Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 2 / 10F (Probability)

For any $x,z\in \{\pm 1\}$ there is exactly one $y=xz$, so $\mathbb{P}(X=x,Z=z)=1/4=\mathbb{P}(X=x)\mathbb{P}(Z=z)$. Similarly $(Y,Z)$ is independent, and $(X,Y)$ is independent by assumption. They are pairwise independent but not jointly independent, because $XYZ=1$ surely.

Using $\max \{X^2,Y^2\}=(X^2+Y^2+|X^2-Y^2|)/2$ and Cauchy-Schwarz inequality,

At most two of $X_1>X_2$, $X_2>X_3$, and $X_3>X_1$ can hold in any outcome. Taking expectations gives $a_{12}+a_{23}+a_{31}\le 2$, and hence $\min \{a_{12},a_{23},a_{31}\}\le 2/3$.