Suppose we have random variable $X=\{x_1,\cdots,x_n\}$ with probability mass function $p$.
The entropy is defined by
$$H(X)=\sum_{i=1}^np(x_i)\log_b(p(x_i)^{-1})$$
where $b$ is any integer $\geq 2$.
Now suppose there is a $k$-partition $X=\bigcup_{j=1}^kX_j$, do we have the following inequality?
$$\frac{1}{n}\sum_{j=1}^k | X_j |\; H(X_j)\leq H(X)$$
i.e, the weighted mean of the entropy of these parts is less than the entropy of the whole.
In general, the inequation is false.
Take for example two sets, the first one with $M\gg 1$ elements and $p(x_i)= \epsilon\ll1/M$, the second with a single value. Then the global entropy $H(X)$ is near zero, while the weighted entropy will be dominated by the entropy of the first set, which is $\log_2 M$
Say: $M=1024$, $\epsilon=10^{-6}$. Then $H(X)=0.0214...$ bits, the weighted average is $9.99...$, and the inequality is false.
What is true is the following:
$$\sum \alpha_i H(x_i) = H(X) -H({\bf \alpha}) \le H(X)$$
where the coefficients of the weighted average ${\bf \alpha} = \{\alpha_i\}$ (with $\alpha_i\ge 0$ and $\sum \alpha_i = 1$) are given by $\alpha_i = \sum_{x_j \in X_i} p(x_j)$, i.e, they are not proportional to the support size of each set but to the accumulated probability of each set.