In this paper by Fiedler (1964) : "RELATIONS BETWEEN THE DIAGONAL ELEMENTS OF TWO MUTUALLY INVERSE POSITIVE DEFINITE MATRICES",
it is written in the proof of Theorem (3.2) on page 6 that $a_{ii}A_{ii} \geq det(A)$ is well-known.
Can someone indicate me where this comes from? I've searched for two hours but must be missing something obvious.
The inequality $a_{ii}\alpha_{ii}\ge1$ follows from Cauchy-Schwarz inequality: $$ \begin{align} a_{ii}\alpha_{ii} =(e_i^\ast Ae_i)(e_i^\ast A^{-1}e_i) &=\|A^{1/2}e_i\|^2\|A^{-1/2}e_i\|^2\tag{$\dagger$}\\ &\ge|\langle A^{1/2}e_i, A^{-1/2}e_i\rangle|^2\\ &=|\langle e_i, A^{1/2}A^{-1/2}e_i\rangle|^2 =1. \end{align} $$ To be honest, I am not sure if the same inequality in the form of $a_{ii}A_{ii}\ge\det(A)$ is really well-known, but it certainly follows easily from the use of Schur complement. Let $A=\pmatrix{a_{11}&v^\ast\\ v&P}$. Then $$ \det(A)=\det(P)(a_{11}-v^\ast P^{-1}v)\le a_{11}\det(P)=a_{11}A_{11}. $$ Remark on the paper's remark. Equality holds in $(\dagger)$ if and only if $A^{-1/2}e_i$ is a nonnegative scalar multiple of $A^{1/2}e_i$, i.e., iff $e_i$ is an eigenvector of $A$ corresponding to a nonnegative eigenvalue. Since $A$ is positive definite, this simply means $e_i$ is an eigenvector of $A$, i.e., all off-diagonal elements on the $i$-th row and column of $A$ are zero.