Prove Inequality: x,y,z - Solution Needed

[JennyPA]

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)

 

[mathman]

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.

 

[JennyPA]

You are right I did that part already and think it's not enough

 

[micromass]

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??

 

[JennyPA]

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

 

[micromass]

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?

 

[checkitagain]

** 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.

 

[mathman]

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.

 

[fbs7]

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.

 

[pwsnafu]

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.

 

[fbs7]

IIRC it was a function had to have a unique global maximum.

Good point: if it's unique, then obviously x*=y*=z*

 

[coolul007]

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.

 

[JennyPA]

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.

 

[Office_Shredder]

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

 

[JennyPA]

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

 

[coolul007]

xy/[(1-x)(1-y)] when you eliminate z, then this expression is always 1 when x+y = 1

 

[JennyPA]

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

 

[coolul007]

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)

 

[JennyPA]

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

 

[coolul007]

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.