1) >=8$

anchit saini

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21 Mar 2008 14:39:47 IST

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well this is the best a question could have brought out!!!
3 geniuses posting 3 different and great answers!!!

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