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 $G=\frac{1}{2k\pi}$.