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X^4 + 1 > x^9 + x, then x<=1

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Let x[tex]\in[/tex][tex]\textbf{R}[/tex]. Prove that if x4+1 > x9+x , then x[tex]\leq[/tex]1

    2. Relevant equations

    [above]

    3. The attempt at a solution

    Sort of clueless on this one. My intuition tells me I should algebraically rearrange the equation such that 1 is on one side of the inequality. I also considered factoring out an x.
     
  2. jcsd
  3. Oct 13, 2009 #2

    Mark44

    Staff: Mentor

    Maybe this will help you get started.
    x4 + 1 > x9 + x
    <==> x9 - x4 + x - 1 < 0
    <==> (x - 1)(x8 + x7 + x6 +x5 + x4 + 1) < 0

    I got this factorization by using synthetic division.

    If x = 1, the product is zero, so the inequality doesn't hold. If x > 1, both factors are positive, so the inequality again doesn't hold.

    Can you say anything about the product if 0 < x < 1?

    The inequality holds if x = 0. Can you say something about the product if x < 0?
     
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