Without computing $A^{-1}$, what is a good analytic upper-bound for $e^\top A^{-1} e$, where $A$ is positive-definite and $e = (1,\ldots,1)\;$?

Question

Let $A$ be an $n \times n$ positive-definite matrix and let $e=(1,\ldots,1) \in \mathbb R^n$. Define $r(A) := e^\top A^{-1} e$. Note that we always have the trivial upper-bound $r(A) \le \lambda_\max(A^{-1})\|e\|^2 = n/\lambda_\min(A) =:r_0(A)$. Note that this bound is attained if $A$ is of the form $\alpha I_n$ with $\alpha>0$.

On the other hand, we always have $$r_0(A) / r(A) \le r_0(A) / \lambda_\min(A^{-1})\|e\|^2 = \lambda_\max(A)/\lambda_\min(A)=:\kappa(A), $$ where $\kappa(A)$ is the condition number of $A$.

Question. Without computing the inverse of $A$, what is a a good analytic upper-bound for $r(A)$ which is better than the trivial upper-bound $r(A)\le r_0(A)$ ?

Related question: Nontrivial lower-bound for $\inf_{x \in \Delta_n} \|Gx\|$