I was stuck for a long time to prove the following trace inequality
\begin{equation}\label{ine1} {\text {tr}} ({\bf A}_1^{-1}) \leq {\text {tr}} ({\bf A}^{-1}), \tag{1} \end{equation}
where ${\bf A} \in {\mathbb C}^{K \times K}$ is a positive definite matrix and ${\bf A}_1$ consists of the diagonal elements of ${\bf A}$, i.e., ${\bf A}_1 = {\text {diag}} ( {\text {diag}} ({\bf A}))$. I have run a lot of simulation using Matlab and this inequality always holds. (1) in the simple case with $K=2$ can be easily proven. However, I can not prove it for the more general $K>2$ case. Some of my efforts are as follows.
Denote ${\bf A}_1 = {\text {diag}} \{ a_1, \cdots, a_K \}$ and the eigenvalues of ${\bf A}$ by $ \{ \lambda_1, \cdots, \lambda_K \}$. Then, it is obvious that
\begin{equation}\label{e1} \sum_{k=1}^K a_k = \sum_{k=1}^K \lambda_k. \tag{2} \end{equation}
In addition, using Hadamard's inequality, we have
\begin{equation}\label{ine2} \prod_{k=1}^K a_k \geq \prod_{k=1}^K \lambda_k. \tag{3} \end{equation}
Since (1) is equivalent to
\begin{equation} \sum_{k=1}^K \frac{1}{a_k} \leq \sum_{k=1}^K \frac{1}{\lambda_k}, \tag{4} \end{equation}
I tried to prove (4) using (2), (3), and the relationship between different means (Harmonic mean, etc.). But they didn't work.
Anyone providing the proof or relevant hints will be much appreciated. Thank you.
