TITLE: Inequality....(RMO $1994$...question $8$)
QUESTION [8 upvotes]: If $a$, $b$, $c$ are positive real numbers such that $a+b+c = 1$, prove that
$$(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)\text{.}$$

REPLY [3 votes]: We can easily prove the inequality: $(a+b)(b+c)(c+a)\ge 8abc$.
Replacing $a$ by $(1-a)$, $b$ by $(1-b)$ and $c$ by $(1-c)$ we get:
$$(2-(a+b))(2-(c+b))(2-(c+a))\geq 8(1-a)(1-b)(1-c)$$
but $a + b + c= 1$ therefore:
$$a+b=1-c,~ b+c=1-a, ~\text{and}~ c+a=1-b$$
substituting these values in LHS we get :
$$(1+a)(1+b)(1+c)\ge8(1−a)(1−b)(1−c)$$
which proves the inequality.