Metric in $\mathbb{P}_2$ I have to prove that $\mathbb{P}_2$ with the function $\delta(P,Q)$ defined by "Sine of the angle between two vector in $\mathbb{R}^3$ such that they correspond respectively to P and Q" is effectively a distance.
I am not able to prove the triangular inequality. I am trying with the brute approach, supposing that the three points are $$(1:0:0), (\cos(a):0:\sin(a)), (\cos(b):\sin(b)\sin(c):\sin(b)\cos(c)),$$
I get this one $$\sin^2(b)\cos^2(c)+(\cos(a)\sin(b)\sin(c)-\sin(a)\cos(b))^2\leq(\sin(a)+\sin(b))^2.$$
Note that the weird point refer to the vector $(\cos(b),0,\sin(b))$ rotated by the angle $-c$ with the axis of rotation corresponding to the z-axis.
Are you able to solve the inequality? Or have you got a different approach? 
 A: Here's a geometric proof.
Let $O$ denote the origin.  If $P$ and $Q$ are two points on the unit sphere, then $\sin(\angle POQ)$ is twice the area of triangle $\triangle POQ$.  Thus, it suffices to prove that
$$
\text{area}(\triangle POQ) \,+\, \text{area}(\triangle QOR) \;\geq\; \text{area}(\triangle POR)
$$
for any three points $P$, $Q$, and $R$ on the unit sphere centered at $O$.
Note first that, by reflecting $P$ or $R$ around $O$ if necessary, we may assume that $\angle POQ$ and $\angle QOR$ are acute.  We may also assume that $\angle POQ$ and $\angle QOR$ are both less than $\angle POR$, since otherwise the inequality is trivially true.
Now, rotate $\triangle POQ$ along $\overline{PO}$ so that it partially covers $\triangle POR$, and similarly rotate $\triangle QOR$ along $\overline{OR}$ so that it partially covers $\triangle POR$.  Then both of these rotated triangles contain the orthogonal projection $Q'$ of $Q$ onto the plane of $\triangle POR$, and hence they overlap.  It follows that they completely cover $\triangle POR$, which proves the inequality.
A: A probably-irrelevant remark: if you define the distance to be the sine of half the angle between the points (required to be in the same hemisphere!), then the triangle inequality is straightforward, because $2 \sin \frac{\theta}{2}$ is exactly the length of the chord from $P$ to $Q$, so the thing you're trying to prove becomes...the Euclidean triangle inequality for the three points $P,Q,R$. 
On the other hand, a quick matlab test of a thousand triples of points on the sphere suggests that the original statement, using sine, is in fact correct, since that test didn't produce any counterexamples. But I don't know a quick and easy proof, alas. 
