How to show $\text{Tr}(AB) \leq \|A\|_F\|B\|_F$?

Question

Let $A,B$ be two $n \times n$ matrices. How to show $\text{Tr}(AB) \leq \|A\|_F\|B\|_F$.

My try:

Using Von Neumann trace inequality we have $$ \text{Tr}(AB) \leq \sum_{i=1}^n \sigma_{A,i}\sigma_{B,i} $$ where $\sigma$ is the singular value which are in order. I cannot go further.

Answer 1

Apply Cauchy-Schwarz: $$\sum_i \sigma_{A,i} \sigma_{B,i} \le \sqrt{\sum_i \sigma_{A,i}^2} \sqrt{\sum_i \sigma_{A,i}^2} = \|A\|_F \|B\|_F$$

Answer 2

By Cauchy-Schwarz, we have \begin{align} \operatorname{Tr}(AB) = \sum^n_{i=1} \sum^n_{j=1} a_{ij}b_{ji} \leq&\ \sum^n_{i=1}\left( \sum^n_{j=1}|a_{ij}|^2\right)^{1/2}\left( \sum^n_{j=1}|b_{ij}|^2\right)^{1/2}\\ \leq&\ \left(\sum^n_{i=1} \sum^n_{j=1}|a_{ij}|^2\right)^{1/2} \left(\sum^n_{i=1} \sum^n_{j=1}|b_{ij}|^2\right)^{1/2}\\ =&\ \|A\|_F\|B\|_F \end{align}