I am solving the following problem:
Fix $0<\varepsilon<1$ and suppose that $A \in \mathbb{R}^{m \times m}$ is symmetric and nonsingular. Show that if $\|A-I\|_F \geq \varepsilon$, then $\left\|A^{-1}-I\right\|_F \geq \frac{\varepsilon}{2}$, where $\|\cdot\|_F$ denotes the Frobenius norm.
Here is my attempt: Since $A$ is real and symmetric, $$ \|A\|_F= \sqrt{\text{trace}(A^* A)} = \sqrt{\text{trace}(A^T A)} = \sqrt{\text{trace}(A^2)} . $$ Hence, we have $$ \|A - I\|_F^2= (a_{11} - 1)^2 + \cdots + (a_{mm} - 1)^2 \geq \varepsilon. $$
But, I cannot go further. Could you give me any hint about this problem?
Let $\lambda_1, \lambda_2, \dots, \lambda_m$ denote the eigenvalues of $A$. We have, \begin{align*} \|A^{-1} - I\|_{F}^2 = \sum_{i = 1}^m \left(\frac{1}{\lambda_i} - 1 \right)^2 = \sum_{i = 1}^m \frac{(\lambda_i - 1)^2}{\lambda_i^2}. \end{align*} If for some $j$, $|\lambda_j| \geq 2$, then, $\displaystyle \|A^{-1} - I\|_{F}^2 \geq \frac{(\lambda_j - 1)^2}{\lambda_j^2} \geq \frac{1}{4} \geq \frac{\varepsilon^2}{4}$ as $\varepsilon \in (0,1)$. This implies $\|A^{-1} - I\|_{F} \geq \dfrac{\varepsilon}{2}$. If no such $j$ exists, then it implies that $|\lambda_i| \leq 2$ for all $i = 1, 2,\dots, m$. Thus, \begin{align*} \|A^{-1} - I\|_{F}^2 = \sum_{i = 1}^m \frac{(\lambda_i - 1)^2}{\lambda_i^2} \geq \frac{1}{4} \sum_{i = 1}^m {(\lambda_i - 1)^2} \geq \frac{1}{4} \|A - I\|_F^2. \end{align*} This implies $\|A^{-1} - I\|_{F} \geq \dfrac{\varepsilon}{2}$ since $\|A - I\|_F \geq \varepsilon$.