An improved Greub-Rheinboldt inequality (Q648932)

Let \(A\) and \(B\) be Hermitian positive definite \(n\times n\) matrices satisfying \(AB=BA\). Denote by~\(\lambda_{(1)}\) and \(\lambda_{(n)}\) the largest and respectively smallest eigenvalue of~\(A\), and by~\(\mu_{(1)}\) and \(\mu_{(n)}\) those of~\(B\). Let \(0\neq x\in\mathbb{C}^n\). \textit{W. Greub} and \textit{W.~Rheinboldt} [Proc. Am. Math. Soc.~10, 407--415 (1959; Zbl 0093.12405)] proved that \[ \frac{(x^*A^2x)(x^*B^2x)}{(x^*ABx)^2}\leq \frac{(\lambda_{(1)}\mu_{(1)}+\lambda_{(n)}\mu_{(n)})^2} {4\lambda_{(1)}\mu_{(1)}\lambda_{(n)}\mu_{(n)}}. \] The matrices \(A\) and \(B\) have common eigenvectors. Let \(\lambda_1,\dots,\lambda_n\) and \(\mu_1,\dots,\mu_n\) be the eigenvalues of \(A\) and \(B\) respectively, orderd according to this correspondence. The present author provides a better inequality \[ \frac{(x^*A^2x)(x^*B^2x)}{(x^*ABx)^2}\leq \frac{(M+m)^2} {4Mm}, \] where \[ M=\max_{i}\frac{\lambda_i}{\mu_i},\quad m=\min_{i}\frac{\lambda_i}{\mu_i}. \]