Consider two square matrices $A, B \in \mathbb{R}^{n \times n}$ and let $\| \cdot\|\_1$ and $\|\cdot\|$ be, respectively, the trace norm (the sum of singular values) and the usual operator norm (the maximum of singular values).


Is there a known bound for the following quantity?
$$
\sup\{\alpha > 0: \; \alpha \, \text{tr}(A^TB) \le \| A+B\|\_1, \, \forall B \; \text{s.t.} \;\|B\| \le 1\}
$$



An inequality involving operator and trace norms
Since I cant comment, I will leave this thought here. Since $||\cdot||\_1$ and $||\cdot||$ (as you defined them) are dual norms, it must be that tr$((A+B)^TX)\leq||A+B||\_1$ for any $X$ such that $||X||\leq 1$. Therefore, tr$(A^TB)\leq ||A+B||\_1 - ||B||^2<||A+B||\_1$ (since tr$(B^TB)\geq||B||^2$).


(edit: fixed typo and replaced $||B||$ with $||B||^2$ in the final step)


