Solution
= Solution
For the distribution $Q_C$ from part (a),
$$
L(x)=-\log_2Q_C(x)-\log_2K.
$$
Therefore
$$
\mathbb E_P L(X)
=H(P)+D(P\Vert Q_C)-\log_2K
\geq H(P),
$$
by <Gibbs inequality> and $K\leq1$.
= Solution
For the distribution $Q_C$ from part (a),
$$
L(x)=-\log_2Q_C(x)-\log_2K.
$$
Therefore
$$
\mathbb E_P L(X)
=H(P)+D(P\Vert Q_C)-\log_2K
\geq H(P),
$$
by <Gibbs inequality> and $K\leq1$.