Suppose $A,B,C$ are all real symmetric and positive definite matrices. Is it possible to lower bound ${\bf tr}(ABC)$ in terms of ${\bf tr}(AB)$.
I tried to follow Von Neumann's trace inequality: letting $\{a_i\}, \{c_i\}$ are the singular values in non-increasing order of $AB$ and $C$ respectively, \begin{align*} {\bf tr} (ABC) \le \sum_{i=1}^n a_i b_i \le \sum_{i=1}^n a_i \|C\|_2, \end{align*} as $a_i$'s are all nonnegative. However I could not relate the sum of $a_i$'s with ${\bf tr}(AB)$.
We can also do following: \begin{align*} {\bf tr} (ABC) \le \|BC\|_2{\bf tr} (A) \le \|B\|_2 \|C\|_2 {\bf tr}(A) \le \frac{\|B\|_2\|C\|_2} {\lambda_{\min}(B) } {\bf tr}(AB). \end{align*}
I am in particular interested in a constant that is not determined by $B$.
$\DeclareMathOperator{Tr}{Tr}$ Let $$\begin{align} D&=A^{1/2}BA^{1/2}\\ E&=\frac{A^{-1/2}CA^{1/2}+A^{1/2}CA^{-1/2}}{2}\text{.} \end{align}$$ Then $D$ is positive definite, $E$ is symmetric (square roots of real p.d. matrices are real), and $$\frac{\Tr ABC}{\Tr AB}=\frac{\Tr DE}{\Tr D}\leq \lambda_{\text{max}}(E)\text{,}$$ equality being approached as $D$ approaches the spectral projector for $\lambda_{\text{max}}(E)$. Therefore $$\Tr ABC \leq \Tr AB\,\lambda_{\text{max}}\left(\tfrac{A^{-1/2}CA^{1/2}+A^{1/2}CA^{-1/2}}{2}\right)\text{.}$$
Note that $\lambda_{\text{max}}(E)$ is not bounded by $\lVert C\rVert_{\infty}$ independent of $A$. For example, let $$\begin{align} A^{1/2}&=1+rX & C&=1+sZ \end{align}$$ where $X$ and $Z$ are Pauli matrices, $0<r<1$, and $0<s<1$. Then $$E=1+\frac{s(1+r^2)}{1-r^2}Z$$ has unbounded spectrum as $r\to 1$.