Suppose we have a function $f(x)$. Let $\delta x$ be a small perturbation of $x$ then we write.
$$ \delta f = f(x +\delta x) - f(x) \tag{1} $$
now the absolute condition number $\hat{\kappa} = \hat{\kappa}(x)$ is defined as
$$ \hat{\kappa}(x) =\lim_{\delta \to 0} \sup_{\| \delta x\| \leq \delta } \frac{\| \delta f\|}{\| \delta x\|} \tag{2}$$
we can simply write this as
$$ \hat{\kappa}(x) = \sup_{ \delta x} \frac{\| \delta f\|}{\| \delta x\|} \tag{3}$$
in comparison the relative condition number $\kappa = \kappa(x)$ is defined as
$$ \kappa(x) =\lim_{\delta \to 0} \sup_{\| \delta x\| \leq \delta } \bigg(\frac{\| \delta f\|}{\| f(x)\|} \bigg/ \frac{\| \delta x\|}{\| x\|} \bigg) \tag{4}$$
in the same way as above
$$ \kappa(x) = \sup_{ \delta x } \bigg(\frac{\| \delta f\|}{\| f(x)\|} \bigg/ \frac{\| \delta x\|}{\| x\|} \bigg) \tag{5}$$
With regards to the formula. If you consider multiplying two numbers as matrix times a vector
$$ \kappa(A) = \|A\| \frac{\| x\|}{\|Ax\|} \tag{6}$$
note for numbers this just becomes the absolute value
$$ \kappa(f(x_{1},x_{2})) = |x_{1}| \frac{|x_{2}|}{|x_{1}x_{2}|}\tag{7}$$
I believe
The statement is false. Consider the max norm, like here. So the counterexample is:
$A := \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is invertible and $B := \begin{pmatrix} 1/4 & 1/2 \\ 1/2 & 1 \end{pmatrix}$ is singular.
$$\|A^{-1}\| \cdot \|A-B\| = 1 \cdot \left\|\begin{pmatrix} 3/4 & -1/2 \\ -1/2 & 0 \end{pmatrix} \right\| = \frac{3}{4} < 1$$
The problem is indeed the absence of sub-multiplicative property. But if a matrix norm have this property then, the statement become true by the following lemma.
Lemma: Let $\|\cdot\|$ be a matrix norm on $\mathbb{R}^{n \times n}$. If $\|\cdot\|$ is sub-multiplicative, then there is a vector norm $\|\cdot\|_*$ on $\mathbb{R}^n$ such that both norms are compatible.
Proof: Fix $\boldsymbol{0} \neq \boldsymbol{y} \in \mathbb{R}^n$. Define $\|\cdot\|_* : \mathbb{R}^n \to [0,\infty)$ such that $\|\boldsymbol{x}\|_* := \|\boldsymbol{x}\boldsymbol{y}^T\|$ for every $\boldsymbol{x} \in \mathbb{R}^n$.
It's easy to check that $\|\cdot\|_*$ is a well defined vector norm because $\boldsymbol{y} \neq \boldsymbol{0}$ and the propierties of $\|\cdot\|$ for being, by hyphothesis, a sub-multiplicative matrix norm. So let's just check compatibility.
Let $A \in \mathbb{R}^{n \times n}$ and $\boldsymbol{x} \in \mathbb{R}^n$. Then by sub-multiplicative property of $\|\cdot\|$ we have
$$\|A\boldsymbol{x}\|_* = \|(A\boldsymbol{x})\boldsymbol{y}^T\| = \|A(\boldsymbol{x}\boldsymbol{y}^T)\| \leq \|A\| \cdot \|\boldsymbol{x}\boldsymbol{y}^T\| = \|A\| \cdot \|\boldsymbol{x}\|_*$$
Hence $\|\cdot\|$ is compatible with $\|\cdot\|_*$. $\blacksquare$
Then we are done since in the edited part of my post I had already demonstrated the result when we have sub-multiplicative and compatibility properties.
Best Answer
You have that $X=AX^{-1}$ and so $|| X|| \le || A ||||X^{-1}||$ that is equivalent to second inequality. The first inequality has the same proof by using $X^{-1}=(AX^{-1})^{-1}$