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.

Question

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.

Answer 1

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 $a,b, c \in \Bbb R$.

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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 $b=c$.

Answer 2

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.

Answer 3

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.