Suppose that $A, \hat A$ are invertible real matrices such that $Ax = y$ and $\hat A\hat x = \hat y$, where $\|A - \hat A\| \leq \epsilon_1$ and $\|\hat y - y\| \leq \epsilon_2 \|y\|$. I'm trying to show that if $\epsilon_1 < \sigma_{\min}(A)$, then
$$\frac{\|x - \hat x\|}{\|x\|} \leq \frac{\epsilon_2 \sigma_\max(A) + \epsilon_1}{\sigma_\min(A) - \epsilon_1}$$.
The thing I tried was to expand, $$ \|x - \hat x\| = \|A^{-1}(y - \hat y) + (A^{-1} - \hat A^{-1})\hat y\| \leq \sigma_\max(A^{-1}) \|y - \hat y\| + \|A^{-1} - \hat A ^{-1}\|\|\hat y\|. $$ Then we also have $\|x\| \geq \sigma_\min(A^{-1})\|y\|$, so putting these bounds together, we have $$ \frac{\|x - \hat x\|}{\|x\|} \leq \frac{\sigma_\max(A^{-1}) \|y - \hat y\| + \|A^{-1} - \hat A ^{-1}\|\|\hat y\|}{\sigma_\min(A^{-1})\|y\|} \leq \frac{\epsilon_2 \sigma_\max(A) + \sigma_\max(A)\sigma_\min(A) \|A^{-1} - \hat A^{-1}\|\|\hat y\|}{\sigma_\min(A)}, $$ which I don't really know what to do with. Any hints/help would be greatly appreciated.
There is an error in the last inequality, which should read $$ \frac{\|x-\hat x\|}{\|x\|} \le \frac{\epsilon_2 \sigma_\max(A) + \epsilon_1 \frac{ \|\hat y\| }{ \|y\|}}{ \sigma_\min(A)} \le \frac{\epsilon_2 \sigma_\max(A) + \epsilon_1 (1+\epsilon_2)}{ \sigma_\min(A)} $$ Also I think the left-hand side of the desired inequality should be $\frac{\|x-\hat x\|}{\|\hat x\|}$ to get to the correct right-hand side.