TITLE: Trace inequality under consideration of definiteness
QUESTION [5 upvotes]: 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.

REPLY [7 votes]: 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.