# Thread: [SOLVED] Proving inequalities

1. ## [SOLVED] Proving inequalities

Hello,

I'm trying to prove problems like the following and I'm having issues since I have never been showed just how to work such a problem.

Prove the following inequality:

$\displaystyle |x| \leq 1 \Rightarrow |x^2-x-2| \leq 3|x+1|$

I'm sure this is quiet an easy problem I just need to have it showed to me since I'm really not used to be thinking "in proofs" just yet!

Thanks!

2. $\displaystyle |x^2-x-2| = |x-2| \cdot |x+1|$.

3. Yeah i got that , problem is that I do not understand the why. I get that you can get:

$\displaystyle |x-2|\cdot|x+1|\leq3|x+1|$

and then

$\displaystyle |x-2|\leq3$

but at this point do we just assume that it's obvious that we can remove the absolute function of x-2 (since it will always be positive due to the initital constraint) or how do we prove that we can remove it ?

This is my proof, either I am doing something wrong or I just can't seem to grasp the why in what I'm doing:

$\displaystyle |x-2|\leq3 \Leftrightarrow |-x+2|\leq3$ Since $\displaystyle |x|\leq1$ we can remove the absolute function and we get

$\displaystyle -x+2\leq3$

and then

$\displaystyle -x\leq1$

and now somehow this equals

$\displaystyle |x|\leq1$

why does $\displaystyle |x|=-x$ and/or why am I messing up?

Help and thanks a lot!

4. Originally Posted by tyro89
Yeah i got that , problem is that I do not understand the why. I get that you can get:
$\displaystyle |x|\leq1$
$\displaystyle |x|\leq1$ means $\displaystyle |x-2|\le|x|+|2|\le 3$

5. So I'm just using a method that was completely wack then

So since $\displaystyle |x|\leq1$ implies that

$\displaystyle |x-2|\leq3$

and then we multiply both sides by $\displaystyle |x+1|$ and boom we got

$\displaystyle |x-2||x+1|\leq3|x+1|$

which is equal to our original problem...

Sweet thanks! I think I just got to do a few of these to get myself into the game