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  Since this is a cubic with negative leading
  coefficient and constant term, it has a negative
  root, so it has at most two roots in $(0, 1)$.
  It follows (using Rolle's theorem) that $f$ has
  at most five roots in $[0, 1]$, up to
  multiplicity.

  But
  \[
    f'(x)
    = 2x(\log (1 - x^2) - \log x^2)
    + \phi(x \log x + (1 - x) \log (1 - x))
    - \phi x(\log(1 - x) - \log x)
  .\]
  So $f'(0) = 0$, so $f$ has a double root at $0$.

  We can also calculate (using $\psi^2 + \psi =
  1$):
  \begin{align*}
    f'(\psi)
    &= 2\psi(\log \psi - 2\log \psi) + \phi(\psi
    \log \psi + 2(1 - \psi)\log \psi)
    - (2\log \psi - \log \psi) \\
    &= -2\psi \log \psi + \log \psi + 2\phi \log
    \psi - 2\log \psi - \log \psi \\
    &= 2\log \psi(-\psi + \phi - 1) \\
    &= 2\phi \log \psi(-\psi^2 + 1 - \psi) \\
    &= 0
  \end{align*}
  So there's a double root at $\psi$.

  Also, note $f(1) = 0$.

  So $f$ is either non-negative on all of $[0, 1]$
  or non-positive on all of $[0, 1]$.

  If $x$ is small,
  \begin{align*}
    f(x)
    &= x^2 \log \frac{1}{x^2}
    + (1 - x^2) \log \frac{1}{1 - x^2}
    - \phi x \left( x \log \frac{1}{x} + (1 - x)
    \log \frac{1}{1 - x} \right) \\
    &= 2x^2 \log \frac{1}{x} - \phi x^2 \log
    \frac{1}{x} + O(x^2)
  \end{align*}
  so there exists $x$ such that $f(x) > 0$.
\end{proof}

\begin{fclemma}[]
  \label{lemma:5.3}
  The function $f(x, y) = \frac{\h(xy)}{x\h(y) +
  y\h(x)}$ is minimised on $(0, 1)^2$ at a point
  where $x = y$.
\end{fclemma}

\begin{proof}
  We can extend $f$ continuously to the boundary
  by setting $f(x, y) = 1$ whenever $x$ or $y$ is
  $0$ or $1$. To see this, note first that it's
  valid if neither $x$ nor $y$ is $0$.
  \begin{center}
    \includegraphics[width=0.15\linewidth]{images/49b9b6ad365d4367.png}
  \end{center}
  If either $x$ or $y$ is small, then
  \begin{align*}
    \h(xy)
    &= -xy (\log x + \log y)
    + O(xy) \\
    x\h(y) + y\h(x)
    &= -x(y \log y + O(y)) - y(x \log x + O(x)) \\
    &= h(x) + O(xy)
  \end{align*}
  So it tends to $1$ again.
  
  One can check that $f \left( \half, \half
  \right) < 1$, so $f$ is minimised somewhere in
  $(0, 1)^2$.

  Let $(x^*, y^*)$ be a minimum with $f(x^*, y^*)
  = \alpha$.

  Let $g(x) = \frac{h(x)}{x}$ and note that
  \[
    f(x, y) = \frac{g(xy)}{g(x) + g(y)}
  .\]
  Also,
  \[
    g(xy) - \alpha (g(x) + g(y)) \ge 0
  \]
  with equality at $(x^*, y^*)$. So the partial
  derivatives of LHS are both $0$ at $(x^*, y^*)$.
  \begin{align*}
    y^* g'(x^* y^*) - \alpha g'(x^*)
    &= 0 \\
    x^* g'(x^* y^*) - \alpha g'(y^*)
    &= 0
  \end{align*}
  So $x^* g'(x^*) = y^* g'(y^*)$. So it's enough
  to prove that $x g'(x)$ is an injection.
  $g'(x) = \frac{\h'(x)}{x} - \frac{\h(x)}{x^2}$,
  so
  \begin{align*}
    xg'(x)
    &= \h'(x) - \frac{\h(x)}{x} \\
    &= \log (1 - x) - \log x + \frac{x\log x + (1
    - x) \log (1 - x)}{x} \\
    &= \frac{\log(1 - x)}{x}
  \end{align*}
  Differentiating gives
  \[
    \frac{-1}{x(1 - x)} - \frac{\log(x - 1)}{x^2}
    = \frac{-x - (1 - x) \log(1 - x)}{x^2(1 - x)}
  .\]
  The numerator differentiates to $-1 + 1 + \log(1
  - x)$, which is negative everywhere. Also, it
  equals $0$ at $0$. So it has a constant sign.
\end{proof}

Combining this with \cref{lemma:5.2}, we get that
\[
  \h(xy)
  \ge \frac{\phi}{2} (x\h(y) + y\h(x))
.\]
This allows us to take $1 - \frac{1}{\phi} = 1 -
\frac{\sqrt{5} - 1}{2} = \frac{3 - \sqrt{5}}{2}$.

\newpage
\section{Entropy in additive combinatorics}

We shall need two ``simple'' results from additive
combinatorics due to Imre Ruzsa.

\begin{fcdefnstar}[Sum set / difference set / etc]
  Let $G$ be an abelian group and let $A, B \subset
  G$.

  The \emph{sumset} $A + B$ is the set $\{x + y :
  x \in A, y \in B\}$.

  The \emph{difference set} $A - B$ is the set
  $\{x - y : x \in A, y \in B\}$.

  We write $2A$ for $A + A$, $3A$ for $A + A + A$,
  etc.
\end{fcdefnstar}

\begin{fcdefnstar}[Ruzsa distance]
  \glssymboldefn{ruzd}%
  The \emph{Ruzsa distance} $d(A, B)$ is
  \[
    \frac{|A - B|}{|A|^{\half} |B|^{\half}}
  .\]
\end{fcdefnstar}

\begin{fclemma}[Ruzsa triangle inequality]
  \label{lemma:6.1}
  $\ruzd AC \le \ruzd AB + \ruzd BC$.
\end{fclemma}

\begin{proof}
  This is equivalent to the statement
  \[
    |A - C| |B|
    \le |A - B| |B - C|
  .\]
  For each $x \in A - C$, pick $a(x) \in A$, $c(x)
  \in C$ such that $a(x) - c(x) = x$. Define a map
  \begin{align*}
    \phi : (A - C) \times B
    &\to (A - B, B - C) \\
    (x, b)
    &\mapsto (a(x) - b, b - c(x))
  \end{align*}
  Adding the coordinates of $\phi(x, b)$ gives
  $x$, so we can calculate $a(x)$ (and $c(x)$)
  from $\phi(x, b)$, and hence can calculate $b$.
  So $\phi$ is an injection.
\end{proof}

\begin{fclemma}[Ruzsa covering lemma]
  \label{lemma:6.2}
  Assuming:
  - $G$ an abelian group
  - $A, B$ finite subsets of $G$
  Then:
  $A$ can be covered by at most $\frac{|A +
  B|}{|B|}$ translates of $B - B$.
\end{fclemma}

\begin{proof}
  Let $\{x_1, \ldots, x_k\}$ be a maximal subset
  of $A$ such that the sets $x_i + B$ are
  disjoint.

  Then if $a \in A$, there exists $i$ such that
  $(a + B) \cap (x_i + B) \neq \emptyset$. Then $a
  \in x_i + B - B$.

  So $A$ can be covered by $k$ translates of $B -
  B$. But
  \[
    |B|k
    = |\ub{\{x_1, \ldots, x_k\} + B}_{\subset A +
    B}|
    \le |A + B|
    .
    \qedhere
  \]
\end{proof}
