An upper bound for $trace(XYXZ)$ in terms of $trace(X^2YZ)$ I want to find an upper bound for  $trace(XYXZ)$ in terms of $trace(X^2YZ)$. $X, Y, Z$ are Hermitian positive definite matrices. 
Any idea/help is appreciated.
Update: Assume that $Y$ and $Z$ commute: $YZ=ZY$.
 A: It is possible for $\text{trace}(X^2 Y Z) = 0$ with $X,Y,Z$ positive definite Hermitian.  Try
$$ X^2 = \pmatrix{2 & -1\cr -1 & 1\cr},\ Y = \pmatrix{1/10 & 0\cr 0 & 1\cr},\ Z = \pmatrix{2 & 1\cr 1 & 7/10}$$
However, this will not work with the added condition that $Y$ and $Z$ commute.
If $Y$ and $Z$ commute, so do $Y$ and $Z^{1/2}$, and $YZ = Z^{1/2} Y Z^{1/2}$ is positive definite.  Then 
$$\text{trace}(X^2 Y Z) = \text{trace}((YZ)^{1/2} X^2 (YZ)^{1/2}) > 0$$
A: Short answer: None of the two has a reason to be larger than the other. Some bounds in terms of the condition numbers of $Y,Z$ are
$$trace(XYXZ) \leq \min(\kappa(Y), \kappa(Z)) \cdot trace(X^2YZ)$$
$$trace(X^2YZ) \leq \min(\kappa(Y), \kappa(Z)) \cdot trace(XYXZ)$$

First, we can simultaneously diagonalize $Y$ and $Z$, say $Y = PY'P^{-1}$, $Z = PZ'P^{-1}$ with $P$ unitary, then we get
$$trace(X'Y'X'Z') \quad, \quad trace(X'^2Y'Z')$$
with $X' = P^{-1}XP$ still hermitian. Thus we may suppose $Y,Z$ are diagonal. Then
$$\begin{align*}
trace(XYXZ) &= \sum_{i,j}|X_{ij}|^2Y_{jj}Z_{ii} \\
trace(X^2YZ) &= \sum_{i,j}|X_{ij}|^2Y_{ii}Z_{ii} 
\end{align*}$$
The $Y_{ii},Z_{ii}>0$ are arbitrary, so all we can say is that one is bounded by $$\kappa(Y) = \max_{i \neq j}|Y_{ii}/Y_{jj}|$$ (or $\kappa(Z) = \max_{i \neq j}|Z_{ii}/Z_{jj}|$) times the other.
