I know that, if $A$ is an $n\times n$ symmetric positive definite matrix, then $A_{ii}(A^{-1})_{ii}\geq1$ for all $i=1,\ldots,n$.
A proof is the following: we can write $A=PDP^T$ and $A^{-1}=PD^{-1}P^T$, where $D$ is diagonal and $P$ orthogonal. Then $$ A_{ii}=\sum_{l=1}^n P_{il}^2 D_{ll},\quad (A^{-1})_{ii}=\sum_{l=1}^n \frac{P_{il}^2}{D_{ll}}.$$ Then, by the Cauchy-Schwarz inequality and the fact that the rows of $P$ have norm $1$, $$ A_{ii}(A^{-1})_{ii}=\left(\sum_{l=1}^n (P_{il}\sqrt{D_{ll}})^2\right)\left(\sum_{l=1}^n\left(\frac{P_{il}}{\sqrt{D_{ll}}}\right)^2\right)\geq \left(\sum_{l=1}^n P_{il}\sqrt{D_{ll}}\frac{P_{il}}{\sqrt{D_{ll}}}\right)^2=1.$$
My question is about when the equality holds. Looking at the above proof, we have to use the fact that there is equality in the Cauchy-Schwarz inequality if and only if the vectors are linearly dependent. So, $A_{ii}(A^{-1})_{ii}=1$ if and only if there is a $\lambda_i\in\mathbb{R}$ such that $$P_{il}\sqrt{D_{ll}}=\lambda_i \frac{P_{il}}{\sqrt{D_{ll}}}.$$
I think that, if $A_{ii}(A^{-1})_{ii}=1$ for all $i=1,\ldots,n$, then $A$ is diagonal, but I am not sure. The idea would be to prove that each row of $P$ has a component $1$ and the rest are zeros. So suppose that, for a row $i$, there are two columns $l_i$ and $k_i$ such that $P_{i,l_{i}}\neq0$ and $P_{i,k_{i}}\neq0$. Then $D_{l_i,l_i}=D_{k_i,k_i}=\lambda_i$, that is, $A$ has two eigenvalues equal. I do not know how to proceed and if this fact is important. Any ideas?
Equality holds for all $i$ if and only if $A$ is diagonal. This is evident if you recall that $(A^{-1})_{ii}$ is the inverse of a Schur complement.
Let $A=\pmatrix{a_{11}&b^\top\\ b&C}$. As $A$ is positive definite, so is $C$. Yet, using Schur complement, we get $(A^{-1})_{11}=(a_{11}-b^\top C^{-1}b)^{-1}\ge a_{11}^{-1}$. So, equality holds if and only if $b^\top C^{-1}b=0$, i.e. iff $b=0$. It follows that $A$ is diagonal if $A_{ii}(A^{-1})_{ii}=1$ for all $i$.