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How do I show $|x|<|x^2+b|$ for real $x$ and $b>\frac{1}{4}$ ?

I mean it's clear for $x=0$ and $|x|>1$ but what is if $0<|x|<1$?

Please help!

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hint : consider $(x-\frac{1}{2})^2$ – Raymond Manzoni Jan 4 '12 at 18:44
up vote 2 down vote accepted

I'm not sure of the immediate simplest way, but the first way that struck me is what I'm writing here.

Separate the positive and negative cases, so that we can ignore the absolute value itself. Then note that $x^2 - x + \frac{1}{4} = (x - 1/2)^2 \geq 0$.

Or use the powers of calculus on the separate cases.

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You only have to prove it for $x\geq 0$, since if true for $x$ it is true for $-x$.

If $x\geq 0$, then $|x|=x$ and $|x^2+b|=x^2+b$ and you only need to prove that $x^2+b>x$.

But $$x^2-x+b = x^2 - x + \frac{1}{4} + (b-\frac{1}{4}) = (x-\frac{1}{2})^2 + (b-\frac{1}{4})$$

Since $b>\frac{1}{4}$, this means that $x^2-x+b > 0$, so $x^2+b > x$.

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One possibility is to compare the graphs $y=|x|$ and $y=x^2+\frac{1}{4}$.

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