Let $x_1,x_2, ..., x_n$ be non-negative real numbers. Solve the system of equations: $$x_k+x_{k+1}=x^2_{k+2}\,\,,\,\,\, (k =1,2,...,n),$$where $x_{n+1} = x_1$, $x_{n+2} = x_2$.
Problem
Source: Bosnia and Herzegovina EGMO TST 2019 p1
Tags: algebra, system of equations
grupyorum
07.10.2022 23:08
Either $x_i=0$ for all $i$ or $x_i=2$ for all $i$. Let $x_i=\textstyle \max_{1\le j\le n}x_j$. Note that $x_{i-2}+x_{i-1}=x_i^2\ge x_{i+1}^2 = x_{i-1}+x_i$ implies $x_{i-2}\ge x_i$. Hence, $x_{i-2}=x_i$ due to maximality. Now, $x_{i-1}^2 = x_{i-2}+x_{i-3}$ and $x_{i-2}^2 = x_{i-3}+x_{i-4}$, together with $x_{i-2}$ being maximum, implies $x_{i-1}\ge x_{i-2}$, that is $x_{i-2}=x_{i-1}=x_i$. Iterating, we find all $x_i$ equal. Finally, setting $x_i=c$ we get $c^2=2c$ thus $c\in\{0,2\}$, as claimed.
bin_sherlo
13.04.2023 19:12
$x_1+x_2=x_3^2,x_2+x_3=x_4^2,...,x_n+x_1=x_2^2$ By adding this eqalities,
$2.\sum{x_i}=\sum{x_i^2}$
By Cauchy Schwarz
$\sum{x_i^2}.n \geq (\sum{x_i})^2=\frac{(\sum{x_i^2)^2}}{4} \implies 4n \geq \sum{x_i^2}=2.\sum{x_i} \implies 2n \geq \sum{x_i}$
Let $a_j$ be the minimum.
$x_{j-2}+x_{j-1}=x_j^2 \leq (\frac{x_{j-2}+x_{j-1}}{2})^2$ Then $
(x_{j-2}+x_{j-1})^2 \geq 4(x_{j-2}+x_{j-1})$
If $x_{j-2}+x_{j-1}=0$ then $x_j=0$ so $x_{j-1}=x_{j-2}=0$ then $x_{j-3}=x_{j-4}=0...$
$(0,0,...,0)$ is a solution.
If $+x_{j-2}+x_{j-1} \neq 0$
$x_j^2=x_{j-2}+x_{j-1} \geq 4$ so the minimum of this sequence $x_j \geq 2$
$x_1,x_2,...,x_k \geq 2$ so $x_1+x_2+...+x_n \geq 2n \geq x_1+x_2+...+x_n$
So $x_1=x_2=...=x_k=2$
$(2,2,...,2)$ is another solution.
Answer:$(0,0,...,0);(2,2,...,2)$.
ismayilzadei1387
22.06.2023 16:36
bin_sherlo wrote:
$x_1+x_2=x_3^2,x_2+x_3=x_4^2,...,x_n+x_1=x_2^2$ By adding this eqalities,
$2.\sum{x_i}=\sum{x_i^2}$
By Cauchy Schwarz
$\sum{x_i^2}.n \geq (\sum{x_i})^2=\frac{(\sum{x_i^2)^2}}{4} \implies 4n \geq \sum{x_i^2}=2.\sum{x_i} \implies 2n \geq \sum{x_i}$
Let $a_j$ be the minimum.
$x_{j-2}+x_{j-1}=x_j^2 \leq (\frac{x_{j-2}+x_{j-1}}{2})^2$ Then $
(x_{j-2}+x_{j-1})^2 \geq 4(x_{j-2}+x_{j-1})$
If $x_{j-2}+x_{j-1}=0$ then $x_j=0$ so $x_{j-1}=x_{j-2}=0$ then $x_{j-3}=x_{j-4}=0...$
$(0,0,...,0)$ is a solution.
If $+x_{j-2}+x_{j-1} \neq 0$
$x_j^2=x_{j-2}+x_{j-1} \geq 4$ so the minimum of this sequence $x_j \geq 2$
$x_1,x_2,...,x_k \geq 2$ so $x_1+x_2+...+x_n \geq 2n \geq x_1+x_2+...+x_n$
So $x_1=x_2=...=x_k=2$
$(2,2,...,2)$ is another solution.
Answer:$(0,0,...,0);(2,2,...,2)$.
Same idea , same inequality