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We're interested in a symmetric and positive definitive matrix $\Sigma^{-1}$ \Sigma$that has been orthogonally diagonalized as$ \Sigma=Q^{T}\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})Q $where$\theta_{1}^{2} \geq \theta_{2}^{2} \geq ... \geq \theta_{n}^{2} > 0$, and we let$\theta_{\max}^{2}=\theta_{1}^{2}$and$\theta_{\min}^{2}=\theta_{n}^{2}$. Also, let$\Theta^{2}=\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})$. We want to show for any nonzero vector$\alpha$,$ \frac{\alpha^{T}\Sigma^{-1}\alpha}{\sqrt{\alpha^{T}\alpha}\sqrt{\alpha^{T}\Sigma^{-2}\alpha}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $Note that you can assume without loss of generality that$\alpha$is of length 1. (just scale the length of alpha out of everything on the left hand side of the inequaility.) Also, by using the substitution$x=Q\alpha$, you can reduce this to a problem about the digonal matrix$\Theta$, and then reduce the matrix-vector products to sums. We then want to show for all vectors$x$of length 1,$ \frac{\sum_{i=1}^{n} x_{i}^{2}\theta_{i}^{-2}} {\sqrt{\sum_{i=1}^{n}x_{i}^{2}\theta_{i}^{-4}}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $Unfortunately, I don't see any easy way to proceed from here. 1 Here's a restatement of the problem for those who don't want to find the paper referenced in the question. We're interested in a symmetric and positive definitive matrix$\Sigma^{-1}$that has been orthogonally diagonalized as$ \Sigma=Q^{T}\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})Q $where$\theta_{1}^{2} \geq \theta_{2}^{2} \geq ... \geq \theta_{n}^{2} > 0$, and we let$\theta_{\max}^{2}=\theta_{1}^{2}$and$\theta_{\min}^{2}=\theta_{n}^{2}$. Also, let$\Theta^{2}=\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})$. We want to show for any nonzero vector$\alpha$,$ \frac{\alpha^{T}\Sigma^{-1}\alpha}{\sqrt{\alpha^{T}\alpha}\sqrt{\alpha^{T}\Sigma^{-2}\alpha}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $Note that you can assume without loss of generality that$\alpha$is of length 1. (just scale the length of alpha out of everything on the left hand side of the inequaility.) Also, by using the substitution$x=Q\alpha$, you can reduce this to a problem about the digonal matrix$\Theta$, and then reduce the matrix-vector products to sums. We then want to show for all vectors$x$of length 1,$ \frac{\sum_{i=1}^{n} x_{i}^{2}\theta_{i}^{-2}} {\sqrt{\sum_{i=1}^{n}x_{i}^{2}\theta_{i}^{-4}}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} \$