Lower bound on $tr(A^{-1})$, where $A\in \mathbb{R}^{n \times n}$ symmetric, positive definite.

Question

Let $A \in \mathbb{R}^{n \times n}$ be a symmetric, positive definite matrix with $n \geq 2$. Does the following inequality then hold? $$ \sum_{i=1}^n \frac{1}{\lambda_i} > \frac{1}{\lambda_\min}, $$ where the $\lambda_i$'s are the eigenvalues of $A$ and $\lambda_\min := \min_{i \in \{1, \cdots, n\}} \left( \lambda_i \right)$.

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Thoughts so far:

A simpler inequality that clearly holds is $$ \sum_{i=1}^n \frac{1}{\lambda_i} \geq \frac{n}{\lambda_\max}, $$ but that doesn't seem useful in this situation.

Also, I realize that the left hand size is $\text{trace}(A^{-1})$, not sure what to do with this.

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Background:

An inequality in some lecture notes that I'm reading seems to implicitly hinge on this claim. Specifically, the following is stated as a portion of a longer discourse: $$ \sigma^2 \sum_{i = 1}^p \frac{1}{\lambda_i} > \frac{\sigma^2}{\lambda_\min}, $$ where $(X^T X)$ is a positive definite matrix, $X \in \mathbb{R}^{n \times p}$.

Answer 1

This should be obvious $$\sum_{i=1}^n \frac{1}{\lambda_i} > \frac{1}{\lambda_\min}$$

Because the eigenvalues are positive and one of the eigenvalues, say $\lambda_j$ is the minimum eigenvalue. Thus $$\sum_{i=1}^n \frac{1}{\lambda_i} = \frac{1}{\lambda_\min} + \sum _{i\ne j} \frac{1}{\lambda_i} > \frac{1} {\lambda_\min}$$

Answer 2

Yes.

Since the eigenvalues are positive, all the terms in the sum are positive. Therefore, the sum is larger than any of its terms.