Is it possible to lower bound ${\bf tr}(XYZ)$ in terms of $X, Y, Z$ provided that $X, Y, Z$ are all symmetric and positive definite?

Question

Suppose $X, Y, Z$ are symmetric and positive definite matrices. I am interested in lower bounding ${\bf tr}(XYZ)$. The trace of product of two matrices can be lower bounded by \begin{align*} {\bf tr} (XY) \ge \lambda_{\min}(X) {\bf tr}(Y). \end{align*} I am wondering whether it is all possible to acquire a similar bound.

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We can also lower bound by \begin{align*} 2{\bf tr}(XYZ) = {\bf tr}(XYZ + XZY) \ge {\bf tr}(X) \lambda_{\min}(YZ+ZY). \end{align*} I could not see how to proceed to bound $\lambda_{\min}(YZ+ZY)$.

Answer 1

A real matrix $A$ is a product of $3$ symmetric $>0$ matrices iff $A$ is congruent to a triangular matrix with $>0$ diagonal.

Then $trace(A)$ may be $<0$.

For example $P=\begin{pmatrix}88&41&70\\-82&91&-32\\-70&29&-1\end{pmatrix},B=\begin{pmatrix}19&80&90\\0&21&-35\\0&0&60\end{pmatrix}$. $tr(P^TBP)=-252771$.