Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
hint : consider $(x-\frac{1}{2})^2$ –  Raymond Manzoni Jan 4 '12 at 18:44

3 Answers 3

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.

share|improve this answer

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 $x^2+b > x$.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.