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

\begin{document}
\title{Shortlist 2006 A5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 6}
\maketitle

\section*{Problem}
If $a$, $b$, $c$ are the sides of a triangle, prove that
\[ \frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}
  + \frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}
  + \frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \le 3. \]

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

\section*{External Link}
\url{https://aops.com/community/p741368}

\newpage

\section*{Solution}
To deal with the long denominators,
let $x = \sqrt b + \sqrt c - \sqrt a$, etc.\ for brevity.
Note that $x > 0$, and similarly for the others.
Then we have that
\[ 0 < b+c-a = \left( \frac{x+y}{2} \right)^2
  + \left( \frac{x+z}{2} \right)^2 - \left( \frac{y+z}{2} \right)^2
  = x^2+xy+xz-yz \]
Thus the problem asks us to prove that
\[ \sum_{\text{cyc}} \frac{\sqrt{x^2+xy+xz-yz}}{x} \le 3\sqrt 2 \]
for positive real numbers $x$, $y$, $z$
satisfying the extra condition that all radicals are positive.

By Cauchy-Schwarz though, we have that
\[
  \left( \sum_{\text{cyc}} \frac{\sqrt{x^2+xy+xz-yz}}{x} \right)^2
  \le 3 \sum_{\text{cyc}} \frac{x^2+xy+xz-yz}{x^2} \]
and expanding fully, we see it suffices to show
\[ \sum_{\text{sym}} xy^2z^3 \le
  (xy)^3 + (yz)^3 + (zx)^3 + 3(xyz)^2 \]
which is Schur's inequality applied to $xy$, $yz$, $zx$.

\end{document}
