Solving Gnarly Inequality: K, I, & Alpha Positive Constants

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Homework Help Overview

The discussion revolves around solving an inequality involving positive constants K and α, and a variable I, specifically focusing on the expression \(\frac{K^2 I^4 + K^2 I^2 + 6KI \alpha + \alpha^2}{K^2 I^2 + 5KI^2 \alpha} \leq 2\). The original poster is seeking to establish limits on K and α that satisfy this inequality, given that I is greater than or equal to 1.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to determine limits on K and α to satisfy the inequality, with some suggesting that the first step involves analyzing the denominator. Others propose manipulating the inequality into a quadratic form for easier analysis. Questions arise regarding whether the inequality should be solved with respect to I or if it is meant to be proven.

Discussion Status

The discussion is ongoing, with various approaches being suggested. Some participants express frustration over the complexity of the problem, while others offer different methods for tackling the inequality, such as graphing or transforming it into a quadratic expression. There is no clear consensus on the best method to proceed.

Contextual Notes

Participants note that the problem originates from a Biomathematics course, where understanding the inequality is crucial for evaluating stability properties related to eigenvalues. The original poster mentions feeling overwhelmed by the complexity of the problem and the limits of the variables involved.

happyg1
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Homework Statement



Hi,
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.

Homework Equations


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.
CC
 
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i don't understand you want to solve this wrt I or you want to prove this IE.
 
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.
 
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.

p.s
this doesn't seem a topic in college maths, in what course youv'e encouterd this question?
 
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.
 
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.
 
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 happened.

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.
 
but K and alpha are positive and I is bigger than 1 as well.
so it's obviously positive.
 
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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