The condition of a differentiable function $f:X\rightarrow Y$ is defined as $$\kappa = \frac{\Vert x\Vert_X}{\Vert f(x)\Vert_Y}\Vert Df(x)\Vert_{\leadsto Y},$$ where $\Vert\cdot\Vert_X$ is the norm defined on $X$, $\Vert\cdot\Vert_Y$ the norm defined on $Y$, and $\Vert\cdot\Vert_{\leadsto Y}$ the norm induced by $\Vert\cdot\Vert_Y$.
The condition of a regular matrix $A$ is defined as $k = \Vert A^{-1}\Vert_{\leadsto X} \Vert A\Vert_{\leadsto X}$. I want to show that this is a special case of the aforementioned general case if $f(x) = Ax$. I have managed to show that $$\kappa\leq k$$ using the submultiplicativity of norms: $$\kappa = \frac{\Vert x\Vert}{\Vert Ax\Vert}{\Vert A\Vert}\leq \frac{\Vert A^{-1}\Vert \Vert Ax\Vert}{\Vert Ax\Vert}\Vert A\Vert = \Vert A^{-1}\Vert \Vert A\Vert = k$$ (ignoring the norm subscripts in the notation). However, I fail to work out the other ineqality: $$\kappa = \frac{\Vert x\Vert}{\Vert Ax\Vert}\Vert A\Vert\geq \qquad?\qquad\geq k$$
Does this inequality even hold? I am aware of Matrix Condition Number Definition, however, in that answer it seems that the two notions of condition numbers $k$ and $\kappa$ are just related and not the same.
Proof
$$\kappa=\frac{\|x\|}{\|Ax\|}\|A\|_{ip}\ge\frac{\|x\|}{\|A\|_{ip}\|x\|}\|A\|_{ip}\\ =1\\ =\|I\|_{ip}\\ =\|A^{-1}A\|_{ip}\\ \ge \|A^{-1}\|\|A\|\\ =k$$