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.
 A: 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!
A: This is a kind of extended comment to what Michael Rozenberg did. His bound of $\frac1{22}$ is already an improvement over the original statement. Can it be improved further? What's the supremum of that expression?
Michael got rid of the need for triangle inequalities using Ravi substitution, by using $x,y,z$ instead of $a,b,c$ as he explained. But he still had $x,y,z\ge0$ as a constraint. To make this algebraically more tractable, use squares $x^2,y^2,z^2$ instead of his $x,y,z$. A square is implicitly non-negative, so in that case any real numbers plugged in for $x,y,z$ would describe a valid triangle, and conversely any triangle can be described by such numbers.
$$a=y^2+z^2\qquad b=z^2+x^2\qquad c=x^2+y^2$$
But as the whole setup is scale-invariant, you can pick one of them without loss of generality, e.g. $x=1$.
Now you have an expression in $y$ and $z$, and are looking for extrema. Which can be done by computing the partial derivatives and looking for common zeros of these. I don't suggest you do this by hand, but use a computer algebra system instead. I used sage. If you had to do this manually, I'd suggest using resultants to find the common solutions.
You will find that there are several solutions. For some of them the expression will be minimal, namely zero. For others it will be maximal, and in all of them either $y=0$ or $z=0$, so they describe a triangle that degenerated to a line. The maximum obtained in all of these cases is
$$\alpha=\frac13\sqrt{253-80\sqrt{10}}\approx0.0444562$$
which is one of the solutions of
$$9\,\alpha^4 - 506\,\alpha^2 + 1 = 0\,.$$
This happens for
$$x=1\qquad y=0\qquad z^8 - 2\,z^6 - 7\,z^4 - 2\,z^2 + 1 = 0$$
or expressed in terms of $z^2$ which corresponds to the $z$ from Michael's answer
$$z^2=\frac12\left(1\pm\sqrt2+\sqrt{15\pm10\sqrt2}\right)$$
using the same sign for both the $\pm$. So you can say that
$$\left\lvert\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right\rvert
\le\frac13\sqrt{253-80\sqrt{10}}<\frac18$$
with equality only achievable if you allow degenerate triangles.
