CHSH optimum from the correlation tensor ($S_{\max }=2\sqrt{{\lambda }_1+{\lambda }_2}$)

For spin-projective measurements on a two-qubit state, let ${\lambda }_1\ge {\lambda }_2\ge {\lambda }_3$ be the eigenvalues of $T^{T}T$, where $T$ is its Pauli correlation tensor. Bob's sum and difference of unit axes are orthogonal and have lengths $2\cos \varphi$ and $2\sin \varphi$. Maximizing Alice's two axes aligns them with the corresponding images under $T$. Maximizing $\varphi$ then gives $2\sqrt{\Vert T\mathbf{u}{\Vert }^2+\Vert T\mathbf{v}{\Vert }^2}$ for orthonormal $\mathbf{u},\mathbf{v}$. The maximum of this quadratic-form sum is ${\lambda }_1+{\lambda }_2$, by diagonalizing $T^{T}T$ and selecting its top two eigenvectors. Thus the displayed expression is both an upper bound and attainable. The same maximum applies to the form with the sum of two absolute values, since local outcome signs can select their positive branches.