Let $\alpha_n \leq \alpha_{n-1} \leq \cdots \alpha_1$ be all eigenvalues of a graph $G$ (where the eigenvalues of $G$ is the eigenvalues of its adjacency matrix $A(G)$).
Let $n = |V(G)|$ be the number of vertices of $G$. Prove that an upper bound of the "energy of the graph $G$", definded by $\mathcal{E}(G):= \sum_{i=1}^{n}|\alpha_i|$, is $\frac{1}{2}n(\sqrt{n} + 1)$, i.e., $$\sum_{i=1}^{n}|\alpha_i|\leq\frac{1}{2}n(\sqrt{n} + 1)$$
I once use the following expression, if $m$ denotes the number of the edges of the graph, then using the Cauchy-Schwatz inequality, with the easy fact that the trace of $(A(G))^2$, $\mathrm{tr}((A(G))^2) = 2m$, we have the following coarse estimate: $$\mathcal{E}(G) \leq \sqrt{n} \sqrt{\sum_{i=1}^{n} \alpha_{i}^{2}}=\sqrt{n} \sqrt{2 m}=\sqrt{2 m n}\leq \sqrt{2\binom{n}{2}n} = n\sqrt{n-1}$$
but compared with $\frac{1}{2}n(\sqrt{n} + 1)$, it is bigger unless $n = 1$ or $2$ and I now face the difficulty.
Can anyone offer some help?