I'm looking for a convex reformulation, if any exists, of the following minimisation problem:
Let $A$ be a symmetric, positive definite $n \times n$ matrix, and $b \in \mathbb{R}^n$. Minimise $f(X) = - b^T (A+X)^{-T} X (A+X)^{-1} b$, where $X$ is a symmetric positive semidefinite $n \times n$ matrix.
The following is a plot of $-f$ for a toy version of the problem: $n=2$; diagonal $X$; $A$ and $b$ left unspecified.
By inspection, $-f$ is not even a quasiconvex function of $(X_{11}, X_{22})$. Therefore, quasiconvex-optimisation techniques would not, in general, be applicable.
Although I haven't proved it, my intuition tells me that $f$ does have a unique minimum. This leads me to believe that a convex reformulation of the problem may exist... but I can't find any.
Can you think of some nonlinear change of variables that transform this nonconvex optimisation problem into a convex one?
Making the substitution $Y=(A+X)^{-1}$, the objective becomes $b^T(Y A Y - Y )b$, which is convex in $Y$. The semidefinite constraint $X\succeq 0$ is equivalent to $0 \prec Y \preceq A^{-1}$.
As a bonus, this transformation allows us to solve the problem in closed form. Denoting the transformed objective by $g(Y)$, its gradient is:
$$\begin{aligned}\nabla g(Y) &= bb^T YA + AYbb^T - bb^T \\ &= b \left(AYb - \frac{1}{2} b\right)^T + \left(AYb - \frac{1}{2} b\right) b ^T\end{aligned}$$
So the gradient is zero and the variable is feasible if and only if the following conditions hold:
$$Yb=\frac{1}{2} A^{-1}b, \quad 0 \prec Y \preceq A^{-1} \quad \Leftrightarrow \quad XA^{-1}b=b, \quad X \succeq 0$$
Since $g$ is convex, these give sufficient conditions for optimality. Furthermore, solutions to these conditions always exist; one example is given by $Y^* = \frac{1}{2}A^{-1} \Leftrightarrow X^* = A$.