IIT JEE , AIEEE Forum - Mathematics , Algebra

nano0101

0<a,b,c<1 and a+b+c=1 , then prove

(1-a)/a.(1-b)/b.(1-c)/c >= 8

nick

the eq. can be written as

(1/a-1)(1/b-1)(1/c-1)

let a+b+c=x

therefore

= (x/a-1)(x/b-1)(x/c-1)

as x=1 (given)

on simplifying it gives

(a+b)(a+c)(b+c)/abc RIGHT!!............................1

NOW

we know

A.M. >=G.M

(a+b)/2>=(ab)^1/2

similarly for

(b+c) and (a+c)

multiplying them we have,

(a+b)(a+c)(b+c)>= 8abc

which can be written as

(a+b)(a+c)(b+c)/abc>=8

from (1)

we get the desired result.....

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amar.gupta

good work nick

nano0101

Thank you !!!!