# Question about Bell's Theorem and hidden variable equations [closed]

I am trying to understand Griffiths explanation of Bell's theorem in chapter 12 of 2nd edition Intro to QM (or the afterword section A.2 in the 1st edition).

He starts with given two detectors measuring the spin component along axes denoted by the vector $\vec{a}$ and right detector along $\vec{b}$, define a function that is the average of the product of the spin of both particles called $P(\vec{a}, \vec{b})$

He then defines two more functions $A(\vec{a},\lambda)=\pm 1$ and $B(\vec{b},\lambda)=\pm 1$ where these two functions express the component of a particle's angular momentum as a function of a hidden variable $\lambda$ and along the above mentioned axes.

For entangled particles, if the detectors are aligned then the spins are anti-correlated such that

$$A(\vec{a},\lambda)=-B(\vec{a},\lambda)$$

Next he defines the average of the product of the spins using the hidden variable with a probability distribution $\rho(\lambda)$ as follows

\begin{align*} P(\vec{a}, \vec{b}) &= \int \rho(\lambda)\, A(\vec{a},\lambda)\, B(\vec{b},\lambda) \, d\lambda\\ &= -\int \rho(\lambda) \, A(\vec{a},\lambda) \, A(\vec{b},\lambda) \, d\lambda \end{align*}

Now given any other unit vector $\vec{c}$, then

\begin{align*} P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{c}) &= -\int \rho(\lambda) \Big[ \, A(\vec{a},\lambda) \, A(\vec{b},\lambda) - A(\vec{a},\lambda) \, A(\vec{c},\lambda) \Big] \, d\lambda\\ &= -\int \rho(\lambda) \Big[ 1 - A(\vec{b},\lambda) \, A(\vec{c},\lambda) \Big] \, A(\vec{b},\lambda) \, A(\vec{c},\lambda) \, d\lambda \hspace{5mm}\textbf{ since } [A(b,\lambda)]^2=1 \end{align*}

The next part is where he loses me. He then states that since $-1 \leq A(\vec{b},\lambda) \, A(\vec{c},\lambda)\leq +1$ and that since $\rho(\lambda) \Big[ 1 - A(\vec{b},\lambda) \, A(\vec{c},\lambda) \Big] \geq 0$ (which I get both of these points) he then jumps to this (which I don't get):

$$|P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{c})| \leq \int \rho(\lambda) \Big[ 1 - A(\vec{b},\lambda) \, A(\vec{c},\lambda) \Big] d\lambda$$