let  prime number $p\equiv 1\pmod 4$,and $x_{i}\ge 0$ such $$x_{1}+x_{2}+\cdots+x_{p}=1$$
show that
$$\sum_{1\le i<j\le p}\left(\dfrac{i-j}{p}\right)x_{i}x_{j}\le\dfrac{p-1}{2p+6}$$

Here $\left(\dfrac{\cdot}{\cdot}\right)$ is the Legendre symbol.

This problem is an inequality that my colleague encountered when he was writing this paper,and We are not going to prove this inequality yet, but it is interesting to find that the right constant seems to be the best and we found that P = 5 can be reached:
$$p=5,x_{1}=x_{2}=0.5,x_{3}=x_{4}=\cdots=0$$