 Solution

Let $V,V^{\prime },U$ be mutually independent, with $V,V^{\prime }$ identically distributed. Entropy submodularity for three independent sums givesIndependent addition cannot decrease discrete entropy, because $H(V+V^{\prime }+U)\ge H(V+V^{\prime }+U\mid U)=H(V+V^{\prime })$. Hence

Take $U$ to have the distribution of $-X$ and take $V,V^{\prime }$ to be independent copies of $X$, all mutually independent. Then $U+V$ and $U+V^{\prime }$ both have the distribution of $X-Y$, whereas $V+V^{\prime }$ has the distribution of $X+Y$ and $H(U)=H(X)$. HenceorThe denominator is nonnegative because conditioning on $Y$ recovers $X$ from $X-Y$, so $H(X-Y)\ge H(X-Y\mid Y)=H(X)$.