I wanted to share with you this little demonstration I got making some homework, it seems to be beautiful for me, also it would be nice to receive some feedback. I would also like to clear out that it isn't, at least, entire mine.
Let f be:
$f(x)=(a_{1}x+b_{1})^2+(a_{2}x+b_{2})^2+...+(a_{n}x+b_{n})^2$
Where $a_{i}$ and $b_{i}$ are the row and column coefficients of A and B both $\in R^{n\times n}$ respectively.
From the polynomial expansion:
$f(x)=a_{1}^2x^2+2a_{1}b_{1}+b_{1}^2+a_{2}^2x^2+2a_{2}b_{2}+b_{2}^2+...+a_{n}^2x^2+2a_{n}b_{n}+b_{n}^2$
$=\sum_{i=1}^{n}a_{i}^2x^2+2\sum_{i=1}^{n}a_{i}b_{i}+\sum_{i=1}^{n}b_{i}^2$
Notice the parable is always above the x axis for all element in the images, so it doesn't have real roots, then its discriminant $\Delta=b^2-4ac$, from $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ is less than 0, so:
$\Delta=4(\sum_{i=1}^{n}a_{i}b_{i})^2-4\sum_{i=1}^{n}a_{i}^2\sum_{i=1}^{n}b_{i}^2\leq0$
$(\sum_{i=1}^{n}a_{i}b_{i})^2-\sum_{i=1}^{n}a_{i}^2\sum_{i=1}^{n}b_{i}^2\leq0$
$(\sum_{i=1}^{n}a_{i}b_{i})^2\leq\sum_{i=1}^{n}a_{i}^2\sum_{i=1}^{n}b_{i}^2$
From here we can sum for all the rows and columns of A and B respectively:
$\sum_{j=1}^{n}\sum_{i=1}^{n}(\sum_{k=1}^{n}{a_{i}}_{k}{b_{k}}_{j})^2\leq\sum_{j=1}^{n}\sum_{i=1}^{n}{{a_{i}}_{j}}^2\sum_{j=1}^{n}\sum_{i=1}^{n}{{b_{j}}_{i}}^2$
Then applying square root both sides:
$\sqrt{\sum_{j=1}^{n}\sum_{i=1}^{n}(\sum_{k=1}^{n}{a_{i}}_{k}{b_{k}}_{j})^2} \leq\sqrt{\sum_{j=1}^{n}\sum_{i=1}^{n}{a_{i}}_{j}^2\sum_{j=1}^{n}\sum_{i=1}^{n}{b_{j}}_{i}^2}$
$\sqrt{\sum_{j=1}^{n}\sum_{i=1}^{n}(\sum_{k=1}^{n}{a_{i}}_{k}{b_{k}}_{j})^2} \leq \sqrt{\sum_{j=1}^{n}\sum_{i=1}^{n}{a_{i}}_{j}^2}\sqrt{\sum_{j=1}^{n}\sum_{i=1}^{n}{b_{j}}_{i}^2}$
Notices that this is precisely:
$||AB||_{F}\leq||A||_{F}||B||_{F}$