I was reading a book about inequalities, in that I found that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}>1\tag{1}$$ is a symmetric inequality in $a$,$b$,$c$. But if I change the order from $(a,b,c)$ to $(a,c,b)$, I am not getting the same inequality which is the basic definition of symmetric inequality. What am I doing wrong?

Here is my attempt: After changing the order to $(a,c,b)$, we get $$\frac{a}{c}+\frac{c}{b}+\frac{b}{a}>1$$ this is clearly not same as $(1)$.

Here is the excerpt from the book which I am studying:

  • $\begingroup$ Are a, b, c positive numbers $\endgroup$ – Darkrai Jan 24 '18 at 12:46
  • $\begingroup$ @Manthanein it is not given in the book that a,b,c are positive numbers $\endgroup$ – user521346 Jan 24 '18 at 12:47
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    $\begingroup$ The cyclic symmetry is (a, b, c) ; (b, c, a) ; (c, a, b) and (a, c, b) is not symmetric with the given sequence $\endgroup$ – Darkrai Jan 24 '18 at 12:49
  • $\begingroup$ @Manthanein I did not arrive, this is given in a book $\endgroup$ – user521346 Jan 24 '18 at 12:49
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    $\begingroup$ @Manthanein I am talking about symmetric, not cyclic $\endgroup$ – user521346 Jan 24 '18 at 12:51

It's not symmetric inequality because the permutation $(a,b,c)\rightarrow(a,c,b)$ gives another inequality: $$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}>1.$$

By the way, the inequality $$a+b+c>abc$$ is symmetric.

  • $\begingroup$ So the book is wrong? $\endgroup$ – user521346 Jan 24 '18 at 12:48
  • $\begingroup$ If the book says that it's a symmetric inequality then the book is wrong. $\endgroup$ – Michael Rozenberg Jan 24 '18 at 12:49
  • $\begingroup$ I agree $a+b+c>abc$ is symmetric, but this is a big claim that the book is wrong, isn't it? $\endgroup$ – user521346 Jan 24 '18 at 12:53
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    $\begingroup$ @MichaelRozenberg This is the book springer.com/in/book/9780387989426 see section 2.7 $\endgroup$ – user521346 Jan 24 '18 at 12:57
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    $\begingroup$ @GaurangTandon You can define "symmetry" for a triplet as $P(a,b,c) = P(a',b',c')$ for every permutation of $(a',b',c')$ in $(a,b,c)$ $$P(a,b,c)=P(a,c,b)=P(b,a,c)=P(b,c,a)=P(c,a,b)=P(c,b,a)$$ $\endgroup$ – Vivek Jan 24 '18 at 14:04

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