Entanglement isn't about interaction or information transfer betweeen entangled particles.
Consider spin-entaglement of two spin-$\frac{1}{2}$ particles:
Let them be in singulet-state relative to an arbitrary axis (say z-axis):
$$ |\Psi \rangle = \frac{1}{\sqrt{2}} (\ |\uparrow_z, \downarrow_z \rangle - |\downarrow_z,\uparrow_z\rangle \ ) $$
The propability $P$ to measure both particles in state $|i,j \rangle$ with $i,j \in \{ \uparrow, \downarrow \}$ where the axis of both measurments enclose the angle $\theta$ is given by:
$$ P_{i,j} = \| \langle i,j | \Psi \rangle \|^2 = \frac{1}{4} (1 - i \cdot j \cdot \cos \theta )$$
if we take $i,j$ to be 1 and -1 for $\uparrow$ and $\downarrow$, respectively.
The reduced propability $p_i$ of measuring only one particle (e.g. if we don't care about the other) is given by:
$$ p_i = \sum_{j \in \{1,-1\}} P_{i,j} = \frac{1}{2} $$
The conditional propability of measuring the other particle (after we already know about the measurment of the first particle) is given by:
$$ \tilde{p}_{j|i} = \frac{P_{i,j}}{p_i} = \frac{1}{2} (1 - i \cdot j \cdot \cos \theta ) $$
This does involve the angle $\theta$ and usually one starts here to argue about non-locality and instantanious actions changing the outcome of experiment when we change the angle $\theta$ at the first measurment apparatus.
This is however not true. If we are talking about conditional propabilities we have already performed a measurment and set the measurment axis of the first measurment. Changing this axis afterwards will not affect the propability as the angle $\theta$ is relative to the measured axis. Changing the axis of the second measurment only changes the propability predicting the outcome of the later measurment for the first observer because he has that extra knowledge.
The propability for the second observer stays the same, as this is the reduced propability (he doesn't know about the first measurment):
$$ p_j = \sum_{i \in \{1,-1\}} P_{i,j} = \frac{1}{2} $$
In short: Without the extra knowledge of the first measurment, entanglement is not important for the second observer. To gain that extra knowledge there must be an additional information transfer to the second observer and this is restricted by means of relativity-causality ($v\le c$ etc.). So entanglement neither breaks causality nor can it transfer any information.
$$$$
Sometimes one comes about the argument that the violation of Bell inequalities shows, that entanglement is still something more than classical perception would allow.
So let's have a look at a certain expectation value. The axis for spin measurment shall be labeled by normalized vectors $\vec{a}$ and $\vec{b}$ such that $\vec{a}\cdot\vec{b} = \cos\theta$. Consider
\begin{equation}
\langle \Psi|\vec{a}\cdot\vec{S_1} \ \ \vec{b} \cdot \vec{S}_2 | \Psi \rangle = -\frac{\hbar^2}{4}\vec{a}\cdot\vec{b} = -\frac{\hbar^2}{4} \cos\theta
\tag{1}
\end{equation}
, which is the expectation value of the product of both measurments results. Here we have $\vec{S} = \frac{\hbar}{2}(\sigma_x, \sigma_y, \sigma_z)^T$ with $\sigma_x, \sigma_y, \sigma_z$ the Pauli matrices.
We now follow the reasoning of John Bell in his original work since other, similar inequalities are based on the same problem.
The argument goes like this: Assume a classical, statistical system with non-hidden and hidden variables all labeled by $\vec{\lambda} = (\lambda_1, \dots, \lambda_n)$ for some $n\in\mathbb N$. Furthermore there exists two functions $A(\vec{a},\vec{\lambda})$ and $B(\vec{b},\vec{\lambda})$ that give the results of spin measurment on particle 1 and 2, respectively. They can only yield $\pm\frac{\hbar}{2}$, since that is the only outcome of experiment. Those functions depend on one measurment axis only, because there shall be no action between measurment apparatus 1 and 2 (this is the assumed locality).
$$$$
Because the system is studied on a statistical basis, there exists a propability density $ \varrho(\vec{\lambda}) $ that is a function of the system parameters $\vec{\lambda}$ and allows calculation of the expectation value
$$ E(\vec{a},\vec{b}) = \int \varrho(\vec{\lambda}) \cdot A(\vec{a},\vec{\lambda}) B(\vec{b},\vec{\lambda}) \ d^n\lambda $$,
which should equal the one from above (1) if it is to be interpreted on a classical, local basis (Note: one can incorporate discrete statistical variables by terms like $\sum_j \alpha_j \cdot \delta(c_j-\lambda_m)$). The malicious assumption here is that $\varrho$ is no function of the axis-vectors $\vec{a}$ and $\vec{b}$. This is, however, quite natural for classical systems with correlation. The point is: Allowing $\varrho(\vec{\lambda}, \vec{a}, \vec{b})$ or even just $\varrho(\vec{\lambda}, \vec{a} \cdot \vec{b})$, the Bell inequalities cannot be derived! Such propability densities can cause violation of the inequality. To understand that, I will now derive them and point out which step is not possible with the modified density:
$$$$
Assume
$$ E(\vec{a},\vec{b}) = -\frac{\hbar^2}{4} \vec{a} \cdot \vec{b} \tag2 $$,
so that quantum mechanical description is in agreement with the classical one. For $\vec{a} = \vec{b}$:
\begin{equation}
\begin{aligned}
-\frac{\hbar^2}{4} & = \int \underbrace{\varrho(\vec{\lambda})}_{\ge 0} \cdot \underbrace{A(\vec{a},\vec{\lambda}) B(\vec{a},\vec{\lambda})}_{\ge -\frac{\hbar^2}{4}} \, d^n\lambda \\ & \Leftrightarrow \\
0 & = \int \underbrace{\varrho(\vec{\lambda})}_{\ge 0} \cdot \left( \underbrace{A(\vec{a},\vec{\lambda}) B(\vec{a},\vec{\lambda}) + \frac{\hbar^2}{4}}_{\ge 0} \right) \, d^n\lambda
\end{aligned}
\end{equation}
because $\varrho$ is a normalized propability density. It follows that
\begin{equation}
\begin{aligned}
A(\vec{a},\vec{\lambda}) B(\vec{a},\vec{\lambda}) = -\frac{\hbar^2}{4}
\end{aligned}
\end{equation}
is a valid equation under the integral with $\varrho$. This can only hold if
\begin{equation}
\begin{aligned}
B(\vec{a},\vec{\lambda}) = - A(\vec{a},\vec{\lambda})
\end{aligned}
\tag3
\end{equation}.
Note that this holds for any vector $\vec{a}$. Now take another normalized vector $\vec{c}$ and do the following calculations:
\begin{align}
\frac{\hbar^2}{4}|(-\vec{a}\cdot\vec{b}) - (-\vec{a}\cdot\vec{c})| & = |E(\vec{a},\vec{b}) - E(\vec{a},\vec{c}) | \\
& = \left| - \int \varrho(\vec{\lambda}) \cdot (A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) - A(\vec{a},\vec{\lambda}) A(\vec{c},\vec{\lambda})) \, d^n\lambda \right| \\
& = \left| \int \varrho(\vec{\lambda}) \cdot A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) \cdot (1 - \frac{4}{\hbar^2}A(\vec{b},\vec{\lambda}) A(\vec{c},\vec{\lambda})) \, d^n\lambda \right| \\
& \le \int | \varrho(\vec{\lambda}) | \cdot | A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) | \cdot |1 - \frac{4}{\hbar^2}A(\vec{b},\vec{\lambda}) A(\vec{c},\vec{\lambda})| \, d^n\lambda \\
& = \int \varrho(\vec{\lambda}) \cdot (\frac{\hbar^2}{4} - A(\vec{b},\vec{\lambda}) A(\vec{c},\vec{\lambda})) \, d^n\lambda \\
& = \frac{\hbar^2}{4} + E(\vec{b},\vec{c}) = \frac{\hbar^2}{4} - \frac{\hbar^2}{4}\vec{b}\cdot\vec{c} \tag4
\end{align}
In the first equality we used (2). In the second we used (3). In the third we used $A(\vec{b},\vec{\lambda})^2 = \frac{\hbar^2}{4}$. The fourth step is the triangle inequality for integrals. In the fifth step we used $A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) = \pm \frac{\hbar^2}{4}$ and $\varrho(\vec{\lambda}) \ge 0$. In the last step we used (2) and the fact that $\varrho$ is normalized.
So we finaly have Bell's inequality
\begin{equation}
\begin{aligned}
|\vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{c}| + \vec{b}\cdot \vec{c} \le 1 \, ,
\end{aligned}
\tag5
\end{equation},
which can be violated for some choise of $\vec{a},\vec{b},\vec{c}$. This usually shows that our first assumption (2) is false. Therefore, no classical, local system should be able to describe the expectation value (1).
$$$$
With the modified probability density the steps in (4) look like this:
\begin{align}
\frac{\hbar^2}{4}|(-\vec{a}\cdot\vec{b}) - (-\vec{a}\cdot\vec{c})| & = |E(\vec{a},\vec{b}) - E(\vec{a},\vec{c}) | \notag \\
& = \left| - \int \varrho(\vec{\lambda}, \vec{a}, \vec{b}) \cdot A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) - \varrho(\vec{\lambda}, \vec{a}, \vec{c}) \cdot A(\vec{a},\vec{\lambda}) A(\vec{c},\vec{\lambda}) \, d^n\lambda \right| \notag \\
& = \left| \int A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) (\varrho(\vec{\lambda}, \vec{a}, \vec{b}) - \varrho(\vec{\lambda}, \vec{a}, \vec{c}) \frac{4}{\hbar^2}A(\vec{b},\vec{\lambda}) A(\vec{c},\vec{\lambda})) \, d^n\lambda \right| \notag \\
& \le \int \frac{\hbar^2}{4} \cdot \left| \varrho(\vec{\lambda}, \vec{a}, \vec{b}) - \varrho(\vec{\lambda}, \vec{a}, \vec{c}) \frac{4}{\hbar^2}A(\vec{b},\vec{\lambda}) A(\vec{c},\vec{\lambda}) \right| \, d^n\lambda
\end{align}
Note that one cannot proceed from here since in general $\varrho(\vec{\lambda}, \vec{a},\vec{b}) \ne \varrho(\vec{\lambda}, \vec{a},\vec{c})$. Also the second equality shouldn't work here anyway since (3) is only vaild when multiplied by $\varrho(\vec{\lambda},\vec{a},\vec{a})$. For instance, when $\varrho(\vec{\lambda},\vec{a},\vec{a}) = 0$ equation (3) can be violated in general. Nevertheless, one could only try to use another triangle equation on the term $|\dots|$, leaving us finally with the inequality
\begin{equation}
\begin{aligned}
|\vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{c}| \le 2 \, ,
\end{aligned}
\end{equation},
which is not to be violated by any choise of $\vec{a},\vec{b},\vec{c}$.
$$$$
In summary: If one allows propability densities $\varrho(\vec{\lambda}, \vec{a}, \vec{b})$, that depend on some parameters of the measurment, the derivation of an inequality which is violated by quantum mechanical expectation values is not possible in the usual way. Above, I already argued that the dependence on $\vec{a}, \vec{b}$ is in general no cause for non-local behaviour as long as the reduced propability of a subsystem is only depended on its own parameters. This problem is inherent to inequalities that are derived on the same arguments like Bell's inequality: see for example the CHSH-inequality on page 527 equation 2, which is frequently used in experiments!
$$$$
So if we would find some functions $A$ and $B$ that satisfy our locality-conditions from above there is no reason to think of the expectation value (1) as a non-local one. Take
\begin{align}
p_{i,j}(\vec{a},\vec{b}) & = \frac{1}{4} (1 - i j \ \vec{a}\cdot\vec{b}) \\
A(i,\vec{a}) & = \frac{\hbar}{2} \ i \\
B(j,\vec{b}) & = \frac{\hbar}{2} \ j
\end{align}
Then we have
$$ E(\vec{a}, \vec{b}) = \sum_{i,j \in \{1,-1 \}} p_{i,j}(\vec{a},\vec{b}) \cdot A(i,\vec{a}) B(j,\vec{b}) = - \frac{\hbar^2}{4} \ \vec{a}\cdot\vec{b} = - \frac{\hbar^2}{4} \ \cos\theta$$,
which equals (1) on a pure, local and classical basis.
