Here's what I have:
\begin{aligned} \|A\|_F &= \sqrt{\sum_{i = 1}^m \|a_i\|_2^2 }& a_i \text{ are column vectors}\\ & \leq \sqrt{m \max_i \|a_i\|_2^2} & \text{algebra} \\ & = \sqrt{m}\max_i\|a_i\|_2 & \\ & \leq \sqrt{m}\max_i \|a_i\|_1 & \|a\|_2\leq\|a\|_1\\ & \leq \sqrt{m} \|A\|_1 & \text{1-norm is greatest col sum} \end{aligned}
My problem is, we know $\|a\|_2 \leq \|a\|_1$, but I don't have it proven that $\max_i \|a_i\|_2 = \max_i \|a_i\|_1$. I feel this is true; if the sum of squared values of a column is the greatest, the sum of absolute values should also be the greatest. But I'm not quite sure how to prove it. Do y'all have any thoughts or suggestions? Should I try another tack?
Okay, I think I got it. Was being silly. Here's what I have now:
\begin{aligned} \|A\|_F^2 &= \sum_{i=1}^m\sum_{j=1}^m a_{ij}^2 & \text{definition} \\ &\leq m\max_i\sum_{i=1}^ma_{ij}^2 & \text{greatest column sum squared, m times} \\ & \leq m\left(\max_i\sum_{i=1}^m|a_{ij}|\right)^2 & \text{sum of squares less than square of sums} \\ & = m\|A\|_1^2 \end{aligned}
Does this make sense?