Help with the following inequality

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In summary, the conversation discusses a complicated inequality and suggests using a common denominator and expanding the polynomials to solve it. The idea of changing variables is also brought up as a potential solution. The conversation ends with a suggestion to use mathematical software to simplify the process.
  • #1
evagelos
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please help with the following inequality:



.....[(x+y)(y+z)(z+x) +xy+xz+yz]/xyz >=
.....(3xy+3xz+3yz+x^2+y^2+z^2)/[(x+y)(y+z)(z+x)]
 
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  • #2


That doesn't look like a very appealing inequality. Is there a condition for this inequality? I guess you could try dumbassing it by assuming x >= y >= z since the inequality is symmetric in x,y,z and then cross multiplying but I haven't figured out a nice solution yet.
 
  • #3


thanks for the quick respond the only condition is that x>0,y>0,z>0
 
  • #4


What have you tried?
 
  • #5


I would suggest, if x,y,z are positive, that you just get a common denominator and multiply the numerators out into expanded polynomials, then match up the terms.
 
  • #6


Yeah just as I had guessed, the inequality is basically just multiplying out, moving everything to one side, and relying on the fact that the sum of positive real numbers is greater than 0.

That is really a nasty inequality. It's symmetric but that doesn't make anything better. It's not homogeneous, so we can't come up with any additional constraints. It doesn't rely on any clever manipulations or other well-known inequalities (unless you count a simple albeit important fact). It's somewhat instructive but taken in perspective it's no where near as instructive as the other inequality you posed.
 
  • #7


so what do we do now?
 
  • #8


Give me a solid proof please
 
  • #9


The change of variables:

[tex]s = x + y + z, t = xyz, w = 1/x + 1/y + 1/z[/tex]

might help. Note that [tex](x + y)(y + z)(x + z) = swt - t[/tex] and [tex]xy + xz + yz = tw[/tex]. The inequality becomes: [tex]sw + w - 1 \geq (s^2 + 2tw)/(stw - t)[/tex], if my calculations aren't off. Also note that [tex]sw \geq 1[/tex].
 
  • #10


cellotim you mixed up worst now
 
  • #11


Does this mean that I made a mistake or that I've confused you more?

If you want to know how I would solve this problem for a class, I would sit down at Maple or Mathematica, cross-multiply the denominators, and have the computer symbolically expand each side (in x,y,z), then I would make sure that all the coefficients on the LHS are bigger than all the coefficients on the RHS. It would be a huge mess, but it would get it done. I don't know if you have access to any maths software, but that is how I would do it. (Of course you can do it all or partially by hand if you need to show that work. It wouldn't take more than 30-1hr.)
 

What is an inequality?

An inequality is a mathematical statement that compares two quantities, expressing that one is greater than, less than, or not equal to the other.

How do I solve an inequality?

To solve an inequality, you need to isolate the variable on one side of the inequality symbol (>, <, ≥, ≤) and simplify the other side. Remember to reverse the inequality symbol if you multiply or divide by a negative number.

What are the different types of inequalities?

There are three types of inequalities: greater than (>), less than (<), and not equal to (≠). You can also have greater than or equal to (≥) and less than or equal to (≤) inequalities.

Can I graph an inequality?

Yes, you can graph an inequality on a number line. Use an open circle for > and < inequalities, and a closed circle for ≥ and ≤ inequalities. Shade the region to the right of > and ≥ inequalities, and to the left of < and ≤ inequalities.

Are there any rules for solving inequalities?

Yes, there are a few rules to keep in mind when solving inequalities. When adding or subtracting a number on both sides, the inequality sign remains the same. However, when multiplying or dividing by a negative number, the inequality sign must be reversed. Additionally, when solving for a variable in an absolute value inequality, you must consider both the positive and negative solutions.

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