Frobenius Norm Inequality; Spectral Radius is smaller than Frobenius Norm Let $A \in \mathbb{R}^{n \times m}$ and $x \in \mathbb{R}^n$. Prove the following inequality. $\left\lVert \cdot \right\rVert_F$ denotes the Frobenius norm and $\left\lVert \cdot \right\rVert_2$ denotes the $p$-norm with $p=2$.
$$\left\lVert Ax \right\rVert_2 \leq \left\lVert A \right\rVert_F \left\lVert x \right\rVert_2$$

tl;dr: I'm essentially stuck at this (or similar) inequality:
$$ \lambda_{max} (A^T \cdot A) \leq \left\lVert A^T \cdot A \right\rVert_F$$
while $\lambda_{max}$ is the largest eigenvalue of $A^T \cdot A$. I know that this inequality holds if the norm was a natural norm, but since frobenius norm isn't induced by a vector norm, I'm not sure how to proceed.

How I got to this point:
$$\left\lVert Ax \right\rVert_2 \leq \left\lVert A \right\rVert_2 \left\lVert x \right\rVert_2$$
So we have to show:
$$\left\lVert A \right\rVert_2 \leq \left\lVert A \right\rVert_F$$
or
$$\left\lVert A \right\rVert_2^2 \leq \left\lVert A \right\rVert_F^{2}$$
We have:
$$\left\lVert A \right\rVert_2^2 = \lambda_{max}(A^TA) \leq \left\lVert A^T A \right\rVert_F$$
The last inequality is the part I can't prove. If I could show it, we have:
$$\left\lVert A^T A \right\rVert_F \leq \left\lVert A^T \right\rVert_F \left\lVert A \right\rVert_F = \left\lVert A \right\rVert_F^2$$
Which was the thing we wanted to show above.

These threads were helpful:
Show that $ \lVert A \rVert_2^2 \leq \lVert A \rVert _1 \lVert A \rVert _ \infty $
The spectral radius of the matrix $A$ is less than or equal any natural norm
Anyways, thank you for your help. It is greatly appreciated.
 A: Recall that the Frobenius norm is also the Schatten $2$-norm, and thus has a characterization in terms of $2$-norm of the singular values.
Letting $B = A^T A$,
$$
\lVert B\rVert_F = \sqrt{\operatorname{Tr} B^T B}
= \sqrt{\sum_{i=1}^n \lambda_i(B)^2} \geq \sqrt{(\max_{1\leq i\leq n}\lambda_i(B) )^2} = \max_{1\leq i\leq n}\lambda_i(B) 
$$
where $(\lambda_i(B))_{1\leq i\leq n}$ are the (real) eigenvalues of the symmetric matrix $B$. This shows
$$
\lVert A^TA \rVert_F \geq \lambda_\max(A^TA)
$$
as desired.
A: Here's another way: First prove the following
Lemma For two matrices $A,B$ we have $\| A B \|_F \le \| A \|_F \| B \|_F$.
Let $( A B )_{i,j} := (c_{i,j})_{i,j}$
Then we have
\begin{align*}
\| A B \|_F^2
& \overset{\text{Def}}{=} \sum_{i,j = 1}^{n} | c_{i,j} |^2
= \sum_{i,j = 1}^{n} | \langle a_{i, \ast}, b_{\ast,j} \rangle |
\overset{\text{(CS)}}{\le} \sum_{i,j = 1}^{n} \| a_{i, \ast} \|_2^2 \cdot \| b_{\ast,j} \|_2^2 \\
& = \sum_{i,j = 1}^{n} \| a_{i, \ast} \|_2^2 \cdot \sum_{i,j = 1}^{n} \| b_{\ast,j} \|_2^2
= \| A \|_F^2 \cdot \| B \|_F^2,
\end{align*}
where (CS) means the Cauchy-Schwarz inequality. 
Now, since the function $x \mapsto \sqrt{x}$ is strictly increasing and monotone, you can take the square root from both sides and the inequality (since it's $\le$ and not $<$ will be preserved).

Now, we can prove $\| A x \|_2 \le \| A \|_F \| x \|_2$.
Using the lemma and putting $B = x$ we have $\| x \|_F = \| x \|_2$ (since the vector $x$ has, viewed as Matrix, rank 1).
Therefore, we have
\begin{align*}
\| A x \|_{2}
= \| Ax \|_F
\overset{\text{(L)}}{\le} \| A \|_F \| x \|_F
= \| A \|_F \| x \|_2,
\end{align*}
where (L) is the inequality from the lemma.
