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/chapter4-5.tex",3 "problem_type": "calculation",4 "problem": "例2. 设 $p, q \\in \\mathbf{R}^{+}, x \\in\\left(0, \\frac{\\pi}{2}\\right)$, 试求\n$$\n\\frac{p}{\\sqrt{\\sin x}}+\\frac{q}{\\sqrt{\\cos x}}\n$$\n的最小值.",5 "solution": "解:由柯西不等式, 得\n$$\n(\\sqrt{p m}+\\sqrt{q m})^2 \\leqslant\\left(\\frac{p}{\\sqrt{\\sin x}}+\\frac{q}{\\sqrt{\\cos x}}\\right)(m \\sqrt{\\sin x}+n \\sqrt{\\cos x}),\n$$\n当且仅当 $\\frac{\\frac{p}{\\sqrt{\\sin x}}}{m \\sqrt{\\sin x}}=\\frac{\\frac{q}{\\sqrt{\\cos x}}}{n \\sqrt{\\cos x}}$ 时, 等号成立.\n故\n$$\n\\tan x=\\frac{n p}{m q} .\n$$\n又 $(m \\sqrt{\\sin x}+n \\sqrt{\\cos x})^2==\\left(\\frac{m}{a} \\cdot a \\sqrt{\\sin x}+\\frac{n}{b} \\cdot b \\sqrt{\\cos x}\\right)^2$\n$$\n\\begin{aligned}\n& \\leqslant\\left(\\frac{m^2}{a^2}+\\frac{n^2}{b^2}\\right)\\left(a^2 \\sin x+b^2 \\cos x\\right) \\\\\n& \\leqslant\\left(\\frac{m^2}{a^2}+\\frac{n^2}{b^2}\\right) \\sqrt{a^4+b^4},\n\\end{aligned}\n$$\n当且仅当 $\\tan x=\\frac{a^2}{b^2}, \\frac{a^2 \\sin x}{\\frac{m^2}{a^2}}=\\frac{b^2 \\cos x}{\\frac{n^2}{b^2}}$ 时, 即 $\\tan x=\\frac{b^4 m^2}{a^4 n^2}=\\frac{a^2}{b^2}$ 时, 等号成立.\n故\n$$\n\\frac{m}{n}=\\frac{a^3}{b^3}, \\tan x=\\left(\\frac{m}{n}\\right)^{\\frac{2}{3}},\n$$\n且 $\\quad m \\sqrt{\\sin x}+n \\sqrt{\\cos x} \\leqslant\\left(m^{\\frac{4}{3}}+n^{\\frac{4}{3}}\\right)^{\\frac{3}{4}}$,\n从而\n$$\n\\begin{aligned}\n& \\left(\\frac{m}{n}\\right)^{\\frac{2}{3}}=\\frac{n p}{m q}, \\\\\n& \\frac{m}{n}=\\left(\\frac{p}{q}\\right)^{\\frac{3}{5}} .\n\\end{aligned}\n$$\n令 $m=p^{\\frac{3}{5}}, n=q^{\\frac{3}{5}}$, 得\n$$\n\\frac{p}{\\sqrt{\\sin x}}+\\frac{q}{\\sqrt{\\cos x}} \\geqslant \\frac{(\\sqrt{p m}+\\sqrt{n q})^2}{\\left(m^{\\frac{4}{3}}+n^{\\frac{4}{3}}\\right)^{\\frac{3}{4}}}=\\left(p^{\\frac{4}{5}}+q^{\\frac{4}{5}}\\right)^{\\frac{5}{4}},\n$$\n当且仅当 $\\tan x=\\left(\\frac{m}{n}\\right)^{\\frac{2}{3}}=\\left[\\left(\\frac{p}{q}\\right)^{\\frac{3}{5}}\\right]^{\\frac{2}{3}}=\\left(\\frac{p}{q}\\right)^{\\frac{2}{5}}$ 时, 等号成立.",6 "remark": "注:这里, 在两次利用柯西不等式时, 引进了参数 $n 、 m 、 a 、 b$.",7 "figures": []8}