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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: $$A^{-1} \preceq B^{-1} - A_{\max}^{-1}B_{\max}^{-1}C_{\min}I$$

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    $\begingroup$ Why are you sure it's true? $\endgroup$ Jan 17, 2021 at 20:00
  • $\begingroup$ @FedorPetrov If $A, B, C$ are scalars, the inversion inequality should be true. That can be easily verified. But for the matrix case, I'm not entirely sure if it's true. I think it's likely to be true because $A, B, C$ are all bounded symmetric positive definite matrices and it's easy to see that $A_{\max}I\succeq B_{\max}I + C_{\min}I$. I wonder if there's a way to do eigendecomposition and convert the problem to 1D case. If you think it is not true, I'd also like to know if there are cases to make it untrue. Thanks. $\endgroup$
    – Evan
    Jan 17, 2021 at 23:21
  • $\begingroup$ ok, but if you are not sure please ask "prove or disprove", not "prove". As stated now, this looks like a homework question which is not welcomed. $\endgroup$ Jan 18, 2021 at 6:57
  • $\begingroup$ I agree. In the future I will pay more attention to the statement. This is something I need to prove in my research. $\endgroup$
    – Evan
    Jan 18, 2021 at 15:24

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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 $\lambda\leqslant \beta$, $\lambda+\gamma\leqslant \alpha$.

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  • $\begingroup$ Thanks a lot, Fedor. I would up vote but I'm new here and don't have enough reputation:( . Just a commont on the proof, when decomposing the matrix $Y$, which is symmetric positive definite, should it be $Y = Y^{\frac{1}{2}}Y^{\frac{\top}{2}}$? $\endgroup$
    – Evan
    Jan 18, 2021 at 18:31
  • $\begingroup$ $Y^{1/2}$ is also symmetric positive definite, and it's square equals $Y$ $\endgroup$ Jan 18, 2021 at 18:49
  • $\begingroup$ I see. You are using positive definite square root here. $\endgroup$
    – Evan
    Jan 18, 2021 at 20:06

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