Let $X$ be a smooth complete variety over an algebraically closed field of dimension $\geq3$. Given a divisor $D_1$ on $X$ with $D_1 \cdot C>0$ for every curve $C \subset X$, and a divisor $D_2$ on $X$ satisfying $D_2^2 \cdot S>0$ for every surface $S \subset X$, does there exist a divisor $D$ on $X$ satisfying $D \cdot C>0$ *and* $D^2 \cdot S>0$ for every curve and surface respectively? I am willing to make any assumption on $X$, except that $X$ be projective.

As I understand it, since $D_1$ is nef, we have that $D_1^2 \cdot S\geq 0$, so even if for $m>>0$ we manage to have $(mD_1+D_2) \cdot C>0$ for every curve, the other requirement becomes $(mD_1+D_2)^2 \cdot S = m^2(D_1^2 \cdot S) + 2m(D_1\cdot D_2 \cdot S) + D_2^2 \cdot S >0$. The first term is non-negative, the last term is positive, but what needs to happen to ensure the middle term is non-negative also?