Let $X$ and $Y$ be discrete random variables. The conditional entropy of $Y$ is:
$$H(Y|X) = \sum_{x}P(x)H(Y|X=x)$$
Assume we quantize $X$ with a quantizer $Q(X)$. How could we prove the resulting conditional entropy is greater than the original conditional entropy?
$$H(Y|X) \leq H(Y|Q(X))$$
Here I just want to elaborate on what @stochasticboy321 commented. We know that $Q(X)$ is conditionally independent of $Y$ given $X$. So we have a Markov chain like this:
$$Y ----- X ----- Q(X)$$
Now, we are able to use The Data Processing Inequality(DPI) which states that given random variables X,Y and Z that form a Markov chain in the order X→Y→Z, then the mutual information between X and Y is greater than or equal to the mutual information between X and Z. Using this we have:
$$I(Y;X) \geq I(Y;Q(X))$$
By expanding the mutual information in terms of entropy we have:
$$H(Y) - H(Y|X) \geq H(Y)-H(Y|Q(X))$$
This leads us to our desired result: $$H(Y|Q(X)) \geq H(Y|X)$$