Bell's Original Paper - Local hidden variable theories correlations smaller than entanglement

Question

I'm having trouble following Bell's derivation of equation 22 in his original paper. Particularly, how to go from

$$ | \overline{P} (\vec{a}, \vec{b}) - \overline{P}(\vec{a}, \vec{c}) | \leq 1 + \overline{P}(\vec{b}, \vec{c}) + \varepsilon + \delta $$

to

$$ | \vec{a} \cdot \vec{c} - \vec{a} \cdot \vec{b} | - 2 (\varepsilon + \delta) \leq 1 - \vec{b} \cdot \vec{c} + 2(\varepsilon + \delta) $$

using

\begin{equation} | \overline{P}(\vec{a}, \vec{b}) + \vec{a} \cdot \vec{b} | \leq \varepsilon + \delta. \qquad(1) \end{equation}

My attempt was

$$ \begin{equation} \begin{split} &\overline{P}(\vec{b}, \vec{c}) + \vec{b} \cdot \vec{c} \leq | \overline{P}(\vec{b}, \vec{c}) + \vec{b} \cdot \vec{c} | \leq \varepsilon + \delta\\ \implies & \overline{P}(\vec{b}, \vec{c}) \leq (\varepsilon + \delta) - \vec{b} \cdot \vec{c}\\ \implies & 1 + (\varepsilon + \delta) + \overline{P}(\vec{b}, \vec{c}) \leq 1 - \vec{b} \cdot \vec{c} + 2(\varepsilon + \delta)\\ \implies& | \overline{P} (\vec{a}, \vec{b}) - \overline{P}(\vec{a}, \vec{c})| \leq 1 - \vec{b} \cdot \vec{c} + 2(\varepsilon + \delta) \end{split} \end{equation} $$

How do I deal with the left hand side, i.e., the absolute value of difference of probabilities using $(1)$?

Edited to add more info

We are dealing with a pair of spin one-half particles in a singlet state moving freely in opposite directions.

$$ P (\vec{a}, \vec{b}) \equiv \int d \lambda \rho(\lambda)A(\vec{a}, \lambda) B(\vec{b}, \lambda) $$

where $\lambda$ is a continuous parameter, $\rho(\lambda)$ is its probability distributions and $A(\vec{a}, \lambda) = \pm 1$, $B(\vec{a}, \lambda) \pm 1$ are the possibles results of measuring spin components $\vec{\sigma}_{1}, \vec{\sigma}_{2}$ along $\vec{a}, \vec{b}$ respectively.

The probabilities $\overline{P}$s are nonnegative, the vectors $\vec{a}, \vec{b}, \vec{c}$ are unit vectors and I'm assuning $\varepsilon, \delta > 0$.

Answer 1

What Equation (1) says is that if you replace $P(\vec x,\vec y)$ by $-\vec x\cdot \vec y$, you incur an error of at most $\pm (\epsilon+\delta)$.

Now do this both on the rhs and the lhs of your first equation:

1. On the lhs of the "$\le$", in the worst case the error will make the value larger, which means you need to put a minus sign to still satisfy the inequality. You do two replacements, thus $-2(\epsilon+\delta)$.

2. On the rhs, in the worst case the error will make the value smaller. Thus, you get an extra $+(\epsilon+\delta)$.

If you want to formalize this, this can be done by the triangle inequality, plus the following variant thereof: If $|a-b|\le \eta$, then $b-\eta \le a \le b+\eta$.