Triangle inequality on the projective space

Question

Given a unit $n$-sphere $\mathbb{S}^n = \{x \in \mathbb{R}^{n+1} : \langle x,x \rangle = 1\}$, we define the set $\mathbb{P}^n = \{[x] : x \in \mathbb{S}^n\}$, where $[x] = \{-x, x\}$, and a function $d: \mathbb{P}^n \to \mathbb{R}$ given by $$d([x],[y]) = \min\{\|x-y\|, \|x+y\|\}.$$

I am to prove that $d$ is a distance function in $\mathbb{P}^n$. The exercise should be simple, but I can't figure out how to show that the triangular inequality holds, that is: $$d([x], [y]) \leq d([x], [z]) + d([z], [y]) \qquad\forall [x],[y],[z] \in \mathbb{P}^n.$$ Given two vectors $u,v \in \mathbb{S}^n$, I can say if $d([u], [v])$ is equal to $\|u-v\|$ or $\|u+v\|$ using the angle between the vectors, but separate the problem in different cases hasn't helped. I think there could be a more direct approach.

Answer 1

1. Let $z$ be any representative of $[z]$.
2. Let $x$ be a representative of $[x]$ such that $d([x],[z])=\|x-z\|$.
3. Let $y$ be a representative of $[y]$ such that $d([z],[y])=\|z-y\|$.

Now the desired inequality becomes $$d([x], [y]) \leq \|x-z\| + \|z-y\|$$ and the rest should be clear.