890

若 $0 < a < b < c < 1$ , $b \geq 2 a$ 或 $a + b \leq 1$ ，则 $max {b - a, c - b, 1 - c}$ 的最小值为_____.

解钱莘芪

首先注意到一个引理：因为 $max {f_{1}, f_{2}, f_{3}} \geq f_{i}, i = 1, 2, 3$ ,从而给定 $a_{1}, a_{2}, a_{3} > 0$ ,有 $max {f_{1}, f_{2}, f_{3}} \geq \frac{a_{1} f_{1} + a_{2} f_{2} + a_{3} f_{3}}{a_{1} + a_{2} + a_{3}} .$

回到题目.如果成立 $b \geq 2 a$ ，则 $\begin{array}{r} max {b - a, c - b, 1 - c} \geq max {\frac{b}{2}, c - b, 1 - c} \geq \frac{b + (c - b) + (1 - c)}{2 + 1 + 1} = \frac{1}{4} . \end{array}$
等号成立当且仅当 $(a, b, c) = (\frac{1}{4}, \frac{1}{2}, \frac{3}{4})$ .
如果成立 $a + b \leq 1$ ,则 $\begin{array}{r} max {b - a, c - b, 1 - c} \geq max {2 b - 1, c - b, 1 - c} \geq \frac{(b - \frac{1}{2}) + (c - b) + (1 - c)}{\frac{1}{2} + 1 + 1} = \frac{1}{5} . \end{array}$
等号成立当且仅当 $(a, b, c) = (\frac{2}{5}, \frac{3}{5}, \frac{4}{5})$ .

因此答案为两者中较小的 $\frac{1}{5}$ .