Inequality involving condition number

Question

How should I even begin to attempt to show that: $$\frac{\|\bf{x} - \tilde{x} \|}{\|\bf{x}\|} \leq \frac{cond(\bf{A})}{1 - \|\bf{A}^{-1} (\bf{A} - \bf{\tilde{A}}) \|} \left( \frac{\|\bf{b} - \bf{\tilde{b}} \|}{\|\bf{b}\|} + \frac{\|\bf{A} - \bf{\tilde{A}} \|}{\|\bf{A}\|} \right)$$ with $\bf{Ax = b}$ and $\bf{\tilde{A}\tilde{x} = \tilde{b}}$ for invertible real $n \times n$ matrices $\bf{A}$ and $\tilde{\bf{A}}$; and the vectors are elements of $\mathbb{R}^n$ ? Any hint is much appreciated.

Note that the norm $\| \:.\|$ is just any (consistent) norm.

Answer 1

I might edit this answer into a proper one once I'm a bit more free, but for now, please accept these screenshots since I don't have enough time to convert my custom LaTeX commands into normal ones.

Answer 2

HINT: From $$\begin{split} x-\tilde{x} &= A^{-1}b-\tilde{A}^{-1}\tilde{b} \\&=A^{-1}b-\tilde{A}^{-1}b+\tilde{A}^{-1}b-\tilde{A}^{-1}\tilde{b} \\&=\tilde{A}^{-1}(\tilde{A}-A)A^{-1}b+\tilde{A}^{-1}(b-\tilde{b}) \\&=\tilde{A}^{-1}[(\tilde{A}-A)x+(b-\tilde{b})], \end{split} $$ and $\|b\|\leq\|A\|\|x\|$ we have that $$ \frac{\|x-\tilde{x}\|}{\|x\|} \leq \|\tilde{A}^{-1}\|\|A\|\left(\frac{\|\tilde{A}-A\|}{\|A\|}+\frac{\|b-\tilde{b}\|}{\|b\|}\right). $$

Now you just need to make up a bound on $\|\tilde{A}^{-1}\|\|A\|$.