2021 IMO Problems/Problem 2


Show that the inequality \[\sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i-x_j|} \le \sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i+x_j|}\] holds for all real numbers $x_1,x_2,\dots,x_n$.


then, since \[\sqrt{x}\geq 0,   \sum_{i=1}^{n}\sum_{j=1}^{n}(\sqrt{x_i-x_j}^4)\leq \sum_{i=1}^{n}\sum_{j=1}^{n}(\sqrt{x_i+x_j}^4)\] then, \[\sum \sum x_i^2+x_j^2-2x_ix_j \leq \sum \sum x_i^2+x_j^2+2x_ix_j \to \sum \sum 4x_ix_j\geq 0,\] therefore we have to prove that \[\sum \sum a_ia_j\geq 0\] for every list $x_i$, and we can describe this to \[\sum \sum a_ia_j=\sum a_i^2 + \sum\sum a_ia_j(i\neq j)\] we know that \[\frac{a_i^2}{2}+\frac{a_j^2}{2} \geq |a_ia_j|\] therefore, \[a_i^2+a_j^2 \geq -(a_ia_j+a_ja_i)\] \[\to \sum a_i^2 + \sum\sum a_ia_j \geq 0\] \[Q.E.D.\] --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]


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