singlet and triplet states could equally be established for any other state of the form (5a) or (5b) for which $|c_1|=|c_2|=1/\sqrt{2}$ and $\vartheta_3 =\pi/2$, thereby showing that for every maximally entangled state there are infinitely many sets of directions giving the maximal violation of the CHSH inequality. As we have already said, such infinity of directions arises due to the [*nonuniqueness*]{} of the Schmidt decomposition of a maximally entangled state.

Concluding remarks
==================

Lastly, a further remark is in order about the relationship between the conditions needed to maximally violate the CHSH inequality and those required by the three-particle state (2) to produce a direct (“all or nothing”) nonlocality contradiction [@GHZ; @GHSZ]. As pointed out by KZ, it is remarkable that the same type of spin measurements involved in the two-particle case also forms the basis of the argument leading to the GHZ contradiction [@GHSZ]. This connection can be best appreciated when we generalize the CHSH inequality to a measurement scheme involving three spin-$\frac{1}{2}$ particles. The appropriate inequality is of the form [@Hardy] $$|\langle B_{\text H} \rangle | \leq 2 \, ,$$ where now the relevant Bell operator is $$\begin{aligned}
B_{\text H} \equiv &[& \sigma(\vec{e}_1)\otimes \sigma(\vec{e}\,^{\prime}_{2}) +
\sigma(\vec{e}\,^{\prime}_{1})\otimes \sigma(\vec{e}_{2})]\otimes
\sigma(\vec{e}\,^{\prime}_{3})  