$a+b+c=1$, prove $(1+a)(1+b)(1+c) \ge 8(1-a)(1-b)(1-c) $, without AM-GM $a,b,c \in \mathbb{R}^{+}$, if $a+b+c=1$, prove that
$$ (1+a)(1+b)(1+c) \ge 8(1-a)(1-b)(1-c) $$
but without AM-GM as the only tool.

Collecting data:
Since $a+b+c=1$, we cannot have one of the variable to be greater than or equal to 1. So
$$ \boxed{0 < a,b,c < 1} $$
Also by AM-GM: $1=a+b+c \ge 3(abc)^{1/3}$, or
$$ \boxed{abc \le \frac{1}{27}} $$
Next,
$$ (1-a)(1-b)(1-c) = [1 - (a+b) + ab](1-c) = \boxed{ 1 - (a+b+c) + (ab + ac + bc) - abc }$$
$$ =\boxed{ (ab+ac+bc)-abc} $$
Next,
$$ (1+a)(1+b)(1+c) = [1 + (a+b) + ab](1+c)  = 1 + (a+b+c) + (ac+ab+bc) + abc$$
$$ = \boxed{2 + (ac+bc+bc) + abc} $$
so the inequality is equivalent with:
$$ 2 + ac+bc+ab + abc \ge 8(ab+ac+bc) - 8abc $$
or
$$ \boxed{ 2 + 9abc \ge 7 (ab+ac+bc) } \:\: \leftarrow \:\: \text{to be proven} $$
Next, $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab+ac+bc) = 1$,
$$ \boxed{ab+ac+bc = \frac{1- (a^{2} + b^{2}+ c^{2})}{2}} $$
which means
$$ 7(ab+ac+bc) = \frac{7}{2} -\frac{7}{2}(a^{2}+b^{2}+c^{2}) = 2 + \frac{3 - 7(a^{2}+b^{2}+c^{2})}{2}$$
and it is left to prove
$$ \frac{3-7(a^{2}+b^{2}+c^{2})}{2} \le 9abc $$
$$ \boxed{3  \le 18 abc  + 7(a^{2}+b^{2}+c^{2})} \:\: \leftarrow \:\: \text{to be proven}$$

Proceeding along the OP:
It suffices to prove that
$$2 + 9abc \ge 7(ab + bc + ca).$$
Degree three Schur inequality yields
$$(a + b + c)^3 - 4(a + b + c)(ab + bc + ca) + 9abc \ge 0$$
that is
$$1 - 4 (ab + bc + ca) + 9abc \ge 0.$$
It suffices to prove that
$$2 + 4(ab + bc + ca) - 1 \ge 7(ab + bc + ca)$$
or
$$1 \ge 3(ab + bc + ca)$$
which is true since $(a + b + c)^2 - 3(ab + bc + ca) = a^2 + b^2 + c^2 - ab - bc - ca$ $= \frac12[(a-b)^2+(b-c)^2 + (c-a)^2]\ge 0$.
We are done.
