\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{JMO 2026/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 176}
\maketitle

\section*{Problem}
Let $a$, $b$, $c$ be distinct positive integers such that $ab+c=c^2$.
Prove that $(a-b)^2 \ge 4c$.



\section*{External Link}
\url{https://aops.com/community/p37578110}

\newpage

\section*{Solution}
Note that the conclusion is equivalent to
\[ (a-b)^2 \ge 4c \iff (a+b)^2 \ge 4c+4ab = 4c^2 \iff a+b \ge 2c. \]
We now solve the problem:
\begin{claim*}
  We have $a+b \ge 2c$.
\end{claim*}
\begin{proof}
  Because of distinctness, we may as well assume WLOG that
  \[ a > c > c-1 > b. \]
  Since the function
  \begin{align*}
    f \colon (\sqrt{c(c-1)}, \infty) & \to \RR \\
    x & \mapsto x + \frac{c(c-1)}{x}
  \end{align*}
  is strictly increasing,
  it follows that
  \[ a + b = f(a) > f(c) = 2c-1 \implies a+b \ge 2c. \qedhere \]
\end{proof}

\end{document}
