A Bell inequality is, in this context, a linear combination of conditional probability distributions that defines an inequality that can be violated by some (set of probabilities produced by a) quantum state, but not by a classical distribution.
More precisely, a linear inequality of the form
$$\sum_{a,b,x,y} c_{a,b,x,y} p(ab|xy) \le C$$
for some set of real coefficients $c_{a,b,x,y}$ and $C$.
A "classical behaviour" is in this context, sticking to a two-party scenario, a conditional probability distribution of the form
$$p(ab|xy)=\sum_\lambda p_\Lambda(\lambda)p_{A|X}(a|x)p_{B|Y}(b|y)$$
for some distributions of hidden variables $\Lambda$ and local conditional probability distributions $p_{A|X}$ and $p_{B|Y}$. Here $a,b$ denote measurement outcomes and $x,y$ measurement choices.
A separable bipartite state is by definition a state of the form
$$\rho = \sum_k p_k \,\rho_k^A\otimes\rho_k^B.$$
From the usual Born rule, the probability distributions that can be produced by this are
$$p(ab|xy) = \operatorname{tr}(\rho(\Pi_{a|x}\otimes \Pi_{b|y}))
= \sum_k p_k \operatorname{tr}(\rho_k^A \Pi_{a|x})\operatorname{tr}(\rho_k^B \Pi_{b|y}).$$
By slightly changing the notation as $\operatorname{tr}(\rho_k^A\Pi_{a|x})=p_{A|X}(a|x)$ and $\operatorname{tr}(\rho_k^B\Pi_{b|y})=p_{B|Y}(b|y)$ you should immediately see that any such distribution is "classical" in the sense outlined before. Hence any separable state is completely classical from this perspective, and won't by definition violate any Bell inequality.
This is all for general separable states. If you want to stick specifically to pure ones, the matter becomes much more trivial because a separable pure state is just a product state, and thus corresponds to a total lack of correlations.