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 $\varepsilon, \delta > 0$.