Is this exotic function actually a metric?

Question

I've got this pretty exotic metric of which I cannot seem to prove the triangle inequality. Given that I already have a metric $\delta$ on the unit ball in $\mathbb{R}^n$, I define a new metric $d(x,y)$ to be zero whenever $x=y$ and $\ln\dfrac{2}{\delta(x,y)}$ whenever this is not the case.

The first two properties are pretty straightforward. As for the triangular inequality, in the case that $x,y$ and $z$ are all different points (the other cases are trivial), I get this: $$\ln\dfrac{2}{\delta(x,y)} \leq \ln\dfrac{2}{\delta(x,z)}+\ln\dfrac{2}{\delta(z,y)} = \ln\dfrac{4}{\delta(x,z)\delta(z,y)} \Leftrightarrow \delta(x,z)\delta(z,y) \leq 2\delta(x,y)$$

I do have the extra condition that none of these three values are zero, nor do they exceed $1$. But now I'm stuck. I'm also not sure that this IS a metric indeed, yet I have failed to find a counterexample. The only other condition I have is that $\delta$ does satisfy the triangular inequality, hence $$\delta(x,y) \leq \delta(x,z)+\delta(z,y)$$

Who can help?

Answer 1

There is a triangle $x,y,z$ in the ball such that $\delta(x,y)$ is arbitrarily close to $0$, and $\delta(x,z),\delta(y,z)\approx 1$. So that defeats your triangle inequality for $d$.

Answer 2

Your can find a triangle whose sides in terms of Euclidean space's natural metric are arbitrarily close to 2, 2 and 0. Then their $\delta $ values would be close to 0 for the first two and arbitrarily large for the third one. Then the triangle inequality is not satisfied.