Concavity alone does not give two-unitary subadditivity

The question is whether, for every real-valued concave function $f$ on
$[0,\infty)$ with $f(0)\ge0$, not assumed monotone, and all positive
semidefinite $A,B$, there are unitaries $U,V$ with
$f(A+B)\le Uf(A)U^*+Vf(B)V^*$. The note shows that this fails in dimension
two. For $f(t)=t-t^2$, let $P=\operatorname{diag}(1,0)$ and let $Q$ be the
orthogonal projection onto the line spanned by $(3,4)^T$. Then
$f(P)=f(Q)=0$, while $f(P+Q)=-(PQ+QP)$ has eigenvalues $6/25$ and $-24/25$,
so it is not dominated by the zero matrix that every right-hand side equals.
The failure persists for the positive definite inputs $P+\frac1{20}I_2$ and
$Q+\frac1{20}I_2$: the largest eigenvalue of $f(A+B)$ is $1/4$, while every
right-hand side is at most $\frac{19}{200}I_2$. The example does not
contradict the theorem for nonnegative concave functions, since a concave
function that is nonnegative on the whole half-line is nondecreasing.