Optimizing Inequality with Constraints

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In summary, the conversation discusses trying to prove an inequality with given constraints and the possibility of adding additional constraints to make the proof easier. The original inequality is shown to be true with the given constraints and an example is provided. The conversation then discusses the effectiveness of the constraints for different values and suggests additional constraints that could make the proof easier.
  • #1
nasshi
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I'm interested in proving the following inequality, given the following constraints. This is for personal gain, and not for homework. I'm not even sure if I can prove this without additional constraints. If you can solve it by adding constraints, please let me know what options I have for adding constraints.

Given:
0<a<1
x>0, y>0, z>0
x>y

Prove:[tex]\frac{x-y}{(1+x)(1+y)} < \frac{z*a}{1+y+z}[/tex]

Thanks so much for your help!
 
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  • #2
I believe it's proven wrong with the example, (a,x,y,z)=(1/2,2,1,1).
 
  • #3
You are indeed correct.

How about we add the constraint:
[tex]y+z>x[/tex]

What about now?
 
  • #4
Well y+z>x really just means z>x in your case, but I'll try to prove it:

[tex]
\frac {x-y}{(1+x)(1+y)}<\frac {z*a}{1+y+z}[/tex]

Fallowing constraints, we set (a,x,y,z) to (1/2,2,1,2)

[tex]

\frac {2-1}{(1+2)(1+1)}<\frac {2*1/2}{1+1+2}[/tex]

[tex]\frac {1}{6}<\frac {1}{4}[/tex]

This proves true.
Assume [tex](a,x,y,z)\longrightarrow(a+1,x+1,y+1,z+1)[/tex]; plug in the original set.

[tex]\frac {(x+1)-(y+1)}{(1+(x+1))(1+(y+1))}<\frac {(z+1)(a+1)}{1+(y+1)+(z+1)}[/tex]

[tex]\frac {x-y}{(x+2)(y+2)}<\frac {(z+1)(a+1)}{y+z+3}[/tex]

Now plug in, and simplify:

[tex]\frac {2-1}{(2+2)(1+2)}<\frac {(2+1)(1/2+1)}{1+2+3}[/tex]

[tex]\frac {1}{12}<\frac {9}{12}[/tex]

I don't know if this is formal, but I'd call it adequate.
 
  • #5
I think it works well for larger values of x,y,z. But when x,y,z are small, for example, between 0 and 1, I think your method won't work.
 
  • #6
I should have set (a,x,y,z) to (d,2d,d,3d) and (1-d,2d,d,3d).
Anyways, you have to fix your constraints...
 
  • #7
I can't exactly "fix my constraints". I'm trying to prove something as generally as possible. Can you think of any constraints I can add that would easily establish the desired result? Because I can't.
 
  • #8
Here are some constraints that I think work:
x>y
0<y<1
a>6
z>2/35
 

1. How do you prove a desired inequality?

To prove a desired inequality, you must first determine the desired inequality and the variables involved. Then, you can use mathematical techniques such as algebra, calculus, or induction to manipulate the equation and show that it is true for all possible values of the variables.

2. What is the importance of proving a desired inequality?

Proving a desired inequality is important because it validates the validity of a mathematical statement or theory. It allows for a deeper understanding of the relationship between variables and can be used to make predictions and solve problems in various fields of science and mathematics.

3. What are some common methods used to prove inequalities?

Some common methods used to prove inequalities include algebraic manipulation, the use of mathematical identities and theorems, and the application of mathematical induction. In some cases, computer simulations or mathematical software may also be used to help prove an inequality.

4. How do you know if your proof is correct?

A correct proof of a desired inequality should logically and mathematically follow from the given information and assumptions. It should also clearly show each step of the proof, with no gaps or errors in reasoning. Double-checking your work and having someone else review your proof can also help ensure its accuracy.

5. Can inequalities be disproven?

Yes, inequalities can be disproven if they are shown to be false for at least one set of variables or if there is a logical flaw in the proof. However, if a desired inequality is proven to be true, it is considered a valid mathematical statement and cannot be disproven.

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