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

\begin{document}
\title{KJMO 2012/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 19}
\maketitle

\section*{Problem}
Let $a$, $b$, $c$ be positive real numbers satisfying $ab+bc+ca=1$.
Prove that
\[
  \frac{a+b}{\sqrt{ab(1-ab)}}
  + \frac{b+c}{\sqrt{bc(1-bc)}}
  + \frac{c+a}{\sqrt{ca(1-ca)}} \le \frac{\sqrt{2}}{abc}
\]

\section*{Video}
\href{https://www.youtube.com/watch?v=lC1g-01Me-w&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/lC1g-01Me-w}}



\newpage

\section*{Solution}
Since $1-ab=c(a+b)$ the inequality rewrites as
\[ \sum_{\text{cyc}} \sqrt{a+b}
  \le \sqrt{\frac{2}{abc}}(ab+bc+ca) \]
or simply
  \[ \sum_{\text{cyc}} \sqrt{\frac{a+b}{2}}
    \le \sum_{\text{cyc}} \sqrt{\frac{ab}{c}} \]
If we let $x = \sqrt{\frac{bc}{a}}$, etc.,
then the inequality rewrites as
\[ \sum_{\text{cyc}} \sqrt{\frac{x(y+z)}{2}}
  \le x+y+z.  \]

This follows by using AM-GM
\[ \sqrt{\frac{x(y+z)}{2}} \le \half\left[ x + \frac{y+z}{2} \right] \]
and summing cyclically.

\end{document}
