\documentclass[../../script.tex]{subfiles}

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\begin{document}
\section{Elementary Inequalities}
\begin{eg}\leavevmode
\begin{itemize}
	\item $x \in \realn \implies x^2 \ge 0$
	\item $x^2 - 2xy + y^2 = (x - y)^2 \ge 0 ~~\forall x, y \in \realn$
	\item $x^2 + y^2 \ge 2xy$
\end{itemize}
\end{eg}

\begin{thm}[Absolute inequalities]\label{thm:abs}
Let $x \in \realn$, $c \in [0, \infty)$. Then
\begin{enumerate}[(i)]
	\item $-|x| \le x \le |x|$
	\item $|x| \le c \iff -c \le x \le c$
	\item $|x| \ge c \iff x \le -c \vee c \le x$
	\item $|x| = 0 \iff x = 0$
\end{enumerate}
\end{thm}

\begin{thm}[Triangle inequality]\label{thm:triangle}
Let $x, y \in \realn$. Then
\[
	|x + y| \le |x| + |y|
\]
\end{thm}
\begin{proof}
From \Cref{thm:abs} follows $x \le |x|$ and $y \le |y|$.
\begin{equation}
	\implies x + y \le |x| + |y|
\end{equation}
However, from the same theorem follows $-|x| \le x$ and $-|y| \le y$.
\begin{align}
	&\implies -|x|-|y| = x + y \\
	&\implies |x + y| \le |x| + |y|
\end{align}
\end{proof}

\begin{cor}
$n \in \natn$, $x_1, \cdots, x_n \in \realn$. Then
\[
	\left| \sum_{i=1}^n x_i \right| \le \sum_{i=1}^n |x_i|
\]
\end{cor}
\begin{proof}
Proof by induction. Let $n = 1$:
\begin{equation}
	|x_1| \le |x_1|
\end{equation}
This statement is trivially true. Now assume the corollary holds for $n \in \natn$. Then
\begin{equation}
\begin{split}
	\left| \sum_{i=1}^{n+1} x_i \right| = \left| \sum_{i=1}^n x_i + x_{n+1} \right| &\le \left| \sum_{i=1}^n  x_n \right| + |x_{n+1}| \\
	&\le \sum_{i=1}^n |x_i| + |x_{n+1}| \\
	&= \sum_{i=1}^{n+1} |x_i|
\end{split}
\end{equation}
\end{proof}

\begin{thm}[Bernoulli inequality]\label{thm:bernoulli}
Let $x \in [-1, \infty)$ and $n \in \natn$. Then
\[
	(1 + x)^n \ge 1 + nx
\]
\end{thm}
\begin{proof}
Proof by induction. Let $n = 1$:
\begin{equation}
	1 + x \ge 1 + 1\cdot x
\end{equation}
This is trivial. Now assume the theorem holds for $n \in \natn$. Then
\begin{equation}
\begin{split}
	(1 + x)^{n+1} = (1+x)^n (1+x) &\ge (1 + nx)(1 + x) \\
	&= 1 + (n+1)x + nx^2 \\
	&\ge 1 + (n+1)x
\end{split}
\end{equation}
\end{proof}
\end{document}