$\left|(a + b)(b + c)(c + a)\right| = \left|(a − b)(b − c)(c − a)\right| \implies | \frac{a}{b} + \frac{b}{c} + \frac{c}{a} | \geq 1$ If $a,b,c \in \mathbb{R}$ satisfy $\left|(a + b)(b + c)(c + a)\right| = \left|(a − b)(b − c)(c − a)\right|$ then $\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right| \geq 1$.
Given what we want to prove, additional assumption of $a,b,c \not= 0$ seems to be necessary.
I can only see kind of 'cyclic symmetry', by which I mean that any $2$ variables can be swapped without changing the equation or inequality. Also, if we could assume $a,b,c > 0$ then the result follows simply from the AM-GM inequality:
$ \frac{1}{3} \left( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right)  \geq \left(\frac{a}{b}  \frac{b}{c}  \frac{c}{a}\right)^\frac{1}{3} = 1 \implies \left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right| \geq 3 \geq 1$
But it doesn't help. Such assumption definitely doesn't follow from the initial condition since for instance $a=b=1$ and $c=-1$ satisfy both the equation and inequality.
Other than that, I have no idea how to get from that equation to the desired inequality. Any hints will be appreciated.
 A: Let us transform the left side of the inequality being proved
$$\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right| = \frac{|ca^2 + ab^2 +bc^2|}{|abc|}.$$
It follows from the equality that two cases are possible.
$\textbf{Case 1.}$
$$(a+b)(b+c)(c+a) = (a-b)(b-c)(c-a) \, \Longleftrightarrow \, a^2b + a^2c + ab^2+2abc + $$
$$+ac^2 + b^2c+bc^2 = -a^2b+a^2c+ab^2-ac^2-b^2c+bc^2 \, \Longleftrightarrow \, $$
$$\, \Longleftrightarrow \, abc = -a^2b -ac^2 -cb^2 \, \Longrightarrow \, |abc| =|a^2b +ac^2 +cb^2|.$$
In this case $\frac{|ca^2 + ab^2 +bc^2|}{|abc|} = \frac{|ca^2 + ab^2 +bc^2|}{|a^2b +ac^2 +cb^2|}$ and we need to prove, that $|ca^2 + ab^2 +bc^2| \ge |a^2b +ac^2 +cb^2|$. Try to prove last inequality, using equalities
$$(ca^2 + ab^2 +bc^2)^2 = c^2a^4 + a^2b^4 + b^2c^4 +2abc(ba^2+cb^2+ac^2),$$
$$ abc = -ba^2 -cb^2 - ac^2$$
and inequality of arithmetic and geometric means
$\textbf{Case 2.}$
$$(a+b)(b+c)(c+a) = -(a-b)(b-c)(c-a) \, \Longleftrightarrow \, a^2b + a^2c + ab^2+2abc + $$
$$+ac^2 + b^2c+bc^2 = a^2b-a^2c-ab^2+ac^2+b^2c-bc^2 \, \Longleftrightarrow \, $$
$$\, \Longleftrightarrow \, abc = -ab^2 - ca^2 - bc^2 \,\Longrightarrow \, |abc| = |ab^2 + ca^2 + bc^2| .$$
So, in this case $\frac{|ca^2 + ab^2 +bc^2|}{|abc|}=1$ and inequality holds.
A: For $abc\neq0$ we obtain:
Case1.$$\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)=\sum_{cyc}(a^2c-a^2b),$$ which gives
$$\sum_{cyc}a^2b=-abc$$ or
$$\sum_{cyc}\frac{a}{c}=-1$$
and by AM-GM we obtain: $$\left(\sum_{cyc}\frac{a}{b}\right)^2=\sum_{cyc}\frac{a^2}{b^2}-2\geq3-2=1,$$ which gives
$$|\sum_{cyc}\frac{a}{b}|\geq1.$$
Case2.
$$\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)=-\sum_{cyc}(a^2c-a^2b),$$ which gives
$$\sum_{cyc}a^2c=-abc$$ or
$$\sum_{cyc}\frac{a}{b}=-1,$$ which gives
$$|\sum_{cyc}\frac{a}{b}|=1\geq1.$$
