Maths Olympiad Prep

Library / UkraineMO /3 of 57

UkraineMO

Algebra Difficulty 4.6 AIME Prove it Ukraine

Let x1,x2,…,xnx_1, x_2, \dots, x_n be positive numbers such that:
x1n−1+x2n−1+⋯+xnn−1=x1x2…xn x_1^{n-1} + x_2^{n-1} + \dots + x_n^{n-1} = x_1 x_2 \dots x_n
Prove the inequality:
(x1−n+1)(x2−n+1)…(xn−n+1)≥1.(Serdiuk Nazar) (x_1 - n + 1)(x_2 - n + 1)\dots(x_n - n + 1) \ge 1. \quad (\text{Serdiuk Nazar})

Solution

Using Cauchy inequality we get ∀i=1,n\forall i=1, n:
x1x2…xn=x1n−1+x2n−1+⋯+xnn−1≥x1n−1+(n−1)x1x2…xn−1xn+1…xn⇒x1x2…xn−1xn+1…xn(xi−n+1)≥xin−1⇒xi−n+1≥xin−1x1x2…xn−1xn+1…xn⇒∏i=1n(xi−n+1)≥∏i=1nxin−1x1x2…xn−1xn+1…xn=1, \begin{align*} x_1 x_2 \dots x_n &= x_1^{n-1} + x_2^{n-1} + \dots + x_n^{n-1} \ge x_1^{n-1} + (n-1) x_1 x_2 \dots x_{n-1} x_{n+1} \dots x_n \\ &\Rightarrow x_1 x_2 \dots x_{n-1} x_{n+1} \dots x_n (x_i - n + 1) \ge x_i^{n-1} \Rightarrow x_i - n + 1 \ge \frac{x_i^{n-1}}{x_1 x_2 \dots x_{n-1} x_{n+1} \dots x_n} \\ &\Rightarrow \prod_{i=1}^n (x_i - n + 1) \ge \prod_{i=1}^n \frac{x_i^{n-1}}{x_1 x_2 \dots x_{n-1} x_{n+1} \dots x_n} = 1, \end{align*}
as needed.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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