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!

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.
