I need some help to prove the following inequality: let $x\in\mathbb{R}$, then $|x|\le\frac12+\frac {x^2}2.$
Estimate for absolute valuue
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real-analysis
inequality
2 Answers
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$y^2\geq 0$
(Consider $y=1\pm x$.)
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1Or, maybe even simpler: $y=|x|-1$. – 2012-04-14
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You may assume that $x\ge 0$. Then the inequality becomes $ x\le {1\over2}+{x^2\over 2}; $ which is equivalent to $ 2x\le 1+x^2, $ or $ x^2-2x+1\ge 0. $ Now note that $x^2-2x+1=(x-1)^2$.
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0@MichaelHard$y$ I wish I had a nickel for every time I had to say or write "e$q$uations (or inequalities) are not *equal* to each other". – 2012-04-14