Q:

Inequality for $AB + BA$ when $A,B\geq0$, reference request

Let $A,B\geq0$ be positive semidefinite matrices of arbitrary size $n\times n$. Denote by $\alpha$ and $\beta$ their largest eigenvalues. 
It is well-known that the eigenvalues of the expression $AB + BA$ are bounded by [F. Zhang, "Matrix Theory", Sec. 7.2] 
$$
-\frac{1}{4}\alpha\beta I \,\,\leq\,\, AB + BA \,\,\leq\,\, 2\alpha\beta I.
$$
I can show that
$$
- \rm{tr}(AB)I -  \rm{tr}(A)B -  \rm{tr}(B)A \,\leq\,  AB + BA \,\leq\,  \rm{tr}(AB)I +  \rm{tr}(B)A + \rm{tr}(A)B \quad (\dagger)
$$
Now normalize to $\rm{tr}(A) = \rm{tr}(B) = 1$. When $\rm{tr}(AB) = 0$, that is $A\perp B$ in terms of the Hilbert-Schmidt inner product, the expression reduces to
$$
-(A+B) \,\,\leq\, AB + BA \,\leq\,\, (A + B) \quad (\ddagger)
$$
These inequalities $(\dagger)$ and $(\ddagger)$ look rather simple and are for $n\geq 3$ even tight. However I failed to find them in the literature, including in Bernstein's book on "Matrix Facts" or in the books by Bhatia. Am I missing something, are these known or can they straightforwardly be derived from other known expressions?
edit: they are weaker than the simple sum-of-squares, $(A-B)^2 \geq 0$; see answer below.

A:

It seems both bounds are much weaker than what can be obtained from a simple sum-of-squares: namely, for all Hermitian matrices $A$ and $B$, one has
$(A-B)^2 \geq 0$. Expand
$$ AB + BA \leq A^2 + B^2 $$
Replacing $A\rightarrow -A$ yields also the lower bound
$$ -(A^2 + B^2) \leq AB + BA $$
Because of $\rm{tr}(A) = \rm{tr}(B)=1$, these are stronger than both $(\dagger)$ and $(\ddagger)$ in the question. 

