Just for fun, here is an answer in the purely algebraic setting.

So, suppose that $X$ is irreducible projective algebraic of dimension $n$, $\alpha=c_1(\mathcal O_X(D))$ is the class of a nef divisor $D$, and $\omega=c_1(\mathcal O_X(A))$ is the class of an ample divisor.

What we want to show is that if there exists an integer $1\le k \le n$ such that $D^k\cdot A^{n-k}=0$, then for every irreducible subvariety $Z\subseteq X$ of dimension $k$, one has $D^k\cdot Z=0$.

By induction on $n$, the case $n=1$ being clear.

If $k=n$, then there is nothing to prove. Otherwise, fix a $k$-dimensional irreducible subvariety $Z\subseteq X$ and, by taking a large multiple $mA$ of $A$, choose a divisor $H\in|mA|$ containing $Z$, say
$$
H=m_1 H_1+\cdots+m_N H_N,
$$
where the $H_j$'s are the reduced irreducible components of $H$, $m_j>0$, and $Z\subseteq H_1$.

Now,
$$
\begin{aligned}
0 & =D^k\cdot A^{n-k}=\frac 1m\, D^k\cdot H\cdot A^{(n-1)-k} \\
& =\frac 1m\sum m_j\, D^k\cdot H_j\cdot A^{(n-1)-k} \\
& =\frac 1m\sum m_j\, D_{H_j}^k\cdot A_{H_j}^{(n-1)-k}
\end{aligned}
$$
where the $D_{H_j}$'s and $A_{H_j}$'s are respectively the restrictions of $D$ and $A$ to $H_j$.

Since $D_{H_j}$ is nef and $A_{H_j}$ is ample, we obtain that each $D_{H_j}^k\cdot A_{H_j}^{(n-1)-k}$ must be zero. In particular, $D_{H_1}^k\cdot A_{H_1}^{(n-1)-k}=0$.

Since $\dim H_1=n-1$, by induction hypothesis we have that $D_{H_1}^k$ intersects in zero each subvariety of $H_1$ of dimension $k$. But then, $0=D_{H_1}^k\cdot Z=D^k\cdot Z$.