2024/050) If a+b+c =1 for positive numbers a,b c prove that $(1+a)(1+b)(1+c) >= 8(1-a)(1-b)(1-c)$

We have 

$a+b+c = 1$

So $1+a + b+c =  2$

Or $1+a = 2 - b-c = (1-b)+(1-c)$

because (1-b) and (1-c) are positive using AM GM inequality we have

$(1-b)(a-c) >= 2\sqrt{(1-b)(1-c)}$

or $(1+a) >= 2\sqrt{(1-b)(1-c)}\cdots(1)$

similarly 

$(1+b) >= 2\sqrt{(1-c)(1-a)}\cdots(2)$

and

$(1+c) >= 2\sqrt{(1-a)(1-b)}\cdots(3)$

Multiply above 3 equations we get

 $(1+a)(1+b)(1+c) >= 8(1-a)(1-b)(1-c)$