No finite two-unitary constant for the maximal symmetric modulus

For a square complex matrix $Z$, the maximal symmetric modulus
$\mathcal M(Z)=\lim_{r\to\infty}(|Z|^r+|Z^*|^r)^{1/r}$, over positive integers
$r$, is the spectral-order join of $|Z|$ and $|Z^*|$. The note shows that no
finite constant $C\ge0$ makes every pair $A,B\in\mathbb M_2(\mathbb C)$ admit
unitaries $U,V$ with
$\mathcal M(A+B)\le C\bigl(U\mathcal M(A)U^*+V\mathcal M(B)V^*\bigr)$; in
particular the constant-one inequality is false. For $A=e_1e_1^*$ and
$B=te_1e_2^*$ one has $\mathcal M(A)=A$, $\mathcal M(B)=tI_2$ and
$\mathcal M(A+B)=\sqrt{1+t^2}\,I_2$, the last because the root sequence is
$\sqrt{1+t^2}$ times $R^{1/r}$ for a positive definite matrix $R$. Every
right-hand side before scaling has eigenvalues $1+t$ and $t$, so testing on a
unit vector orthogonal to $Ue_1$ forces $C\ge\sqrt{1+t^2}/t$, which is
unbounded as $t\downarrow0$. The construction also answers the preceding
dimension-dependent finite-constant question negatively.