2023 CMO Problems/Problem 2
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Problem
Find the largest real number 
such that for any positive integer 
and any real numbers 
, the following inequality holds:
![\[\sum_{i=1}^n \sum_{j=1}^n(n-|j-i|) x_i x_j \geq C \sum_{i=1}^n x_i^2\]](http://latex.artofproblemsolving.com/9/b/0/9b064612e98d033faee6fa01a15cd663eb4c1140.png)
Solution 1
Define the 
matrix 
as follows:
![\[A_n=\left(\begin{array}{cccccc} n & n-1 & n-2 & \cdots & 2 & 1 \\ 1 & n & n-1 & \cdots & 3 & 2 \\ 2 & 1 & n & \cdots & 4 & 3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n-2 & n-3 & n-4 & \cdots & n & n-1 \\ n-1 & n-2 & n-3 & \cdots & 1 & n \end{array}\right)\]](http://latex.artofproblemsolving.com/7/b/1/7b18997f1e970086cc0f15649cb581e897aa71d3.png)
The problem simplifies to finding the smallest real number such that for all 
and any vector 
, the following inequality holds:
![\[x^T A_n x \geq C x^T x\]](http://latex.artofproblemsolving.com/9/6/8/968049c1b791004d6e4a0a87895176c9ac9d5212.png)
In other words, find the real number such that:
![\[C \leq \inf _{n \in \mathbb{N}^*} \inf _{\substack{x \in \mathbb{R}^n \\ x \neq 0}} \frac{x^T A_n x}{x^T x}\]](http://latex.artofproblemsolving.com/3/d/2/3d2e61e90d784129e2cff251b5609ac44b4ef69c.png)
Given that is not easily invertible directly, but is invertible (as it is a sparse matrix):
![\[A_n^{-1}=\left(\begin{array}{cccccc} \frac{n+2}{2 n+2} & \frac{1}{2} & 0 & \cdots & 0 & \frac{1}{2 n+2} \\ -\frac{1}{2} & 1 & \frac{1}{2} & \cdots & 0 & 0 \\ 0 & -\frac{1}{2} & 1 & \frac{1}{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & -\frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2 n+2} & 0 & \cdots & 0 & -\frac{1}{2} & \frac{n+2}{2 n+2} \end{array}\right)\]](http://latex.artofproblemsolving.com/d/9/b/d9b9d2bc2760d8a96033f6ac795e03b9a6862b73.png)
Since the inverse has non-zero entries:
![\[\inf _{\substack{x \in \mathbb{R}^n \\ x \neq 0}} \frac{x^T A_n x}{x^T x}=\left(\sup _{\substack{x \in \mathbb{R}^n \\ x \neq 0}} \frac{x^T A_n^{-1} x}{x^T x}\right)^{-1}\]](http://latex.artofproblemsolving.com/d/a/3/da3e80969ef7ba6722b29de31c3306a74400780f.png)
For the eigenvalues 
of :
![\[\frac{1}{2}+\frac{1}{2 n+2} \leq \mu-1 \leq \frac{1}{2}+\frac{1}{2 n+2}\]](http://latex.artofproblemsolving.com/e/a/8/ea853aa670803bda9679382c5cf004bc76327987.png)
Thus:
![\[0 \leq \mu \leq 2\]](http://latex.artofproblemsolving.com/4/d/1/4d1ca35fdddf39087f194873973ca2828ff287b3.png)
Given that 
is invertible, 
, therefore:
![\[\begin{gathered} 0<\mu \leq 2 \\ \sup _{\substack{x \in \mathbb{R}^n \\ x \neq 0}} \frac{x^T A_n^{-1} x}{x^T x} \leq 2 \end{gathered}\]](http://latex.artofproblemsolving.com/4/d/0/4d0bdf3077f7dc144b265c60f5a060af1185ede7.png)
For a specific ![$y \in \mathbb{R}^n, y=[1,-1,1,-1, \ldots]^T \in \mathbb{R}^n$](http://latex.artofproblemsolving.com/a/a/1/aa1ee09ca353fbdf42bf070b84cbd4e3afab2fd7.png)
, we have:
![\[\begin{gathered} y^T y=n \\ y^T A_n^{-1} y= \begin{cases}2(n-2)+\frac{2(n+2)}{n+1} & \text { if } n \text { is even } \\ 2(n-1) & \text { if } n \text { is odd }\end{cases} \\ \sup _n \sup _{\substack{x \in \mathbb{R}^n \\ x \neq 0}} \frac{x^T A_n x}{x^T x} \geq \sup _n \frac{y^T A_n y}{y^T y} \geq 2 \end{gathered}\]](http://latex.artofproblemsolving.com/c/9/b/c9b7a24f0b4122452656b6fd1e8efa2a61063064.png)
In conclusion:
![\[\inf _{\substack{x \in \mathbb{R}^n \\ x \neq 0}} \frac{x^T A_n x}{x^T x}=\inf _n\left(\sup _{\substack{x \in \mathbb{R}^n \\ x \neq 0}} \frac{x^T A_n^{-1} x}{x^T x}\right)^{-1}=\left(\sup _{\substack{n \\ n \in \mathbb{R}^n \\ x \neq 0}} \frac{x^T A_n^{-1} x}{x^T x}\right)^{-1}=\frac{1}{2}\]](http://latex.artofproblemsolving.com/4/7/0/470bdc8363e495beff43e8dc4c70f164b0f0a724.png)
Therefore, the maximum value of the real number is 
.
~dalian xes|szm
See also
| 2023 CMO(CHINA) (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All CMO(CHINA) Problems and Solutions | ||
