On an inequality involving operator norm of matrices and singular value Let $A, E \in M_n(\mathbb C)$ be as in this question On invertibility of $A+E$ where $||E||_2<$ smallest singular value of $A$ and $||A^{-1}E||_2<1$  .
How to prove that $\dfrac {||A^{-1}b-(A+E)^{-1}b||_2}{||A^{-1}b||_2}\le \dfrac {||E||_2||A^{-1}||_2}{1-\frac {||E||_2}{\sigma_\min}}$ ? 
 A: I assume that $b\in Bbb C^n$ is an arbitrary non-zero vector and $$\sigma_\min(M)=\inf \{\|Mx\|_2: x\in\Bbb C^n\mbox{ and }\|x\|_2=1\}$$ for each $M\in\Bbb M_n(\Bbb C)$. Put  $c=(A+E)^{-1}b$.  Then $A^{-1}b=A^{-1}(A+E)c=c+A^{-1}Ec$. Now we have to prove that 
$$\dfrac {\|A^{-1}Ec \|_2}{\| c+A^{-1}Ec \|_2}\le \dfrac {||E||_2||A^{-1}||_2}{1-\frac {||E||_2}{\sigma_\min}}.$$ 
Put $d=\frac{c}{\|c\|_2}$. It suffices to prove that 
$$\dfrac {\|A^{-1}Ed \|_2}{\| d+A^{-1}Ed \|_2}\le \dfrac {||E||_2||A^{-1}||_2}{1-\frac {||E||_2}{\sigma_\min}}.$$ 
Since $\|A^{-1}Ed \|_2\le ||E||_2||A^{-1}||_2$, it suffices to prove that
$$1-\frac {||E||_2}{\sigma_\min}\le \| d+A^{-1}Ed \|_2.$$
Since $\| d+A^{-1}Ed \|_2\ge \| d\|_2- \|A^{-1}Ed \|_2=1-\| A^{-1}Ed \|_2$, it suffices to prove that
$${\sigma_\min}\| A^{-1}Ed \|_2\le ||E||_2.$$
Since $$\| A^{-1}Ed \|_2\le \| A^{-1} \|_2\|Ed \|_2\le \| A^{-1} \|_2\|E \|_2\|d\|_2=\| A^{-1} \|_2\|E \|_2,$$ 
It suffices to check that $${\sigma_\min}\| A^{-1}\|_2\le 1.$$
Indeed, 
$$\sigma_\min\| A^{-1}\|_2=\sigma_\min\sup \{\|A^{-1}x\|_2: x\in\Bbb C^n\mbox{ and }\|x\|_2=1\}.$$
Let $x\in\Bbb C^n$ and $\|x\|_2=1$. We have $$1=\|x\|_2=\|AA^{-1}x\|_2\ge \sigma_{\min} \|A^{-1}x\|_2,$$ so  $\sigma_\min\| A^{-1}\|_2\le 1$.
