已知 ${a_1} + {a_2} + {a_3} = {b_1} + {b_2} + {b_3}$，${a_1}{a_2} + {a_2}{a_3} + {a_3}{a_1} = {b_1}{b_2} + {b_2}{b_3} + {b_3}{b_1}$．
若已知 $\min\{a_{1},a_{2},a_{3}\}\leqslant \min\{b_{1},b_{2},b_{3}\}$，求证：$\max\{a_{1},a_{2},a_{3}\}\leqslant \max\{b_{1},b_{2},b_{3}\}$．
若已知 $\min\{a_{1},a_{2},a_{3}\}\leqslant \min\{b_{1},b_{2},b_{3}\}$，求证：$\max\{a_{1},a_{2},a_{3}\}\leqslant \max\{b_{1},b_{2},b_{3}\}$．
【难度】
【出处】
2008年北京大学自主招生保送生测试
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【答案】
略
【解析】
设 ${a_1} + {a_2} + {a_3} = {b_1} + {b_2} + {b_3} $ $ = \alpha $，$ {a_1}{a_2} + {a_2}{a_3} + {a_3}{a_1} = {b_1}{b_2} + {b_2}{b_3} + {b_3}{b_1}$ $ = \beta $，构造函数 $ f(x)$ 与 $ g(x)$：\[\begin{split} &f\left( x \right) = \left( {x - {a_1}} \right)\left( {x - {a_2}} \right)\left( {x - {a_3}} \right) = {x^3} - \alpha {x^2} + \beta x - {a_1}{a_2}{a_2},\\&g\left( x \right) = \left( {x - {b_1}} \right)\left( {x - {b_2}} \right)\left( {x - {b_3}} \right) = {x^3} - \alpha {x_2} + \beta x - {b_1}{b_2}{b_3}.\end{split}\]不妨设 $ {a_1} \leqslant {a_2} \leqslant {a_3} $，$ {b_1} \leqslant {b_2} \leqslant {b_3} $，则 $ {a_1} \leqslant {b_1} $，需要证明 $ {a_3} \leqslant {b_3}$．
根据已知，$$g\left( {{a_1}} \right) - f\left( {{a_1}} \right) = \left( {{a_1} - {b_1}} \right)\left( {{a_1} - {b_2}} \right)\left( {{a_1} - {b_3}} \right) \leqslant 0.$$若 ${a_3} > {b_3} \geqslant {b_2} \geqslant {b_1}$，则$$g\left( {{a_3}} \right) - f\left( {{a_3}} \right) = \left( {{a_3} - {b_1}} \right)\left( {{a_3} - {b_2}} \right)\left( {{a_3} - {b_1}} \right) > 0,$$矛盾．因此 ${a_3} \leqslant {b_3}$．
根据已知，$$g\left( {{a_1}} \right) - f\left( {{a_1}} \right) = \left( {{a_1} - {b_1}} \right)\left( {{a_1} - {b_2}} \right)\left( {{a_1} - {b_3}} \right) \leqslant 0.$$若 ${a_3} > {b_3} \geqslant {b_2} \geqslant {b_1}$，则$$g\left( {{a_3}} \right) - f\left( {{a_3}} \right) = \left( {{a_3} - {b_1}} \right)\left( {{a_3} - {b_2}} \right)\left( {{a_3} - {b_1}} \right) > 0,$$矛盾．因此 ${a_3} \leqslant {b_3}$．
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