How to prove $\|AB\|_F \le \|A\|_F\|B\|_2$? $A,B$ are matrices (not necessarily square). $\|\cdot\|_F$ and $\|\cdot\|_2$ are the matrix Frobenius-norm and 2-norm, respectively.
It suffices to show $\|AB\|_F^2 \le \|A\|_F^2\|B\|_2^2$. Let's denote $a_i^T$ the $i^{th}$ row of $A$, $b_j$ the $j^{th}$ column of $B$, and $e_j$ the $j^{th}$ standard basis. I can only show the following:
\begin{eqnarray}
\|AB\|_F^2 & = & \sum_i\sum_j \|a_i^Tb_j\|_2^2 \\
& = & \sum_i\sum_j \|a_i^T B e_j\|_2^2 \\
& \le & \sum_i\sum_j \|a_i^T\|_2^2 \|B e_j\|_2^2 \\
& \le & \left(\sum_i \|a_i^T\|_2^2\right) \left(\sum_j \frac{\|B e_j\|_2^2}{\|e_j\|_2^2}\right) \\
& \le & \|A\|_F^2 \cdot n\|B\|_2^2
\end{eqnarray}
where $n$ is the number of columns of $B$.
I know that I need to make a vector $x_j$ that is different from $e_j$ such that the 2-norm of it will be $1/n$, which then cancels out the $n$ factor in front of $\|B\|_2^2$. But I tried several vectors, say, $x_j=(1/n,\cdots,1/n)^T$, doesn't seem to work.
Any hint?

Hint:


*

*$\|X\|_F^2 =\sum_i \|x_{i\ast}^T\|_2^2$ ($x_{i\ast}$ denotes the $i$-row of $X$),

*$\color{red}{(XB)_{i\ast}=X_{i\ast}B}$, and

*$\|v^TB\|_2\le\|v^T\|_2\|B\|_2$.

