Find the largest real number $c$ such that $$\sum_{i=1}^{n}\sum_{j=1}^{n}(n-|i-j|)x_ix_j \geq c\sum_{j=1}^{n}x^2_i$$for any positive integer $n $ and any real numbers $x_1,x_2,\dots,x_n.$
Problem
Source: Nov 28, 2023
Tags: inequalities, algebra, CMO, China
28.11.2023 15:59
\[\begin{aligned}\sum_{i=1}^n\sum_{j=1}^n(n-|i-j|)x_ix_j&=\sum_{i=1}^{n}(x_1+x_2+\cdots+x_i)^2+\sum_{i=2}^n(x_i+x_{i+1}+\cdots+x_n)^2 \\&=\frac{x_1^2+x_n^2}{2}+\sum_{i=1}^{n-1}\frac{1}{2}((x_1+\cdots+x_i)^2+(x_1+\cdots+x_{i+1})^2) \\&+\sum_{i=1}^{n-1}\frac{1}{2}((x_i+\cdots+x_n)^2+(x_{i+1}+\cdots+x_n)^2) \\&\geq\frac{x_1^2+x_n^2}{2}+\sum_{i=1}^{n-1}\frac{1}{4}x_{i+1}^2+\sum_{i=1}^{n-1}\frac{1}{4}x_i^2 \\&\geq\frac{1}{2}\sum_{i=1}^nx_i^2. \end{aligned}\]letting $n=2k+1,x_1=1,x_2=-2,x_3=2,x_4=-2,\ldots,x_{2k}=-2,x_{2k+1}=1$ gives $c\leq\frac{1}{2}$. so $c_{max}=\frac{1}{2}.$ also an exercise using Linear Algebra(Rayleigh-Ritz)
28.11.2023 16:28
The expression on the LHS seems quite similar to that in ISL 2015 A3 and hence the beginning of the above solution is common for the both problems.
28.11.2023 16:34
28.11.2023 17:57
The skill is very popular so it is not very difficult now, The main step is $$\sum_{1\le i<j\le n}(j-i)x_ix_j=\sum_{1\le i\le k<j\le n}x_ix_j=\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k}x_i\right)\left(\sum_{j=i+1}^{n}x_j\right)$$And let $S_j=x_1+x_2+\ldots +x_j$, the rest is just Cauchy.
29.11.2023 17:48
Let $s=\sum_{i=1}^{n}x_i$ and denote $y_k=\dfrac12\left(-\sum_{l=1}^{k}x_l+\sum_{l=k+1}^{n}x_l\right)$ with $y_0=-y_n=\dfrac s2$. \begin{align*} \sum_{i,j}(n-|i-j|)x_ix_j=&s^2+\sum_{k=1}^{n-1}\left((\sum_{l=1}^{k}x_l)^2+(\sum_{l=k+1}^{n} x_l)^2\right)\\=&s^2+\sum_{k=1}^{n-1}(2y_k^2+\dfrac{s^2}2)\\=&\dfrac{n-1}2s^2+\sum_{k=1}^{n}(y_{k-1}^2+y_k^2)\\\ge&0+\dfrac12\sum_{k=1}^{n}(y_{k-1}-y_k)^2=\dfrac12\sum_{k=1}^{n}x_k^2\end{align*}with equality case same as above.
29.11.2023 18:04