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

\begin{document}
\title{Putnam 1979 B6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 66}
\maketitle

\section*{Problem}
For $k=1,2,\dots, n$ let $z_k = x_k + iy_k$,
where the $x_k$ and $y_k$ are real and $i=\sqrt{-1}$.
Let $r$ be the absolute value of the real part of
\[ \pm\sqrt{z_1^2 + z_2^2 + \dotsb + z_n^2}. \]
Prove that $r\leq |x_1| + |x_2| + \dotsb + |x_n|$.

\section*{Video}
\href{https://www.youtube.com/watch?v=F1-jL2dSQJc&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/F1-jL2dSQJc}}

\section*{External Link}
\url{https://aops.com/community/p24910033}

\newpage

\section*{Solution}
If the square root is $a+bi$, and
$S = \sum x_k^2 \ge 0$ then
\begin{align*}
  (a^2-b^2) + 2ab \cdot i &= \sum_k (x_k^2-y_k^2) + \sum_k 2x_k y_k i \\
  a^2-b^2 &= \sum_k x_k^2 - y_k^2 \\
  ab &= \sum_k x_k y_k \le \sqrt{ \sum_k x_k^2 \sum_k y_k^2} \\
  &= \sqrt{\left( \sum_k x_k^2 \right)\left( \sum_k x_k^2 - (a^2-b^2) \right)} \\
  a^2b^2 &\le S^2 - (a^2-b^2)S \\
  0 &\le (S-a^2)(S+b^2).
\end{align*}
This means
\[ a^2 \le \sum_k x_k^2
  \implies a \le \sqrt{\sum_k x_k^2} \le \sum_k |x_k|. \]

\end{document}
