A generalized singular value inequality for Heinz means (Q2800493)

Let \(A\) and \(B\) be Hermitian positive semidefinite \(n\times n\) matrices, and let \(s_j(\cdot)\) denote the \(j\)th largest singular value of a given matrix. \textit{K. M. R. Audenaert} [Linear Algebra Appl. 422, No. 1, 279--283 (2007; Zbl 1116.15014)] proved that if \(0\leq\nu\leq 1\), then NEWLINE\[NEWLINEs_j(A^\nu B^{1-\nu}+A^{1-\nu}B^\nu)\leq s_j(A+B), \tag{1}NEWLINE\]NEWLINE \(j=1,\dots,n\). In order to extend this, the present author claims that if \(0\leq\nu\leq\frac{3}{2}\) and \(0\leq t\leq 1\), then NEWLINE\[NEWLINE2\sqrt{t(1-t)}s_j(tA^\nu B^{1-\nu}+(1-t)A^{1-\nu}B^\nu)\leq s_j(tA+(1-t)B), \tag{2}NEWLINE\]NEWLINE \(j=1,\dots,n\). However, (1) cannot be extended to \(\nu>1\). To see why, simply take \(a>0\), \(A=aI\), \(B=I\), and denote \(\nu=1+\delta\). Then, by (1), \(a^{1+\delta}+a^{-\delta}\geq a+1\), which is false if \(a\) is large enough. The proof of (2) is questionable also for \(0\leq\nu\leq 1\).