# Positive semidefinite relationship after partial trace

Let $$\rho_{ABC}$$ and $$\sigma_{C}$$ be arbitrary quantum states and $$\lambda\in \mathbb{R}$$ be minimal such that

$$\rho_{ABC}\leq \lambda \rho_{AB}\otimes\sigma_C$$

We assume there are no issues with support in the above statement to avoid infinities. Now, one traces out the $$B$$ register. Let $$\mu\in \mathbb{R}$$ be minimal so that

$$\rho_{AC}\leq \mu\rho_A\otimes \sigma_C$$

Clearly, $$\lambda\geq\mu$$ since partial tracing is a completely positive quantum operation but in this case, since the traced out register had the same state on both the lhs and rhs, is $$\lambda = \mu$$?

No, not necessarily. For example, take $$\rho$$ to be a GHZ state and let $$\sigma$$ be the completely mixed state of one qubit. We then have $$\lambda=4$$ and $$\mu=2$$.