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

\begin{document}
\title{Twitch 037.3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 37}
\maketitle

\section*{Problem}
If $x+y=2$ for real numbers $x,y > 0$, prove that $\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$.

\section*{Video}
\href{https://www.youtube.com/watch?v=Rsr1cKKDljs&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/Rsr1cKKDljs}}



\newpage

\section*{Solution}
Square both sides and rewrite as
\begin{align*}
  \sqrt{x^2+3} + \sqrt{y^2+3} &\ge 6-\sqrt{xy+3} \\
  \iff x^2+y^2+6 + 2\sqrt{(x^2+3)(y^2+3)} &\ge 39+xy - 12\sqrt{xy+3} \\
  \iff x^2+y^2 + 12\sqrt{xy+3} + 2\sqrt{(x^2+3)(y^2+3)} &\ge 33+xy \\
  \iff 12\sqrt{xy+3} + 2\sqrt{(x^2+3)(y^2+3)} &\ge 29+3xy \\
  \iff 12\sqrt{xy+3} + 2\sqrt{x^2y^2 - 6xy + 21} &\ge 29+3xy.
\end{align*}
Let $t = xy$, so want $12\sqrt{t+3} + 2\sqrt{t^2-6t+21} \ge 29+3t$.
Our constraint on $t$ is $t \le 1$.
Squaring both sides, this is
\[ 144(t+3) + 4(t^2-6t+21) + 24\sqrt{(t^2-6t+21)(t+3)}
  \ge 9t^2 + 174t + 841 \]
which is equivalent to
\[ 48\sqrt{(t^2-6t+21)(t+3)} \ge 5t^2 + 54t + 325. \]
Squaring both sides again, we find this is equivalent to
\begin{align*}
  & 2304t^3 - 6912 t^2 + 6912t + 145152 \\
  &\qquad \ge 25t^4 + 540t^3 + 6166t^2 + 35100t + 105625 \\
  \iff
  0 &\ge 25t^4 - 1764 t^3 + 13078 t^2 + 28188 t - 39527 \\
  &= (t-1)(25t^3 - 1739t^2 + 11339 t + 39527)
\end{align*}
which is obviously true for $t \le 1$.

\end{document}
