math-ai/BlueMO
BlueMO 🚀 BlueMO: A Comprehensive Collection of Challenging Mathematical Olympiad Problems from the Little Blue Book Series BlueMO is a comprehensive and challenging dataset comprising mathematical olympiad problems paired with detailed solutions, meticulously curated from the esteemed "Little Blue Book" (小蓝书) series (Second Edition)—a vital resource for Chinese students training for national and international olympiad math competitions. Designed to… See the full description on the dataset page: https://huggingface.co/datasets/math-ai/BlueMO.
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1{2 "source_file": "./raw_volume-zh/volume4/exercise2.tex",3 "problem_type": "calculation",4 "problem": "问题16. 设 $a>0, x_1, x_2, \\cdots, x_n \\in[0, a](n \\geqslant 2)$ 且满足\n$$\nx_1 x_2 \\cdots x_n=\\left(a-x_1\\right)^2\\left(a-x_2\\right)^2 \\cdots\\left(a-x_n\\right)^2 .\n$$\n求 $x_1 x_2 \\cdots x_n$ 的最大值.",5 "solution": "由平均值不等式, 得 $\\left(x_1 x_2 \\cdots, x_n\\right)^{\\frac{1}{2 n}}=\\left[\\left(a-x_1\\right)\\left(a-x_2\\right) \\cdots(a-\\right. \\left.\\left.x_n\\right)\\right]^{\\frac{1}{n}} \\leqslant a-\\frac{x_1+\\cdots+x_n}{n} \\leqslant a-\\left(x_1 \\cdots x_n\\right)^{\\frac{1}{n}}$. 令 $y=\\left(x_1 x_2 \\cdots x_n\\right)^{\\frac{1}{2 n}} \\geqslant 0$, 则有 $y \\leqslant a-y^2$, 即 $y^2+y-a \\leqslant 0$. 解不等式得 $0 \\leqslant y \\leqslant \\frac{-1+\\sqrt{4 a+1}}{2}$, 故 $x_1 x_2 \\cdots x_n$ 的最大值为 $\\left(\\frac{-1+\\sqrt{4 a+1}}{2}\\right)^{2 n}$.",6 "remark": "",7 "figures": []8}