Let $x_1,x_2,\dots,x_n$ be vectors of unit norm in a Euclidean space. Show that the number of unordered pairs $(i,j)$ for which $|x_i + x_j| < 1$ is at most $\lfloor{n^2\over4}\rfloor$
A hint to its answer suggests to
Show that if $x_1,x_2,x_3$ all have unit norm then $|x_i + x_j|\geqslant1$ for some $1\leqslant i<j\leqslant3$
I still can't seem to figure it out.
Consider a graph $G$ whose vertices are the given vectors $x_1,\dots, x_n$ and vertices $x_i$ and $x_j$ are adjacent iff $|x_i+x_j|<1$. The hint claim implies that the graph $G$ has no cycles of length $3$. Then by Turán's or even Mantel's theorem, the graph $G$ has at most $\lfloor{n^2\over4}\rfloor$ egdes, which is equivalent to the required claim. Its bound is attained, for instance, when $x_1=\dots=x_{ \lfloor{n\over 2}\rfloor}=e$ and $ x_{ \lfloor{n\over 2}\rfloor}=\dots=x_n$ where $e$ is a fixed vector of the unit length.
Proof of the hint claim. Assume the converse, that is that $|x_i + x_j|<1$ for each $1\leqslant i<j\leqslant3$. This means
$(x_i+x_j, x_i+x_j)<1$
$(x_i, x_i)+ (x_j x_j)+2(x_i, x_j)<1$
$(x_i, x_j)<-1/2$.
Since $(x_i, x_j)=\|x_i\|\|x_i\|\cos \alpha_{ij}=\cos \alpha_{ij}$, where $\alpha_{ij}$ is an angle between vectors $x_i$ and $x_j$, we see that $\alpha_{ij}>2\pi/3$ for each $1\leqslant i<j\leqslant3$. That is $\alpha_{12}>2\pi/3$, $\alpha_{13}>2\pi/3$, and $\alpha_{23}>2\pi/3$, which is impossible, because $\alpha_{12}+\alpha_{13}+\alpha_{23}\le 2\pi$ or because $$\sum (x_i,x_i)-2\sum_{i<j} (x_i,x_j) =\left(\sum x_i, \sum x_i\right)>0. $$