Let $X$ be a smooth projective complex K3 surface and $L, D$ two effective divisors, $L^2\geq0$ and $D^2\geq0$. 


(Q1). do we have $L\cdot D\geq0$ ?


If either one has positive self-intersection, the answer is yes by index theorem. Assume $L^2=D^2=0$. If either one is basepoint free then it is a multiple of an elliptic curve and so the answer is yes. 


It remains to consider the case when $L$ and $D$ are both not basepoint free. 


I did not manage to show this, so I tried to construct a simple counterexample where $L$ and $D$ both share the same rational curve as fixed part, but I could not find any. Is there any? 



Positivity question on K3 surfaces
Let me prove that $(L,D)\geq 0$.


Consider the set of all (1,1)-classes which satisfy 
$\eta^2\geq 0$. It is a union of two components $P\_+$ and $P\_-$, 
intersecting in 0. If $\eta, \eta'$ are in the same component
($P\_+$ or $P\_-$), they intersect non-negatively, by Hodge index 
theorem. Let $P\_+$ be a component containing a Kahler class. 
It remains to prove that any effective class that satisfies 
$\eta^2=0$ lies in $P\_+$.


However, for each $a\in P\_-$ and $b\in P\_+$,
one has $(a,b) \leq 0$, because vectors in $P\_+$
intersect positively by Hodge index theorem.
Therefore, an intersection of any class
in $P\_-$ and the Kaehler class is non-positive.
Therefore, $P\_-$ does not contain effective classes.


