TITLE: Optimisation: $H(X_1) + H(X_2) +H(X_3) - H(X_1+X_2+X_3)$
QUESTION [7 upvotes]: Consider the following optimisation.
$$\max [H(X_1) + H(X_2) +H(X_3) - H(X_1+X_2+X_3)],$$
where H denotes the Shannon entropy, + denotes addition over real numbers, and the maximum is taken over all the random variables $X_1$, $X_2$, $X_3$ that are independent Bernoulli.
Question: How to prove that the maximum is achieved when $X_1$, $X_2$, and $X_3$ follow a uniform distribution?

Edit (as requested): For a discrete random variable $X$ with probability mass function $p$ with support $\mathcal{X}$, $H(X) = - \sum_{x \in \mathcal{X}} p(x) \log p(x)$.

REPLY [2 votes]: Too long for a comment. Note that $\bar{\bar{x}}=1-\bar{x}=1-(1-x)=x$. Similar for $y$ and $z$. Denote $u:=(x,y,z)$ then $\bar{u}=(\bar{x},\bar{y},\bar{z})$. So your problem can be rewritten as
$$\max_{u\in(0,1)^3}f(u)+f(\bar{u})$$ 
Note that if $u^*\in\arg\max_{u\in(0,1)^3}f(u)+f(\bar{u})$ then so is $\bar{u}^*$ because 
$$f(\bar{u}^*)+f(\bar{\bar{u}}^*)=f(\bar{u}^*)+f(u^*)=f_{\max}$$
It suffices therefore to prove that $f(u)+f(\bar{u})$ is a strictly concave function of $u\in(0,1)^3$ since in this case if a maximum exists it would be unique. This would force $u^*=\bar{u}^*$ or equivalently $x^*=1-x^*,y^*=1-y^*,z^*=1-z^*$ implying $x^*=y^*=z^*=1/2$.