Trace of an inverse inequality $\text{Tr}(A^{-1}) \ge n^2 \text{Tr}(A)^{-1}$ Let $A \in \mathbb{R}^n$ be a positive definite matrix. Then, it is well known that 
\begin{align}
\text{Tr}(A^{-1}) \ge n^2  \text{Tr}(A)^{-1}
\end{align}
The proof follows by using the fact that trace is and a sum of eigenvalues and using AMGM inequality. 

My question: Does this inequality hold with equality iff and only if $A$ is a diagonal matrix?

I know also that this inequality holds with equality iff eigenvalues of $A$ are identical.   But not sure of this implies that $A$ is a diagonal matrix.
All positive-definite matrices are unitary diagonalizable $A=UDU^T$ and
$$
\operatorname{Tr}A=\operatorname{Tr}UDU^T=\operatorname{Tr}DU^TU=\operatorname{Tr}D.
$$
Similarly, $\operatorname{Tr}A^{-1}=\operatorname{Tr}D^{-1}$. Thus, it makes no restriction to assume that $A$ is diagonal. Of course, it is not going to be equality for a diagonal matrix in general (otherwise, it would be equality for all positive-definite matrices). It is equality iff all eigenvalues are equal, that is, iff $A=cI$, a scalar multiple of the identity matrix.

P.S. Just for completeness: the proof of the inequality follows immediately from Cauchy-Schwarz
$$
\left(\sum\sqrt{\lambda_i}\cdot\frac{1}{\sqrt{\lambda_i}}\right)^2\le \sum\lambda_i\cdot\sum\frac{1}{\lambda_i}
$$
with equality iff $(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n})$ and $(\frac{1}{\sqrt{\lambda_1}},\ldots,\frac{1}{\sqrt{\lambda_n}})$ are parallel, i.e. all $\lambda_i$ are equal.
Edit : I missed the assumption that the matrix should be symmetric. 



There is equality if and only if all the eigenvalues of $A$ are equal. This does not imply that $A$ is diagonal. For instance, take $$A = \begin{pmatrix} 1&amp;1\\0&amp;1\end{pmatrix} \, . $$ 
It is not diagonal but saturates your inequality.