$\newcommand\Legendre{\genfrac(){}{}}$Let prime number $p\equiv 1\pmod 4$,and $x_{i}\ge 0$ such $$x_{1}+x_{2}+\dotsb+x_{p}=1.$$ 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 is an inequality that my colleague encountered when he was writing this paper,and We are can't prove this inequality , so I ask it,and we found the right constant seems to be the best because when $p = 5$ can be reached: $$p=5,x_{1}=x_{2}=0.5,x_{3}=x_{4}=\dotsb=0.$$

$\newcommand\Legendre{\genfrac(){}{}}$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: $$p=5,x_{1}=x_{2}=0.5,x_{3}=x_{4}=\dotsb=0.$$

$\newcommand\Legendre{\genfrac(){}{}}$Let prime number $p\equiv 1\pmod 4$,and $x_{i}\ge 0$ such $$x_{1}+x_{2}+\dotsb+x_{p}=1.$$ 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 is an inequality that my colleague encountered when he was writing this paper,and We are can't prove this inequality , so I ask it,and we found the right constant seems to be the best and we found that $p = 5$ can be reached: $$p=5,x_{1}=x_{2}=0.5,x_{3}=x_{4}=\dotsb=0.$$

$\newcommand\Legendre{\genfrac(){}{}}$Let prime number $p\equiv 1\pmod 4$,and $x_{i}\ge 0$ such $$x_{1}+x_{2}+\dotsb+x_{p}=1.$$ 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 is an inequality that my colleague encountered when he was writing this paper,and We are can't prove this inequality , so I ask it,and we found the right constant seems to be the best because when $p = 5$ can be reached: $$p=5,x_{1}=x_{2}=0.5,x_{3}=x_{4}=\dotsb=0.$$

$\newcommand\Legendre{\genfrac(){}{}}$Let prime number $p\equiv 1\pmod 4$,and $x_{i}\ge 0$ such $$x_{1}+x_{2}+\dotsb+x_{p}=1.$$ 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 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: $$p=5,x_{1}=x_{2}=0.5,x_{3}=x_{4}=\dotsb=0.$$

$\newcommand\Legendre{\genfrac(){}{}}$Let prime number $p\equiv 1\pmod 4$,and $x_{i}\ge 0$ such $$x_{1}+x_{2}+\dotsb+x_{p}=1.$$ 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 is an inequality that my colleague encountered when he was writing this paper,and We are can't prove this inequality , so I ask it,and we found the right constant seems to be the best and we found that $p = 5$ can be reached: $$p=5,x_{1}=x_{2}=0.5,x_{3}=x_{4}=\dotsb=0.$$