每日一题[867]“存在”的反面

已知$a_1,a_2,\cdots,a_{10}$为大于零的正实数，且

$${a_1} + {a_2} + \cdots + {a_{10}} = 30,{a_1}{a_2} \cdots {a_{10}} < 21,$$ 求证：$a_1,a_2,\cdots,a_{10}$中必有一个数在$(0,1)$之间．

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分析与解　用反证法，假设$\forall {a_i}\left( {i = 1,2, \cdots ,10} \right)$， ${a_i} \geqslant 1$．

令${a_i} = 1 + {b_i}\left( {i = 1,2, \cdots ,10} \right)$，则${b_i} \geqslant 0$，且${b_1} + {b_2} + \cdots + {b_{10}} = 20$．
所以

$$\begin{split}{a_1}{a_2} \cdots {a_{10}}& = \left( {1 + {b_1}} \right)\left( {1 + {b_2}} \right) \cdots \left( {1 + {b_{10}}} \right)\\&= 1 + {b_1} + {b_2} + \cdots + {b_{10}} + {b_1}{b_2} + {b_2}{b_3} + \cdots \\& = 21 + {b_1}{b_2} + {b_2}{b_3} + \cdots \\&\geqslant 21\end{split}$$ 与${a_1}{a_2} \cdots {a_{10}} < 21$矛盾．

所以$\exists {a_i}\left( {i = 1,2, \cdots ,10} \right)$，使$0<{a_i} < 1$．

注　事实上，不需要$a_i$是正实数的条件就可以证明存在一个数$a_i$小于$1$．