Let $A$ and $B$ be positive definite matrices. How to show that $(A+B)^{-2}< A^{-2}$?
2026-03-27 02:02:48.1774576968
Inequality involving negative powers of positive definite matrices
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You can't because it isn't true. Counterexample: $(A+B)^{-2}<A^{-2}$ is equivalent to $(A+B)^2-A^2>0$, but \begin{aligned} &\left[\pmatrix{1&5\\ 5&26}+\pmatrix{2&10\\ 10&51}\right]^2-\pmatrix{1&5\\ 5&26}^2\\ =&\pmatrix{3&15\\ 15&77}^2-\pmatrix{1&5\\ 5&26}^2\\ =&\pmatrix{234&1200\\ 1200&6154}-\pmatrix{26&135\\ 135&701}\\ =&\pmatrix{208&1065\\ 1065&5453} \end{aligned} is not positive definite because its determinant is $-1$.