A Matrix Inequality

An exercise from Braess FEM book: for $A, B$ symmetric, positive definite matrices, let $A \le B$. We want to show that the inverses satisfies a similar property $B^{-1} \le A ^{-1}$.

The book actually gave quite a lot of hints for this one.

\begin{aligned}x^TB^{-1}x &= x^T A^{-1/2} A^{1/2} B^{-1} x \\&\le \sqrt{x^T A^{-1} x} \sqrt{x^T B^{-1} A B^{-1} x}.\end{aligned}

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Then from the hypothesis, $A \le B \implies x^T(B - A)x \ge 0 \implies y^T(B^{-1} - B^{-1}AB^{-1})y \ge 0$.
So we can plug this in to our equation above to find that $x^TB^{-1}x \le \sqrt{x^T A^{-1} x}\sqrt{x^T B^{-1}x}$.

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