Given an integer $n \geq 3$, determine the maximum value of product of $n$ non-negative real numbers $x_1,x_2, \ldots , x_n$ when subjected to the condition \begin{align*} \sum_{k=1}^n \frac{x_k}{1+x_k} =1 \end{align*}
Problem
Source: Balkan MO ShortList 2011 A2
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VicKmath7
01.07.2020 15:17
Solution?
XbenX
01.07.2020 17:25
Let $\frac{x_i}{x_i+1}=a_i$, then $a_1+a_2+\ldots +a_n=1$ and $x_i=\frac{a_i}{1-a_i}$. By Am-Gm we have $$\prod_{i=1}^{n}x_i=\frac{\prod_{i=1}^{n}a_i}{\prod_{i=1}^{n} \sum_{j\neq i} a_j}\leq \frac{\prod_{i=1}^{n}a_i}{\prod_{i=1}^{n}(n-1)\sqrt[n-1]{\prod_{j\neq i}a_j}}=\frac{1}{(n-1)^{n}}.$$
Psyduck909
31.08.2020 09:21
XbenX wrote: Let $\frac{x_i}{x_i+1}=a_i$, then $a_1+a_2+\ldots +a_n=1$ and $x_i=\frac{a_i}{1-a_i}$. By Am-Gm we have $$\prod_{i=1}^{n}x_i=\frac{\prod_{i=1}^{n}a_i}{\prod_{i=1}^{n} \sum_{j\neq i} a_j}\leq \frac{\prod_{i=1}^{n}a_i}{\prod_{i=1}^{n}(n-1)\sqrt[n-1]{\prod_{j\neq i}a_j}}=\frac{1}{(n-1)^{n}}.$$ veryy beautiful
IMO2022Goldinshallah
10.08.2021 07:23
Answer (n-1)^(-n)