Does the tensor product respect semidefinite ordering in this way? I'll use $\succeq$ to denote the positive semidefinite ordering: for square matrices $X,Y$, one has $X \succeq Y$ iff $X - Y$ is positive semidefinite. It's a well known fact that if $X, Y \succeq 0$ then $X \otimes Y \succeq 0$. However, if one has two pairs of matrices with $X \succeq Y$ and $X' \succeq Y'$, it isn't necessarily the case that $X \otimes X' \succeq Y \otimes Y'$ (for instance, if $X = X' = I$ and $Y = Y' = -2I$).
My question is: if we add that $Y, Y' \succeq 0$, so our assumptions are
  $$ X \succeq Y \succeq 0,\ \ \ X' \succeq Y' \succeq 0 $$
Is it necessarily true that $X \otimes X' \succeq Y \otimes Y'$?
I personally feel that this should be true (and I could swear I've seen this result before but can't find it anywhere), but I'm struggling to prove it. Does anyone have a reference for this fact, and/or know how to prove (or disprove) it?

\begin{aligned}
&X\otimes X'-Y\otimes Y'\\
&=\left[(X-Y)+Y\right]\otimes\left[(X'-Y')+Y'\right]-Y\otimes Y'\\
&=\left[(X-Y)\otimes(X'-Y')+Y\otimes(X'-Y')+(X-Y)\otimes Y'+Y\otimes Y'\right]-Y\otimes Y'\\
&=(X-Y)\otimes(X'-Y')+Y\otimes(X'-Y')+(X-Y)\otimes Y'.
\end{aligned}
Now the result follows because $X-Y,\,X'-Y',\,Y$ and $Y'$ are all positive semidefinite (and this is where $Y,Y'\succeq0$ is essential).
