Inequality

Problem : 18 ) 

Prove the inequality (1 + a) (1 + b) (1 + c) ≥ 8 (1 - a) (1 - b) (1 - c)

a, b, c are positive numbers and their sum is 1. Also indicate when equality occurs.

Solution : 

Given : a, b, c > 0 and a + b + c = 1
=> (1 − a) > 0 ;(1 − b) > 0 ; (1 − c) > 0

Therefore, a = (1 − b − c ) ==> 1 + a = (1 − b) + (1 − c)
and. . . . . b = (1 − a − c ) ==> 1 + b = (1 − a) + (1 − c)
similarly, . c = (1 − a − b ) ==>1 + c = (1 − a) + (1 − b)

Using A.M.- G.M. Inequality ;
(1 + a) = (1 − b) + (1 − c) ≥ 2 √[(1 − b) × (1 − c)] - - -(1)
(1 + b) = (1 − a) + (1 − c) ≥ 2 √[(1 − a) × (1 − c)] - - - -(2)
(1 + c) = (1 − a) + (1 − b) ≥ 2 √[(1 − a) × (1 − b)] - - - - (3)

Multiplying (1), (2) and (3) we get,

(1 + a)(1 + b)(1 + c) ≥ 8 (1 − a)(1 − b)(1 − c)

Equality holds true if a = b = c = 1/3

Link to Y/A