What is Tsirelson's bound for the original bell's inequality

The folklore says the upper bound is 3/2 but I don't remember a reference. I am going to reproduce some old notes of mine demonstrating that bound.

First, Bell's original inequality is eqn (15) of [1]:

$$1+P(vec{b},vec{c})ge|P(vec{a},vec{b})-P(vec{a},vec{c})|,$$

where $a,b,c$ are unit vectors. The associated quantum prediction is obtained by using eqn (3) which states that

$$P(vec{x},vec{y})=-vec{x}cdotvec{y}.$$

So the quantum version of the inequality can be written

$$underbrace{|vec{a}cdotvec{b}-vec{a}cdotvec{c}|+vec{b}cdotvec{c}}_{displaystyle R}le 1.$$

Your question would then be how badly $R$ can violate that upper bound of 1. By a suitable choice of basis, we can always write

$$begin{align} vec{c}&=(cosvarphi,sinvarphi,0), vec{b}&=(cosvarphi,-sinvarphi,0), vec{a}&=(cosphisintheta,sinphisintheta,costheta), end{align}$$

with $varphi,phiin[0,2pi]$ and $thetain[0,pi]$.

Then

$$begin{align} vec{a}cdotvec{b}&=cos(phi+varphi)sintheta, vec{a}cdotvec{c}&=cos(phi-varphi)sintheta, vec{b}cdotvec{c}&=cos^2varphi-sin^2varphi, end{align}$$

and

$$R=2|sinvarphisinphisintheta|+1-2sin^2varphi.$$

This is a second-order polynomial in $|sinvarphi|$ whose maximum is reached for $|sinvarphi|=frac{1}{2}|sinphisintheta|$ and the maximum value is $frac{1}{2}|sinphisintheta|+1.$ This latter bound is then maximum for $phi=theta=pi/2$ and the maximum value is $3/2$. It is therefore reached for $varphi=pi/4$.

[1] John S. Bell. On the Einstein Podolsky Rosen paradox. Physics, 1(3):195–200, 1964. (pdf)

Tsirelson's bound for CHSH works only with matrix sum. If one uses Kronecker sum instead then the eigenvalues are -4,-2,0,2,4. So the average of CHSH can reach 4 in tensor quantum mechanics.