\documentclass[../../script.tex]{subfiles} % !TEX root = ../../script.tex \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}