Let x, y, z be real numbers and x y z 12 . If x+y+z= 2 , then the minimum value of x y z is

Step-by-step solution

Given x+y+z= 12 and x y z 12 Now, Take x siny z . aligned & 1 2 (2 x y z)= 1 2 ( x( (y+z)+ (y-z))) & 1 2 ( x+ y+2) aligned Here, y + z = 2 - x , 1 2 ( x ( 2 -x ) ) 1 2 ^2 x . If we take minimum value of y=z= 12 , then x= 3 So, 1 2 ^2 x = 1 2 ( 3 )^2= 1 2 1 4 = 1 8 Therefore, Minimum value of the given expression is 1 8