Is there a shorter way to prove triangular inequality for the metric SNCF?

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I'm trying to solve this problem.

Let $E=\mathbb{R}^{2}$ with the Euclidian metric $d_{2}$. Fix a point $P \in E$ and define a map $d : E \times E \to \mathbb{R}$ by $$d(x, y) = \begin{cases} d_{2}(x, y) & \text{if } x, y, P \text{ are on the same line } \\ d_{2}(x, P) + d_{2}(y, P) & \text{otherwise} \end{cases}$$

Prove that $d$ is a metric on $E$.

I found that to prove the triangular inequality by case-by-case ($x,y,z$ are on the same line, v.v..) is rather boring.

I would like to ask if I can prove $d(x,y) \le d(x,z) + d(y,z)$ more directly. Please shed me some light!


My attempt:

We have $$d(x,z) \ge \min\{d_{2}(x, z), d_{2}(x, P) + d_{2}(z, P)\}$$ $$d(y,z) \ge \min\{d_{2}(y, z), d_{2}(y, P) + d_{2}(z, P)\}$$ $$d(x,y) \le \max\{d_{2}(x, y), d_{2}(x, P) + d_{2}(y, P)\}$$

I fail to manipulate to get the desire inequality.