For $a, b, c$ is the length of three sides of a triangle. Prove that $\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}$

Question

For $a, b, c$ is the length of three sides of a triangle. Prove that $$\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}$$

Answer 1

Let $a=y+z$, $b=x+z$ and $c=x+y$.

Hence, $x$, $y$ and $z$ be positives and we need to prove that $$(2x+y+z)(2y+x+z)(2z+x+y)\geq8\sum_{cyc}(y-x)(2x+y+z)(2y+x+z)$$ or $$\sum_{cyc}\left(2x^3+15x^2y-x^2z+\frac{16}{3}xyz\right)\geq0,$$ which is obvious because $x^3+y^3+z^3\geq x^2z+y^2x+z^2y$ by Rearrangement.

Answer 2

HINT: use that we get for the sum $$\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=\frac{(a-b) (a-c) (b-c)}{(a+b) (a+c) (b+c)}$$