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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.

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2 Answers 2

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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.

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  • $\begingroup$ $|ca^2 + ab^2 +bc^2| \geq |abc|$ is equivalent to $(ca^2 + ab^2 +bc^2)^2 \geq (abc)^2$ since $f(x)=x^2$ is injective on $[0,\infty)$. So $$(ca^2 + ab^2 +bc^2)^2 = c^2a^4 + a^2b^4 + b^2c^4 +2abc(ba^2+cb^2+ac^2) = c^2a^4 + a^2b^4 + b^2c^4 - 2(abc)^2 \geq (abc)^2$$ which simplifies to $$\frac{1}{3}(c^2a^4 + a^2b^4 + b^2c^4) \geq (abc)^2 = (c^2a^4 \cdot a^2b^4 \cdot b^2c^4)^{\frac{1}{3}}$$ which is just AM-GM inequality. Thank you, that was very clear. $\endgroup$
    – Figment
    Jan 22 at 11:31
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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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