有n個數值，由小到大排列是 a 1 ≦ a 2 ≦ a 3 ⋯ ≦ a n {\displaystyle a_{1}\leqq a_{2}\leqq a_{3}\cdots \leqq a_{n}} ，則必然 a 1 ≦ ∑ k = 1 n a k n ≦ a n {\displaystyle a_{1}\leqq {\frac {\sum _{k=1}^{n}a_{k}}{n}}\leqq a_{n}} 。 請問這是為什麼？謝謝！
∵ a 1 ⩽ a 2 {\displaystyle a_{1}\leqslant a_{2}} a 1 ⩽ a 3 {\displaystyle a_{1}\leqslant a_{3}} a 1 ⩽ a 4 {\displaystyle a_{1}\leqslant a_{4}} …… a 1 ⩽ a n {\displaystyle a_{1}\leqslant a_{n}} ∴ a 1 + a 1 + a 1 + . . . + a 1 ⩽ a 2 + a 3 + a 4 + . . . + a n {\displaystyle a_{1}+a_{1}+a_{1}+...+a_{1}\leqslant a_{2}+a_{3}+a_{4}+...+a_{n}} ∴ a 1 + ( a 1 + a 1 + a 1 + . . . + a 1 ) ⩽ a 1 + ( a 2 + a 3 + a 4 + . . . + a n ) {\displaystyle a_{1}+(a_{1}+a_{1}+a_{1}+...+a_{1})\leqslant a_{1}+(a_{2}+a_{3}+a_{4}+...+a_{n})} 即 n a 1 ⩽ ∑ k = 1 n a k {\displaystyle na_{1}\leqslant \sum _{k=1}^{n}a_{k}} ∴ a 1 ⩽ ∑ k = 1 n a k n {\displaystyle a_{1}\leqslant {\frac {\sum _{k=1}^{n}a_{k}}{n}}} 同理可证， a n ⩾ ∑ k = 1 n a k n {\displaystyle a_{n}\geqslant {\frac {\sum _{k=1}^{n}a_{k}}{n}}}
我的作法不太一樣，用反證法： 假設 a 1 > ∑ k = 1 n a k n {\displaystyle a_{1}>{\frac {\sum _{k=1}^{n}a_{k}}{n}}} ，則 a 1 > ∑ k = 1 n a k n {\displaystyle a_{1}>{\frac {\sum _{k=1}^{n}a_{k}}{n}}} a 2 > ∑ k = 1 n a k n {\displaystyle a_{2}>{\frac {\sum _{k=1}^{n}a_{k}}{n}}} (因為 a 2 ≧ a 1 > ∑ k = 1 n a k n {\displaystyle a_{2}\geqq a_{1}>{\frac {\sum _{k=1}^{n}a_{k}}{n}}} ) a 3 > ∑ k = 1 n a k n {\displaystyle a_{3}>{\frac {\sum _{k=1}^{n}a_{k}}{n}}} ⋯ {\displaystyle \cdots } a n > ∑ k = 1 n a k n {\displaystyle a_{n}>{\frac {\sum _{k=1}^{n}a_{k}}{n}}} 全部加起來， a 1 + a 2 + a 3 + ⋯ + a n > n × ∑ k = 1 n a k n {\displaystyle a_{1}+a_{2}+a_{3}+\cdots +a_{n}>n\times {\frac {\sum _{k=1}^{n}a_{k}}{n}}} 即 ∑ k = 1 n a k > ∑ k = 1 n a k {\displaystyle \sum _{k=1}^{n}a_{k}>\sum _{k=1}^{n}a_{k}} (自己大於自己) 矛盾，故 a 1 ≦ ∑ k = 1 n a k n {\displaystyle a_{1}\leqq {\frac {\sum _{k=1}^{n}a_{k}}{n}}} 另一邊同理。
