If $X$, $Y$, $Z$ positive, does $Tr\{XZ\}\le Tr\{YZ\}$ hold? If $X\le Y$, that is, $Y-X$ is positive semi-definite, does it imply that $Tr\{XZ\}\le Tr\{YZ\}$?
If it's true, how should one go about proving it? Is it a result from some other theorem?
*edit: $Z$ is also positive, all matrices are Hermitian.
 A: If the matrices are not hermitian, here is an example to prove that the statement is false.
The statement is equivalent to is $\text{Tr}((Y-X)Z) \geq 0$
For instance, take 
$Y-X = \left(\begin{array}{c c c c}
1 & -1 & -1 & -1 \\
0 & 1 & -1 & -1 \\
0 & 0 & 1 & -1 \\
0 & 0 & 0 & 1
\end{array} \right)$
$Z = \left(\begin{array}{c c c c}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 \\
1 & 1 & 1 & 1 \\
\end{array} \right)$
Clearly, $Y>X$ and $Z>0$, however the product is
$(Y-X)Z = \left(\begin{array}{c c c c}
-2 & -1 & 0 & 1 \\
-3 & -1 & 0 & 1 \\
-2 & -2 & 0 & 1 \\
-1 & -1 & -1 & 1 \\
\end{array} \right)$
and the trace of the above is negative and hence $\text{Tr}(XZ) \leq \text{Tr}(YZ)$ is false.
However, the fact that the matrices are Hermitian might turn out to be helpful in proving that the statement is true. My hunch is that the statement is true but I do not have a proof now.
A: Yes.
Hints: 


*

*You can write $Y - X = S^2$, $Z = T^2$ with $S,T \geq 0$.

*Recall $\text{Tr}(AB) = \text{Tr}(BA)$ and $\text{Tr}(C^{\ast}C) \geq 0$.

