How to prove this inequality, that: $$Tr(\mathbf{X})\geq \frac{N^2}{Tr(\mathbf{X}^{-1})}$$ where $\mathbf{X}\in \mathbb{R}^{N\times N}$ is an arbitrary positive definite matrix.
Horn R A, Johnson C R. Matrix analysis[M]. Cambridge university press, 2012.
On
As copper.hat suggests in the comments, use the fact that the eigenvalues of $X^{-1}$ are $1/\lambda_k$.
Then $$\newcommand\Tr{\operatorname{Tr}}\Tr(X)\Tr(X^{-1}) = \sum_{i=1}^N\lambda_i \sum_{j=1}^N\lambda_j^{-1} =\sum_{i=1}^N \frac{\lambda_i}{\lambda_i} + \sum_{i> j} \left(\frac{\lambda_i}{\lambda_j} + \frac{\lambda_j}{\lambda_i}\right)$$ Then for any $t>0$, $t+\frac{1}{t} \ge 2$. This yields $$\Tr(X)\Tr(X^{-1}) \ge N + \sum_{i > j} 2 = N + 2\frac{N(N-1)}{2} = N+N^2-N = N^2,$$ as desired.
Let $\lambda_1, \ldots, \lambda_N$ be the eigenvalues of $X$, which are positive real numbers by assumption. Then the eigenvalues of $X^{-1}$ are $\lambda_i^{-1}$. Clearing denominators and expressing trace in terms of eigenvalues, The inequality becomes
$$\left(\sum \lambda_i\right)\left(\sum 1/\lambda_i\right)\geq N^2.$$
Define $v$ to be the vector whose $i$th component is $\lambda_i^{1/2}$ and $w$ the vector whose $i$th component is $\lambda_i^{-1/2}$. The inequality is then Cauchy-Schwarz applied to $v$ and $w$.