I have tried to prove the following inequality for smooth min-entropies
\begin{equation} H_{min}^{\epsilon}(XY|K) \geq H_{min}^{\epsilon}(X|K) \end{equation}
I started trying to prove it for the non-smooth min-entropy. The definition of the min-entropy is: $H_{min}(X|K) \geq -\log( p_{max}^{(x)})$, where
\begin{equation} p_{max}^{(x)} = \max_{x,k} P_{x|k}(x|k) \end{equation}
and $H_{min}(XY|K) \geq -\log( p_{max}^{(xy)})$, where
\begin{equation} p_{max}^{(xy)} = \max_{xy,k} P_{x,y|k}(x,y|k). \end{equation}
Then we can write the first probability distribution as the marginal
\begin{equation} P_{x|k}(x|k) = \sum_yP_{xy|k}(x,y|k) \end{equation}
From this, since probabilities are non-negative, we know that each term in $\sum_yP_{xy|k}(x,y|k)$ is less or equal to $P_{x|k}(x|k)$, so for each term we have
\begin{equation} P_{x,y|k}(x|k) \leq P_{x|k}(x|k) \end{equation}
Then it follows that
\begin{equation} p_{max}^{(x)} \geq p_{max}^{(xy)} \end{equation}
and from this
\begin{equation} H_{min}(XY|K) \geq H_{min}(X|K) \end{equation}
Can someone please confirm if this calculation is correct, and if this can be extended to smooth min-entropies?
