Difference between revisions of "2021 IMO Problems/Problem 2"
(→Video solutions) |
m (→Solution) |
||
| (7 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
| − | ==Problem== | + | == Problem == |
Show that the inequality | Show that the inequality | ||
<cmath>\sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i-x_j|} \le \sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i+x_j|}</cmath> | <cmath>\sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i-x_j|} \le \sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i+x_j|}</cmath> | ||
holds for all real numbers <math>x_1,x_2,\dots,x_n</math>. | holds for all real numbers <math>x_1,x_2,\dots,x_n</math>. | ||
| − | ==Video solutions== | + | == Solution == |
| + | then, since <cmath>\sqrt{x}\geq 0, | ||
| + | \sum_{i=1}^{n}\sum_{j=1}^{n}(\sqrt{x_i-x_j}^4)\leq \sum_{i=1}^{n}\sum_{j=1}^{n}(\sqrt{x_i+x_j}^4)</cmath> | ||
| + | then, | ||
| + | <cmath>\sum \sum x_i^2+x_j^2-2x_ix_j \leq \sum \sum x_i^2+x_j^2+2x_ix_j | ||
| + | \to \sum \sum 4x_ix_j\geq 0,</cmath> | ||
| + | therefore we have to prove that | ||
| + | <cmath>\sum \sum a_ia_j\geq 0</cmath> for every list <math>x_i</math>, | ||
| + | and we can describe this to | ||
| + | <cmath>\sum \sum a_ia_j=\sum a_i^2 + \sum\sum a_ia_j(i\neq j)</cmath> | ||
| + | we know that | ||
| + | <cmath>\frac{a_i^2}{2}+\frac{a_j^2}{2} \geq |a_ia_j|</cmath> | ||
| + | therefore, <cmath>a_i^2+a_j^2 \geq -(a_ia_j+a_ja_i)</cmath> | ||
| + | <cmath>\to \sum a_i^2 + \sum\sum a_ia_j \geq 0</cmath> | ||
| + | <cmath>Q.E.D.</cmath> | ||
| + | --[[User:Mathhyhyhye|Mathhyhyhy]] 13:29, 6 June 2023 (EST) | ||
| + | |||
| + | == Video solutions == | ||
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems] | https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems] | ||
https://youtu.be/akJOPrh5sqg [uses integral] | https://youtu.be/akJOPrh5sqg [uses integral] | ||
| + | |||
| + | https://www.youtube.com/watch?v=P9Ge8HAf6xk | ||
| + | |||
| + | == See also == | ||
| + | {{IMO box|year=2021|num-b=1|num-a=3}} | ||
Latest revision as of 05:11, 24 April 2024
Contents
Problem
Show that the inequality
![\[\sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i-x_j|} \le \sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i+x_j|}\]](http://latex.artofproblemsolving.com/6/2/6/6264c77345de1aea2a587da1100b58af528389ae.png)
holds for all real numbers 
.
Solution
then, since ![\[\sqrt{x}\geq 0, \sum_{i=1}^{n}\sum_{j=1}^{n}(\sqrt{x_i-x_j}^4)\leq \sum_{i=1}^{n}\sum_{j=1}^{n}(\sqrt{x_i+x_j}^4)\]](http://latex.artofproblemsolving.com/4/a/5/4a5f541df30d5df60b52ad7dee594f769c8008d2.png)
then,
![\[\sum \sum x_i^2+x_j^2-2x_ix_j \leq \sum \sum x_i^2+x_j^2+2x_ix_j \to \sum \sum 4x_ix_j\geq 0,\]](http://latex.artofproblemsolving.com/c/6/7/c67fc8a9bc1b1ce4747d189f3c74f195ee2125f2.png)
therefore we have to prove that
![\[\sum \sum a_ia_j\geq 0\]](http://latex.artofproblemsolving.com/8/3/c/83cec7b6308bfa46b4fdb55946477d7d7c82ea87.png)
for every list 
,
and we can describe this to
![\[\sum \sum a_ia_j=\sum a_i^2 + \sum\sum a_ia_j(i\neq j)\]](http://latex.artofproblemsolving.com/1/c/e/1ce4fb46048728bb733357570892c32578c0c0b8.png)
we know that
![\[\frac{a_i^2}{2}+\frac{a_j^2}{2} \geq |a_ia_j|\]](http://latex.artofproblemsolving.com/6/1/f/61f0dcafdcd7e863aa1465d38f9316d5c1075684.png)
therefore, ![\[a_i^2+a_j^2 \geq -(a_ia_j+a_ja_i)\]](http://latex.artofproblemsolving.com/a/e/8/ae8807e864a6aa0854a87b56573cee3c92ed4081.png)
![\[\to \sum a_i^2 + \sum\sum a_ia_j \geq 0\]](http://latex.artofproblemsolving.com/e/d/e/edebc20ea086aeb91fcf55781d4996aa928388d0.png)
![\[Q.E.D.\]](http://latex.artofproblemsolving.com/b/c/7/bc779cb200fbc15c503e7049874b39ad6f0d1f8a.png)
--Mathhyhyhy 13:29, 6 June 2023 (EST)
Video solutions
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]
https://youtu.be/akJOPrh5sqg [uses integral]
https://www.youtube.com/watch?v=P9Ge8HAf6xk
See also
| 2021 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
