Suppose $A, B, C\in\mathbb{R}^{n\times n}$ are all symmetric positive definite matrices, and they satisfy the inequality $A \succeq B + C$. Assume also that all of the three matrices are bounded, i.e., $A\_{\min}I \preceq A \preceq A\_{\max}I$, $B\_{\min}I \preceq B \preceq B\_{\max}I$ and $C\_{\min}I \preceq C \preceq C\_{\max}I$, where $A\_{\min}, A\_{\max}, B\_{\min}, B\_{\max}, C\_{\min}, C\_{\max}\in\mathbb{R}^+$. How to prove the following inequality:
$$A^{-1} \preceq B^{-1} - A\_{\max}^{-1}B\_{\max}^{-1}C\_{\min}I$$



Matrix inversion inequality
Well, this is true. We consequently have the following:


**Lemma 1.** If $X\succeq I$, then $X^{-1}\preceq I$.


**Proof.** Write $X$ in the diagonal basis.


**Lemma 2.** If $X\succeq Y\succ 0$ then $X^{-1}\preceq Y^{-1}$.


**Proof.** We have $X=Y+Z=Y^{1/2}(I+Y^{-1/2}ZY^{-1/2})Y^{1/2}$ for $Z\succeq 0$, then $$X^{-1}=Y^{-1/2}(I+Y^{-1/2}ZY^{-1/2})^{-1}Y^{-1/2}\\=Y^{-1/2}(I-W)Y^{-1/2}=Y^{-1}-Y^{-1/2}WY^{-1/2}\leqslant Y^{-1}$$ for certain $W\succeq 0$ by Lemma 1.


Now denote $\gamma=C\_{\min}$, $\alpha=A\_{\max}$, $\beta=B\_{\max}$. We have
$A\succeq B+C\succeq B+\gamma I$, thus by Lemma 2 we have $A^{-1}\preceq (B+\gamma I)^{-1}$, and it suffices to prove $(B+\gamma I)^{-1}\preceq B^{-1}-\frac{\gamma}{\alpha\beta}I$ which reduce to the 1-dimensional case if we write $B$ in the diagonal basis: we need to show that $(\lambda+\gamma)^{-1}\leqslant \lambda^{-1}-\frac{\gamma}{\alpha\beta}$ for any eigenvalue $\lambda$ of $B$, this is equivalent to
$\frac{\gamma}{\alpha\beta}\leqslant \lambda^{-1}-(\lambda+\gamma)^{-1}=\frac{\gamma}{\lambda(\lambda+\gamma)}$ that is true since $\lambda\leqslant \beta$, $\lambda+\gamma\leqslant \alpha$.


