Q:

If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression

If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression:
$$\min \{a(1-c)+b(1-a)+c(1-b)\}$$
The given expression can be rewritten as:
$$a+b+c-ab-ac-bc = S_1-S_2  \tag{1}$$
Then from the given condition:
$$\begin{align}
abc &= 1-(a+b+c)+ab+ac+bc-abc \\
S_3 &= 1-S_1+S_2-S_3\\
S_1-S_2 &= 1-2S_3 \tag{2}
\end{align}$$
Using $AM \ge GM$
$$\frac{1-S_3 -S_3}{3} \ge\sqrt[3]{S_3^2}$$
for extreme value I set them equal to get:
$$\begin{align}
\frac{(1-2S_3)^3}{27}-S^2_3 &= 0 \\
8S_3^3 +15S_3^2 +6S_3-1 &=0 \\
(S_3 +1)(8S_3^2+7S_3-1) &= 0\\
(S_3+1)^2\left(S_3-\frac{1}{8}\right) &= 0
\end{align}$$
From here we get for minimum value, $S_3 = \frac{1}{8}$. We ignore $S_3 = -1$  because all $a,b,c \ge 0$.
So we can say:
$$\min \{1-2S_3\} = \frac{3}{4}$$
Q. Is there error in any argument (highly probable)
Q. Please provide a better method for inequality, if possible!
NOTE : 
I hope you understand my terminology, cannot put it in a better way:
$$S_i = \sum_{cyc} a_1 a_2 a_3 ... ('i'\text{ factors})$$

A:

We'll prove that $\frac{3}{4}$ is a minimal value.
Let $c=0$. 
Hence, $(1-a)(1-b)=0$ and $\sum_{cyc}a(1-c)=a+b-ab=1>\frac{3}{4}.$
Now, let $abc\neq0$, $\frac{a}{1-a}=x$, $\frac{b}{1-b}=y$ and $\frac{c}{1-c}=z$.
Thus, $xyz=1$, $a=\frac{x}{1+x},$ $b=\frac{y}{1+y}$, $c=\frac{z}{1+z}$ and we need to prove that 
$$\sum_{cyc}\frac{x}{1+x}\left(1-\frac{z}{1+z}\right)\geq\frac{3}{4}$$ or
$$\sum_{cyc}\frac{x}{(1+x)(1+z)}\geq\frac{3}{4}$$ or 
$$4\sum_{cyc}x(1+y)\geq3\prod_{cyc}(1+x)$$ or
$$4(x+y+z)+4(xy+xz+yz)\geq3+3(x+y+z)+3(xy+xz+yz)+3xyz$$ or
$$x+y+z+xy+xz+yz\geq6,$$ which is true by AM-GM.
The equality occurs for $x=y=z=1$, which says that $\frac{3}{4}$ is the answer.

