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: $$prod_{i = 1}^{n}left(1 + {1 over 3i - 2} ight) > sqrt[3]{3n + 1}.$$ 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. $$prod_{i = 1}^{n}left(1 + {1 over 3i - 2} ight) > sqrt[3]{3n + 1}$$ $$prod_{i = 1}^{n}left(1 + {1 over 3i - 2} ight)^3 > 3n + 1$$ $$prod_{i = 1}^{n}left({3i -1 over 3i - 2} ight)^3 > 3n + 1$$ $$prod_{i = 1}^{n}{(3i -1)^3 over (3i - 2)^2} > (3n + 1)prod_{i = 1}^{n}(3i - 2)$$ $$prod_{i = 1}^{n}{(3i -1)^3 over (3i - 2)^2} > (3n + 1)cdot1cdot4cdotscdot(3n-2)$$ $$prod_{i = 1}^{n}{(3i -1)^3 over (3i - 2)^2} > prod_{i = 1}^{n + 1}(3i - 2)$$ $$prod_{i = 1}^{n}{(3i -1)^3 over (3i - 2)^2} > prod_{i = 1}^{n}(3i + 1)$$ $${(3i -1)^3 over (3i - 2)^2} > 3i + 1$$ $$(3i- 1)^3 > (3i + 1)(3i - 2)^2$$ $$3i - 1 > sqrt[3]{(3i + 1)(3i - 2)^2}$$ 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, $$1 + {1over 3cdot 1 - 2} = 2 = sqrt[3]{8} > sqrt[3]{3cdot1 + 1} = sqrt[3]{4}.$$ 假设 $n = j$ 时成立, 即 $$prod_{i = 1}^{j}left(1 + {1 over 3i - 2} ight) > sqrt[3]{3j + 1}.$$ 最后证明 $n = j+1$ 时亦成立: $$prod_{i = 1}^{j+1}left(1 + {1 over 3i - 2} ight) = prod_{i = 1}^{j}left(1 + {1 over 3i - 2} ight)cdotleft(1 + {1 over 3j + 1} ight)$$ $$> sqrt[3]{3j+1}cdot{3j+2 over 3j+1}= {3j+2 over sqrt[3]{(3j+1)^2}}$$ 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.