Prove |A^2 - B^2| \leq \frac{1}{2} \{2|B|+\frac{1}{2}\}

[bullpup]

There are many questions i can't get. This is just one. Can anyone give me a hint on what to do? It's probably really simple :(

If A and B are real numbers such that |A - B| < \frac{1}{2}, show that:

|A^2 - B^2| \leq \frac{1}{2} \{2|B|+<br /> \frac{1}{2} \}

 

[arildno]

Welcome to PF!
Try to rewrite |A^{2}-B^{2}| as a product of two expressions..
(It's a well known identity)

 

[bullpup]

Is this what you meant?

|(A+B) (A-B)| \leq |B| + \frac{1}{4}

If so, i already had that but didn't know where to go from there...already wasted a lot of paper going no where

Am i supposed to use -\frac{1}{2} &lt; A - B &lt; \frac{1}{2} somewhere?

thanks

 

[arildno]

First of all:
You are to PROVE that inequality, not ASSUME it!
Hence, what you start with, is the following EQUALITY:
|A^{2}-B^{2}|=|A-B||A+B|
(Ok so far?)
Now, use the inequality you've been given to derive:
|A^{2}-B^{2}|=|A-B||A+B|\leq\frac{1}{2}|A+B|
Agreed?
Now you'll need to use your given inequality to provide an estimate of |A+B|!

 

[arildno]

PS:
You'll need to use the triangle inequality as well..