The Frobenius norm is submultiplicative

Idea. Every entry of $AB$ is a dot product; Cauchy-Schwarz controls it by the lengths of the row and the column involved. Summing over all entries reassembles the Frobenius norms.

Step 1: one entry. Let $r_i$ be the $i$-th row of $A$ and $c_j$ the $j$-th column of $B$, both viewed in $\mathbb{R}^n$. Then $(AB)_{ij}=\sum_ka_{ik}b_{kj}=\langle r_i,c_j\rangle$, and by Cauchy-Schwarz
$$(AB)_{ij}^2\le|r_i|^2|c_j|^2.$$

Step 2: sum over $i,j$.
$$\|AB\|_F^2\le\sum_{i,j}|r_i|^2|c_j|^2=\Big(\sum_i|r_i|^2\Big)\Big(\sum_j|c_j|^2\Big)=\|A\|_F^2\|B\|_F^2.$$
The two sums factor because $i$ and $j$ vary independently, and $\sum_i|r_i|^2=\sum_{i,k}a_{ik}^2=\|A\|_F^2$ (similarly for $B$).

Conclusion. Taking square roots gives $\|AB\|_F\le\|A\|_F\|B\|_F$.

Remark. Together with completeness this makes $(M_n(\mathbb{R}),\|\cdot\|_F)$ a Banach algebra. Compare with the max-entry norm, which is *not* submultiplicative (next problems). A useful corollary is $\|A^k\|_F\le\|A\|_F^k$.