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