Let $A\in\mathbb{R}^{n\times n}$ be a matrix having eigenvalues with strictly negative real part (in other words, $A$ is supposed to be Hurwitz stable). Let $\mathrm{tr}(\cdot)$ denote the trace operator and $I$ the identity matrix. Let $>$ be the standard partial order in the cone of positive definite matrices.

Consider the following optimization problem $$\tag{$\star$}\label{prob} m(A) :=\sup_{\substack{X\in\mathbb{R}^{n\times n}\\ X>0,\ \mathrm{tr}(X)=1}} \mathrm{tr}\left(X(I+P)^{-1}\right), $$ where $P$ is the (unique) positive definite solution of the following Lyapunov equation $AP+PA^\top=-X$.

My question.Suppose that $A$ is, does the following inequality $$ m(A)\ge \frac{2\,\mathrm{tr}\,A}{2\,\mathrm{tr}\,A-1} $$ always hold true?upper triangular

**A special case.** Notice that if $A+A^\top<0$, then the answer is in the affirmative. To see this, just pick $P^\star= -\frac{1}{2\mathrm{tr}(A)}I$ and observe that $X^\star=-(AP^\star+P^\star A^\top)$ is positive definite and satisfies $\mathrm{tr}(X^\star)=1$.

**Numerical evidences.** Quite surprisingly, after runnning an extensive number of numerical simulations, it seems that the answer is in the affirmative for ** any** (upper triangular Hurwitz stable) $A$. More precisely it seems that (modulo numerical errors of magnitude $\sim 10^{-6}$) the conjectured inequality actually holds with equality, that is
$$
m(A) = \frac{2\,\mathrm{tr}\,A}{2\,\mathrm{tr}\,A-1}.
$$

However, this fact does not seem trivial to prove (I spent quite some time thinking about this, but I didn't manage to prove it); so any help in clarifying this conjecture is greatly appreciated. Thanks!

**An (perhaps useful?) equivalent formulation.** By plugging $X=-AP-PA^\top$ into the trace functional in \eqref{prob}, the latter can be rewritten as
\begin{align}\tag{$\star\star$}
m(A) &=\sup_{\substack{X\in\mathbb{R}^{n\times n}\\ X>0,\ \mathrm{tr}(X)=1}} -\mathrm{tr}\left((AP+PA^\top)(I+P)^{-1}\right)\notag \\
&=\sup_{\substack{X\in\mathbb{R}^{n\times n}\\ X>0,\ \mathrm{tr}(X)=1}} -2\mathrm{tr}\left(A(I+P^{-1})^{-1}\right)\\
&=-2\inf_{\substack{X\in\mathbb{R}^{n\times n}\\ X>0,\ \mathrm{tr}(X)=1}} \mathrm{tr}\left(A(I+P^{-1})^{-1}\right)\notag\\
&=-2\inf_{\substack{P\in\mathbb{R}^{n\times n}\\ P>0,\ \mathrm{tr}(AP)=-\frac{1}{2}\\ AP+PA^\top<0}} \mathrm{tr}\left(A(I+P^{-1})^{-1}\right)\\
&\overset{(\#)}{=}-2\inf_{\substack{P\in\mathbb{R}^{n\times n}\\ P>0,\ \mathrm{tr}(AP)=-\frac{1}{2}\\ AP+PA^\top<0}} \mathrm{tr}\left(A-A(I+P)^{-1}\right) \\
&=-2\,\mathrm{tr}\,A + 2\sup_{\substack{P\in\mathbb{R}^{n\times n}\\ P>0,\ \mathrm{tr}(AP)=-\frac{1}{2}\\ AP+PA^\top<0}} \mathrm{tr}\left(A(I+P)^{-1}\right), \\\label{prob-eq}
\end{align}
where in (#) I used the Woodbury matrix identity.