maximum value of $H(X_1+X_2)$ Let $X_1$ & $X_2$ be two independent binary r.v.s, taking values in {0,1}; let $Y \triangleq X_1+X_2$ be their sum, i.e. a ternary r.v. taking values in {0,1,2}.  How do we prove that the maximum value of $H(Y)$ is 1.5 (bits)?  Note that with $X_1$ & $X_2$ being i.i.d. Bern($\frac{1}{2}$), a simple caculation shows that the entropy of the sum is 1.5 bits.  However, I'm having difficulty proving that this is indeed its maximum value, and would appreciate some help.  Thanks a lot!
 A: One approach is to formulate it as an analysis problem, as Max also suggested above.  A few tricks & observations will simplify the solution.
Let $X_1 \sim Bern(a), X_2 \sim Bern(b),$ and $Y \triangleq X_1+X_2,$ where $a,b \in [0,1].$  Defining $f(a,b) \triangleq H(Y),$ it follows that
$$f(a,b) = -(1-a)(1-b)log(1-a)(1-b) - ab\ log(ab) -(a+b-2ab)log(a+b-2ab)$$
First of all, $f$ is a continuous function on $S \triangleq [0,1]\times[0,1]$, which is a compact set.  So a maximum of $f$ must exist on $S.$
Secondly note that if $a = 0$ or $1$, then $f(a,b)=H(Y)=H(X_2)\leq 1$ bit.  Similarly, if $b=0$ or $1$, then $f(a,b)\leq 1$ bit.  So a maximum of $f$ does not occur on the boundary of $S$, because $f(0.5,0.5)=1.5$ bits$>1$ bit.  It must occur in $int\ S,$ i.e.$(0,1)\times(0,1)$.
To simplify the analysis,let's use natural log from this point on.  Since $f \in C'$ on $int\ S$, with 
$$f_a(a,b)=(1−b)log(1−a)(1−b)−(1−2b)log(a+b−2ab)−b\ log(ab),$$
$$f_b(a,b)=(1−a)log(1−a)(1−b)−(1−2a)log(a+b−2ab)−a\ log(ab).$$
It follows that a maximum must satisfy $\triangledown f = [0,0].$  Now note that
$$f_a(a,b)=0 \Leftrightarrow (1-b)log\frac{(1-a)(1-b)}{a+b-2ab}=b\ log\frac{ab}{a+b-2ab}$$
$$f_b(a,b)=0 \Leftrightarrow (1-a)log\frac{(1-a)(1-b)}{a+b-2ab}=a\ log\frac{ab}{a+b-2ab}$$
To solve this set of equations, it helps to note that the log's above are well-defined on $int\ S$.  Moreover, they cannot both be zero, because $X_1$ & $X_2$ are independent.  As a result, neither of them can be zero; otherwise the equalities above cannot hold.  This observation greatly simplifies the solution, because it implies that
$$\frac{a}{1-a}=\frac{b}{1-b},$$
which in turn implies $a=b.$
Substituting $a=b$ into $f_a(a,b)=0$, with some simplification, we have
$$g(a)\triangleq log\frac{1-a}{2a}+2a\ log2=0.$$
Note that $a=0.5$ is clearly a solution.  To prove that this is the only solution, it suffices to show that $g(a)$ is strictly decreasing on $(0,1).$  This can be easily seen by differentiating $g(a)$, and noting 
$$g'(a)=2\ log2-\frac{1}{a(1-a)}< 0,$$
for all $a \in (0,1).$  We therefore conclude $a=b=0.5$ is the only solution of $\triangledown f=[0,0]$ on $int\ S$. So $f(0.5,0.5) = 1.5$ (bits) is indeed the maximum.  This completes the proof.
