Prove Cycle Graphs are never hyper-energetic. The energy of a graph G, is defined as $$\varepsilon(G)=\sum_{i=1}^{n}\left|\lambda_i\right|$$
where $\lambda_i$ are the eigenvalues associated with the adjacency matrix of the Graph.
The energy of a Complete Graph $K_n$ is $\varepsilon(K_n)=2(n-1)$.
A Hyper-energetic graph is defined as a graph $G$ of order $n$ with $\varepsilon(G) \geq 2(n-1)$.
It is known that the eigenvalues of the Cycle Graph on $n$ vertices ($n \geq 3$), $C_n$, is $2\cos(\frac{2\pi k}{n})$ for $k=0,1,2,\ldots, n-1$. Thus the energy of $C_n$ is $$\varepsilon(C_n)=\sum_{k=0}^{n-1}\left|2\cos\left(\frac{2\pi k}{n}\right)\right|$$
Prove that for $n \geq 3$ $$\sum_{k=0}^{n-1}\left|2\cos\left(\frac{2\pi k}{n}\right)\right| \leq 2(n-1)$$

I know the sum is bounded by $2n$, and I believe using the roots of unity might have a play in it. But I'm not sure how to go about.
 A: $\left|\,\cos(x)\,\right|$ is a Lipschitz-continuous function on $[0,2\pi]$ with Lipschitz constant $1$, hence:
$$ \frac{2\pi}{n}\sum_{k=0}^{n-1}\left|\cos\left(\frac{2\pi k}{n}\right)\right|\leq\frac{2\pi}{n}+\int_{0}^{2\pi}\left|\,\cos(x)\,\right|\,dx=\frac{2\pi}{n}+4 $$
and by multiplying both sides by $\frac{n}{\pi}$ we get:

$$ \sum_{k=0}^{n-1}\left|2\cos\left(\frac{2\pi k}{n}\right)\right|\leq \color{red}{2+\frac{4}{\pi}\,n}$$

that is stronger than the given inequality for any $n\geq 6$. 
One just have to check by hand the cases $n\in\{3,4,5\}$.
As an alternative, we may exploit the Fourier cosine series of $\left|\,\cos(x)\,\right|$:
$$ \left|\,\cos(x)\,\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{m\geq 1}\frac{(-1)^m \cos(2mx)}{4m^2-1} $$
together with the fact that:
$$ \sum_{k=0}^{n-1}\cos\left(m\cdot\frac{2\pi k}{n}\right) = \text{Re}\sum_{k=0}^{n-1}\exp\left(m\cdot\frac{2\pi i k}{n}\right)=\left\{\begin{array}{rcl}n & \text{if} & m\mid n\\0 & \text{if} & m\nmid n \end{array}\right. $$
to compute the exact energy:
$$ \sum_{k=0}^{n-1}\left|\cos\left(\frac{2\pi k}{n}\right)\right|=\frac{2n}{\pi}-\frac{4n}{\pi}\sum_{m\geq 1}\frac{(-1)^{nm}}{4n^2 m^2-1}$$
that is:

$$ \sum_{k=0}^{n-1}\left|\cos\left(\frac{2\pi k}{n}\right)\right|=\left\{\begin{array}{rcl}\frac{1}{\sin\frac{\pi}{2n}}&\text{if}& n\text{ is odd}\\ \frac{1}{\tan\frac{\pi}{2n}}&\text{if}& n \text{ is even}.\end{array}\right.$$

