You Meany! Topic: Algebra/Calculus. Level: Olympiad.

Problem: Problem-Solving Through Problems - 6.6.9) Let $ f(x) $ be differentiable on $ [0,1] $ with $ f(0) = 0 $ and $ f(1) = 1 $. For each positive integer $ n $ and arbitrary given positive numbers $ k_1, k_2, \ldots, k_n $, show that there exist distinct $ x_1, x_2, \ldots, x_n $ such that

$ \displaystyle \sum_{i=1}^n \frac{k_i}{f^{\prime}(x_i)} = \sum_{i=1}^n k_i $.

Solution: For convenience, define $ S_m = k_1+k_2+\cdots+k_m $ for $ m = 1, 2, \ldots, n $. Let $ a_i $, $ i = 1, 2, \ldots, n $, be a sequence of positive reals defined by: $ a_i $ is the smallest positive real satisfying $ f(a_i) = \frac{S_i}{S_n} $. The existence of each $ a_i $ is guaranteed by the Intermediate Value Theorem since $ 0 < a_i \le 1 $ for all $ i $. In addition, let $ a_0 = 0 $.

By the Mean Value Theorem, there exist $ x_i $ for $ i = 1, 2, \ldots, n $ such that

$ f^{\prime}(x_i) = \frac{f(a_i)-f(a_{i-1})}{a_i-a_{i-1}} = \frac{\frac{S_i-S_{i-1}}{S_n}}{a_i-a_{i-1}} = \frac{k_i}{S_n(a_i-a_{i-1})} $

because $ S_i-S_{i-1} = k_i $. Then

$ \frac{k_i}{f^{\prime}(x_i)} = S_n(a_i-a_{i-1}) $.

So

$ \displaystyle \sum_{i=1}^n \frac{k_i}{f^{\prime}(x_i)} = S_n(a_n-a_{n-1}+a_{n-1}-a_{n-2}+\cdots+a_1-a_0) = S_n(a_n-a_0) = S_n $

as desired, since $ a_n = 1, a_0 = 0 $. QED.

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Comment: Probably the best application of the Mean Value Theorem and Intermediate Value Theorem together I've ever seen. This problem is a really good exercise in thinking creatively to achieve the desired result.

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Practice Problem: (Problem-Solving Through Problems - 6.6.2) Let $ f: \mathbb{R} \to \mathbb{R} $ be a function such that for all reals $ x $ and $ y $, $ |f(x)-f(y)| \le (x-y)^2 $. Prove that $ f $ is a constant function.