Let $n$ be a positive integer. If $x_1, x_2, \ldots, x_n \geq 0$, $x_1+x_2+\ldots+x_n=1$ and, assuming $x_{n+1}=x_1$, find the maximal value of $$\sum_{k=1}^n \frac{1+x_k^2+x_k^4}{1+x_{k+1}+x_{k+1}^2+x_{k+1}^3+x_{k+1}^4}.$$
Problem
Source: CGMO 2024 P7
Tags: inequalities
sqing
15.08.2024 02:08
Let $n$ be a positive integer. If $x_1, x_2, \ldots, x_n \geq 0, x_1+x_2+\ldots+x_n=1.$ Find the maximum value and the minimum value of $$\sum_{k=1}^n \frac{1+x_k^2+x_k^4}{1+x_{k+1}+x_{k+1}^2+x_{k+1}^3+x_{k+1}^4}.$$Where $x_{n+1}=x_1.$
a_507_bc wrote:
Let $n$ be a positive integer. If $x_1, x_2, \ldots, x_n \geq 0$, $x_1+x_2+\ldots+x_n=1$ and, assuming $x_{n+1}=x_1$, find the maximal value of $$\sum_{k=1}^n \frac{1+x_k^2+x_k^4}{1+x_{k+1}+x_{k+1}^2+x_{k+1}^3+x_{k+1}^4}.$$
invt
01.11.2024 17:00
does anyone have solution?
lbh_qys
03.12.2024 11:19
Since \[ (5 + 10a - 4b)(1+b+b^2+b^3+b^4) - 5(1+a^2 + a^4) = b+b^2+b^3+b^4-4b^5 + 5a(2b+2b^2+2b^3+2b^4 + 2-a^2-a^4) \geq 0, \] we have \[ \sum_{k=1}^n \frac{1+x_k^2 + x_k^4}{1+x_{k+1}+x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4} \leq \sum_{k=1}^n \left( 1 + 2x_k - \frac{4}{5}x_{k+1} \right) = n + 2 - \frac{4}{5} = n + \frac{6}{5}. \] Equality holds when \( n \geq 2\), \(x_1 = 1\) and \(x_2 = \cdots = x_n = 0\). And for \( n = 1 \) the maximal is \( \frac{3}{5} \).