if $\sum_{i=1}^{n}\sum_{j=1}^{n}|1-x_{i}x_{j}|=\sum_{i=1}^{n}\sum_{j=1}^{n}|x_{i}-x_{j}|$ show that $\sum_{i=1}^{n}x_{i}=n$

Question

let $n>1$ is give postive integers,and $x_{i}>0,i=1,2,\cdots,n$,and such $$\sum_{i=1}^{n}\sum_{j=1}^{n}|1-x_{i}x_{j}|=\sum_{i=1}^{n}\sum_{j=1}^{n}|x_{i}-x_{j}|$$

show that $$\sum_{i=1}^{n}x_{i}=n$$ when $n=2$, since $$|1-x^2_{1}|+2|1-x_{1}x_{2}|+|1-x^2_{2}|=2|x_{1}-x_{2}|$$ (1):if $x_{1}>x_{2}>1$, then we have $$x^2_{1}-1+2(x_{1}x_{2}-1)+x^2_{2}-1=2(x_{1}-x_{2})$$ then $$x^2_{1}+x^2_{2}+2x_{1}x_{2}+2x_{2}=2x_{1}+4$$ it clear $$LHS=x^2_{1}+x^2_{2}+2x_{1}x_{2}+2x_{2}>2x_{1}+4$$ other case $x_{1}>1>x_{2}$ it seem not easy deal it,so The problem How to solve?Thanks

Answer 1

Thank the Gabriel Romon give this links,because this have equality: $x_{i},x_{i}\ge 0$,then $$\sum_{1\le i,j\le n}\min(x_{i}x_{j},y_{i}y_{j})\le \sum_{1\le i,j\le n}\min(x_{i}y_{j},x_{j}y_{i})$$ take $y_{i}=1$, and use $\min{(a,b)}=\dfrac{a+b-|a-b|}{2}$ we have $$2n\sum_{i=1}^{n}-\sum_{1\le i,j\le n}|x_{i}-x_{j}|\ge n^2+\left(\sum_{i=1}^{n}x_{i}\right)^2-\sum_{1\le i,j\le n}|1-x_{i}x_{j}|$$ takeing in to account that $$\sum_{1\le i,j\le n}|1-x_{i}x_{j}|=\sum_{1\le i,j\le n}|x_{i}-x_{j}|$$ we obtain $$2n\sum_{i=1}^{n}x_{i}\ge n^2+\left(\sum_{i=1}^{n}x_{i}\right)^2$$ which can be rewritten as $$\left(\sum_{i=1}^{n}x_{i}-n\right)^2\le 0$$ therefore $$\sum_{i=1}^{n}x_{i}=n$$First of all, let's admit that this is a pretty good solution, but if we don't know the inequality, then we can't solve it, I guess there's got to be some other transmutation technique we can use? or maybe this problem have exist other methods?Thanks