a monotone relation for s-numbers   Assume $A, B$ are self-ajoint 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 s-numbers?
 A: *

*$s_n(A+iB) \not\le s_n(2A+iB)$, for example by setting
\begin{equation*}
   A = \begin{bmatrix}
   -2 & -6\\\\
    -6 & -2
   \end{bmatrix}\qquad B = 
   \begin{bmatrix}
   10 & 4\\\\
    4 & 16
   \end{bmatrix}.
  \end{equation*}
In this case, $s_2(A+iB) = 7.7192$ and $s_2(2A+iB) = 4.9433$.

*$\prod_{k=1}^n s_k(A+iB) \not\le \prod_{k=1}^n s_k(2A+iB)$ for the same example as above. The lhs is 156, while the rhs is 129.2440.

*The operator norm version also does not hold (as shown in the nice counterexample by Gerald Edgar)

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