The $a_n,b_n$ and $c_n$ are Fourier coefficients. I start by expressing $a_n$ and $b_n$ in terms of $c_n$ as follows: since for every complex number $z$, $|z|^2=z\overline{z}$ we have that
\begin{align}|a_n|^2+|b_n|^2&=a_n\overline{a_n}+b_n\overline{b_n}=(c_n+c_{-n})(\overline{c_n}+\overline{c_{-n}})+i(c_n-c_{-n})(-i)(\overline{c_n}-\overline{c_{-n}})\\ &=2c_n\overline{c_n}+2c_{-n}\overline{c_{-n}}=2\left(|c_n|^2+|c_{-n}|^2\right), \end{align}
so
$$\sum_{n\in\mathbb{Z}}|a_n|^2+|b_n|^2=2\sum_{n\in\mathbb{Z}}\left(|c_n|^2+|c_{-n}|^2\right).\tag1$$
Here I'm stuck. How do I get the $4|c_0|^2$ infront and how do I get rid of the terms $|c_{-n}|^2$ in the sum?
Assuming $a_n, b_n$ are the usual cosine and sine coefficients, they are defined only for n non-negative, with $a_0=2c_0, b_0=0$ and then the computation works - on the left n non-zero is positive, so on the right you group n and -n to get the sum over the non-zero integers, while the 0-coefficient has the right factor as above.