Assume $A, B$ are selfajoint compact operators. Is it true that $\A+iB\\le \2A+iB\$? Do we have a stronger inequality $\prod_{k=1}^ns_k(A+iB)\le \prod_{k=1}^ns_k(2A+iB)$ or even stronger one $s_n(A+iB)\le s_n(2A+iB)$, $n=1, 2, \ldots$, where $s_n$ are snumbers?

I just tried a few random matrices... $$ \begin{align} &A=\begin{bmatrix} 0.1 & 0.4\cr 0.4&0\end{bmatrix}, \qquad B=\begin{bmatrix} 1.5 & 0.5 + i\cr 0.5  i& 3.5\end{bmatrix}, \cr &\A+iB\ = \frac{7}{2}+\frac{\sqrt{61}}{10}\approx 4.28 \cr &\2A+iB\ = \frac{7}{2}+\frac{\sqrt{29}}{10}\approx 4.04 \end{align} $$ 


EDIT If, however, $A$ and $B$ are assumed to be positive definite, then these inequalities probably hold. As a hint why they might hold (I have not had the time to check any of the other cases), consider $C=A+iB$ and $D=2A+iB$ with $A,B \ge 0$. Then, \begin{equation*} \begin{split} \prod_{j=1}^n s_j(C) = \det C &= \det(A)\prod_{j=1}^n[1+ s_j(A^{1/2}BA^{1/2})^2]^{1/2}\\\\ \prod_{j=1}^n s_j(D) = \det D &= \det(A)\prod_{j=1}^n2\left[1+ \frac{s_j(A^{1/2}BA^{1/2})^2}{4}\right]^{1/2}\\\\ &= \det(A)\prod_{j=1}^n[4+ s_j(A^{1/2}BA^{1/2})^2]^{1/2} \ge \det(C). \end{split} \end{equation*} 

