Sides $\frac{|b - c|}{\sqrt{(b^2 + 1)(c^2 + 1)}}, \frac{|c - a|}{\sqrt{(c^2 + 1)(a^2 + 1)}}, \frac{|a - b|}{\sqrt{(a^2 + 1)(b^2 + 1)}}$ of a triangle. 
Prove that for all pairwise distinct $a, b, c \in \mathbb R$, $$\frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}}, \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}, \frac{|a - b|}{\sqrt{a^2 + 1}\sqrt{b^2 + 1}}$$ are always sides of a triangle.

For all $\triangle MNP$ where $m = MP, n = PM, p = MN$, we have that $$n + p > m, p + m > n, m + n > p$$
We need to obtain that $$\frac{|a - b|}{\sqrt{a^2 + 1}\sqrt{b^2 + 1}} + \frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}} > \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}$$
First attempt, we have that $$\frac{(a - b)^2}{|a - b|\sqrt{a^2 + 1}\sqrt{b^2 + 1}} + \frac{(b - c)^2}{|b - c|\sqrt{b^2 + 1}\sqrt{c^2 + 1}}$$
$$ \ge \frac{(c - a)^2}{\sqrt{b^2 + 1} \cdot \left(|b - c|\sqrt{c^2 + 1} + |a - b|\sqrt{a^2 + 1}\right)}$$
and $$\left(|b - c|\sqrt{c^2 + 1} + |a - b|\sqrt{a^2 + 1}\right)^2 \le \left[(b - c)^2 + (a - b)^2\right] \cdot (c^2 + a^2 + 2)$$
It is needed to prove that $$\sqrt{\left[(b - c)^2 + (a - b)^2\right] \cdot (b^2 + 1)(c^2 + a^2 + 2)} < |c - a|\sqrt{c^2 + 1}\sqrt{a^2 + 1}$$
Second attempt, it is to prove that $$|a - b|\sqrt{c^2 + 1} + |b - c|\sqrt{a^2 + 1} > |c - a|\sqrt{b^2 + 1}$$
According to the Cauchy - Schwarz inequality, we have that $$\left(|a - b|\sqrt{c^2 + 1} + |b - c|\sqrt{a^2 + 1}\right)^2 \ge 2|(a - b)(b - c)|\sqrt{(c^2 + 1)(a^2 + 1)}$$
What needs to be established is $$2|(a - b)(b - c)|\sqrt{(c^2 + 1)(a^2 + 1)} > (c - a)^2(b^2 + 1)$$
Third attempt, let $a = \tan\alpha, b = \tan\beta, c = \tan\gamma$ $\left(\alpha, \beta, \gamma \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\right)$, it could be easily deducted that $$\frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}} = \frac{|\tan\gamma - \tan\alpha|}{\sqrt{\tan\gamma^2 + 1}\sqrt{\tan\alpha^2 + 1}} = \frac{\left|\dfrac{\sin(\gamma - \alpha)}{\cos\gamma\cos\alpha}\right|}{\dfrac{1}{\cos\gamma\cos\alpha}} = \pm\sin(\gamma - \alpha)$$
For all of the above attempts, there need to be considered multiple cases of $a, b, c$, whether they're positive and negative, and their arrangements from littlest to greatest.

Let $a>b>c\geq0$.
Thus, easy to see that $$\frac{a-c}{\sqrt{(a^2+1)(c^2+1)}}>\frac{a-b}{\sqrt{(a^2+1)(b^2+1)}}$$ and
$$\frac{a-c}{\sqrt{(a^2+1)(c^2+1)}}>\frac{b-c}{\sqrt{(b^2+1)(c^2+1)}}$$ because $$a-c>\frac{a}{b}(b-c)$$ and it's enough to prove that:
$$\frac{a-b}{\sqrt{(a^2+1)(b^2+1)}}+\frac{b-c}{\sqrt{(b^2+1)(c^2+1)}}>\frac{a-c}{\sqrt{(a^2+1)(c^2+1)}}$$ or
$$(a-b)\sqrt{c^2+1}+(b-c)\sqrt{a^2+1}>(a-c)\sqrt{b^2+1}$$ or
$$(a-b)(b-c)\sqrt{(a^2+1)(c^2+1)}>(a-b)(b-c)(ac+1),$$ which is true by C-S.
The equality in C-S here does not occur because our variables are different.
Since the given expressions are not changed after substitution $a$ at $-a$, $b$ at $-b$ and $c$ at $-c$,
it remains to assume $a>b\geq0>c,$ which we can end by the similar way.
