$X$ rational surface, $\mathcal{F}$ nef, then $h^2(X,\mathcal{F})=0$ I'm reading this paper on anticanonical surfaces, i.e. surfaces such that $-K_X$ is effective.
Throughout the paper $X$ is a smooth projective surface over an algebraically closed field.
In the proof of Lemma II.2, he only assumes $X$ is rational (not yet anticanonical) and claims that the following result is elementary:

(c) If $\mathcal{F}$ is a nef divisor, then $h^2(X,\mathcal{F})=0$ and $\mathcal{F}^2\geq 0$.

I know that $\mathcal{F}^2\geq 0$ is a classical result on nef divisors (it's true even if $X$ is not rational), but I can't see why $h^2(X,\mathcal{F})=0$ and why $X$ being rational should matter.
The only thing I can think of is Serre duality, so that $h^2(X,\mathcal{F})=h^0(X,K_X-\mathcal{F})$, but I don't know what to make of this.
 A: Actually the solution was much simpler than I thought. We use the following fact:
Suppose that $X$ is a smooth rational surface, then there exists a smooth, irreducible, rational curve $C \subset X$ such that $C^2 \geq 0$ and $K_{X}.C \leq -2$.
proof of fact. Since $X$ is rational, it is either a blow-up of $\mathbb{CP}^2$ or a Hirzebruch surface in a sequence of points. If $X$ is a blow-up of $\mathbb{CP}^2$ let $C$ by a line missing all the blow-up points. If $X$ is a blow-up of a Hirzebruch surface then let $C$ be a $\mathbb{P}^1$-fibre missing all of the blow-up points. Lines have normal bundle $\mathcal{O}(1)$ and $\mathbb{P}^1$-fibres have normal bundle $\mathcal{O}$, so the result follows from a computation.
Proof of Lemma: Take a curve $C$ as in the above fact. Suppose for a contradiction that $h^{2}(\mathcal{F}) = h^{0}(K_X - \mathcal{F}) \neq 0$. Then $K_{X}-\mathcal{F}$ is effective. It follows that, $$C. (K_{X}-\mathcal{F}) \geq 0. \;\;\;\;(*)$$
This is the general fact that an irreducible curve with positive self-intersection always intersects an effective divisor non-negatively. This is exactly "Useful remark 3.5" in the book "Complex Algebraic Surfaces" by A. Beauville.
But $\mathcal{F}$ is nef so $C . \mathcal{F} \geq 0$ and from above $K_{X}. C \leq -2$, the contradicts the inequality (*).
