Let $A$ be an Hurwitz stable matrix (i.e. the real part of the eigenvalues of $A$ lie in the left-half plane) and $Q\ne 0$ be a positive semidefinite matrix ($Q\ge 0$, for short). Let $P\ge 0$ be the solution of the following Lyapunov equation
$$
AP+PA^\top = -Q,
$$
and define $f(Q) := \mathrm{tr}(PQ)$, where $\mathrm{tr}(\cdot)$ denotes the trace operator.

>**My question.** Let $\lambda_{\max}(P)$ denote the largest eigenvalue of $P\ge 0$. It is quite easy to see that
$$\tag{$\star$}\label{star}
f(Q) \ge \frac{\mathrm{tr}(Q)}{\lambda_{\max}(P)}.
$$
However, does there *always* (i.e., for any choice of $A$ Hurwitz stable) exist a $Q\ge 0$, $Q\ne 0$, for which \eqref{star} is attained with equality?