# 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.

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.