a,b and c are three non-zero POSITIVE REAL numbers subjected to the condition that each of a,b and c are STRICTLY than 1 and a+b+c=2. Prove that [a/(1-a)].[b/(1-b)].[c/(1-c)]>=8.

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