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Math Help - [SOLVED] Proving inequalities

  1. #1
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    [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:

    |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!
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  2. #2
    Senior Member Sampras's Avatar
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     |x^2-x-2| = |x-2| \cdot |x+1| .
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  3. #3
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    Yeah i got that , problem is that I do not understand the why. I get that you can get:

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

    and then

    |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:

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

    -x+2\leq3

    and then

    -x\leq1

    and now somehow this equals

    |x|\leq1

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

    Help and thanks a lot!
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  4. #4
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    Quote Originally Posted by tyro89 View Post
    Yeah i got that , problem is that I do not understand the why. I get that you can get:
    |x|\leq1
    |x|\leq1 means |x-2|\le|x|+|2|\le 3
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  5. #5
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    So I'm just using a method that was completely wack then

    So since |x|\leq1 implies that

    |x-2|\leq3

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

    |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
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