EGMO 2023 | Problem 1 | Solution - 17 April 2023

Problem 1.
There are $n \ge 3$ positive real numbers $a_1, a_2, \dots, a_n$ . For each $1 \le i \le n$ we let $b_i = \frac{a_{i-1} + a_{i+1}}{a_i}$ (here we define $a_0$ to be $a_n$ and $a_{n+1}$ to be $a_1$ ). Assume that for all $i$ and $j$ in the range $1$ to $n$ , we have $a_i \le a_j$ if and only if $b_i \le b_j$ .
Prove that $a_1 = a_2 = \dots = a_n$ .

Solution
From the given condition we get that $(b_{i+1}-b_i)(a_{i+1}-a_{i})\geq 0$ for $i=1,..,n$ .
Hence $\sum_{i=1}^n(b_{i+1}-b_i)(a_{i+1}-a_{i})\geq 0\Leftrightarrow$
$\Leftrightarrow\sum_{i=1}^nb_{i+1}a_{i+1}-\sum_{i=1}^nb_{i+1}a_i-\sum_{i=1}^nb_ia_{i+1}+\sum_{i=1}^na_ib_i\geq 0$
$\Leftrightarrow 2\sum_{i=1}^na_i\dfrac{a_{i-1}+a_{i+1}}{a_i}\geq \sum_{i=1}^na_i\dfrac{a_i+a_{i+2}}{a_{i+1}}+\sum_{i=1}^na_{i+1}\dfrac{a_{i-1}+a_{i+1}}{a_i}\Leftrightarrow$
$\Leftrightarrow 4\sum_{i=1}^n a_i\geq \sum_{i=1}^n \dfrac{a_i^2+a_ia_{i+2}}{a_{i+1}}+\sum_{i=1}^n\dfrac{a_{i+2}^2+a_ia_{i+2}}{a_{i+1}}\Leftrightarrow$
$\Leftrightarrow 4\sum_{i=1}^n a_i\geq \sum_{i=1}^n \dfrac{(a_i+a_{i+2})^2}{a_{i+1}}\overset{C-S}{\geq }\dfrac{4\left (\displaystyle \sum_{i=1}^n a_i \right )^2}{\displaystyle \sum_{i=1}^n a_i}=4\sum_{i=1}^n a_i$
so $\dfrac{a_i+a_{i+2}}{a_{i+1}}$ must be constant, i.e $b_1=b_2=..=b_n$ .
Now since $b_i\geq b_j$ and $b_j\geq b_i$ for all $i,j$ we get $a_i\geq a_j$ and $a_j\geq a_i$ for all $i,j$ and the result follows.