The easiest way to prove this is using variational calculus. You have to put
$$
  \delta I(G(\omega))=0.
$$
The calculation is quite straigthforward and provides the condition
$$
  \delta G(\omega)=0
$$
and so the extremum is for $G(\omega)=G=constant$. Finally, from the condition you have to set
$$
  \int_{-k\pi}^{k\pi}G(\omega)=2k\pi G=1.
$$
This gives the value of the extremum $G=\frac{1}{2k\pi}$.

**Expanded on OP request**: The idea behind  functional calculus (calculus of variations) is to consider a class of functionals, as in your case, that can be amenable to a generalized differentiation. You can find all the rules and the definition of a functional derivative [here][1] but for a more serious approach some lectures as the ones I pointed out in the comment area are needed. Your case is particularly simple as one is left in each term with the variation of $G(\omega)$ and this must be zero to find an extremum.


  [1]: http://en.wikipedia.org/wiki/Functional_derivative