Nefness on a K3 surface Let $D$ be a divisor on a (complex) K3 surface.
Suppose $D^2\geq0$. In general, $D$ is nef if $D\cdot C\geq0$ for all irreducible curves on the surface.
Is it sufficient in our case  to check this for smooth rational curves (i.e. the (-2) curves) ?
 A: First off, you obviously have to assume something about $-D$ not being effective, because otherwise you could take a negative ample class. 
The cone of curves of a K3 surface is pretty well described in this paper. And there is a newer version of it that works in positive characteristic as well here.
Here is what you get out of this:


*

*It is possible that there are no $(-2)$-curves on a K3 surface, but in this case for every divisor with $D^2\geq 0$ either $D$ or $-D$ is both nef and effective.

*If the Picard number is $2$, it is possible that there is only one $(-2)$-curve. In this case there are actually (effective) curves with non-negative and even positive self-intersection which are not nef. However, they are negative on the sole $(-2)$-curve. (I think I will leave this for the reader for now).

*In all other cases the $(-2)$-curves generate a cone which is dense in the cone of curves, so any divisor that is non-negative on the $(-2)$-curves is non-negative on every effective curve.


So, it actually looks like that what you want is true.
A: Yes, as long as $D$ is effective, it's enough to check this for $C$ running over all smooth rational curves. See Corollary 1.7 of Chapter 8 of Huybrechts's notes http://www.math.uni-bonn.de/people/huybrech/K3Global.pdf and the discussion around it.
