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Gnarly inequality

  1. May 3, 2007 #1
    1. The problem statement, all variables and given/known data

    I have to solve this thing:
    [tex]\frac{K^2 I^4 + K^2 I^2+6KI \alpha +\alpha^2}{K^2 I^2+5KI^2\alpha}\leq 2[/tex]

    I know that [tex]I \geq 1 [/tex] and that K and [tex]\alpha[/tex] are positive constants.

    2. Relevant equations

    3. The attempt at a solution
    I have just been going in circles all day. I can't come up with any set limits on K and [tex]\alpha[/tex] that make this true. I've gotten mental burn out on it.
    Please give me a push in the right direction.
    Last edited: May 3, 2007
  2. jcsd
  3. May 3, 2007 #2
    i don't understand you want to solve this wrt I or you want to prove this IE.
  4. May 3, 2007 #3
    I have to figure out what limits on K and alpha make that IE true.
    The solution I need will look something like when Kis greater than .5 and alpha is less than 1, then it's true. The I is my variable, everything else is a constant. I can't seem to solve it wrt I and get anything meaningful, nor I can find the limits on K and alpha to force it to be true.
  5. May 3, 2007 #4
    i don't intend to crack my head over this but obviously you first need to decide the limits of the denominator, afterwarsd i would multiply by the square of the denominator and hopefully something will be eliminated there.
    but probably you did it already.

    this doesnt seem a topic in college maths, in what course youv'e encouterd this question?
  6. May 3, 2007 #5
    This is from Biomathematics. I found the steady states and am evaluating the trace and determinant find out what the stability properties are, which requires me to figure out that inequality. If that IE is true then I have complex eigenvalues and if not, they are real, so it's a stable node vs. a stable spiral.
    I've been studying for so long it just looks like greek at this point. (and some of it IS!
    Finding the limits of the denominator isn't as easy as I thought it would be. I goes from 1 to infinity and K is just positive, so is alpha. looks like bigger infinity over smaller infinity to me, so when's that less than 2?
    I can't even look at this anymore.
    It's just not getting any clearer.
  7. May 3, 2007 #6

    Tom Mattson

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    There's no good reason to leave this as an inequality involving a rational expression. Multiply both sides by [itex]K^2I^2+5KI^2\alpha[/itex], then collect all the terms to the left hand side. You will then have:

    (a quadratic expression in K) [itex]\leq[/itex] 0.

    That will be much easier for you to work with.
  8. May 4, 2007 #7


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    Unfortunately, multiplying both sides by [itex]K^2I^2+ 5KI^2\alpha[/itex] will [itex]reverse[/itex] the inequality if that is negative, which may well happend.

    Happyg1, I recommend graphing, in the K,[itex]\alpha[/itex] plane, the curves on which the numerator and denominator, independently, are 0. A rational function can change signs only where either the numerator or denominator is 0 so those curves separate areas of different signs.
  9. May 4, 2007 #8
    but K and alpha are positive and I is bigger than 1 as well.
    so it's obviously positive.
  10. May 4, 2007 #9

    Tom Mattson

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    Yes, exactly. I still think that manipulating it into [itex]A_1K^2+A_2K+A_3\leq 2[/itex] would be easier. Actually, I think that bringing it to [itex]A_1K^2+A_2+(A_3-2)=\leq 0[/itex] would be even easier than that. You can examine the roots of that quadratic and then try to determine the conditions for which it is positive. I haven't done it in detail, but that's how I would attack it.
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