No finite two-unitary constant for the arithmetic symmetric modulus

For a square complex matrix $X$, let $\mathcal S(X)=(|X|+|X^*|)/2$ be its
arithmetic symmetric modulus. The note shows that there is no finite constant
$C\ge0$ such that every pair $X,Y\in\mathbb M_3(\mathbb C)$ admits unitaries
$U,V$ with $\mathcal S(X+Y)\le C\bigl(U\mathcal S(X)U^*+V\mathcal S(Y)V^*\bigr)$
in the Loewner order; in particular the conjectured constant $\sqrt2$ fails.
The witnesses are the real rank-one matrices $X=e_1(e_1+te_2)^*$ and
$Y=-(e_1+te_3)e_1^*$. Both symmetric moduli have eigenvalues $(s\pm1)/2$ and
$0$, where $s=\sqrt{1+t^2}$, while $\mathcal S(X+Y)\ge(t/2)I_3$. Testing the
inequality on a unit vector orthogonal to top eigenvectors of both rotated
summands forces $C\ge(\sqrt{1+t^2}+1)/(2t)$, which is unbounded as
$t\downarrow0$. The choice $t=3/4$ gives an explicit rational counterexample
to the constant $\sqrt2$.