Does $d(x,y) = \log(1 + | x - y |)$ define a metric on $\mathbb{R}$? Help would be much appreciated here! I got as far as evaluating the following:


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*$d(x,y) \geq 0$. Argument is that $(1 + |x - y|)$ doesn't produce negative values, since the $|x-y| \geq 0$ itself and, at most, it would be added to $1$. So this condition is satisfied!

*$d(x,y) = d(x,y)$. Since we have that $\log(1 + |x - y|) = \log(1 + |y - x|)$ by commutative of real numbers.
Now, here is where I am stuck. How can I demonstrate this algebraically?


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*$d(x,y) = 0$ implies that $x = y$.


What I could do is $\log(1+|x - y|) = 0$. It seems that the only way this log can be zero, is if $1 + |x - y| = 1$. Therefore, it would $|x - y| = 0$ would rise naturally. Does it make sense?


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*$d(x, y) \leq d(x,z) + d(z,y)$. What is left is add a zero to the inside of the modulo function and then try and break it into the triangle inequality. That is not working so well thus far, but here it goes: 
$\log(1 + |x - y|) = \log(1 + |x - z + z - y|)$ but then it turns out that:
$\log(1 + |x - z + z - y|) \leq \log(1 + |x - z| + |z - y|)$

You have the right idea about how to show that $d(x,y)=0$ implies $x=y$.
Hints for the triangle inequality:


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*\begin{aligned}d(x,z)+d(z,y)&=\log(1+|x-z|)+\log(1+|z-y|) \\&= \log(1+|x-z|+|z-y|+|x-z||z-y|)
\end{aligned}

*The function $d': \mathbb R \times \mathbb R \to \mathbb R$ given by $d'(x,y)=|x-y|$ is a metric on $\mathbb R$.

*$\log(x)$ is nondecreasing on $[1, \infty)$.
