miniliv4

2022-10-11

How is symmetry in an inequality determined?

I was reading a book about inequalities, in that I found that

$$\begin{array}{}\text{(1)}& \frac{a}{b}+\frac{b}{c}+\frac{c}{a}>1\end{array}$$

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?

I was reading a book about inequalities, in that I found that

$$\begin{array}{}\text{(1)}& \frac{a}{b}+\frac{b}{c}+\frac{c}{a}>1\end{array}$$

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?

Zara Pratt

Beginner2022-10-12Added 12 answers

It's not symmetric inequality because the permutation $(a,b,c)\to (a,c,b)$ gives another inequality:

$$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}>1.$$

By the way, the inequality

$$a+b+c>abc$$

is symmetric.

$$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}>1.$$

By the way, the inequality

$$a+b+c>abc$$

is symmetric.

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