Inequalities
(Canadian Mathematical Olympiad 2021 P2, AoPS) Let \(n \geq 2\) be some fixed positive integer and suppose that \(a_1, a_2, \ldots, a_n\) are positive real numbers satisfying \[ a_1 + a_2 + \cdots + a_n = 2^n - 1. \] Find the minimum possible value of \[ \frac{a_1}{1} + \frac{a_2}{1 + a_1} + \frac{a_3}{1 + a_1 + a_2} + \dots + \frac{a_n}{1 + a_1 + a_2 + \cdots + a_{n-1}}. \]
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Note that
\(\begin{align*} & \frac{a_1}{1} + \frac{a_2}{1 + a_1} + \frac{a_3}{1 + a_1 + a_2} + \dots + \frac{a_n}{1 + a_1 + a_2 + \cdots + a_{n-1}} \\ & = \frac{1 + a_1}{1} + \frac{1 + a_1 + a_2}{1 + a_1} + \frac{1 + a_1 + a_2 + a_3}{1 + a_1 + a_2} \\ & \quad + \dots + \frac{1 + a_1 + a_2 + \cdots + a_{n}}{1 + a_1 + a_2 + \cdots + a_{n - 1}} - n \\ & \geq n \sqrt[n]{1 + a_1 + a_2 + \cdots + a_n} - n\\ & = n \sqrt[n]{2^n} - n \\ & = n. \end{align*} \\)
This shows that the minimum possible value of the given expression is at least \(n\). Also note that if \(a_i = 2^{i - 1}\) for all \(1 \leq i \leq n\), then \(a_1 + a_2 + \cdots + a_n = 2^n - 1\) holds, and
\(\begin{align*} & \frac{a_1}{1} + \frac{a_2}{1 + a_1} + \frac{a_3}{1 + a_1 + a_2} + \dots + \frac{a_n}{1 + a_1 + a_2 + \cdots + a_{n-1}} \\ & = \frac{1 + a_1}{1} + \frac{1 + a_1 + a_2}{1 + a_1} + \frac{1 + a_1 + a_2 + a_3}{1 + a_1 + a_2} \\ & \quad + \dots + \frac{1 + a_1 + a_2 + \cdots + a_{n}}{1 + a_1 + a_2 + \cdots + a_{n - 1}} - n \\ & = 2n - n \\ & = n. \end{align*} \\)
This proves that the minimum possible value of the given expression is equal to \(n\).
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