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

\begin{document}
\title{MR J560}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 91}
\maketitle

\section*{Problem}
Show that for all positive reals $a,b,c$,
\[ \frac{2}{a^2}+\frac{5}{b^2}+\frac{45}{c^2}
  > \frac{16}{(a+b)^2}+\frac{24}{(b+c)^2}+\frac{48}{(c+a)^2}. \]

\section*{Video}
\href{https://www.youtube.com/watch?v=xrAkxr6mB9w&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/xrAkxr6mB9w}}



\newpage

\section*{Solution}
By Holder inequality we generally have
\[ \frac{u}{x^2} + \frac{v}{y^2}
    \ge (u^{1/3} + v^{1/3})^3 \cdot \frac{1}{(x+y)^2}
  \]
for any $u$, $v$, $x$, $y$.
Therefore, we will be done if we can find choices
$0 \le r \le 2$, $0 \le s \le 5$, $0 \le t \le 45$
such that
\begin{align*}
  r^{1/3} + (5-s)^{1/3} &\ge 2\sqrt[3]{2} = \sqrt[3]{16} \\
  s^{1/3} + (45-t)^{1/3} &\ge 2\sqrt[3]{3} = \sqrt[3]{24} \\
  t^{1/3} + (2-r)^{1/3} &\ge 2\sqrt[3]{6} = \sqrt[3]{48}
\end{align*}
with at least one inequality being strict.

The official solution uses the choices $(r,s,t)=(1,1,36)$
in which case each of the inequalities above is true by AM-GM.
\begin{remark*}
More blunt approaches exist by simply randomly guessing, say:
\begin{align*}
  r &= 2 \cdot 0.9^3 = 1.458 \\
  s &= 5 - (1.1 \cdot \sqrt[3]{2})^3 = 5 - 2.662 = 2.338 \\
  t &= 45 - (\sqrt[3]{24} - \sqrt[3]{2.338})^3 \\
  &= 45 - 24 + 2.338 + 3\sqrt[3]{24^2 \cdot 2.338} - 3\sqrt[3]{24 \cdot 2.338^2} \\
  &= 23.338 + 3\sqrt[3]{576 \cdot 2.338} - 3\sqrt[3]{24 \cdot 2.338^2} \\
  &> 23.338 + 3\sqrt[3]{1346.688} - 3\sqrt[3]{138.24} \\
  &> 23.338 + 33 - 16 = 39.338 \\
  t^{1/3} + (2-r)^{1/3} &= \sqrt[3]{39.338} + \sqrt[3]{0.542}.
\end{align*}
So we wish to show
\[ \sqrt[3]{39.338} + \sqrt[3]{.542} > \sqrt[3]{48} \]
but $\sqrt[3]{39.338} > 3.4$ since $3.4^3 = 11.56 \cdot 3.4 = 39.304$,
and $\sqrt[3]{.542} > 0.8$,
while plainly $\sqrt[3]{48} < 4$.
\end{remark*}

\end{document}
