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this one is good que. it include am and gm inequality (BUT SIR ACCORDING CONDITION a,b,c should smaller than 1 AND GREATER THAN 0)
let x=1-a, y=1-b, z=1-c
so (1-x)+(1-y)+(1-z)=2
x+y+z=1
so,(1-x/x)(1-y/y)(1-z/z)=(y+z)(x+z)(y+x)/xyz=(a/1-a)(b/1-b)(c/1-c)
so by AM GM inq.
AM=>GM
y+z=>2(y.z)^1/2 1)
x+y=>2(x.y)^1/2 2)
x+z=>2(x.z)^1/2 3)
from (1).(2).(3)
we get,
(x+y)(y+z)(x+z)=>8(x.x.y.y.z.z)^1/2
(x+y)(y+z)(x+z)/x.y.z=>8
so sir, (a/1-a)(b/1-b)(c/1-c)=>8
prooved
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