84 lines
2.1 KiB
TeX
84 lines
2.1 KiB
TeX
\documentclass[../../script.tex]{subfiles}
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% !TEX root = ../../script.tex
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\begin{document}
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\section{Elementary Inequalities}
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\begin{eg}\leavevmode
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\begin{itemize}
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\item $x \in \realn \implies x^2 \ge 0$
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\item $x^2 - 2xy + y^2 = (x - y)^2 \ge 0 ~~\forall x, y \in \realn$
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\item $x^2 + y^2 \ge 2xy$
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\end{itemize}
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\end{eg}
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\begin{thm}[Absolute inequalities]\label{thm:abs}
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Let $x \in \realn$, $c \in [0, \infty)$. Then
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\begin{enumerate}[(i)]
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\item $-|x| \le x \le |x|$
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\item $|x| \le c \iff -c \le x \le c$
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\item $|x| \ge c \iff x \le -c \vee c \le x$
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\item $|x| = 0 \iff x = 0$
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\end{enumerate}
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\end{thm}
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\begin{thm}[Triangle inequality]\label{thm:triangle}
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Let $x, y \in \realn$. Then
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\[
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|x + y| \le |x| + |y|
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\]
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\end{thm}
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\begin{proof}
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From \Cref{thm:abs} follows $x \le |x|$ and $y \le |y|$.
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\begin{equation}
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\implies x + y \le |x| + |y|
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\end{equation}
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However, from the same theorem follows $-|x| \le x$ and $-|y| \le y$.
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\begin{align}
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&\implies -|x|-|y| = x + y \\
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&\implies |x + y| \le |x| + |y|
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\end{align}
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\end{proof}
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\begin{cor}
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$n \in \natn$, $x_1, \cdots, x_n \in \realn$. Then
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\[
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\left| \sum_{i=1}^n x_i \right| \le \sum_{i=1}^n |x_i|
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\]
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\end{cor}
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\begin{proof}
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Proof by induction. Let $n = 1$:
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\begin{equation}
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|x_1| \le |x_1|
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\end{equation}
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This statement is trivially true. Now assume the corollary holds for $n \in \natn$. Then
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\begin{equation}
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\begin{split}
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\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}| \\
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&\le \sum_{i=1}^n |x_i| + |x_{n+1}| \\
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&= \sum_{i=1}^{n+1} |x_i|
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\end{split}
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\end{equation}
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\end{proof}
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\begin{thm}[Bernoulli inequality]\label{thm:bernoulli}
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Let $x \in [-1, \infty)$ and $n \in \natn$. Then
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\[
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(1 + x)^n \ge 1 + nx
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\]
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\end{thm}
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\begin{proof}
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Proof by induction. Let $n = 1$:
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\begin{equation}
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1 + x \ge 1 + 1\cdot x
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\end{equation}
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This is trivial. Now assume the theorem holds for $n \in \natn$. Then
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\begin{equation}
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\begin{split}
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(1 + x)^{n+1} = (1+x)^n (1+x) &\ge (1 + nx)(1 + x) \\
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&= 1 + (n+1)x + nx^2 \\
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&\ge 1 + (n+1)x
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\end{split}
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\end{equation}
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\end{proof}
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\end{document} |