Bound on diagonal entries of inverse matrix

Question

If a complex hermitian matrix $A$ is invertible, and all its diagonal entries are real and $\ge 1$, is it true that all the diagonal entries of $A^{-1}$ are real and $\le 1$?

Answer 1

No. Let $Q$ be a real $n\times n$ orthogonal matrix whose first column is $\frac1{\sqrt{n}}(1,1,\ldots,1)^T$. Then $A=Q(n\oplus\epsilon I_{n-1})Q^T$ is a counterexample when $\epsilon>0$ is sufficiently small.

Conceptually, the inverse of a positive definite but nearly singular matrix must have a large positive eigenvalue. Hence its trace is also large and in turn at least one of its diagonal entry is large.