I am trying to prove this for my numerical analysis class. This is from chapter 4.4 of Kincaid and Cheney's book. So far I haven´t got any good idea.
I have tried $$ \|A\| \|A^{-1}\| = \sup \|A\frac{u}{\|v\|}\|\cdot \|A^{-1}\frac{v}{\|u\|}\|,\quad \|u\|=\|v\| $$
But I coudn't advance any further. What confuses me the most is that there is not $||A^{-1}||$ in the equation. As far as I know there is not relation between a matrix norm and its inverse norm besides $\|A^{-1}\| \geq (\|A\|)^{-1}$ which haven't been useful. Please help.
For any $x,y$ with $\|x\|=\|y\|$ we have, with $z=Ay$, $$ \frac{\|Ax\|}{\|Ay\|}=\frac{\|Ax\|}{\|x\|}\,\frac{\|y\|}{\|Ay\|} =\frac{\|Ax\|}{\|x\|}\,\frac{\|A^{-1}z\|}{\|z\|}\leq\|A\|\,\|A^{-1}\|. $$ Now fix $\varepsilon>0$. Then there exist $x,z$ such that $$\frac{\|Ax\|}{\|x\|}>\|A\|-\varepsilon\,\qquad\qquad\frac{\|A^{-1}z\|}{\|z\|}>\|A^{-1}\|-\varepsilon. $$ By rescaling $z$ if needed, we may assume that $\|A^{-1}z\|=\|x\|$. With $y=A^{-1}z$, $$ \frac{\|Ax\|}{\|Ay\|}=\frac{\|Ax\|}{\|x\|}\,\frac{\|A^{-1}z\|}{\|z\|} >(\|A\|-\varepsilon)(\|A^{-1}\|-\varepsilon). $$ As $\|x\|=\|y\|$ and this can be done for each $\varepsilon>0$, this shows that $$ \|A\|\,\|A^{-1}\|=\sup\Big\{\frac{\|Ax\|}{\|Ay\|}:\ \|x\|=\|y\|\Big\}. $$