math-ai/BlueMO

1{2    "source_file": "./raw_volume-zh/volume4/exercise2.tex",3    "problem_type": "calculation",4    "problem": "问题17. 设 $n \\geqslant 2, x_1, x_2, \\cdots, x_n$ 为实数, 且 $\\sum_{i=1}^n x_i^2+\\sum_{i=1}^{n-1} x_i x_{i+1}=1$, 对每个给定的正整数 $k, 1 \\leqslant k \\leqslant n$, 求 $\\left|x_k\\right|$ 的最大值.",5    "solution": "由已知条件, 得 $2 \\sum_{i=1}^n x_i^2+2 \\sum_{i=1}^{n-1} x_i x_{i+1}=2$. 即 $x_1^2+\\left(x_1+x_2\\right)^2+ \\left(x_2+x_3\\right)^2+\\cdots+\\left(x_{n-2}+x_{n-1}\\right)^2+\\left(x_{n-1}+x_n\\right)^2+x_n^2=2$. 对给定的正整数 $k$, $1 \\leqslant k \\leqslant n$, 由平均值不等式, 得 $\\sqrt{\\frac{x_1^2+\\left(x_1+x_2\\right)^2+\\cdots+\\left(x_{k-1}+x_k\\right)^2}{k}} \\geqslant \\frac{\\left|x_1\\right|+\\left|x_1+x_2\\right|+\\cdots+\\left|x_{k-1}+x_k\\right|}{k} \\geqslant \\frac{\\left|x_1-\\left(x_1+x_2\\right)+\\cdots+(-1)^{k-1}\\left(x_{k-1}+x_k\\right)\\right|}{k}=\\frac{\\left|x_k\\right|}{k}$. 所以 $\\frac{x_1^2+\\left(x_1+x_2\\right)^2+\\cdots+\\left(x_{k-1}+x_k\\right)^2}{k} \\geqslant \\frac{x_k^2}{k^2}$, 即 $x_1^2+\\left(x_1+x_2\\right)^2 +\\cdots+\\left(x_{k-1}+x_k\\right)^2 \\geqslant \\frac{x_k^2}{k}$. 同理, 可得 $\\left(x_k+x_{k+1}\\right)^2+\\cdots+\\left(x_{n-1}+x_n\\right)^2+x_n^2 \\geqslant \\frac{x_k^2}{n-k+1}$. 将以上两式相加, 得 $\\left|x_k\\right| \\leqslant \\sqrt{\\frac{2 k(n+1-k)}{n+1}}(k=1,2, \\cdots, n)$, 当且仅当 $x_1=-\\left(x_1+x_2\\right)=\\left(x_2+x_3\\right)=\\cdots=(-1)^{k-1}\\left(x_{k-1}+x_k\\right)$ 及 $x_k+x_{k+1}=-\\left(x_{k+1}+x_{k+2}\\right)=\\cdots=(-1)^{n-k} x_n$ 时, 等号成立, 即当且仅当 $x_i= x_k(-1)^{i-k} \\frac{i}{k}(i=1,2, \\cdots, k-1)$ 且 $x_j=x_k(-1)^{j-k} \\frac{n+1--j}{n-k+1}(j=k+1$, $k+2, \\cdots, n)$ 时, $\\left|x_k\\right|=\\sqrt{\\frac{2 k(n+1-k)}{n+1}}$. 于是 $\\left|x_k\\right|_{\\text {max }}= \\sqrt{\\frac{2 k(n+1-k)}{n+1}}$.",6    "remark": "",7    "figures": []8}