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Let $G \in \mathbb{R}^{3 \times 3}$ a symmetric, but indefinite matrix and $U \in \mathbb{R}^{3\times 3}$ a symmetric and positive definite matrix. I would like to prove the inequality

$$ \text{Tr} \left( G^2 \right) \leq \text{Tr} \left( GUGU^{-1} \right). $$

If $U$ and $G$ commute, both sides of the inequality are obviously equal. However for more general cases, I have tried to rearrange the inequality to

$$ \text{Tr}(\underbrace{[UG-GU]}_{\text{skew-symmetric}}\ GU^{-1}) \leq 0 $$

and then using the Cauchy-Schwarz inequality. Unfortunately, I have not found a solution yet.

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    $\begingroup$ You may assume U is diagonal. Then write everything out in components. $\endgroup$
    – Michael Renardy
    Commented Jun 28, 2020 at 19:08
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    $\begingroup$ Do you have ample computational evidence that the inequality holds? For instance, have you tried 1000 random-generated examples on a computer? $\endgroup$
    – Federico Poloni
    Commented Jun 28, 2020 at 19:23
  • $\begingroup$ I have tried it with matlab and and a lot of variations, all of them were true. I have forgotten to say that both sides are of course larger than zero $\endgroup$
    – TARS
    Commented Jun 28, 2020 at 19:36
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    $\begingroup$ Yes I Have tried it with random matrices which fulfill the restrictions stated above. All of them were fulfilled. $\endgroup$
    – TARS
    Commented Jun 28, 2020 at 20:17

1 Answer 1

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Write $$G=\left( \begin{array}{ccc} a & b & c \\ b & d & e \\ c & e & f \\ \end{array} \right) $$ and, without loss of generality, $$ U=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & u & 0 \\ 0 & 0 & v \\ \end{array} \right),$$ where $u,v>0$. Then $$\text{Tr}(GUGU^{-1})-\text{Tr}(G^2) =\frac{b^2 (u-1)^2 v+c^2 u (v-1)^2+e^2 (u-v)^2}{u v},$$ which is manifestly $\ge0$, as desired.

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    $\begingroup$ Moreover, this works for $n \times n$ matrices, not just $3\times 3$. You get $$ \text{Tr}(GUGU^{-1} - G^2) = \sum_{i<j} \frac{(u_i - u_j)^2 g_{ij}^2}{u_i u_j}$$ $\endgroup$
    – Robert Israel
    Commented Jun 28, 2020 at 21:33
  • $\begingroup$ @RobertIsrael : Good point! $\endgroup$
    – Iosif Pinelis
    Commented Jun 28, 2020 at 21:35

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