Prove Inequality: x,y,z - Solution Needed

  • Thread starter JennyPA
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In summary: So the maximum for xy/(1-x)(1-y) is 1.I did not work on this problem for a few days, so I may have missed something.In summary, the problem is to prove that the expression sqrt(xy/(z+xy))+sqrt(yz/(x+yz))+sqrt(xz/(y+xz)) is less than or equal to 3/2, given the conditions x+y+z=1 and 0<x,y,z<1. After discussing various methods, including using calculus and eliminating variables, it is suggested that the maximum for the term xy/(1-x)(1-y) is 1. However, it is pointed out that this does not provide a complete proof for all other combinations of
  • #1
JennyPA
7
0
I found this problem the other day, seems interesting but I am still not sure about the solution
Anybody can help

x, y, z are numbers with

x+y+z=1 and 0<x,y,z<1

prove that

sqrt(xy/(z+xy))+sqrt(yz/(x+yz))+sqrt(xz/(y+xz))<=3/2 ("<=" means less or equal)
 
Last edited:
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  • #2
I haven't worked it through. However, it looks like the max is assumed when x=y=z=1/3. In that case each term in the sum = 1/2.
 
  • #3
You are right I did that part already and think it's not enough
 
  • #4
JennyPA said:
You are right I did that part already and think it's not enough

What do you mean, it is not enough?? Why do you think that??
 
  • #5
It should be proved for all other combinations of x,y and z.
Proving only for x,y and z equal 1/3 is incomplete
 
  • #6
JennyPA said:
It should be proved for all other combinations of x,y and z.
Proving only for x,y and z equal 1/3 is incomplete

But the expression reaches a maximum when x=y=z=1/3, so doesn't that tell you anything?
 
  • #7
micromass said:
** But the expression reaches a maximum when x=y=z=1/3, so doesn't that tell you anything?




Devil's Advocate:

** You claim that the maximum is reached when x = y = z = 1/3, but
it has not been shown by anyone in this thread that it is (or referenced to
another place as already known to be).

So . . . . . I am not convinced that the expression could not be greater than 3/2.


mathman stated that "it looks like the max..." <---- Not concrete

When JennyPA stated that she "did that part already" in response
to mathman, she may have just referred to plugging in 1/3 for each
variable to see that particular sum is 3/2 , but had no knowledge of
whether that selection makes a maximum or not.
 
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  • #8
I suggest that someone use some elementary calculus and see of the max is where it appears to be. To avoid partial derivatives, hold x constant and see if the max is at y=z. If that works, the rest should be easy enough.
 
  • #9
There was a previous discussion on this forum about this hypothesis:

If f(x,y,z) is continuous and symmetric, that is, f(x,y,z) = f(y,z,x) = f(z,x,y), then if that function has a global maximum at f(x*,y*,z*), then x*=y*=z*

That discussion didn't include a proof of that, though.
 
  • #10
fbs7 said:
There was a previous discussion on this forum about this hypothesis:

If f(x,y,z) is continuous and symmetric, that is, f(x,y,z) = f(y,z,x) = f(z,x,y), then if that function has a global maximum at f(x*,y*,z*), then x*=y*=z*

That discussion didn't include a proof of that, though.

IIRC it was a function had to have a unique global maximum.
 
  • #11
pwsnafu said:
IIRC it was a function had to have a unique global maximum.

Good point: if it's unique, then obviously x*=y*=z*
 
  • #12
Just a little algebra and the each term can eliminate a variable: xy/(z + xy) = xy/[(1-x)(1-y)]
Likewise for the other 2 terms.
 
  • #13
coolul007 said:
Just a little algebra and the each term can eliminate a variable: xy/(z + xy) = xy/[(1-x)(1-y)]
Likewise for the other 2 terms.

I am still working on this problem and has no clue how to proceed from here.
 
  • #14
The best suggestion kind of got lost in the noise here. Take derivatives and use Lagrange multipliers/whatever calculus method you feel most comfortable with to find the maximum of the function
 
  • #15
Office_Shredder said:
The best suggestion kind of got lost in the noise here. Take derivatives and use Lagrange multipliers/whatever calculus method you feel most comfortable with to find the maximum of the function

The problem is I am not suppose to use calculus for this problem. This is the algebra course
 
  • #16
xy/[(1-x)(1-y)] when you eliminate z, then this expression is always 1 when x+y = 1
 
  • #17
coolul007 said:
xy/[(1-x)(1-y)] when you eliminate z, then this expression is always 1 when x+y = 1

how you eliminate z?
you can replace z with 1-x-y but can't see what you did
 
  • #18
JennyPA said:
how you eliminate z?
you can replace z with 1-x-y but can't see what you did

That's what I did, Then 1-x-y+xy =(1-x)(1-y)
 
  • #19
coolul007 said:
That's what I did, Then 1-x-y+xy =(1-x)(1-y)

I understand that part I did that but can't get the statement x+y=1
 
  • #20
since z is no longer a part of this term, I looked at values for x and y to see where the max for the term is. The restrictions of x + y + z = 1 then x + y can't exceed 1.
 

1. What is an inequality?

An inequality is a mathematical statement that compares two quantities and states that one is larger or smaller than the other. It is denoted by the symbols ">" (greater than), "<" (less than), ">=" (greater than or equal to), "<=" (less than or equal to), or "≠" (not equal to).

2. How do you prove an inequality?

To prove an inequality, you must manipulate the given expressions using algebraic operations to show that one side is always greater or smaller than the other. You can also use properties of inequalities, such as the transitive property, to simplify the expressions and make the comparison easier.

3. What is the difference between an equality and an inequality?

An equality is a mathematical statement that shows that two quantities are exactly equal, while an inequality shows a relationship between two quantities where one is larger or smaller than the other. Inequalities allow for a range of values, while equalities only have one specific value.

4. Can you use substitution to prove an inequality?

Yes, substitution can be used to prove an inequality. By substituting a value for a variable in the expression, you can simplify the inequality and show that one side is always greater or smaller than the other. However, it is important to choose a value that satisfies the given conditions of the inequality.

5. Are there any special rules for proving inequalities?

Yes, there are some important rules to keep in mind when proving inequalities. These include the addition and subtraction rules, the multiplication and division rules, and the rules for dealing with negative numbers. It is also important to remember to reverse the inequality sign when multiplying or dividing both sides by a negative number.

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