Q.3. Let a, b, c be three real numbers such that 0<a, b, c<1n and a+b+c=2. Prove that

Solution:

Let 1 - a = x

thus  a = 1 - x

 

1 - b = y

thus  b = 1 - y

 

1 - c = z

thus  c = 1 - z

 

since a + b + c = 2

thus  1 - x + 1 - y + 1 - z = 2

thus  x + y + z = 1

 

thus  a/(1-a) * b/(1-b) * c/(1-c)  =  (1-x)/x * (1-y)/y * (1-z)/z  

                                                  =  (z+y)/x * (x+z)/y * (x+y)/z

 

applying AM >= GM

 

z+y >= 2*(y*z)^(1/2)

 

x+y >= 2*(y*x)^(1/2)

 

z+x >= 2*(x*z)^(1/2)

 

thus  (z+y)*(y+x)*(x+z)/x*y*z  >=  { 2*(y*z)^(1/2) * 2*(y*z)^(1/2) * 2*(y*z)^(1/2) }/x*y*z

 

                                                >=  2*2*2 = 8

 

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