If $A$ is an $m \times n$ matrix. Then $\|A\| \leq \sqrt{\sum_{i,j}a_{i,j}^2} \leq \sqrt{n} \|A\|$
so $Ax$ is going to be an $m \times 1$ vector of which we take the norm in this case it will be the unit vector that gives maximum norm. If we said $m = 2$ it will be $\sqrt{a_1^2 + a_2^2}$. Now assuming I understood the notation correctly $\sqrt{\sum_{i,j}a_{i,j}^2} = \sqrt{a_{1,1}^2 + a_{1,2}^2 + a_{2,1}^2 + a_{2,2}^2 }$. That part of the inequality makes sense.
But the second bit is throwing me off because if I take the same $\|A\|$ I still have the same $m \times 1$ vector I'm taking the…