I've been researching some operator spaces and have stumbled upon the following problem. Suppose $x,y\in\mathcal{B}(\mathcal{H})$ (for $\mathcal{H}$ separable) such that $0\leq x\leq y\leq 1$. Further let $t\in[0,1]$ and $\rho$ be a density matrix (i.e. $\rho\geq 0$, $\mathrm{Tr}(\rho)=1$). Is it true that $$ \mathrm{Tr}\big(\vert \rho^{1-t}x\rho^t\vert\big)\leq \mathrm{Tr}\big(\vert \rho^{1-t}y\rho^t\vert\big) \leq 1? $$ Obviously this holds for $t=\frac{1}{2}$, but I'm not really sure how to approach it for other values. I've tried playing around with the polar decomposition but haven't gotten anywhere.
I'm convinced that the inequality is actually true at least in the case of $2\times 2$ matrices as I've run a randomized computer search and have not found any counterexamples, but even there trying to prove it directly by writing out all the matrices is very ugly.
EDIT: I've verified the inequality numerically on matrices of sizes up to $5\times 5$ with $10\,000$ random samples per size (hopefully the code is correct, generating random unitaries is a bit tricky).
It seems my code was indeed incorrect.
The first inequality is not true.
For a counterexample, consider $t = 1$, $\rho = v v^{*}$ for a unit vector $v$. Then one checks that $\textrm{Tr}(|x \rho|) = \|x(v)\|$, so the inequality is equivalent to $$ \langle x^{2}(v), v\rangle = \|x(v)\|^{2} \leq \|y(v)\|^{2} \leq \langle y^{2}(v), v\rangle, $$ i.e. $x^2 \leq y^{2}$. It is however not true in general that $0 \leq x \leq y \leq 1$ implies $x^2 \leq y^2$, as can be seen by considering say $x = u u^{*}$ and $y = x + \varepsilon w w^{*}$ for small enough $\varepsilon$ and $u, w \in \mathcal{H}$ not parallel or orthogonal.
The second inequality is true and follows from Hölder's inequality.