Difference between revisions of "2003 IMO Problems/Problem 5"

(Solution)
 
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==Solution==
 
==Solution==
We have
+
{{solution}}
 +
have
 
\begin{align*}\left(\sum_{i,j=1}^{n}|x_i-x_j|\right)^2 &=\left(2\sum_{1\le i\le j\le n}(x_j-x_i)\right)^2 \
 
\begin{align*}\left(\sum_{i,j=1}^{n}|x_i-x_j|\right)^2 &=\left(2\sum_{1\le i\le j\le n}(x_j-x_i)\right)^2 \
 
&= \left((2n-2)x_n+(2n-6)x_{n-1}+\dots +(2-2n)x_1\right)^2 \
 
&= \left((2n-2)x_n+(2n-6)x_{n-1}+\dots +(2-2n)x_1\right)^2 \

Latest revision as of 03:08, 26 March 2024

Problem

Let $n$ be a positive integer and let $x_1 \le x_2 \le \cdots \le x_n$ be real numbers. Prove that

\[\left( \sum_{i=1}^{n}\sum_{j=i}^{n} |x_i-x_j|\right)^2 \le \frac{2(n^2-1)}{3}\sum_{i=1}^{n}\sum_{j=i}^{n}(x_i-x_j)^2\]

with equality if and only if $x_1, x_2, ..., x_n$ form an arithmetic sequence.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. have (i,j=1n|xixj|)2=(21ijn(xjxi))2=((2n2)xn+(2n6)xn1++(22n)x1)2((2n2)2+(2n6)2+(2n10)2++(22n)2)(x12+x22++xn2)=4(n1)(n)(n+1)3(x12+x22++xn2)=2(n21)32(nx12+nx22++nxn2)=2(n21)32((n1)(i=1nxi2)+(i=1nxi)221i<jnxixj)=2(n21)32(1i<jn(xixj)2)=2(n21)3i,j=1n(xixj)2

See Also

2003 IMO (Problems) • Resources
Preceded by
Problem 4
1 2 3 4 5 6 Followed by
Problem 6
All IMO Problems and Solutions