<s>[INST] Show $|x|<|x^2+b|$ for real $x$ and $b>\frac{1}{4}$ 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!
 [/INST] 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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