Variant versions of the Lewent type determinantal inequality

Document Type: Original Article

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Abstract

‎In this paper‎, ‎we present a refinement of the Lewent determinantal inequality and then‎, ‎we show that the following inequality holds‎
‎\begin{align*}‎
‎ &\det\frac{I_{\mathcal{H}}+A_1}{I_{\mathcal{H}}-A_1}+\det\frac{I_{\mathcal{H}}+A_n}{I_{\mathcal{H}}-A_n}-\sum_{j=1}^n\lambda_j \det\left(\frac{I_{\mathcal{H}}+A_j}{I_{\mathcal{H}}-A_j}\right)\\‎
‎ & \ge \det\left[\left(\frac{I_{\mathcal{H}}+A_1}{I_{\mathcal{H}}-A_1}\right)\left(\frac{I_{\mathcal{H}}+A_n}{I_{\mathcal{H}}-A_n}\right)\prod_{j=1}^n \left(\frac{I_{\mathcal{H}}+A_j}{I_{\mathcal{H}}-A_j}\right)^{-\lambda_j}\right]\,‎,
‎\end{align*}‎
‎where $A_j\in\mathbb{B}(\mathcal{H})$‎, ‎$0\le A_j < I_\mathcal{H}$‎, ‎$A_j's$ are trace class operators and $A_1 \le A_j \le A_n~(j=1,\cdots,n)$ and $\sum_{j=1}^n\lambda_j=1,‎~ ‎\lambda_j \ge 0‎~ ‎(j=1,\cdots,n)$‎. In addition‎, ‎we present some new versions of the Lewent type determinantal inequality‎.

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