let$x_{i},y_{i}(i=1,2,\cdots,n)$ be real numbers.and such $x_{1}\le x_{2}\le\cdots\le x_{n}\le y_{1}\le y_{2}\le\cdots\le y_{n}$ and $$\sum_{i=1}^{n}(x_{i}+y_{i})=0$$ prove or disprove $$\sum_{i=1}^{n}x_{i}y_{i}\le 0$$
I 've been thinking about it for a long time, and so far I haven't found any counterexamples, probably the Abel transformation, but it doesn't seem to be an easy solution.Of course, it could be very simple. A simple deformation, but I didn't think of it.
Case $n=1$ then $x_{1}=-y_{1}$,so $x_{1}y_{1}=-x^2_{1}\le 0$
Case $n=2$ then $x_{1}+x_{2}=-(y_{1}+y_{2})$ then we have $$2x_{1}x_{2}+x^2_{1}+x^2_{2}=2y_{1}y_{2}+y^2_{1}+y^2_{2}$$ then we have $$2(x_{1}x_{2}+y_{1}y_{2})=4y_{1}y_{2}+y^2_{1}+y^2_{2}-x^2_{1}-x^2_{2}$$ last I found when n=2 can't prove it.too.Now I wonder if there are any counterexample at this case –
ADD it:
tired comment that $\sum_{i=1}^{n}\sum_{j=1}^{n}x_{i}y_{j}\le 0$ because we have $$ 2\sum_{i=1}^{n}x_{i}\le\sum_{i=1}^{n}(x_{i}+y_{i})=0,2\sum_{i=1}^{n}y_{i}\ge\sum_{i=1}^{n}(x_{i}+y_{i})=0$$ so $$\sum_{i,j=1}^{n}x_{i}y_{j}=\sum_{i=1}^{n}x_{i}\sum_{j=1}^{n}y_{j}\le0$$
Substituting $$ x_1 = - y_1 - \sum_{i=2}^n (x_i + y_i) $$ we get $$ \sum_{i=1}^{n}x_{i}y_{i} = -y_1^2 - y_1 \sum_{i=2}^n (x_i + y_i) + \sum_{i=2}^{n}x_{i}y_{i} \\ = -n y_1^2 + \sum_{i=2}^n (x_i-y_1)(y_i-y_1) \le 0 $$
From here it is not difficult to see that equality holds if and only if $$ x_{k} = \ldots = x_n = 0 \\ y_1 = \ldots = y_{k-1} = 0 $$ for some $k$, $2 \le k \le n$, i.e. if and only if $x_iy_i=0$ for all $i$.