\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{USAMO 2021/5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 65}
\maketitle
\section*{Problem}
Let $n \ge 4$ be an integer.
Find all positive real solutions to the following
system of $2n$ equations:
\begin{align*}
a_1 &= \frac{1}{a_{2n}} + \frac{1}{a_{2}}, & a_2 &= a_1 + a_3, \\[1ex]
a_3 &= \frac{1}{a_{2}} + \frac{1}{a_{4}}, & a_4 &= a_3 + a_5, \\[1ex]
a_5 &= \frac{1}{a_{4}} + \frac{1}{a_{6}}, & a_6 &= a_5 + a_7, \\[1ex]
&\vdotswithin{=} &&\vdotswithin{=} \\
a_{2n-1} &= \frac{1}{a_{2n-2}} + \frac{1}{a_{2n}}, & a_{2n} &= a_{2n-1} + a_1.
\end{align*}
\section*{Video}
\href{https://www.youtube.com/watch?v=9WNgDETHOlI&t=1700}{\texttt{https://youtu.be/9WNgDETHOlI}}
\section*{External Link}
\url{https://aops.com/community/p21498967}
\newpage
\section*{Solution}
The answer is that the only solution is
$(1,2,1,2,\dots,1,2)$ which works.
We will prove $a_{2k}$ is a constant sequence,
at which point the result is obvious.
\paragraph{First approach (Andrew Gu).}
Apparently, with indices modulo $2n$, we should have
\[ a_{2k} = \frac{1}{a_{2k-2}}
+ \frac{2}{a_{2k}} + \frac{1}{a_{2k+2}} \]
for every index $k$ (this eliminates all $a_{\text{odd}}$'s).
Define
\[ m = \min_k a_{2k} \qquad\text{and}\qquad M = \max_k a_{2k}. \]
Look at the indices $i$ and $j$
achieving $m$ and $M$ to respectively get
\begin{align*}
m &= \frac2m + \frac{1}{a_{2i-2}} + \frac{1}{a_{2i+2}}
\ge \frac2m + \frac1M + \frac1M = \frac2m + \frac2M \\[1ex]
M &= \frac2M + \frac{1}{a_{2j-2}} + \frac{1}{a_{2j+2}}
\le \frac2M + \frac1m + \frac1m = \frac2m + \frac2M.
\end{align*}
Together this gives $m \ge M$, so $m = M$.
That means $a_{2i}$ is constant as $i$ varies,
solving the problem.
\paragraph{Second approach (author's solution).}
As before, we have
\[ a_{2k} = \frac{1}{a_{2k-2}}
+ \frac{2}{a_{2k}} + \frac{1}{a_{2k+2}} \]
The proof proceeds in three steps.
\begin{itemize}
\ii Define
\[ S = \sum_k a_{2k},
\quad\text{and}\quad
T = \sum_k \frac{1}{a_{2k}}. \]
Summing gives $S = 4T$.
On the other hand, Cauchy-Schwarz says $S \cdot T \ge n^2$,
so $T \ge \half n$.
\ii On the other hand,
\[ 1 = \frac{1}{a_{2k-2} a_{2k}}
+ \frac{2}{a_{2k}^2} + \frac{1}{a_{2k} a_{2k+2}} \]
Sum this modified statement to obtain
\[
n = \sum_k \left( \frac{1}{a_{2k}} + \frac{1}{a_{2k+2}} \right)^2
\overset{\text{QM-AM}}{\ge}
\frac 1n \left( \sum_k \frac{1}{a_{2k}} + \frac{1}{a_{2k+2}} \right)^2
= \frac 1n \left( 2T \right)^2
\]
So $T \le \half n$.
\ii Since $T \le \half n$ and $T \ge \half n$,
we must have equality everywhere above.
This means $a_{2k}$ is a constant sequence.
\end{itemize}
\begin{remark*}
The problem is likely intractable over $\CC$,
in the sense that one gets a high-degree polynomial
which almost certainly has many complex roots.
So it seems likely that most solutions must involve
some sort of inequality,
using the fact we are over $\RR_{>0}$ instead.
\end{remark*}
\end{document}