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

\begin{document}
\title{MR U552}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 85}
\maketitle

\section*{Problem}
Find all polynomials with real coefficients $P$ which satisfy
\begin{align*}
  P(P(a+b))-2ab(2P(a+b)-ab) &\geq P(a^2)+P(b^2) \\
  &\geq P(a^2+b^2)-P(ab\sqrt{2})
\end{align*}
for all $a,b \in \RR$.

\section*{Video}
\href{https://www.youtube.com/watch?v=i1atXI9YdxE&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/i1atXI9YdxE}}



\newpage

\section*{Solution}
The answers are:
\begin{itemize}
  \ii $P(x) = cx^2$ for $-1 \le c \le 0$;
  \ii $P(x) = x^2$;
  \ii $P(x) = x^2 + \ell$ for $\ell \ge 1$.
\end{itemize}

\begin{claim*}
  We have $\deg P \le 2$.
\end{claim*}
\begin{proof}
  Assume that $\deg P > 2$.
  Comparing the leading terms when $a=b$, we find that
  the leading coefficient of the dominating $P(P(a+b))$ is positive.
  However, plugging in $a=-b$ now gives a contradiction.
\end{proof}

\begin{claim*}
  The $x^1$ coefficient of $P$ is zero and $P(0) \ge 0$.
\end{claim*}
\begin{proof}
  Let $P(x) = cx^2 + kx + \ell$.
  Expand the eastern inequality:
  \[ ca^4+ka^2+\ell + cb^4+kb^2+\ell \ge c(a^4+b^4) + k(a^2+b^2) -
  kab\sqrt2. \]
  Mass cancellation boils this down to
  \[ 2\ell \ge -kab \sqrt 2. \]
  Since this is valid for all $a,b \in \RR$, this means $k = 0$
  and $\ell \ge 0$.
\end{proof}

Finally, put $P(x) = cx^2 + \ell$
and the western inequality reads
\[ c^3(a+b)^4 + 2\ell c^2(a+b)^2 + \ell^2 c + 2a^2b^2 \ge
  c(a^4+b^4)+\ell +4c \cdot ab(a+b)^2 + 4ab \cdot \ell. \]
We make some deductions:
\begin{itemize}
  \ii Let $b=0$ and $a \to \infty$ to get $c^3 \ge c$.
  \ii Let $a=-b=x$ gives
  \[ P(P(0)) + 4x^2P(0) + 2x^4 \ge 2P(x^2) \]
  which means $c \le 1$.
  \ii Combining these two, we get $c \in [-1,0] \cup \{1\}$.
  \ii Let $a=b=0$ to get $\ell^2 c \ge \ell$.
  Since $\ell \ge 0$ is known, either $\ell = 0$ or $\ell c \ge 1$.
  The latter requires $c > 0$, or $c = 1$.
\end{itemize}
Hence, we obtain that either
\[ \left( c \in [-1,0] \cup \{1\} \text{ and } \ell = 0 \right)
  \text{ or } \left( c = 1 \text{ and } \ell \ge 1 \right) \]
which gives the answer set.
One can manually check this works by plugging it in and bashing some
more (details left to excessively bored reader).

\end{document}
