Let $A\in\mathbb{R}^{n\times n}$ be a positive semidefinite matrix and $D\in\mathbb{R}^{n\times n}$ be a diagonal positive definite matrix. Let $\mathrm{diag}(X)\in\mathbb{R}^{n\times n}$ denote the diagonal part of a matrix $X\in\mathbb{R}^{n\times n}$.



> 
> **My question:** Does the following inequality hold true
>  $$
> \mathrm{diag}\left[(A+D)^{-1}\right] \ge \left[\mathrm{diag}(A)+D\right]^{-1}\ \ ?
> $$
> 
> 
> 


For the sake of clarity, I describe below a simple example. Consider
$$
A=\begin{bmatrix}2 & 1 \\ 1 & 3\end{bmatrix}, \quad D=\begin{bmatrix}0.5 & 0 \\ 0 & 2\end{bmatrix}.
$$
We have
$$
\mathrm{diag}\left[(A+D)^{-1}\right] = \begin{bmatrix}0.2927 & 0\\0 & 0.1341 \end{bmatrix}, \quad \left[\mathrm{diag}(A)+D\right]^{-1} = \begin{bmatrix}0.1818 & 0 \\0 & 0.0833 \end{bmatrix},
$$
so that $\mathrm{diag}\left[(A+D)^{-1}\right] \ge \left[\mathrm{diag}(A)+D\right]^{-1}$. 


(If true, the above inequality perhaps is just a well-know inequality but I wasn't able to find it in any "classical" matrix theory textbooks.)



$\mathrm{diag}\left[(A+D)^{-1}\right] \ge \left[\mathrm{diag}(A)+D\right]^{-1}$?
This is true. As Darij suggests, denote $B=A+D$, let $e=e\_i$ be a basic unit vector. We have to prove $(B^{-1}e,e)\geqslant (Be,e)^{-1}$. Denote further $C=B^{1/2}$, this rewrites $\|Ce\|\cdot \|C^{-1}e\|\geqslant 1$. This follows from Cauchy-Bunyakovsky-Schwarz, as $(Ce,C^{-1}e)=(e,e)=1$.


