# Why is the cross-entropy always more than the entropy?

I understand intuitively why cross-entropy is always bigger. However, could someone show that mathematically?

Let's say you have two distributions $$p$$ and $$q$$. Cross entropy is: $$H(p,q)=-\sum_x{p(x)\log{q(x)}}$$. First, you'll manipulate it to obtain the very well known form: $$H(p,q)=H(p)+D_{KL}(p||q)$$, where $$D_{KL}(p||q)$$ is called the KL distance.
$$H(p,q)=-\sum_x{p(x)\log{\left(\frac{q(x)p(x)}{p(x)}\right)}}=-\sum_xp(x)\log{\left(\frac{q(x)}{p(x)}\right)}-\sum_x{p(x)\log{p(x)}}=D_{KL}(p||q)+H(p)$$
Then, it only remains to prove that $$D_{KL}(p,q)\geq 0$$, which can be done in various ways. The page I shared uses $$\log(x)\leq x-1$$: $$D_{KL}(p||q)\geq \sum_x{p(x)\left(1-\frac{q(x)}{p(x)} \right)}=\sum_x{p(x)}-\sum_x{q(x)}=1-\sum_x{q(x)}\geq 0$$
From the beginning, we assume that $$x$$ is in the support set of $$p(x)$$, i.e. $$p(x)$$ is non-zero. In the wikipedia entry, it says $$\sum_x{q(x)}=1$$, but I disagree with it, since support set of $$q(x)$$ can be different.