 Solution

By asymptotic Riemann–Roch and the preceding bounds for higher sheaf cohomology,A nef divisor satisfies $D^n\ge 0$: for a fixed ample divisor $A$, the divisors $D+\epsilon A$ are ample for positive rational $\epsilon$, and continuity of the intersection product gives the inequality as $\epsilon \to 0$. If $h^0(X,mD)=0$ eventually, the left side is nonpositive. Dividing by $m^n$ forces $D^n=0$, since $h^1\ge 0$ and $D^n\ge 0$. The same identity then gives $h^1(X,mD)=O(m^{n-1})$.