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Let $A$ be a $n\times n$ real (symmetric) positive definite matrix with spectrum contained in $[m, M]$ and also let $X$ be an $n\times p$ matrix such that $X'X=I_p$. It is known that $$\frac{(M+m)^2}{4Mm}(X'AX)^{-1}-X'A^{-1}X \tag{$\star$}$$ is positive semidefinite. Is it true that $$\frac{(M+m)^4}{16M^2m^2}(X'AX)^{-2}-(X'A^{-1}X)^2 \tag{$\dagger$}$$ is also positive semidefinite?

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Yes, you comment is right. We know generally, for psd of $S,T$, psd of $S-T$ does not imply psd of $S^2-T^2$. –  Sunni Aug 24 '11 at 20:04

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From maximizing the given condition, denote it by $(\star)$, we obtain, $$ \left( \frac{(M+m)^2}{4Mm}\frac{1}{m}-\frac{1}{M}\right) I \succeq (\star)\succeq 0 $$ and thus, $(M+m)^2-4m^2\geq0$ which gives us nothing other than $M\geq m$. On the other hand, minimizing the second expression, say $(\dagger)$, we obtain $$ \frac{(M+m)^4}{16M^2m^2}\frac{1}{M^2}-\frac{1}{m^2} = \frac{(M+m)^4-(2M)^4}{16M^4m^2} $$ and this expression is positive semi-definite if and only if $m\geq M$. Therefore combining both, if $M=m$ then $(\dagger)$ is positive semi-definite.

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