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 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.
Update on OP request: Let us introduce the following functional $$ Z_m[G]=\int_{-k\pi}^{k\pi}\frac{A}{G(\omega)+A}e^{-im\omega}d\omega $$ The functional we are considering takes the form $$ I[G]=Z_0[G]-\frac{Z_1^*[G]Z_1[G]}{Z_0[G]}. $$ Now we have $$ \delta Z_m[G]=-\int_{k\pi}^{-k\pi}\frac{A}{(G(\omega)+A)^2}\delta G(\omega)e^{-im\omega}d\omega. $$ Chain rule applies also to functionals and we can evaluate $\delta I[G]$ immediately to give $$ \delta I[G]=\delta Z_0[G]-\frac{Z_1^*[G]Z_1[G]\delta Z_0[G]-Z_0[G]\delta(Z_1^*[G]Z_1[G])}{Z_0^2[G]} $$ and we see that the condition $\delta G(\omega)=0$ sets the variation to zero. This solution is consistent with the given constraint provided $G=\frac{1}{2k\pi}$. The application of the constarint a posteriori fixes the value of the constant.
Further clarification for OP: I will show that a functional that does not depend from at least a first derivative is a constant in one dimension. Let us consider the functional $$ S=\int_a^bL(q(t),q'(t),t)dt. $$ The condition for the extremum just gives $\delta S=0$ yielding Euler-Lagrange equation $$ \frac{d}{dt}\frac{\partial L}{\partial q'(t)}=\frac{\partial L}{\partial q(t)}. $$ Then, if there is no dependence on derivative we are left with $\frac{\partial L}{\partial q(t)}=0$ that implies immediately $L=L(t)$ and $q(t)=constant$.