数学奥林匹克问题解答: 新西兰数学奥林匹克第一题解答

Posted 2020-08-14 赵胤数学课

新西兰数学奥林匹克总计3道题目, 考试时间为90分钟.

 

证明不等式: $$\prod_{i = 1}^{n}\left(1 + {1 \over 3i - 2}\right) > \sqrt[3]{3n + 1}.$$ 解答一:

采用分析法证明之. 核心想法是先在不等式右边增加连乘符号, 然后两边同时去掉连乘, 最后使用均值不等式完成证明. $$\prod_{i = 1}^{n}\left(1 + {1 \over 3i - 2}\right) > \sqrt[3]{3n + 1}$$ $$\prod_{i = 1}^{n}\left(1 + {1 \over 3i - 2}\right)^3 > 3n + 1$$ $$\prod_{i = 1}^{n}\left({3i -1 \over 3i - 2}\right)^3 > 3n + 1$$ $$\prod_{i = 1}^{n}{(3i -1)^3 \over (3i - 2)^2} > (3n + 1)\prod_{i = 1}^{n}(3i - 2)$$ $$\prod_{i = 1}^{n}{(3i -1)^3 \over (3i - 2)^2} > (3n + 1)\cdot1\cdot4\cdots\cdot(3n-2)$$ $$\prod_{i = 1}^{n}{(3i -1)^3 \over (3i - 2)^2} > \prod_{i = 1}^{n + 1}(3i - 2)$$ $$\prod_{i = 1}^{n}{(3i -1)^3 \over (3i - 2)^2} > \prod_{i = 1}^{n}(3i + 1)$$ $${(3i -1)^3 \over (3i - 2)^2} > 3i + 1$$ $$(3i- 1)^3 > (3i + 1)(3i - 2)^2$$ $$3i - 1 > \sqrt[3]{(3i + 1)(3i - 2)^2}$$ 最后一步成立可由 $AM-GM$ 不等式证明: $$3i - 1 = {(3i + 1) + (3i - 2) + (3i - 2) \over 3} > \sqrt[3]{(3i + 1)(3i - 2)^2}.$$

解答二:

采用数学归纳法证明之.

当 $n = 1$ 时, $$1 + {1\over 3\cdot 1 - 2} = 2 = \sqrt[3]{8} > \sqrt[3]{3\cdot1 + 1} = \sqrt[3]{4}.$$ 假设 $n = j$ 时成立, 即 $$\prod_{i = 1}^{j}\left(1 + {1 \over 3i - 2}\right) > \sqrt[3]{3j + 1}.$$ 最后证明 $n = j+1$ 时亦成立: $$\prod_{i = 1}^{j+1}\left(1 + {1 \over 3i - 2}\right) = \prod_{i = 1}^{j}\left(1 + {1 \over 3i - 2}\right)\cdot\left(1 + {1 \over 3j + 1}\right)$$ $$> \sqrt[3]{3j+1}\cdot{3j+2 \over 3j+1}= {3j+2 \over \sqrt[3]{(3j+1)^2}}$$ 即需要证明 $${3j+2 \over \sqrt[3]{(3j+1)^2}} > \sqrt[3]{3j+4}.$$ 这可由 $AM-GM$ 不等式证明之: $$3j + 2 = {(3j+1) + (3j+1) + (3j+4) \over 3} > \sqrt[3]{(3j+1)^2(3j+4)}.$$

 

Q.E.D.