Mantel's Theorem proof verification

Question

I found the following proof for Mantel's theorem in Lecture 1 of David Conlon's "Extremal graph theory" course. I cannot understand how the equality that I have highlighted in the proof quoted below was arrived at. I would appreciate some assistance.

Theorem 1 (Mantel's theorem) *If a graph $G$ on $n$ vertices contains no triangle then it contains at most $\frac {n^2}{4}$ edges.

First proof Suppose that $G$ has $m$ edges. Let $x$ and $y$ be two vertices in $G$ which are joined by an edge. If $d(v)$ is the degree of a vertex $v$, we see that $d(x)+d(y)\leq n$. This is because every vertex in the graph $G$ is connected to at most one of $x$ and $y$. Note now that $$\bbox[5px,border:2px solid red]{ \sum_x d^2(x)=\sum_{x,y\in E} \big( d(x)+d(y)\big) } \leq mn.$$ On the other hand, since $\sum_x d(x)=2m$, the Cauchy-Schwarz inequality implies that $$\sum_x d^2(x)\geq\frac{\big(\sum_x d(x)\big)^2}{n} \geq \frac{4m^2}{n}.$$ Therfore $$\frac{4m^2}{n} \leq mn,$$ and the result follows. $\tag*{$\square$}$

The part I don't understand is: $$\sum_x d^2(x)=\sum_{x,y\in E} \big( d(x)+d(y)\big).$$

Answer 1

Pick any vertex $x \in V$ of degree $k$ (that is, let $k = d(x)$). Then $G$ contains $k$ edges of the form: $$ xy_1, xy_2, \ldots, xy_k $$ where $y_1,y_2,\ldots,y_k \in V$. Hence, in the summation: $$ \sum_{xy \in E}(d(x) + d(y)) $$ we know that the term $d(x)$ will appear exactly $k$ times. In other words, the vertex $x$ will contribute an amount of: $$ \underbrace{d(x) + d(x) + \cdots + d(x)}_{k \text{ times}} = k \cdot d(x) = d(x) \cdot d(x) = d^2(x) $$ Thus, since $x$ was arbitrary, it follows that: $$ \sum_{xy \in E}(d(x) + d(y)) = \sum_{x \in V} d^2(x) $$ as desired.

Answer 2

In the sum $S:=\sum_{xy\in E}\bigl(d(x)+d(y)\bigr)$ the ends $x$ and $y$ of each edge $e=xy$ get $d(x)$, resp. $d(y)$, points each. This means that every vertex $x\in V$ gets $d(x)$ points for each edge having an end at $x$. It follows that $S=\sum_{x\in V}d^2(x)$.