Showing that Frobenius Norm is Submultiplicative Considering the Frobenius norm $\|.\|_{F}$, how can we prove that it's submultiplicative?
N.b: I noticed there is a user who asked a similar question but proof looked like directing away from the question. I hope anyone can help me.
 A: \begin{align*}\|AB\|_{F}^{2}=
\left\|\left(\begin{array}{c}
\widehat{a}_{0}^{H} \\
\widehat{a}_{1}^{H} \\
\vdots \\
\widehat{a}_{m-1}^{H}
\end{array}\right)\left(\begin{array}{llll}
b_{0} & b_{1} & \cdots & b_{n-1}
\end{array}\right)\right\|_{F}^{2}
&=\left\|\left(\begin{array}{c|c|c|c}
\widehat{a}_{0}^{H} b_{0} & \widehat{a}_{0}^{H} b_{1} & \cdots & \widehat{a}_{0}^{H} b_{n-1} \\
\hline \hat{a}_{0}^{H} b_{0} & \widehat{a}_{0}^{H} b_{1} & \cdots & \widehat{a}_{0}^{H} b_{n-1} \\
\vdots & \vdots & & \vdots \\
\hline \widehat{a}_{m-1}^{H} b_{0} & \widehat{a}_{m-1}^{H} b_{1} & \cdots & \widehat{a}_{m-1}^{H} b_{n-1}
\end{array}\right)\right\|_{F}^{2}\\ \\&=\sum_{i} \sum_{j}\left|\widehat{a}_{i}^{H} b_{j}\right|^{2} \\ 
&\leq \sum_{i} \sum_{j}\left\|\hat{a}_{i}^{H}\right\|_{2}^{2}\left\|b_{j}\right\|^{2} \quad \text { (Cauchy-Schwartz) } \\
&=\left(\sum_{i}\left\|\widehat{a}_{i}\right\|_{2}^{2}\right)\left(\sum_{j}\left\|b_{j}\right\|^{2}\right)\\ &=\left(\sum_{i} \widehat{a}_{i}^{H} \widehat{a}_{i}\right)\left(\sum_{j} b_{j}^{H} b_{j}\right) \\
&\leq\left(\sum_{i} \sum_{j}\left|\widehat{a}_{i}^{H} \widehat{a}_{j}\right|\right)\left(\sum_{i} \sum_{j}\left|b_{i}^{H} b_{j}\right|\right)\\&=\|A\|_{F}^{2}\|B\|_{F}^{2}
\end{align*}
Now, to conclude this proof, we shall take the square root of both sides of the inequality  :$$\|AB\|_{F}^{2}\leq\|A\|_{F}^{2}\|B\|_{F}^{2}$$
And hence, this proves that Frobenius norm is submultiplicative.
