On an eigenvalue inequality involving the Hadamard product

Let A, B be n×n positive definite matrices. Then for 0≤t≤1∏i=knλi(A∘B)≥∏i=knλi((A♯tB)(A♯1−tB))≥∏i=knλi(AB),k=1,…,n. This gives a weighted extension of a result of Ando [1]. The case where the eigenvalues λi are replaced with the singular values σi is also considered.