For $a$, $b$, $c$ the sides of a triangle, prove $\left| \frac{a}b-\frac{b}a+\frac{b}c-\frac{c}b+\frac{c}a-\frac{a}c\right|< 1$

Question

The inequality I have to prove that if $a$, $b$, and $c$ are the sides of a triangle, then

$$\left| \frac{a}b-\frac{b}a+\frac{b}c-\frac{c}b+\frac{c}a-\frac{a}c\right|< 1$$

I initially thought of changing it in terms of sine using the law of sines, which gives me

$$\left| \frac{\sin A}{\sin B}-\frac{\sin B}{\sin A}+\frac{\sin B}{\sin C}-\frac{\sin C}{\sin B}+\frac{\sin C}{\sin A}-\frac{\sin A}{\sin C}\right| < 1$$

However, from there I'm not sure how to proceed. Any tips?

Answer 1

Hint: Using the substitution $a = x+y, b=y+z, c = z+x$ (if $a, b, c$ are sides of a triangle then $x, y, z$ always exist and are positive) we get $$\left| \sum_{cyc} \left( \frac{a}b-\frac{b}a \right)\right| = \left|\frac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)} \right| < 1$$

Answer 2

By an inequality of the triangle $\left|\sum\limits_{cyc}\left(\frac{a}{b}-\frac{a}{c}\right)\right|=\frac{|\sum\limits_{cyc}(a^2c-a^2b)|}{abc}=\frac{\prod\limits_{cyc}|a-b|}{abc}=\prod\limits_{cyc}\frac{|a-b|}{c}<1$