|
|
Line 9: |
Line 9: |
| ==Solution== | | ==Solution== |
| {{solution}} | | {{solution}} |
− | \[
| + | have |
− | \text{We first make use of symmetry to rewrite the inequality as} | + | \begin{align*}\left(\sum_{i,j=1}^{n}|x_i-x_j|\right)^2 &=\left(2\sum_{1\le i\le j\le n}(x_j-x_i)\right)^2 \ |
− | \]
| + | &= \left((2n-2)x_n+(2n-6)x_{n-1}+\dots +(2-2n)x_1\right)^2 \ |
− | \[
| + | &\le ((2n-2)^2+(2n-6)^2+(2n-10)^2+\dots + (2-2n)^2)(x_1^2+x_2^2+\dots + x_n^2) \ |
− | \left(\sum_{1\le i<j\le n}|x_i-x_j|\right)^2\le\frac{n^2-1}3<br="">\left(\sum_{1\le i<j\le n}|x_i-x_j|^2\right)<br="">\] | + | &= \frac{4(n-1)(n)(n+1)}{3}(x_1^2+x_2^2+\dots + x_n^2) \ |
− | \[
| + | &= \frac{2(n^2-1)}{3}\cdot 2(nx_1^2 + nx_2^2 + \dots + nx_n^2) \ |
− | \text{WLOG that } x_1\le x_2\le\dots\le x_n \text{ and let } x_{i-1}-x_i=a_i.
| + | &= \frac{2(n^2-1)}{3}\cdot 2\left((n-1)\left(\sum_{i=1}^{n}{x_i^2}\right) + \left(\sum_{i=1}^{n}{x_i}\right)^2 - 2\sum_{1\le i<j\le n}x_ix_j\right) \ |
− | \]
| + | &= \frac{2(n^2-1)}{3}\cdot 2\left(\sum_{1\le i<j\le n}(x_i-x_j)^2\right) \\ |
− | \[
| + | &= \frac{2(n^2-1)}{3}\sum_{i,j=1}^{n}(x_i-x_j)^2 |
− | \text{The inequality is equivalent to}
| + | \end{align*} |
− | \]
| |
− | \[
| |
− | \left(\sum_{1\le i<j\le n}\left(a_i+\dots+a_{j-1}\right)="" \right)^2\le\frac{n^2-1}3<br="">\left(\sum_{1\le i<j\le n}\left(a_i+\dots+a_{j-1}\right)^2\right)="" <br="">\]
| |
− | \[
| |
− | \text{for all } a_1,\dots,a_{n-1}. \text{But this can be rewritten as}
| |
− | \]
| |
− | \[
| |
− | \left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right) \right)^2\le\frac{n^2-1}3
| |
− | \left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right)^2\right) | |
− | \]
| |
− | \text{By Cauchy-Schwarz:}
| |
− | \begin{align*}
| |
− | \left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right)^2\right)\left(\sum_{l=1}^{n-1}\sum_{j-i=l}l^2\right)&\ge\left(\sum_{l=1}^{n-1}\sum_{j-i=l}l\left(a_i+\dots+a_j\right)\right)^2\ &=\left(\sum_{l=1}^{n-1}l\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2
| |
− | \end{align*} | |
− | \text{We claim that } | |
− | \[
| |
− | \sum_{j-i=l}(a_i+\dots+a_j)=\sum_{j-i=(n-l)}(a_i+\dots+a_j).
| |
− | \]
| |
− | \text{Indeed, we may consider the } <math>l\times(n-l)</math> \text{ matrix:}
| |
− | \[
| |
− | \left( \begin{array}{cccc} | |
− | a_1 & a_2 & \dots & a_l \
| |
− | a_2 & a_3 & \dots & a_{l+1} \
| |
− | \vdots & \vdots & \ddots & \vdots\
| |
− | a_{n-l} & a_{n-l+1} & \dots & a_n
| |
− | \end{array} \right)
| |
− | \]
| |
− | \text{The first sum corresponds to summing the matrix row by row, and the second corresponds to summing it column by column. Thus the two sums are equal, as claimed.}
| |
− | \text{Hence:}
| |
− | \begin{align*}
| |
− | \left(\sum_{l=1}^{n-1}l\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2&=\left(\sum_{l=1}^{n-1}\frac n2\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2\ &=\frac{n^2}4\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2 | |
− | \end{align*} | |
− | \text{We may also check that }
| |
− | \[
| |
− | \sum_{l=1}^{n-1}\sum_{j-i=l}l^2=\sum_{l=1}^{n-1}(n-l)l^2=\frac{n^4-n^2}{12}.
| |
− | \]
| |
− | \text{Thus we have proven that } | |
− | \[
| |
− | \frac{n^4-n^2}{12}\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right)^2\right)\ge\frac{n^2}4\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2
| |
− | \]</j\le></j\le></j\le></j\le> | |
| | | |
| ==See Also== | | ==See Also== |
| | | |
| {{IMO box|year=2003|num-b=4|num-a=6}} | | {{IMO box|year=2003|num-b=4|num-a=6}} |
Revision as of 15:02, 3 December 2023
Problem
Let be a positive integer and let be real numbers. Prove that
with equality if and only if 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
See Also