We need to proove that
\[\begin{align}H(W) \geq 0 \end{align}\]Consider two auxiliary functions,
\[\begin{align} f(x) &= log(x) \\ g(x) &= x - 1 \end{align}\]Looking at the plots of the two equations,
Lets consider three situations, \(x = 1\), \(x > 1\) and \(0< x < 1\)
For \(x = 1\),
\[\begin{align} f(x) &= log(1) \\ &= 0 \\ g(x) &= 1 -1 \\ &= 0 \end{align} \\\]For \(x > 1\),
\[\\ \begin{align} g(x) & \ldots \text{is increasing along with x and greater than 0} \\ f(x) &\ldots \text{is increasing, but slower than $g(x)$ but still greater than 0} \end{align} \\\]For \(0 < x < 1\),
\[\begin{align} g(x) & \ldots \text{is decreasing along with x} \\ f(x) &\ldots \text{is decreasing, but faster than $g(x)$} \end{align}\]so,
\[\\ \begin{align} g(x) & \geq f(x) \ \space \ldots \text{ $\forall x > 0$}\\ x-1 & \geq log(x) \space \ldots \text{ $\forall x > 0$} \\ 1-x & \leq -log(x) \space \ldots \text{ $\forall x > 0 \ldots (1)$} \end{align} \\\]Now,
\[\\ \begin{align} H(W) &= - \sum_{i=1}^{W} p(w_{i}) \cdot log(p(w_{i}))\\ &= \sum_{i=1}^{W} p(w_i) \cdot (-log(p(w_{i)))}\\ &= \sum_{i=1}^{W}\underbrace{p(w_{i})}_\text{$ \geq 0$} \cdot \underbrace{(1 - p(w_{i}))}_\text{$ \geq 0$} \ldots \text{ from $(1)$} \end{align} \\\]