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Problem 1253

AIME late

Algebra Difficulty 5.0 Prove it China Girls' Mathematical Olympiad · China

Let $a_1, a_2, \dots, a_n$ be $n$ non-negative real numbers.

$\frac{1}{1+a_1} + \frac{a_1}{(1+a_1)(1+a_2)} + \dots + \frac{a_1a_2\dots a_{n-1}}{(1+a_1)(1+a_2)\dots(1+a_n)} \le 1.$

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $a_0 = 1$ . We prove the following identity:
$\sum_{k=1}^{n} \prod_{j=1}^{k} \frac{a_{j-1}}{1+a_j} = 1 - \prod_{j=1}^{n} \frac{a_j}{1+a_j} \qquad \textcircled{1}$
by induction on $n$ .

It is evident that $\textcircled{1}$ is true for $n = 1$ . Suppose that $\textcircled{1}$ is true for $n-1$ , $n \ge 2$ , then for $n$ ,
$\begin{aligned} \sum_{k=1}^{n} \prod_{j=1}^{k} \frac{a_{j-1}}{1+a_j} &= \sum_{k=1}^{n-1} \prod_{j=1}^{k} \frac{a_{j-1}}{1+a_j} + \prod_{j=1}^{n} \frac{a_{j-1}}{1+a_j} \\ &= 1 - \prod_{j=1}^{n-1} \frac{a_j}{1+a_j} + \prod_{j=1}^{n} \frac{a_{j-1}}{1+a_j} \\ &= 1 - \prod_{j=1}^{n} \frac{a_j}{1+a_j}. \end{aligned} \quad \square$

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.