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I need some help to prove the following inequality: let $x\in\mathbb{R}$, then $$|x|\le\frac12+\frac {x^2}2.$$

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$y^2\geq 0$

(Consider $y=1\pm x$.)

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    Or, 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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    The phrase "is equivalent to" reminds me of a time when I was explaining to a class how to prove a trigonometric identity. When I said something is equivalent to something else, someone asked: "Does equivalent to mean 'equal to'?". People are full of surprises. (Just in case anyone is unsure: _No_. It does _not_ mean "equal to".)2012-04-14
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    @MichaelHardy I wish I had a nickel for every time I had to say or write "equations (or inequalities) are not *equal* to each other".2012-04-14