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．
【难度】
【出处】
2012年北京大学保送生试题
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【答案】
略
【解析】
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 $0<{a_i} < 1$．
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 $0<{a_i} < 1$．
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