I'm reading about Bell's inequalities and hidden variables theory. And I wonder where does the CSHS inequality really come from ? I read the paper and the inequality we find is $$\left|P(a,b)-P(a,b')\right|+P(a',b)+P(a',b')\leq 2 \tag{1}$$
while the most common form I found elsewhere is $$P(a,b)-P(a,b')+P(a',b)+P(a',b')\leq 2 \tag{2}$$
Where does the absolute value disappear ? I don't understand how (2) could be contained in (1)
Note that, in this paper, the notation $P(a,b)$ is a (joint) expectation value : $$P(a,b) = \int d\lambda \rho(\lambda)A(a, \lambda) B(b, \lambda)= \langle AB\rangle_{ab}$$, where $A$ and $B$ can only take values $\pm1$
You have :
$$-2 \leq A(a, \lambda) (B(b, \lambda) - B(b', \lambda)) + A(a', \lambda) (B(b, \lambda) + B(b', \lambda))\leq 2 \tag{1}$$ because if one of the 2 terms is $2$ or $-2$, the other term is zero, and these are the only possibilities.
Summing on the $\lambda$ with the density $\rho(\lambda)$, we get :
$$-2 \leq P(a,b)-P(a,b')+P(a',b)+P(a',b') \leq 2 \tag{2}$$
Now, considering : $|A(a, \lambda) (B(b, \lambda) - B(b', \lambda))| + A(a', \lambda) (B(b, \lambda) + B(b', \lambda))$, if the first term is zero, the second term is $\leq 2$, and if the first term is $2$, then the second term is zero, so finally:
$$|A(a, \lambda) (B(b, \lambda) - B(b', \lambda))| + A(a', \lambda) (B(b, \lambda) + B(b', \lambda)) \leq 2\tag{3}$$
Summing on the $\lambda$, we get :
$$\int d\lambda \rho(\lambda)|A(a, \lambda) (B(b, \lambda) - B(b', \lambda)) |+P(a',b)+P(a',b') \leq 2\tag{4}$$
But we have :
$$|P(a,b)-P(a,b')| = |\int d\lambda \rho(\lambda)A(a, \lambda) (B(b, \lambda) - B(b', \lambda))| \\ \leq\int d\lambda \rho(\lambda)|A(a, \lambda) (B(b, \lambda) - B(b', \lambda)) |\tag{5}$$
So, finally,
$$|P(a,b)-P(a,b')|+P(a',b)+P(a',b') \leq 2\tag{6}$$
