For a given natural number $n > 3$, the real numbers $x_1, x_2, \ldots, x_n, x_{n + 1}, x_{n + 2}$ satisfy the conditions $0 < x_1 < x_2 < \cdots < x_n < x_{n + 1} < x_{n + 2}$. Find the minimum possible value of \[\frac{(\sum _{i=1}^n \frac{x_{i + 1}}{x_i})(\sum _{j=1}^n \frac{x_{j + 2}}{x_{j + 1}})}{(\sum _{k=1}^n \frac{x_{k + 1} x_{k + 2}}{x_{k + 1}^2 + x_k x_{k + 2}})(\sum _{l=1}^n \frac{x_{l + 1}^2 + x_l x_{l + 2}}{x_l x_{l + 1}})}\] and find all $(n + 2)$-tuplets of real numbers $(x_1, x_2, \ldots, x_n, x_{n + 1}, x_{n + 2})$ which gives this value.
Problem
Source: China TST 2001, problem 4
Tags: algebra unsolved, algebra
Jumbler
28.10.2006 06:54
Anybody get this?
Rust
28.10.2006 09:01
Let $a_{i}=\frac{x_{i+1}}{x_{i}},i=1,2,...,n+1$, then \[X=\frac{s(s+b)}{(2s+b)(2t+c)},\] were $s=\sum_{i=1}^{n}a_{i}, \ \ t=\sum_{i=1}^{n}\frac{1}{a_{i}}, b=a_{n+1}-a_{1},c=\frac{1}{a_{n+1}}-\frac{1}{a_{1}}.$ We can chose $a_{1}=a_{n+1}$, then b=c=0 and $X=\frac{s}{4t}$. Minimal value for X we get when $a_{i}=a$, then $X=\frac{1}{4}a^{2}$, because $a\ge 1$, minimal value $X=\frac{1}{4}$.
MathPanda1
04.04.2016 22:58
If it holds for all $a_i$ equal, shouldn't the minimum be 1? And why can we choose $a_1=a_{n+1}$?
P-H-David-Clarence
17.10.2016 17:11
Using Rust 's notation,in fact we are looking for the minimum value of $$\dfrac {1}{\left( \sum ^{n}_{k=1}\dfrac {1}{\dfrac {1}{a_{k}}+\dfrac {1}{a_{k+1}}}\right) (\dfrac {1}{\sum ^{n}_{l=1}a_{l}}+\dfrac {1}{\sum ^{n}_{l=1}a_{l+1}})}$$for all $a_{i} > 1$. And I think it's not so easy.
P-H-David-Clarence
17.10.2016 17:27
And I think it's not so easy!
Skravin
17.10.2016 17:28
wow.... what a formula!