Consider two positive semidefinite and symmetric matrices, namely $\bf{A}$ and $\bf{B}$. Denote $\Vert\cdot\Vert$ as the operator norm. If we have another positive semidefinite matrix $\bf{B}^*$ that satisfies $\Vert\bf{B}^*\Vert\geqslant\Vert\bf{B}\Vert$, can we get $$\Vert\bf{A}+\bf{B}\Vert\leqslant\Vert\bf{A}+\bf{B}^*\Vert$$? I have worked out that for general $\bf{A}$, $\bf{B}$ that is not positive semidefinite, we cannot get the target result since the eigenvalues may be negative. However, I cannot prove the case for positive semidefinite matrices.
This problem is important for me because I am trying to use a concentration method to control operator norm of summation of dependent matrices. If we have independent $\mathbf{X}_1,...,\mathbf{X}_n\in\mathbb{R}^{m_1\times m_2}$, we can use basic concentration inequalities (such as matrix Bernstein and matrix Hoeffding) to control $\Vert\sum_{i}\mathbf{X}_i\Vert$, which fails in the case when $\mathbf{X}_i$ are dependent. Suppose we have independent $\mathbf{X}_i^*$, it is easy to control $\Vert\sum_{i}\mathbf{X}_i^*\Vert$. So if we have $\Vert\sum_{i}\mathbf{X}_i\Vert\leqslant\Vert\sum_{i}\mathbf{X}_i^*\Vert$ given that $\mathbf{X}_i$ and $\mathbf{X}_i^*$ satisfy some conditions, then we can deal with $\Vert\sum_{i}\mathbf{X}_i\Vert$ for dependent $\mathbf{X}_i$.
If $B^\ast \geq B$, then the result is true since then $A + B \leq A + B^\ast$, which implies $\|A + B\| \leq \|A + B^\ast\|$.
If you’re only assuming $\|B^\ast\| \geq \|B\|$, then the result is not necessarily true. Just choose $A = \begin{pmatrix} 0 & 0\\0 & 2 \end{pmatrix}$, $B = \begin{pmatrix} 0 & 0\\0 & 1 \end{pmatrix}$, and $B^\ast = \begin{pmatrix} 2 & 0\\0 & 0 \end{pmatrix}$, say. Then $\|B^\ast\| = 2 > 1 = \|B\|$ but $\|A + B\| = 3 > 2 = \|A + B^\ast\|$.