Diagonal entries of inverse larger than 1 over those for the original matrix A friend asked me if, given $A \in \mathcal{S}^n_{++}$ (positive definite), is it true $\forall$ j=1,...,n that
$$(A^{-1})_{jj} \geq \frac{1}{A_{jj}}$$
Not sure if it's true but we haven't found counterexamples yet. 
Attempt (assuming true): Let $A_{-ij} \in \mathbb{R}^{n-1 \times n-1}$ be A w/ the ith row and jth column removed.  It's equivalent to show $A_{jj}(A^{-1})_{jj} \geq 1$. Using  this , it's equivalent to show
$$\frac{A_{jj} det(A_{-jj})}{\sum_{i=1}^n (-1)^{i+j}A_{ij}det(A_{-ij})} \geq 1$$
Note the numerator and denominator both have terms $A_{jj} det(A_{-jj})$ so it's equivalent to show $\sum_{i \neq j}(-1)^{i+j}A_{ij}det(A_{-ij}) \leq 0$, and here is where I'm stuck.
I'm not sure if the attempt is the best approach, so alternate proposals are also much appreciated.
 A: Answer assuming $A$ is symmetric:
If $A$ is symmetric and positive definite, then there exists a symmetric positive definite $B$ such that $B^2 = A$. Note that $A_{jj} = e_j^TAe_j = e_j^TB^TBe_j = \|Be_j\|^2$ and similarly $(A^{-1})_{jj} = \|B^{-1}e_j\|^2$. By Cauchy-Schwarz,
$$\langle Be_j, B^{-1}e_j\rangle^2 \le \|Be_j\|^2\|B^{-1}e_j\|^2 = A_{jj}(A^{-1})_{jj}.$$
But $\langle Be_j, B^{-1}e_j\rangle = e_j^T(B^{-1})^TBe_j = e_j^TB^{-1}Be_j = e_j^Te_j = 1$. Thus $A_{jj}(A^{-1})_{jj}\ge 1$, as desired.
A: Since $A$ is symmetric positive definite, then there exists $Q$ orthogonal matrix such that
\begin{align}
A = Q^TDQ
\end{align}
where $D$ is a diagonal matrix with positive entries. Then it follows
\begin{align}
A^{-1} = Q^TD^{-1}Q.
\end{align}
In particular, we see that
\begin{align}
(A^{-1})_{ij} = \sum_{k, \ell} (Q^T)_{i\ell}(D^{-1})_{\ell k}(Q)_{kj} = \sum_{k, \ell} (Q^T)_{i\ell}(Q)_{kj}\frac{1}{\lambda_k}\delta_{\ell, k} = \sum_{k} (Q)_{ki}(Q)_{kj}\frac{1}{\lambda_k}
\end{align}
which means
\begin{align}
(A^{-1})_{ii}= \sum_{k}\frac{1}{\lambda_k}(Q)_{ki}(Q)_{ki}=\sum_{k}\frac{1}{\lambda_k}(Q)_{ki}^2.
\end{align}
Likewise, we could show
\begin{align}
(A)_{ii} = \sum_{k}\lambda_k(Q)_{ki}^2.
\end{align}
Next, observe
\begin{align}
(A^{-1})_{ii}(A)_{ii} = \sum_k\lambda_k (Q)_{ki}^2\sum_\ell \frac{1}{\lambda_\ell}(Q)_{\ell i}^2 \geq \sum_k (Q)_{ki}^2
\end{align}
where the last inequality is a Cauchy-Schwarz. Using the unit property of the columns and rows of $Q$, we have our desired result
\begin{align}
(A^{-1})_{ii}(A)_{ii} \geq 1.
\end{align}
A: Write $A=P^TDP$, so $A^{-1}=P^TD^{-1}P$. Now $A_{ii} = P_i^T D P_i$ (where $i$ indexes the columns), while $(A^{-1})_{ii} = P_i^T D^{-1} P_i$
Your question boils down to whether $P_i^T D^{-1} P_i \geq \frac{1}{P_i^T D P_i}$, or, equivalently, to whether $\sum_{j} \frac{P_{ij}P_{ij}}{D_{jj}} \geq \frac{1}{\sum_{j} P_{ij}P_{ij}D_{jj}}$. This is the celebrated mean vs. harmonic mean inequality.
A: For a statistical approach to this question, note that $A$ can be interpreted as the covariance matrix of an $n$-vector of Gaussian random variables $X \sim \mathcal N(0, A)$. Now, some basic facts about such random vectors: 


*

*$A_{ii}$ is the marginal variance of $X_i$. 

*$1/(A^{-1})_{ii}$ is the conditional variance of $X_i$ given $X_{-i} = (X_1, \ldots, X_{i-1}, X_{i+1}, \ldots, X_n)$. 

*The conditional variance of $X_i$ does not depend on the value of $X_{-i}$. Hence, from the conditional variance formula $\mbox{Var}(X_i) = \mbox{Var}(X_i | X_{-i}) + \mbox{Var}(E(X_i \mid X_{-i}))$. 


Combining these facts we immediately get $A_{ii} \ge 1/(A^{-1})_{ii}$, or equivalently 
$$
(A^{-1})_{ii} \ge \frac{1}{A_{ii}}
$$
as desired. 
