Triangle inequality for hyperbolic metric of logarithm of cross ratio.

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Consider the Poincaré Half-Plane model of the Hyperbolic Space $ \mathbb{H}^2 $. I need to proof that the following d function is a metric.$ d:\mathbb{H}^2\rightarrow\mathbb{R}, d(A,B) = \big{|} log \frac{\bar{AA_{\infty}}.\bar{BB_{\infty}}}{\bar{BA_{\infty}}.\bar{AB_{\infty}}} \big{|}$ is indeed a metric. The argument of log is also called cross ratio of $A,B,A_{\infty}$ and $B_{\infty}$.

(Here $log$ is the logarithm in basis $e$, $\bar{AB}$ means the euclidean measure of the euclidean segment AB, $A_{\infty}$ is the extreme of the hyperbolic support line through the hyperbolic segment AB closer (euclideanaly) to A. Look at the image)enter image description here

My problem is how to proof that d satisfies $d(A,C) \leq d(A,B) + d(B,C)$ . I prooved that the function d satisfies the other metric axioms without very much work, but this is going hard to me and I'm stucked.

Thanks!

P.S.: I know that there is already a post similar to this, but the answer it got was to change to Klein model, which I can't use.

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Suppose we have 3 points $p, q, r\in\mathbb{H}^2$. We want to show that $$d(p, q)+d(q, r)≥d(p, r).$$

enter image description here

How hyperbolic circles are also Euclidean circles, then we can move the picture by a symmetry so that $p=i, q=\alpha+ai, r=bi$.

Let $Q=ai$ be the point on the $Y$ axis which is on the same horizontal line as $q$.

Here $a$ and $b$ and $\alpha$ are constants which depend on the distances involved. We have $$d(p, Q)=\ln(a), \quad d(Q, r)=\ln\left(\frac{b}{a}\right), \quad d(p, r)=\ln(b).$$

In particular $$d(p, Q)+d(Q, r)=\ln(a)+\ln\left(\frac{b}{a}\right)=\ln(b)=d(p, r).$$

So $p, Q, r$ satisfy the triangle inequality. However, we want to prove this for $q$ rather than $Q$.

Let $C$ be the circle centered at $p$ (in the hyperbolic sense) which contains $q$. By symmetry, the $Y$ axis is a diameter of $C$.

The point $Q$ is contained in the disk bounded by $C$. Therefore $$d(p, q)≥d(p, Q).$$

Similarly, $$d(r, q)≥d(r, Q).$$ Then $$d(p, q)+d(q, r)≥d(p, Q)+d(Q, r)=d(p, q).$$