Show that if $A \mathbf{x} = \mathbf{b}$, then $\text{cond}_p(\mathbf{A}) \geq \frac{\|A\|_p\|x\|_p}{\|b\|_p}$

Question

I have this simple little question to show that the above inequality holds with respect to the p-norm (i.e. the norm defined: by $\|x\|_p = \left( \sum_{i=1}^{n} |x_i|^p \right)^{\frac{1}{p}}$).

I just don't know how to go from the definition of $\text{cond}_p(\mathbf{A}) = \|A\|_p \cdot \|A^{-1}\|_p$ to obtain the inequality.

Any tips for how to link the two sides are appreciated in advance!