Question concerning inequality of norms of matrices Prove that, for $A \in M_{m \times n}(\mathbb{R})$, we have $$|A| \leq \sqrt{\sum_{i,j}a_{ij}^2}\leq \sqrt{n}*|A|$$
where $\displaystyle |A| = \max_{\|{x}\|=1}{\|{Ax}\|}$ for $x \in \mathbb{R}^n$.
Over several hours, our study group made little progress on this. We tried using the property of $|A|$, ending up trying to tie in other inequalities such as Cauchy-Schwartz and Triangle, but this got nowhere. We also suggested various geometric proofs, none of which were really "rigorous."
Any and all help is appreciated. Thank you kindly!
 A: Here's a proof of the first inequality:
Let $A_i$ be the $i$th row of $A$. Then, for any $x \in \mathbb{R}^n$, we have $$||Ax||_2^2 = \sum_i\left|\sum_ja_{ij}x_j\right|^2=\sum_i\left|\langle A_i,x\rangle \right|^2 \\\leq \sum_i\langle A_i,A_i\rangle\langle x,x\rangle = \|x\|_2^2\sum_{i,j}a_{ij}^2,$$ 
where the inequality comes from Cauchy-Schwarz. Then, taking the square-root of each side we get, $$\|Ax\|_2 \leq \|x\|_2 \sqrt{\sum_{i,j}a_{ij}^2}.$$
Thus, $$\max_{\|x\|_2=1}\|Ax\|_2 \leq \max_{\|x\|_2=1}\|x\|_2 \sqrt{\sum_{i,j}a_{ij}^2} = \sqrt{\sum_{i,j}a_{ij}^2}.$$
Note, that the Frobenius norm is defined as $\|A\|_F = \sqrt{\sum_{i,j}a_{ij}^2}.$
Therefore, for part two, we want to show $$\|A\|_F \leq \sqrt{n}|A|.$$ Now, note $$\|A\|_F^2 = \sum_i\|Ae_i\|_2^2 \leq \sum_i\|A\|_2^2\|e_i\|_2^2 = |A|^2\sum_i1 = n|A|^2,$$ where $e_i$ is the vector of all zeros except for the $i$th position which has a $1$.
Taking square-roots gives the desired result. Note, the inequality here comes from the "compatibility" of induced norms. That is, for an vector norm and its induced matrix norm $\|\cdot\|$, we know that $$\|Ax\| = \frac{\|Ax\|}{\|x\|}\|x\| \leq \max_{y\neq 0}\left\{\frac{\|Ay\|}{\|y\|}\right\}\|x\| \equiv\max_{\|y\|=1}\{\|Ay\|\}\|x\| =:\|A\|\cdot\|x\|.$$
