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

Question

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.

Answer 1

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.