Consider symmetric positive definite matrices of same size $A_1, A_2, \ldots, A_m$. Is the following statement true?
$$ Tr(\prod_{i=1}^m A_i) \leq \prod_{i=1}^m Tr(A_i)$$
My guess is it's true based on the following fact from Wikipedia:
If $A, B > 0$ and of same size, then
$$ 0 \leq [tr(AB)]^2 \leq tr(A^2)tr(B^2) \leq [tr(A)]^2 [tr(B)]^2$$
First of all, we can easily extend your theorem by repeatedly applying the inequality like this (I'm a bit too lazy to formally work out the formal induction, let me know if you need help for that):
$tr(ABC)^2 \leq tr(AB)^2 tr(B)^2 \leq tr(A)^2 tr(B)^2 tr(C)^2$
Secondly, the trace of any matrix is equal to the sum of its eigenvalues. Since all eigenvalues in a positive definite matrix are positive, the sum of eigenvalues (hence the trace) is also positive.
So since $tr(ABC...)^2 \leq tr(A)^2 \cdot tr(B)^2 \cdot tr(C)^2...$ and since we know that all individual matrix traces are positive, we can just remove the squares (doesn't matter whether $tr(ABC...)$ is negative or positive).
Hence $tr(ABC...) \leq tr(A)\cdot tr(B) \cdot tr(C) ...$, which proves your theorem.