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

\begin{document}
\title{Twitch 118.1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 118}
\maketitle

\section*{Problem}
Prove that for real numbers $a \ge 0$,
\[ \sqrt{a^2 + 5 - \sqrt{a^2 + 6a}} + \sqrt{a^2 + 7a + 8} \geq 5. \]

\section*{Video}
\href{https://www.youtube.com/watch?v=W5EgJeHEIec&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/W5EgJeHEIec}}



\newpage

\section*{Solution}
Two approaches.

\paragraph{First solution (via Cauchy).}
Rearranging:
\begin{align*}
  \iff a^2 + 5 - \sqrt{a^2 + 6a} &\geq \left( 5 -\sqrt{a^2+7a+8} \right)^2 \\
  &= 25 - 10\sqrt{a^2+7a+8} + (a^2+7a+8) \\
  \iff 10\sqrt{a^2+7a+8} &\ge 28 + 7a + \sqrt{a^2+6a}.
\end{align*}
The latter claim follows by Cauchy-Schwarz in the form
\[
  \left( 49 + 1 \right)\left( (a+4)^2 + (a^2+6a) \right)
  \ge \left( 7(a+4) + \sqrt{a^2+6a} \right)^2.
\]

\paragraph{Second solution (geo).}
Let $b = \sqrt{a^2+6a}$; then this reads
\[ \sqrt{(a-3)^2+(b-1)^2} + \sqrt{(a+4)^2+b^2} \geq \sqrt{50} \]
which follows from a geometric argument.

\end{document}
