Diagonal entries of inverse larger than 1 over those for the original matrix

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.

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}

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.