Question

If three positive real numbers, a, b, c are such that a + b + c = 1, then the minimum value of \(\frac{(1-a)(1-b)(1-c)}{abc}\) is

(a) 2 

(b) 3 

(c) 9 

(d) 8

Answer 1

(d) 8

a + b + c = 1 ⇒ b + c = 1 – a 

a + c = 1 – b 

a + b = 1 – c 

Now, a > 0, b > 0, c > 0 ⇒ AM > GM

∴ Minimum value is 8.