<s>[INST] $\left|(a + b)(b + c)(c + a)\right| = \left|(a − b)(b − c)(c − a)\right| \implies | \frac{a}{b} + \frac{b}{c} + \frac{c}{a} | \geq 1$ 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.
 [/INST] 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
$$|\sum_{cyc}\frac{a}{b}|=1\geq1.$$
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