prove or disprove $\sum_{i=1}^{n}x_{i}y_{i}\le 0$

Question

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 $$\sum_{i,j=1}^{n}x_{i}y_{j}=\sum_{i=1}^{n}x_{i}\sum_{j=1}^{n}y_{j}\le0$$

In your case $n=2$, shouldn't you be worrying about $x_1y_1 + x_2y_2$? — Naweed G. Seldon, May 05 '18 at 12:46
last I found when $n=2$ can't prove it.too.Now I wonder if there are any counterexample at this case — math110, May 05 '18 at 13:00
The inequality is alright, so you won’t get any counter examples... — Macavity, May 05 '18 at 13:15
@functionsug do you want to check for $\sum_{i,j}x_iy_j$ or $\sum_i x_i y_i$ ? — Abr001am, May 05 '18 at 13:43
An observation: since $x_i<y_i$
$2 \sum_i x_i < \sum_i(x_i+y_i)=0$

so $\sum_i x_i <0 $ and reverting the argument $\sum_i y_i >0$

$so \sum_i x_i \sum_j y_j <0$ — tired, May 05 '18 at 13:54
@functionsug that's interesting question,not simple to prove at all. — Abr001am, May 05 '18 at 14:58

Martin R · Accepted Answer · 2018-05-06T15:46:37.803

7

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 $i$.

edited May 06 '18 at 15:46

answered May 05 '18 at 15:23

Martin R

113,040

pretty (+1) ... – tired May 05 '18 at 15:29
some one helps to understand where $ny_1^2$ came from? thanks. I just want to understand the wit behind this. – Abr001am May 05 '18 at 15:46
@Abra001: Expanding the products in the last line gives a term $(n-1)y_1^2$ which must be subtracted again to make the equality hold. – Martin R May 05 '18 at 15:47
@MartinR that's imaginarily deft (+1) – Abr001am May 05 '18 at 15:50
1

Thanks all, but it was also a bit of luck. I got the result for $n=2$ first, and then (magically) it worked for arbitrary $n$ :) – Martin R May 05 '18 at 15:53

prove or disprove $\sum_{i=1}^{n}x_{i}y_{i}\le 0$

1 Answers1