Frobenius norm inequality alternative proof

Question

Is there any direct and "natural" way to show this property of Frobenius norm : Let $A,B \in \mathbb R^{n \times n}$ $$\lVert AB \rVert _F \leq \lVert A \rVert _F \lVert B \rVert _F$$
I started by doing the following : $$\lVert AB \rVert _F =\sqrt{\sum_{ij}(AB)_{ij}^2}=\sqrt{\sum_{ij}\big(\sum_{k=1}^nA_{ik}B_{kj}\big)^2} $$
I feel like I can/must use Cauchy-Schwarz inequality somewhere but I am stuck here

Answer 1

Yes, you use Cauchy-Schwarz at the only place where it makes sense: \begin{align} \lVert AB \rVert _F^2 &={\sum_{ij}|(AB)_{ij}|^2}={\sum_{ij}\bigg|\sum_{k=1}^nA_{ik}B_{kj}\bigg|^2}\\[0.3cm] &\leq{\sum_{ij}\sum_{k=1}^n|A_{ik}|^2\sum_{k=1}^n|B_{kj}|^2}\\[0.3cm] &={\sum_{i}\sum_{k=1}^n|A_{ik}|^2\sum_j\sum_{k=1}^n|B_{kj}|^2}\\[0.3cm] &={\|A\|_F^2\,\|B\|_F^2} \end{align}