$\newcommand\Legendre{\genfrac(){}{}}$Let  $p\equiv 1\pmod 4$ be a prime number, and $x_{i}\ge 0$ be such that $$x_{1}+x_{2}+\dotsb+x_{p}=1.$$
Show that
$$\sum_{1\le i<j\le p}\Legendre{i-j}{p}x_{i}x_{j}\le\dfrac{p-1}{2p+6}$$

Here $\Legendre\cdot\cdot$ is the Legendre symbol.

This problem was encountered by a  colleague of mine when he was writing a paper, and we couldn't  prove this inequality. So I ask it. We found  the constant in the right-hand side seems to be the best one because when  $p = 5$ it can be reached:
$$p=5,x_{1}=x_{2}=0.5,x_{3}=x_{4}=\dotsb=0.$$