An inequality for \(t\)-geometric means (Q2805358)

Let \(A_1,\dots,A_m\) and \(B_1,\dots,B_m\) be Hermitian positive definite \(n\times n\) matrices, and let \(\|\cdot \|\) be a unitarily invariant norm. The author proves that NEWLINE\[NEWLINE\begin{multlined} \|\sum_{i=1}^m(A_i\sharp_tB_i)^r\|\leq \|\big[(\sum_{i=1}^mB_i)^\frac{trs}{2}(\sum_{i=1}^mA_i)^{(1-t)rs} (\sum_{i=1}^mB_i)^\frac{trs}{2}\big]^\frac{1}{s}\|\\ \leq \|\big[(\sum_{i=1}^mA_i)^\frac{(1-t)rs}{2} (\sum_{i=1}^mB_i)^\frac{trs}{2}\big]^\frac{1}{s}\|. \end{multlined}NEWLINE\]NEWLINE Here, \(0\leq t\leq 1\), \(r\geq 1\), \(s>0\), and NEWLINE\[NEWLINE A_i\sharp_tB_i=A_i^\frac{1}{2}(A_i^{-\frac{1}{2}}B_iA_i^{-\frac{1}{2}})^tA_i^\frac{1}{2}, NEWLINE\]NEWLINE the \(t\)-geometric mean of \(A_i\) and \(B_i\). Consequently, if \(A_i\) and \(B_i\) commute, \(i=1,\dots,m\), then NEWLINE\[NEWLINE \|\sum_{i=1}^mA_iB_i\|\leq\|(\sum_{i=1}^mA_i^\frac{1}{2}B_i^\frac{1}{2})^2\|\leq \|(\sum_{i=1}^mA_i)(\sum_{i=1}^mB_i)\|, NEWLINE\]NEWLINE proved previously by \textit{K. M. R. Audenaert} [Electron. J. Linear Algebra 30, 80--84 (2015; Zbl 1326.15030)].NEWLINENEWLINEReviewer's remark: ``Applying Theorem~2.2 for \dots'' on p.~768 should read: ``Applying Theorem~2.2 and its proof for \dots''.