math-ai/BlueMO

1{2    "source_file": "./raw_volume-zh/volume13/chapter1.tex",3    "problem_type": "calculation",4    "problem": "例5. 给定整数 $n \\geqslant 3$, 实数 $a_1, a_2, \\cdots, a_n$ 满足 $\\min _{1 \\leqslant i<j \\leqslant n}\\left|a_i-a_j\\right|=1$, 求 $\\sum_{k=1}^n\\left|a_k\\right|^3$ 的最小值.",5    "solution": "解:妨设 $a_1<a_2<\\cdots<a_n$, 则对 $1 \\leqslant k \\leqslant n$, 有\n$$\n\\left|a_k\\right|+\\left|a_{n-k+1}\\right| \\geqslant\\left|a_{n-k+1}-a_k\\right| \\geqslant|n+1-2 k|,\n$$\n所以 $\\sum_{k=1}^n\\left|a_k\\right|^3=\\frac{1}{2} \\sum_{k=1}^n\\left(\\left|a_k\\right|^3+\\left|a_{n+1-k}\\right|^3\\right)$\n$$\n=\\frac{1}{2} \\sum_{k=1}^n\\left(\\left|a_k\\right|+\\left|a_{n+1-k}\\right|\\right)\\left(\\frac{3}{4}\\left(\\left|a_k\\right|-\\left|a_{n+1-k}\\right|\\right)^2+\\right.\n$$\n$$\n\\begin{aligned}\n& \\left.\\frac{1}{4}\\left(\\left|a_k\\right|+\\left|a_{n+1-k}\\right|\\right)^2\\right) \\\\\n\\geqslant & \\frac{1}{8} \\sum_{k=1}^n\\left(\\left|a_k\\right|+\\left|a_{n+1-k}\\right|\\right)^3 \\\\\n\\geqslant & \\frac{1}{8} \\sum_{k=1}^n|n+1-2 k|^3 .\n\\end{aligned}\n$$\n当 $n$ 为奇数时, $\\sum_{k=1}^n|n+1-2 k|^3=2 \\cdot 2^3 \\cdot \\sum_{i=1}^{\\frac{n-1}{2}} i^3=\\frac{1}{4}\\left(n^2-1\\right)^2$;\n当 $n$ 为偶数时, $\\sum_{k=1}^n|n+1-2 k|^3=2 \\sum_{i=1}^{\\frac{n}{2}}(2 i-1)^3=2\\left(\\sum_{j=1}^n j^3-\\right. \\left.\\sum_{i=1}^{\\frac{n}{2}}(2 i)^3\\right)=\\frac{1}{4} n^2\\left(n^2-2\\right)$.\n所以, 当 $n$ 为奇数时, $\\sum_{k=1}^n\\left|a_k\\right|^3 \\geqslant \\frac{1}{32}\\left(n^2-1\\right)^2$;\n当 $n$ 为偶数时, $\\sum_{k=1}^n\\left|a_k\\right|^3 \\geqslant \\frac{1}{32} n^2\\left(n^2-2\\right)$, 等号均在 $a_i=i-\\frac{n+1}{2}, i= 1,2, \\cdots, n$ 时成立.\n因此, $\\sum_{k=1}^n\\left|a_k\\right|^3$ 的最小值为 $\\frac{1}{32}\\left(n^2-1\\right)^2$ ( $n$ 为奇数), 或者 $\\frac{1}{32} n^2\\left(n^2-2\\right)$ ( $n$ 为偶数).",6    "remark": "",7    "figures": []8}