TITLE: A matrix inequality involving trace norm of a matrix and its inverse
QUESTION [13 upvotes]: Let $A,B \succeq 0$ be two positive semidefinite matrices. Can we get a closed form expression for the following quantity?
$$ \inf_{X \succ 0} \mathrm{tr}(XA) + \mathrm{tr}(X^{-1}B) $$
We assume all matrices involved are symmetric.

REPLY [5 votes]: We assume that $B$ symmetric $\geq 0$ and $A,X$ symmetric $>0$. Indeed, if $\det(A)=0,\det(B)\not= 0$, then the lower bound is not reached. For example, when $A=0$, the lower bound $0$ is "reached" for $X$ s.t. $||X||=\infty$.
$\textbf{Part 1}$. Consider the function $f:X>0\rightarrow tr(XA)+tr(X^{-1}B)$. The tangent space to $S_n^{++}$ is $S_n$, the vector space of symmetric matrices. Then 
$Df_X:H\in S_n\rightarrow tr(HA)-tr(X^{-1}HX^{-1}B)=tr(H(A-X^{-1}BX^{-1}))$.
If $X$ reaches a local minimum of $f$, then $Df_X=0$, that is, $A-X^{-1}BX^{-1}$ is skew symmetric; that is equivalent to 
$A=X^{-1}BX^{-1}$ or $(*)$ $XAX-B=0$.
The Riccati equation $(*)$ has a unique symmetric $\geq 0$ solution $X_0$; since the term of degree $1$ in $X$ does not exist, there is a closed form 
$X_0=A^{-1/2}(A^{1/2}BA^{1/2})^{1/2}A^{-1/2}$.
Note that $f(X_0)=2tr((A^{1/2}BA^{1/2})^{1/2})$.
$\textbf{Part 2}$. $D^2f_X:(H,K)\in S_n\times S_n\rightarrow tr(X^{-1}KX^{-1}HX^{-1}B)+tr(X^{-1}HX^{-1}KX^{-1}B)$,
and $D^2f_X(H,H)=2tr((X^{-1}HX^{-1}HX^{-1})B)$.
The matrix $S=X^{-1}HX^{-1}HX^{-1}$ is symmetric $\geq 0$ because
$y^TSy=(HX^{-1}y)^TX^{-1}(HX^{-1}y)\geq 0$.
Then $D^2f_X(H,H)=2tr(SB)\geq 0$ and $f$ is convex (recall that $S_n^{++}$ is convex). Finally, $f(X_0)$ is the required minimum because (for a convex function $f$), $Df_{X_0}=0$ implies that $inf(f)=f(X_0)$.