A triangle inequality for angles

Question

Let $M$ be a complete Riemannian manifold with nonnegative curvature and $x,y,z,p$ four points on $M$. We denote by $\theta(x,y),\theta(y,z),\theta(z,x)$, respectively, the angles at $\tilde p$ of the triangles in $\mathbb R^2$ with the edge lengths:$\{d(x,y),d(x,p),d(y,p)\}$, $\{d(y,z),d(y,p),d(z,p)\}$ and $\{d(z,x),d(z,p),d(x,p)\}$. Then is it true that $$ \theta(x,y)+\theta(y,z) \ge \theta(z,x) $$

Answer 1

No, this is false. As a counterexample, let $M$ be the sphere of radius $1$, which I'll think of as embedded as the set of points a distance $1$ from the origin in $\mathbb{R}^3$. Let $p = (0,0,1)$ be the north pole; let $x = (-1,0,0), y = (0,1,0),$ and $z = (1,0,0)$. Thus $x, y, z$ are three points on the equator of the sphere such that $x$ and $z$ are antipodal and $y$ lies halfway between them. Then $d(x, z) = \pi$, and the distance between any other pair of points is $\pi/2$, so by your definitions $$ \theta(x,z) = \pi, \qquad \theta(x,y) = \theta(y,z) = \pi/3, $$ so your triangle inequality fails.