<s>[INST] Prove that $\left | \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a} \right | < \frac{1}{8}.$ 
Let $a,b,$ and $c$ be the lengths of the sides of a triangle. Prove that $$\left | \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} \right | < \dfrac{1}{8}.$$

The best idea I had was to expand the fractions to get something nicer. So we get $\dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} = \dfrac{a^2 b-a^2 c-a b^2+a c^2+b^2 c-b c^2}{(a+b) (a+c) (b+c)}.$ Then I would try to relate this to side lengths of a triangle to prove the desired inequality but that is the step where I am unsure.
 [/INST] I'll prove a stronger inequality.

Let $a,b,$ and $c$ be the lengths-sides of a triangle. Prove that $$\left | \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} \right | < \dfrac{1}{22}$$

Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, 
$$\sum_{cyc}\frac{a-b}{a+b}=\frac{\sum\limits_{cyc}(a-b)(a+c)(b+c)}{\prod\limits_{cyc}(a+b)}=\frac{\sum\limits_{cyc}(a-b)(c^2+ab+ac+bc)}{\prod\limits_{cyc}(a+b)}=$$
$$=\frac{\sum\limits_{cyc}(a-b)c^2}{\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{\prod\limits_{cyc}(a+b)}=\frac{(y-x)(z-x)(z-y)}{\prod\limits_{cyc}(2x+y+z)}.$$
Thus, we need to prove that
$$(2x+y+z)(2y+x+z)(2z+x+y)\geq22\left|(x-y)(x-z)(y-z)\right|$$
Since the last inequality is symmetric, we can assume 
$x\leq y\leq z$ and $y=x+u$, $z=x+v$, $v=ku$, 
where $x>1$ (if $k=1$ so our inequality is obviously true). 
Id est, we need to prove that 
$$(4x+u+v)(4x+2u+v)(4x+u+2v)\geq22uv(v-u)$$
and since $$(4x+u+v)(4x+2u+v)(4x+u+2v)>(u+u)(2u+v)(u+2v),$$
it remains to prove that 
$$(k+1)(2k+1)(k+2)\geq22(k-1)k$$ or
$$2k^3-15k^2+29k+2\geq0,$$
which is true by AM-GM:
$$2k^3+29k+2>2k^3+29k\geq2\sqrt{2\cdot29}k^2>15k^2.$$
Done!
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