How will you prove that d(x,y)>=0 in a metric space.
That is part of the definition of a metric space, not something to be proved about metric spaces.
Unless you've been given a specific set and distance function on that set, and need to prove that the distance function is non-negative, in which case we'd need to see the details before we can help.
We can define a metric on a set $\displaystyle A$ as a map $\displaystyle d:A\times A\rightarrow \mathbb{R}$ satisfying:
$\displaystyle M.1\quad d(x,y)=0 \Leftrightarrow x=y$
$\displaystyle M.2\quad d(x,y)+d(x,z)\geq d(y,z)$
From $\displaystyle M.1$ and $\displaystyle M.2$ it is easy to prove:
$\displaystyle (i)\quad d(x,y)=d(y,x)$
$\displaystyle (ii)\;\; d(x,y)\geq 0$
For example, to prove $\displaystyle (ii)$ take $\displaystyle z=y$ .
Fernando Revilla