# Quantum channel cannot increase Holevo information of an ensemble

I need to prove the fact that a quantum channel (a superoperator) cannot increase the Holevo information of an ensemble $$\epsilon = \{\rho_x, p_x\}$$. Mathematically expressed I need to prove

\begin{align} \chi(\(\epsilon)) \leq \chi(\epsilon) \end{align} \tag{1}\label{1}

where $$\$$ represents a quantum channel (a positive trace preserving map) which works on the ensemble as $$\ \epsilon = \{ \\rho_x, p_x \}$$. This needs to be done with the property that a superoperator $$\$$ cannot increase relative entropy (some may remember my previous question about this):

\begin{align} S(\\rho || \\sigma) \leq S(\rho || \sigma) \end{align}\tag{2}\label{2}

with relative entropy defined as $$S(\rho || \sigma) = \mathrm{tr}(\rho \log \rho - \rho \log \sigma).$$

The Holevo information is defined as (if someone would not know)

\begin{align} \chi(\epsilon) = S(\sum_x p_x \rho_x ) - \sum_x p_x S(\rho_x) \end{align}.

Does anyone know how we get from equation $$\eqref{2}$$ to equation $$\eqref{1}$$? Maybe what density operators to fill in $$\eqref{2}$$? Or how do we start such a proof?

Suppose that $$\mathsf{X}$$ is a register that can store each possible choice for $$x$$, as a classical state, while $$\mathsf{Y}$$ is a register that can store each possible state $$\rho_x$$. It is then natural to associate the classical-quantum state $$\rho = \sum_x p_x |x\rangle \langle x| \otimes \rho_x$$ with the ensemble $$\{(p_x,\rho_x)\}$$.
1. Verify that the Holevo information of the ensemble $$\{(p_x,\rho_x)\}$$ is just another name for the mutual information $$\text{I}(\mathsf{X}:\mathsf{Y})_\rho$$ between $$\mathsf{X}$$ and $$\mathsf{Y}$$.
2. Verify, for an arbitrary state $$\rho$$ of the pair $$(\mathsf{X},\mathsf{Y})$$, that the mutual information $$\text{I}(\mathsf{X}:\mathsf{Y})_\rho$$ can be expressed as $$\mathrm{I}(\mathsf{X}:\mathsf{Y})_\rho = \mathrm{S}\bigl(\rho\,\big\|\, \rho^{\mathsf{X}}\otimes \rho^{\mathsf{Y}}\bigr).$$