Prove that the function $d:\Bbb R \times \Bbb R \rightarrow \Bbb R$, defined by
$$d(x,y):=\ln(1+|x-y|)$$
Here's what I said.
It's not hard to show the first two criteria are met ;
i)
let $r \in \Bbb R^+$, $\ln(r)\geq 0 \forall r \in \Bbb R^+$
also $1+|x-y| \geq 0$.
$\therefore \ln(1+|x-y|)\geq0$, with equality iff x=y as then we have $\ln(1)$ which is equal to zero.
ii) $d(x,y)=\ln(1+|x-y|)=\ln(1+|y-x|)=d(y,x)$
The triangle inequality I'm not as sure about.
I think we should begin by exponentiating
$d(x,z)=\ln(1+|x-z|)$ which after exponentiation yields $e^{1+|x-z|}=e^1e^{|x-z|}$
$d(x,y)=\ln(1+|x-y|)$ which after exponentiation yields $e^{1+|x-y|}=e^1e^{|x-y|}$
$d(y,z)=\ln(1+|y-z|)$ which after exponentiation yields $e^{1+|y-z|}=e^1e^{|y-z|}$
but what I'm really wondering about here is can we take that $x$ is a smaller number than $y$ and $y$ is a smaller number than $z$ and then use this to prove the necessary inequality or else how can we prove the triangle inequality here?
We know $|x+y|+|y-z|\geq|x-z|$ by the triangle equality on $\mathbb{R}$. It follows that $$1+|x-y|+|y-z|+|x-y||y-z|\geq 1+|x-z|.$$ Taking log of both sides will not change the sign of the inequality since the both sides are $\geq 1$. Taking log of both sides gives $$\ln(1+|x-y|+|y-z|+|x-y||y-z|)\geq\ln(1+|x-z|),$$ or equivalently, $\ln(1+|x-y|)+\ln(1+|y-z|)\geq\ln(1+|x-z|)$, as required.