How tight is this trace inequality?

Question

I would like to know how tight the following trace inequalities are for real symmetric $A$ and real symmetric $B \succeq 0$

$$\mbox{trace} (AB) \leq \lambda_{\max} (A) \cdot \mbox{trace} (B) $$

or $$\mbox{trace} (BA) \leq \lambda_{\max} (B) \cdot \mbox{trace} (A) $$

since it is said below eq (1) of

https://www.researchgate.net/profile/Qingling_Zhang4/publication/3032531_Eigenvalue_Inequalities_for_Matrix_Product/links/0c96052cc720dc4b9d000000.pdf

the second line holds. I tried to run some simulations (for various $A,B$ obeying the above), but there seems to be a big difference between the two upper bounds and the upper bounds are pretty far away from $\mbox{trace} (AB)$. Any theoretical insights on the tightness of this would be much appreciated. Thanks!

Answer 1

They are tight, in the sense that we have $\text{trace}(AB) = \lambda_{max}(A)\; \text{trace}(B)$ if $A = I$. Similarly in the second one if $B=I$.