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If `a ,b ,c` are real numbers such that `0 < a < 1,0 < b < 1,0 < c < 1,a+b+c=2,` then prove that `a/(1-a)b/(1-b)c/(1-c)geq8`

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Put `a = y + z , b = z + x , c = x + y` , so that `x + y + z = 1`.
As `x = 1 -a , y = 1 - b , z = 1 - c ` and `0 lt a 1 lt 1 , 0 lt a lt 1 , 0 lt b lt 1 , 0 lt c lt 1`
it follows that `x , y , z gt0` , Now ,
`(a)/(1-a) (b)/(1-b) (c)/(1-c) = ([(y+ z) (z+x) (x+ y)])/(xyz) " " (1)`
Now , `A.M. ge G.M`
`implies (y+z)/(2) ge sqrt(yx) , (z+x)/(2) ge sqrt(zx) , (x+y)/(2) ge sqrt(xy)`
Multiplying , we get
`([(y+z) (z+x) (x + y)])/((xyz)) ge 8 " " (2)`
From (1) and (2) , we get
`(a)/(1-a) (b)/(1-b) (c)/(1-c) ge 8`.
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