Why can't separable pure states violate some Bell inequality?

In ch 4 of Preskill's lecture notes on quantum information, he tells us that any entangled pure state of two qubits violate some Bell inequality, whereas no such violation occurs for separable pure states. I do not understand why this is the case for separable pure states. What is the physical motivation behind this, and how do I show this?

Link to lecture notes: http://theory.caltech.edu/~preskill/ph229/

Local measurements on separable pure states generate independent (and hence local) random variables.

To see this take any POVM $$\{M_a\}$$ for Alice and any POVM $$\{N_b\}$$ for Bob and suppose they measure them on a separable pure state $$|\phi\rangle_{AB} = |\psi_A\rangle \otimes |\psi_B\rangle$$. Then the outcome probability distribution takes the form \begin{aligned} \mathbb{P}[A=a,B=b] &= \langle \phi|M_a\otimes N_b|\phi\rangle \\ &= (\langle\psi_A| \otimes \langle\psi_B|) M_a\otimes N_b (|\psi_A\rangle \otimes |\psi_B\rangle) \\ &= \langle\psi_A|M_a|\psi_A\rangle \langle\psi_B|N_b|\psi_B\rangle \\ &= \mathbb{P}[A=a]\,\mathbb{P}[B=b] \end{aligned} i.e., the joint distribution is always a product distribution! In other words the outcomes of the measurements of the different parties are always independent and therefore are local.

Separable mixed states also generate local distributions however the measurement outcomes need not be independent.

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.

You can make separable pure states via local operations and classical communications. The measurement results on them can be simulated locally with access to some shared randomness only and no further communication. This exactly described the situation of a local hidden variable model. The whole point of a Bell inequality is that it shouldn't be violated by a local hidden variable model.

What do you need to prove? Prove the existence of the local hidden variable model that describes all possible (joint but separable) measurement results on the two qubits.