Theorem: $e_2\leq (y-1)[\frac{n}{2}-y+1+i]$ Help with this proof:
DEF: Let G be a $(3,l)$-graph. We let $v_i=l−1−i$, and we define $s_i$ to be the number of vertices of G of degree $v_i$.
Let $G$ be a graph and $p$ a point of $G$. By $H_1$ we will mean the graph spanned by all points of $G$ which are joined to $p$ by an edge. $H_2$ will denote the graph spanned by all points different from $p$ which are not joined to $p$ by an edge. We will indicate this disection of $G$ by stating that $p$ is the preferred point.
Let $e_k$ be the number of edges in $H_k$ (for k = 1, 2).
DEF: Let $G$ be an $(x,y)$-graph and let $v$ be the minimum valence among all points of $G$; define $\sigma(G)=y-1−v$.
DEF: $G$ is an $(x,y)$-graph if $x>C(G)$ and $y>I(G)$, where, $I(G)$, the independence number of the graph G, is the maximum number of points of $G$ that can be chosen so that no two are joined by an edge and, C(G), the clique number of the graph G, is the maximum number of points in any complete subgraph of G.

Theorem:
If $G$ is a $(3, y)$-graph on $n$ points, $p$ a preferred point
of valence $v_i$, then
$$e_2=(y-1)[\frac{n}{2}-y+1+i]+\sum_{j=1}^{\sigma(G)}j(t_j-\frac{s_j}{2})$$
where $t_j$ the number of points in $H_1$ which have valence $v_j$ in G.

Theorem:
Let $G$ be a $(3, y)$-graph with two points of valence $v_i$
joined by an edge. Then by preferring one of these points we can obtain
$$e_2\leq (y-1)[\frac{n}{2}-y+1+i]$$
proof:
Get $p$ is one of the two given points. Let $p'$ be the other point and $t_j'$ be the number of points of valence $v_j$ in $H_1$ when $p'$ is preferred.
Since $p$ and $p'$ are joined by an edge, they cannot have edges to a
common point. Hence $t_j +t_j' \leq s_j$ for all $j$. Therefore
$$\sum_{j=1}^{\sigma(G)}j(t_j-\frac{s_j}{2})+\sum_{j=1}^{\sigma(G)}j(t_j'-\frac{s_j}{2})=\sum_{j=1}^{\sigma(G)}j(t_j+t_j'-s_j)\leq 0$$
The proposition follows.
I dont undestand why the proof ends here! 
the author assumes that $\sum_{j=1}^{\sigma(G)}j(t_j-\frac{s_j}{2})\leq 0$ 
But I do not understand why.. 
someone who can explain that part?
 A: To make the dependence on the preferred point explicit, write $e_2(p)$ for the number of edges in the graph $H_2$ obtained when $p$ is chosen as the preferred point. Let $M = (y-1)[\frac{n}{2}-y+1+i]$ be the upper bound we want to prove.
Then, assuming you're happy with the first theorem (the equation for $e_2(p)$) and with the inequality $t_j + t_j' \le s_j$, we have:
\begin{align}
   e_2(p) &= M + \sum_{j=1}^{\sigma(G)}j(t_j-\frac{s_j}{2}) \\
   e_2(p') &= M + \sum_{j=1}^{\sigma(G)}j(t_j'-\frac{s_j}{2})
\end{align}
and so by summing these two equations, 
\begin{align}
 e_2(p) + e_2(p') &= 2M + \sum_{j=1}^{\sigma(G)} \left(j(t_j-\frac{s_j}{2}) + j(t_j'-\frac{s_j}{2}) \right) \\
     &= 2M + \sum_{j=1}^{\sigma(G)} j \cdot (t_j + t_j' - s_j) \\
     &\le 2M + \sum_{j=1}^{\sigma(G)} j \cdot 0 = 2M.
\end{align}
where the inequality comes from the inequality $t_j + t_j' - s_j \le 0$.
Since $e_2(p) + e_2(p') \le 2M$, we must have either $e_2(p) \le M$ or $e_2(p') \le M$. This is an argument by contradiction: if we had $e_2(p) > M$ and $e_2(p') > M$, then we'd have $e_2(p) + e_2(p') > 2M$ by adding these inequalities.
