فهرست نشریات علمی دسترسی آزاد ایران

‎Let $\lambda_1,\dots,\lambda_n$  be positive real numbers such that $\sum_{k=1}^n \lambda_k=1$. In this paper, we prove that for any positive operators $a_1,a_2,\ldots, a_n$ in semifinite von Neumann algebra $M$ with faithful normal trace that $\t(1)<\infty$, $$\prod_{k=1}^n(\det a_k)^{\lambda_k}\,\le\,\det (\sum_{k=1}^n \lambda_k a_k),$$where $\det a=exp(\int_0^{\t(1)} \mu_a(t)\,dt)$. If furthermore $\t(a_i)<\infty$ for every $1\le i\le n$ and $ \prod_{k=1}^n(\det a_k)^{\lambda_k}\neq 0$, then equality holds if and only if $a_1=a_2=\cdots =a_n$. A log-majorisation version of Young inequality are given as well.