Inspired by parv mor (and M. Fourier)

Since we are only interested in the integral of $f$ over $[0,2\pi]$ , we may as well assume (by repeating the values of $f$ for $x$ between $0$ and $2\pi$ ) that $f$ is periodic of period $2\pi$ , putting us squarely in M. Fourier's territory. This may mean that $f$ is no longer continuous at $0$ (and $2\pi$ ), but no matter. If we define the Fourier coefficients $c_n \; = \; \frac{1}{2\pi}\int_0^{2\pi} f(x) e^{-inx}\,dx \qquad \qquad n \in \mathbb{Z}$ then Parseval's Identity tells us that $\frac{1}{2\pi}\int_0^{2\pi} |f(x)|^2\,dx \; = \; \sum_{n \in \mathbb{Z}}|c_n|^2$ The conditions on the function $f$ tell us that $c_0=1$ , $c_1 = 1-i$ and $c_{-1} = 1+i$ (indeed, since $f$ is real-valued, $c_{-n} = c_n^\star$ for all integers $n$ ). It is clear that $\frac{1}{2\pi}\int_0^{2\pi} |f(x)|^2\,dx \; \ge \; |c_{-1}|^2 + |c_0|^2 + |c_1|^2 \; = \; \boxed{5}$ with this minimum value achieved when $f(x) \; = \; 1 + (1-i)e^{ix} + (1+i)e^{-ix} \; = \; 1 + 2\cos x + 2\sin x \;.$

For the sake of variety, let's discuss a solution using real Fourier Series. It is also possible to solve this problem without (explicitly) using Fourier series at all, as @Aditya Kumar suggests, but using Fourier analysis makes it conceptually clearer.

If $f(t)$ satisfies condition (G), then its Fourier polynomial $f_1(t)$ is $1+2\cos(t)+2\sin(t);$ see the coefficients $a_0,a_1,b_1$ here . Note that $\frac{1}{2\pi}\int_{0}^{2\pi}(f_1(t))^2dt=1+2+2=5$ . Now, by Pythagoras, $\frac{1}{2\pi}\int_{0}^{2\pi}(f(t))^2dt=\frac{1}{2\pi}\int_{0}^{2\pi}(f_1(t))^2dt+\frac{1}{2\pi}\int_{0}^{2\pi}(f(t)-f_1(t))^2dt\geq \boxed{5}$

We have $(f(t)-2\sin (t) -2\cos(t)-1)^2 \geq 0, \forall t \in \mathbb{R}$ Expanding the above, integrating the result in the range $0$ to $2 \pi$ , and using the given constraints, we directly arrive at the lower-bound.