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I am trying to show that if $A$ is a bounded, self-adjoint and positive operator on a Hilbert space $H$, $0 \in \rho(A)$, the following inequality holds for all $x \in H$ with $\|x\| = 1$: \begin{align*} 4 \langle x, Ax \rangle \langle x, A^{-1}x \rangle \geq 1. \end{align*}

Since $0 \in \rho(A)$, we know that $A:X \to X$ is bijective, $A^{-1}:X \to X$ is bounded and $A^{-1}$ self-adjoint.
But I don't know how to continue. Thanking you in advance for any help!

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Unless I am missing something, it seems to me that we can get this inequality without the 4 by writing $$1 = \|x\|^2 = \langle A^{1/2} x, A^{-1/2} x \rangle$$ and using Cauchy-Schwarz.

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