prove inequality


by JennyPA
Tags: inequalities, number theory, polynomial
JennyPA
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#1
Jan4-12, 01:05 PM
P: 7
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)
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mathman
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#2
Jan4-12, 03:46 PM
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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
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#3
Jan4-12, 11:00 PM
P: 7
You are right I did that part already and think it's not enough

micromass
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#4
Jan4-12, 11:18 PM
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prove inequality


Quote Quote by JennyPA View Post
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
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#5
Jan4-12, 11:28 PM
P: 7
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
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#6
Jan4-12, 11:29 PM
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Quote Quote by JennyPA View Post
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
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#7
Jan5-12, 12:43 AM
P: 99
Quote Quote by micromass View Post
** 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
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#8
Jan5-12, 03:22 PM
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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
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#9
Jan5-12, 09:11 PM
P: 127
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
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#10
Jan5-12, 10:18 PM
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Quote Quote by fbs7 View Post
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
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#11
Jan5-12, 11:50 PM
P: 127
Quote Quote by pwsnafu View Post
IIRC it was a function had to have a unique global maximum.
Good point: if it's unique, then obviously x*=y*=z*
coolul007
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#12
Jan6-12, 09:02 AM
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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
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#13
Feb25-12, 06:58 PM
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Quote Quote by coolul007 View Post
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
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#14
Feb25-12, 08:27 PM
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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
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#15
Feb25-12, 08:45 PM
P: 7
Quote Quote by Office_Shredder View Post
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
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#16
Feb25-12, 11:54 PM
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xy/[(1-x)(1-y)] when you eliminate z, then this expression is always 1 when x+y = 1
JennyPA
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#17
Feb26-12, 11:44 AM
P: 7
Quote Quote by coolul007 View Post
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
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#18
Feb26-12, 11:59 AM
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Quote Quote by JennyPA View Post
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)


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