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) ?



Nefness on a K3 surface
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](http://link.springer.com/article/10.1007/BF01450509). And there is a newer version of it that works in positive characteristic as well [here](http://link.springer.com/chapter/10.1007%2F978-1-4614-6482-2_8#page-1).


Here is what you get out of this:


1. 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.
2. 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).
3. 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.


