Positivity of the anti-commutator of two positive operators implies commutativity? This is a generalization of question Positivity of the anti-commutator of two positive operators .
note: by positive operator, I mean positive semidefinite (i.e. $\ge 0$, not necessary $>0$).
Let $A$ and $B$ two positive operators on a Hilbert space (I'm interested in the finite-dimensional case, but I think the question is interesting also in infinite dimension). The anti-commutator of $A$ and $B$ is defined as $\{A,B\} = AB + BA$.
If $A$ and $B$ commute, then it's easy to show that $\{A,B\} = 2 AB $ is a positive operator.
If $A$ and $B$ don't commute, we have a counterexample that shows that $\{A, B\}$ can be not positive, e.g. $A = \begin{pmatrix} 1 & 0 \\ 0 & 0\\ \end{pmatrix} $ and $B = \begin{pmatrix} 1 & 1 \\ 1 & 1\\ \end{pmatrix} $.
Question:
If $\{ A, B \}$ is positive, does it imply that $A$ and $B$ must commute? Or do exist non-commuting positive $A$ and $B$ such that $\{A,B\}$ is positive?

Edit: (previous answer didn't consider $A\geq0$, $B\geq0$)
Let
$$
A=\begin{bmatrix}2&0\\0&4\end{bmatrix}, \ \ B=\begin{bmatrix}2&1\\1&2\end{bmatrix}.
$$
Then
$$
AB+BA=\begin{bmatrix}4&2 \\4&8 \end{bmatrix}
+\begin{bmatrix} 4&4\\2&8\end{bmatrix}=\begin{bmatrix} 8&6\\6&16\end{bmatrix}\geq0.
$$
