I have a metric on the space $\Bbb{P}(\Bbb{R}^n)$ of lines through the origin in $\Bbb{R}^n$, that is that $d(L_1, L_2) = \sqrt{1 - \frac{|<v, w>|}{\lVert v \rVert \lVert w \rVert}}$, where $v$ and $w$ are any non-zero vectors in $L_1$ and $L_2$ respectively.
I can see that the Cauchy-Schwartz inequality shows that this metric is well-defined, and that the equality condition for Cauchy-Schwartz ensures positivity. Symmetry is obvious. I'm struggling with the triangle inequality, and after looking for quite a while I haven't succeeded in finding a simple proof of this, although I may just be looking in the wrong place!
I'd like to construct a proof directly if possible - could anyone give me a prod in the right direction?
In situations like that, one trick is to pass to the sphere where things are more intuitive and we can reduce to the usual Euclidean norms.
Namely, observe that since we only care about the directions, it doesn't matter which points on the lines we pick, so we can WLOG consider points on the unit sphere. Then the distance you define reduces to $$d(L_1,L_2) = \sqrt{1-\langle u,v\rangle}= \sqrt{1-\cos(u,v)}$$ Now recall that $\cos2\alpha = 1-2\sin^2\alpha$, thus $$d(L_1,L_2) = \sqrt{2}\sin \frac{\angle(u,v)}{2}.$$ The last observation we need is that the sine of the half-angle between $u$ and $v$ is proportional to the distance $\|u-v\|$ in Euclidean space, by considering the altitude in the isosceles triangle with vertices at $u,v$ and the origin. This reduces your triangle inequality to the triangle inequality in Euclidean space.