Let $M = (m_{ij})$ be $n \times n$ symmetric positive definite matrix. Then it can be proven that $$ M^{1/2}A M^{1/2} \succeq M^{1/2}D M^{1/2}\succ 0 $$ so $$ \lambda_{\min}(M^{1/2}A M^{1/2}) \geq \lambda_{\min}(M^{1/2}D M^{1/2}), $$ where D is diagonal matrix with diagonal elements equal to $\frac{1}{m_{ii}}$ and A is matrix with diagonal elements equal to $$a_{ii} = \frac{1}{n-1}\sum_{j = 1,i\neq j}^n \frac{m_{jj}}{m_{ii}m_{jj} - m_{ij}m_{ji}} $$ and non-diagonal elements equal to $$a_{ij} = \frac{-1}{(n-1)} \frac{m_{ij}}{m_{ii}m_{jj} - m_{ij}m_{ji}}. $$
But in reality, ratio of these two minimal eigenvalues can be arbitrary large, for instance $$ M = \begin{bmatrix} 1& 0.99 \\[0.3em] 0.99 & 1 \end{bmatrix}$$ adding $9s$ to $0.99$, ratio $\frac{\lambda_{\min}(M^{1/2}A M^{1/2})}{\lambda_{\min}(M^{1/2}D M^{1/2})}$ is not bounded. For $n = 2$, answer is that $M$ should be close to singular. And their ratio is only equal to $1$ only for diagonal matrix. My intuition is that similar structure with $2 \times 2$ submatrices (close to singular) should apply for higher dimensions. Also my experiments suggest so, but I can't come up with some theoretical reasoning.