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Triangle inequality problem

  1. Sep 16, 2006 #1
    Hello all, I am having some difficulties with a question. Hope you guys can help shine some light on the situation.

    Use the triangle Inequality and the fact that 0< |a| < |b| + => 1/|b| <
    1/|a| to establish the following chains of inequalities.

    |x-2 / x^2+9| "Is less than and equal to" |x| + 2 / 9

    Sorry about the display of the question, I really dont know other way to correctly type it.

    Please explain your answers, and the reason you use such method to answer this question.

    Thanks alot for your much needed help.
  2. jcsd
  3. Sep 17, 2006 #2
    So the triangle inequality is: [tex] |x+y| \leq |x| + |y| [/tex]. We are also given that [tex] 0 < |x| < |y| \rightarrow \frac{1}{|y|} < \frac{1}{|x|} [/tex] and want to establish [tex] |\frac{x^{2}}{x^{2}+9}|\leq \frac{|x|+2}{9} [/tex]
    Last edited: Sep 17, 2006
  4. Sep 17, 2006 #3


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    There's a little trick called "parentheses."
  5. Sep 17, 2006 #4
    Thanks for making the question much more understandable.... any help?o:)
  6. Sep 18, 2006 #5
    Please guys need some advice.... Thanks.
  7. Sep 19, 2006 #6
    is it |x-2| or |x^2| /{....} ?
  8. Sep 19, 2006 #7


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    How big can [itex]x^2/(x^2+9)[/itex] get?
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