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Write $p=(1+\mu {)}^{-1}$. Since $X$ and the geometric random variable $Z$ have the same expected value $\mu$,This is also the entropy deficit relative to the maximum entropy distribution on the nonnegative integers.

The random variables $X$ and $-Z$ are independent, so part a gives $H(X-Z)\ge H(-Z)=H(Z)$. Consequentlywhich is the required bound in terms of the Entropic Ruzsa distance.

For random variables $A,B$ in a finite additive group, take independent copies $A^{\prime },B^{\prime }$ with the same respective distributions and define the Entropic Ruzsa distance byFor the independent variables in the question, expansion givesPart iii, applied to the independent variables $X,Y,-Z$, saysSubtracting $H(X-Y)$ from both sides proves

Because the Entropic Ruzsa distance depends only on marginal distributions, take $X,Y,Z$ independent with the required marginals. Since $X-Z=(X-Y)+(Y-Z)$ is a function of $(X-Y,Y-Z)$, the data processing inequality for mutual information yieldsThe map $(X,X-Y,Y-Z)\mapsto (X,Y,Z)$ is a bijection. Using independence and the chain rule for information entropy, the left side iswhereas the right side is $H(X-Z)-H(Z)$. Hencewhere the final step is subadditivity of information entropy. Substituting this inequality into the definition of $d_R$ gives the Entropic Ruzsa triangle inequality

Let $X_1,X_2,Y$ be independent, with $X_1,X_2$ distributed as $X$. Apply part (b) to $X_1,-Y,X_2$:Adding an independent random variable cannot decrease information entropy, soIn terms of Entropic Ruzsa distance, this is $d_R(X,-X)\le 2d_R(X,Y)$. The Entropic Ruzsa triangle inequality and invariance under simultaneous negation now givewhich is the Entropic Ruzsa sum-difference inequality.