Let $\mathcal{T}_n$ be the set of ordered binary trees that have n leaves.
$d_T(v)$ means the node $v$'s depth in the tree T.
Prove: for any $T\in \mathcal{T}_n$ , for any $\{c_1,c_2,...c_n\}$ , $c_i >0$ , $S=\sum_i c_i$
$$ -\sum_{i=1}^n(c_i\cdot\log_2(c_i/S)) \le \sum_{i=1}^{n} (c_i\cdot d_T(v_i)) $$
$v_i$ means the i-th leaf in tree T.
My primary thought:
Since it has log2, so maybe 2^x can be a direction. Then 2 leaves have the same parent, their depth are the same. $2^{-depth1} + 2^{-depth} = 2^{-parent's depth}$