This proof is from the book "The Probabilistic Method" Chapter 9 and has to do with quadratic residue tournaments but there are a few lines I cannot seem to follow.
The proof is at the bottom of page 153 here Lemma 9.1.2 http://nguyen.hong.hai.free.fr/EBOOKS/SCIENCE%20AND%20ENGINEERING/MATHEMATIQUE/PROBABILITY/The_Probabilistic_Method.pdf
The lines I am having the most trouble with are the first and third. I do not understand how they use the Cauchy Schwarz inequality on the double summation.A bit more detail on the proof overall would be greatly appreciated.
Let us denote \begin{align} a(i) =\sum_{j \in B}d_{ij} \end{align} Then observe we have \begin{align} \left(\sum_{i\in A}\sum_{j \in B} d_{ij}\right)^2 =&\ \left(\sum_{i \in A}a(i)\cdot 1 \right)^2 = \left(\left(\sum_{i \in A} a(i)^2\right)^{1/2}\left( \sum_{i \in A} 1\right)^{1/2}\right)^2 \\ =&\ |A| \sum_{i \in A}a(i)^2\\ \leq&\ |A| \sum_{i \in GF(p)} a(i)^2. \end{align}
Next, using the fact that $d_{ij}^2 = \chi(i-j)^2 = 1$ if $i\neq j$ and $0$ if $i=j$, then we see that \begin{align} a(i)^2 =&\ \left(\sum_{j \in B} d_{ij} \right) \left(\sum_{l \in B} d_{il} \right)= \sum_{j, l \in B} d_{ij}d_{il} = \sum_{j \in B} d_{ij}^2+ \sum_{l<j}d_{ij}d_{il}+\sum_{j<l}d_{ij}d_{il}\\ \leq&\ |B|+2\sum_{l<j}d_{ij}d_{il}. \end{align}
Using Fact 1, we have \begin{align} |A| \sum_{i \in GF(p)} a(i)^2 \leq&\ |A| \sum_{i \in GF(p)}\left(|B|+2\sum_{l<j}d_{ij}d_{il}\right) = |A||B|p+2|A|\sum_{j<l \in B}\sum_{i \in GF(p)}d_{ij}d_{il}\\ =&\ |A||B|p + 2|A|\sum_{j<l \in B} (-1). \end{align}
Lastly, observe \begin{align} \sum_{j<l \in B} 1 = \frac{|B|(|B|-1)}{2} \end{align} then it follows \begin{align} |A||B|p + 2|A|\sum_{j<l \in B} (-1) \leq |A||B|p - |A||B|(|B|-1) \leq |A||B| p. \end{align}