Von Neumann's trace inequality states that $|tr(AB)| \le \sum_{i=1}^{n} \sigma_i(A) \sigma_i(B)$ where $A, B$ are general square matrices with singular values $(\sigma_i(A)), (\sigma_i(B))$, respectively.
Question: does the analogous inequality hold for 3 (or more) matrices?
Related: If it were true, I would expect to find this generalization elsewhere. However, it seems to be proven in a slightly different form in Theorem 2 of [Kristof 69]. When checking the reference, note (1) that $\lambda_i(A A^*) = \sigma_i(A)^2$ and (2) that $tr(ABC) \le tr(Z_1 diag(\sigma(A)) Z_2 diag(\sigma(B)) Z_3 diag(\sigma(C)))$ where $Z_i$ are unitary. The second claim can be easily argued via SVD and trace cycling. I would like somebody more knowledgable to point out my mistake to me. I haven't checked the proof in the paper.
