# 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.

For complex numbers $$z, w \in \Bbb C$$ is $$d(z, w) = \frac{|z - w|}{\sqrt{|z|^2 + 1}\sqrt{|w|^2 + 1}}$$ (apart from a constant factor) the “spherical distance” of $$z$$ and $$w$$, that is the euclidean distance of the stereographic projections of $$z, w$$ onto a sphere. See for example A metric in $\mathbb{C}^{\infty}$ or What is this metric called?.

$$d$$ is a metric on $$\Bbb C$$. It follows that $$d(a, c) < d(a, b) + d(b, c)$$ for (all permutations of) pairwise distinct $$a, b, c \in \Bbb C$$, with strict inequality because three distinct points on a sphere cannot be collinear. This implies that $$d(a, b)$$, $$d(b, c)$$, $$d(c, a)$$ are the side length of a non-degenerate plane triangle.

In particular this holds for pairwise distinct $$a,b, c \in \Bbb R$$.

Your last approach works also: According to Metric $d(x,y)=\frac{|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}}$ on $\mathbb{R}$ we have $$d(a, b) = \frac{|a - b|}{\sqrt{a^2 + 1}\sqrt{b^2 + 1}} = |\sin(\arctan(a) - \arctan(b))|$$ for $$a, b \in \Bbb R$$. It follows that $$d(a, c) = |\sin(\arctan(a) - \arctan(b) + \arctan(b) - \arctan(c))| \\ \le |\sin(\arctan(a) - \arctan(b))| + |\sin(\arctan(b) - \arctan(c))| \\= d(a, b) + d(b, c)$$ since $$|\sin(x+y)| \le |\sin(x)| + |\sin(y)|$$. Equality holds only if $$x=0$$ or $$y= 0$$, that is if $$a=b$$ or $$b=c$$.

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.

Hint.

Given the three sides $$l_1,l_2,l_3$$ we habe

$$\cos\theta_1 = \frac{l_2^2+l_3^2-l_1^2}{2l_2l_3}$$

now making

$$\cases{ l_1^2 = \frac{(b-c)^2}{\left(b^2+1\right) \left(c^2+1\right)}\\ l_2^2 = \frac{(c-a)^2}{\left(a^2+1\right) \left(c^2+1\right)}\\ l_3^2 = \frac{(a-b)^2}{\left(a^2+1\right) \left(b^2+1\right)} }$$

we have

$$\cos\theta_1 = \frac{b c + 1}{\sqrt{(b^2+1)(c^2+1)}}$$

and

$$-1\lt \cos\theta_1 \lt 1$$

as expected.

NOTE

If $$\sin^2\theta_1 = 1-\cos^2\theta_1 = \frac{(b-c)^2}{\left(b^2+1\right) \left(c^2+1\right)}$$ then

$$\frac{l_k^2}{\sin^2\theta_k}=1$$

so the sinus law is observed as well.