Prove Triangle Inequality for complicated trigonometric function

Question

I have the function \begin{equation*} d : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}_{0}^{+} : (x,y) \mapsto \sin(\tan^{-1}|x - y|) \end{equation*} for which I need to prove the Triangle Inequality (as I am proposing it as a metric). Now, I know of the identity \begin{equation*} \sin(\tan^{-1}x) = \frac{x}{\sqrt{1 + x^2}} \end{equation*} but I don't think I need to state how disgusting that makes the following. \begin{equation*} \frac{|x-z|}{\sqrt{1 + (x-z)^2}} \leq \frac{|x-y|}{\sqrt{1 + (x-y)^2}} + \frac{|y-z|}{\sqrt{1 + (y-z)^2}} \end{equation*} Firstly, do I need to go messing around with identites to prove this, or is there an easier way based simply on the properties the fucntions have? If I do need to, is this expression provable in this form, or is there some other identity I can use to prove this? Thanks in advance.

Answer 1

Denote by $f(x) = \frac{x}{\sqrt{1 + x^2}}$ and consider the following "wishful" at the moment proof of the triangle inequality:

$$ f(d(x,z)) \leq f(d(x,y) + d(y,z)) \leq f(d(x,y)) + f(d(y,z)). $$

What should $f$ satisfy in order for this proof to work? For the first inequality, it is enough to require that $f$ is weakly increasing. For the second inequality, it is enough to require that $f$ is subadditive. Let us check if this holds:

1. We have $f'(x) = (1+x^2)^{-\frac{3}{2}} \geq 0$ and so $f$ is weakly increasing.
2. We have $f''(x) = -3x(1+x^2)^{-\frac{5}{2}} \leq 0$ and so $f$ is concave and together with $f(0) = 0 \geq 0$ this implies that $f$ is subadditive (see the proof here in item 10).