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

\begin{document}
\title{CAMO 2021/6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 60}
\maketitle

\section*{Problem}
If $a,b,c > 0$ and $a+b+c=6$ prove that
\[\frac{a^2-4}{4a^2-9a+6}+\frac{b^2-4}{4b^2-9b+6}+\frac{c^2-4}{4c^2-9c+6}\le0.\]

\section*{Video}
\href{https://www.youtube.com/watch?v=te7Lc344aIs&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/te7Lc344aIs}}



\newpage

\section*{Solution}
First, note that
\begin{align*}
\sum_{\text{cyc}} \left[ \frac{3a^2-12}{4a^2-9a+6} + (2-a) \right]
&= \sum_{\text{cyc}} \frac{(2-a)(4a^2-9a+6)+3a^2-12}{4a^2-9a+6} \\
&= \sum_{\text{cyc}} \frac{-4a^3 + 20a^2 -24 a}{4a^2-9a+6} \\
&= \sum_{\text{cyc}} \frac{-4a(a-2)(a-3)}{4a^2-9a+6}
\end{align*}
So, the inequality amounts to saying that
\[
  \sum_{\text{cyc}} \frac{a(a-2)(a-3)}{4a^2-9a+6} \ge 0
  \qquad \forall a+b+c=6.
\]

Let $f(x) = \frac{x(x-2)(x-3)}{4x^2-9x+6}$,
and define the piecewise function $g \colon [0,6] \to \RR$ by
\[ g(x) = \begin{cases}
    x & x \in [0,2/3] \\
    -\half(x-2) & x \in [2/3, 16/7] \\
    \frac15(x-3) & x \in [16/7, 6].
  \end{cases}
\]
The graph of $f$ and $g$ are drawn below in red and blue respectively.
\begin{center}
\begin{asy}
  import graph;
  real f(real x) { return x*(x-2)*(x-3)/(4*x*x-9*x+6); }
  dot("$0$", (0,0), dir(225));
  dot("$2$", (2,0), dir(225));
  dot("$3$", (3,0), dir(-45));
  draw(graph(f, 0, 6, operator ..), red);
  draw( (0,0)--(2/3,2/3)--(2,0)--(16/7,-1/7)--(3,0)--(6,3/5), blue);
  draw( (0,3)--(0,0)--(6,0), Arrows);
\end{asy}
\end{center}
\begin{claim*}
  We have $f(x) \ge g(x)$ for all $x \in [0,6]$.
\end{claim*}
\begin{proof}
  Direct calculation.
\end{proof}
Hence, it suffices to show $g(a) + g(b) + g(c) \ge 0$ for all $a+b+c=6$.
The proof is divided in cases:
\begin{itemize}
  \ii If $\min(a,b,c) \ge \frac23$,
  then Jensen finishes.
  \ii If $a \le \frac23$ and $b,c \ge \frac23$,
  then by Jensen on the convex part of $g$, it is enough to
  show that $g(a) + 2g(3-a/2) \ge 0$, which is easy.
  \ii If $a,b \le \frac23$, then $c \ge 3$, so all terms are nonnegative.
\end{itemize}
This finishes the proof.

\end{document}
