Let $A$ be $m \times r$ and $B$ be $r \times n$. A better bound here is
$$
\| A B \|_F \le \|A\| \|B\|_F \quad (*)
$$
where $\|A\|$ is the $\ell_2$ operator norm:
$$
\|A\| = \max_{\|x\|_2\, \le\, 1} \|A x\|_2.
$$
It is also equal to the largest singular value of A. From this definition $\|A x\|_2 \le \|A\| \|x\|_2$ for any vector $x$ in $\mathbb R^r.$
Since $\|A\| \le \|A\|_F$, inequality (*) is a strictly better inequality than the sub-multiplicative inequality for the Frobenius norm.
To see the inequality, let $B = [b_1 \mid b_2 \mid \cdots \mid b_n]$ be the column decomposition of $B$. Then, $A B = [Ab_1 \mid A b_2 \mid \dots \mid Ab_n]$ is the column decomposition of $AB$. It follows that
\begin{align*}
\| A B \|_F^2 = \sum_{j=1}^n \|A b_j\|^2 \le \|A\|^2 \sum_j \|b_j\|^2 = \|A\|^2 \|B\|_F^2.
\end{align*}
EDIT in response to the question in the comments, ``Is there a lower bound for the Frobenius norm of the product of two matrices?''. In general, no, except for the obvious lower bound of zero. Consider the following two matrices
\begin{align}
A = \begin{pmatrix}
a & b \\ 0 & 0
\end{pmatrix}, \quad
B = \begin{pmatrix}
-b & 0 \\ a & 0
\end{pmatrix}.
\end{align}
Then $\|A\|_F = \|B\|_F = \sqrt{a^2 + b^2}$, while $\|AB\|_F = 0$.
What if the two matrices are symmetric? Consider
\begin{align}
A = \begin{pmatrix}
a & b \\ b & a
\end{pmatrix}, \quad
B = \begin{pmatrix}
-b & a \\ a & -b
\end{pmatrix}, \quad
A B = \begin{pmatrix}
0 & a^2-b^2 \\ a^2-b^2 & 0
\end{pmatrix}.
\end{align}
Then, $\|A\|_F^2 = \|B\|_F^2 = 2(a^2 + b^2)$ while $\|AB\|_F^2 = 2(a^2 - b^2)^2$ which can be made arbitrarily smaller than either of $\|A\|_F^2$ or $\|B\|_F^2$. For example, take $a=b$.