I've been trying that distance using haversine formula is metric space. I can prove first and second conditon, but i have problem with prove that for haversine is true that $ d(x,y)+d(y,z)>= d(x,z) $
I am stuck, I have continued on different paths and they all seem to give nothing, please help.
We assume we have a sphere $\,S\,$ and measure distances between points $\,x\,$ and $\,y\,$ on $\,S\,$ using the angle they subtend when viewed from the center of $\,S\,$. That is, we denote this angle by $\,d(x,y).\,$ The great-circle distance is $\,d(x,y)\,$ multiplied by the radius of $\,S\,$. Given points $\,x,y,z\,$ on $\,S\,$ forming a spherical triangle define the side angles by $$ a:=d(x,y),\; b:=d(y,z),\; c:=d(x,z).$$ Denote the angle of the spherical triangle at $\,y\,$ by $\,C.\,$
The spherical law of cosines, the cosine addition theorem, and using haversines gives the equation $$ \text{hav}(a+b) = \text{hav}(c) + \sin(a)\sin(b) \cos(C/2)^2. $$ The haversine function is strictly monotone increasing on the interval $\,[0,\pi],\,$ and the sines of the sides are positive. This then implies that $\,c \le a+b\,$ which is the triangle inequality for spherical triangles.
EDIT: Compare this with the Wikipedia article Haversine formula.