Show that for two matrices $A,B\in \mathbb{R}^{m\times n}$ we have $\langle A,B\rangle\leq n\|A\|_2\|B\|_2$.
I was hinted to use the SVD decomposition of $A$. I have $\|A\|_2 := \sup_{\|x\|=1}\|Ax\|_2$ and $\langle A, B \rangle := tr(AB^T) = tr(U\Sigma V^T B^T)$ but I don't know how to continue.
Let $r= \min(m,n)$, and $\|M\|_F =\sqrt{tr (MM^\top)}$ denote the Frobenius norm of $M\in\Bbb R^{m\times n}$. Then it holds $$\langle A,B \rangle \leq \|A\|_F \|B\|_F \leq \sqrt{r}\|A\|_2\,\sqrt{r}\|B\|_2=r\|A\|_2\|B\|_2,$$ where the first inequality follows from Cauchy-Schwartz. The second inequality follows from the equivalence between $\|.\|_2$ and $\|\cdot \|_F$.
Proof of $\|M\|_F \leq \sqrt{r}\|M\|_2$:
Let $M\in\Bbb R^{m\times n}$ and $U,\Sigma, V$ an SVD of $M$, then we have $$\|M\|_F =\sqrt{tr (U\Sigma\Sigma^\top U^T)}=\sqrt{tr (\Sigma\Sigma^\top U^TU)}=\sqrt{\sum_{i} \Sigma_{ii}^2}\\\leq \sqrt{r\max_i\Sigma_{ii}^2}=\sqrt{r}\|M\|_2.$$