Frobenius Norm inequality Is there anyway the following inequality can be proved without using SVD and Unitary matrices properties of the norms? 
$||AB||_F \le ||A||_2 ||B||_F$

Use the fact that $\|Ax\|_2\leq\|A\|_2\|x\|_2$ and that if $B=[b_1,\ldots,b_n]$ then $\|B\|_F^2=\sum\limits_{i=1}^n\|b_i\|_2^2$. We have
$$
\|AB\|_F^2=\sum_{i=1}^n\|Ab_i\|_2^2\leq\|A\|_2^2\sum_{i=1}^n\|b_i\|_2^2=\|A\|_2^2\|B\|_F^2.
$$
