# On some matrix inequality

Suppose we have as symmetric positive definite matrix $A$ such that \begin{align} 0 \preceq A \preceq ( C+ D)^{-1} \end{align} where $C$ and $D$ are both symetric positive definite.

My question does there exists a matrix $B$ such that \begin{align} A= ( B+ D)^{-1} \end{align}

In the scalar case, this is trivial. If we have that \begin{align} 0 \le a \le \frac{1}{c+d} \end{align} then we can always find some $b \ge c$ such that \begin{align} a=\frac{1}{b+d} \end{align}

The scalar case generalises to the matrix case. Note that $A=(B+D)^{-1}$ iff $B=A^{-1}-D$. Since $(C+D)^{-1}\succeq A\succ0$, we have $A^{-1}\succeq C+D\succ0$ and hence $B=A^{-1}-D\succeq C\succ0$.