# Prove inequality

1. Jan 4, 2012

### 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)

Last edited: Jan 4, 2012
2. Jan 4, 2012

### 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.

3. Jan 4, 2012

### JennyPA

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

4. Jan 4, 2012

### micromass

What do you mean, it is not enough?? Why do you think that??

5. Jan 4, 2012

### 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

6. Jan 4, 2012

### micromass

But the expression reaches a maximum when x=y=z=1/3, so doesn't that tell you anything?

7. Jan 5, 2012

### checkitagain

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.

Last edited: Jan 5, 2012
8. Jan 5, 2012

### 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.

9. Jan 5, 2012

### 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.

10. Jan 5, 2012

### pwsnafu

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

11. Jan 5, 2012

### fbs7

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

12. Jan 6, 2012

### 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.

13. Feb 25, 2012

### JennyPA

I am still working on this problem and has no clue how to proceed from here.

14. Feb 25, 2012

### Office_Shredder

Staff Emeritus
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. Feb 25, 2012

### JennyPA

The problem is I am not suppose to use calculus for this problem. This is the algebra course

16. Feb 25, 2012

### coolul007

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

17. Feb 26, 2012

### JennyPA

how you eliminate z????
you can replace z with 1-x-y but can't see what you did

18. Feb 26, 2012

### coolul007

That's what I did, Then 1-x-y+xy =(1-x)(1-y)

19. Feb 26, 2012

### JennyPA

I understand that part I did that but can't get the statment x+y=1

20. Feb 26, 2012

### 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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook