Suppose we have two positive definite symmetric matrices $A,B\in \mathbb{R}^{n\times n}$ and $(X)_{ij}$ denotes the $ij$-th entry of a matrix $X$. Then do we have the following inequality?
$\frac{1}{2}\left((A^{-1})_{11}+(B^{-1})_{11}\right)\ge \frac{2}{(A)_{11}+(B)_{11}}$
If yes, how to prove it?
Write $$A=\begin{bmatrix}a_{11} & u^\top\\ u & A'\end{bmatrix},$$ then $$(A^{-1})_{11}=\frac{det(A')}{a_{11}det(A'-uu^\top/a_{11})}\ge\frac{1}{a_{11}},$$ since $A'\ge A'-uu^\top/a_{11}$ and $det(\cdot)$ is monotone(or using minimax theorem). Therefore Jensen's inequality gives for any probability measure $\mathbb{E}$, we have $$\mathbb{E}[(A^{-1})_{11}]\ge \mathbb{E}[\frac{1}{a_{11}}]\ge 1/\mathbb{E}[a_{11}].$$ And the proposed case is just this $\mathbb{E}$ being Bernoulli.