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Prove inequality 
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#1
Jan412, 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) 


#2
Jan412, 03:46 PM

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P: 6,035

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
Jan412, 11:00 PM

P: 7

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



#4
Jan412, 11:18 PM

Mentor
P: 18,027

Prove inequality



#5
Jan412, 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 


#6
Jan412, 11:29 PM

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P: 18,027




#7
Jan512, 12:43 AM

P: 99

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. 


#8
Jan512, 03:22 PM

Sci Advisor
P: 6,035

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
Jan512, 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. 


#10
Jan512, 10:18 PM

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#11
Jan512, 11:50 PM

P: 127




#12
Jan612, 09:02 AM

P: 234

Just a little algebra and the each term can eliminate a variable: xy/(z + xy) = xy/[(1x)(1y)]
Likewise for the other 2 terms. 


#13
Feb2512, 06:58 PM

P: 7




#14
Feb2512, 08:27 PM

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P: 4,500

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
Feb2512, 08:45 PM

P: 7




#16
Feb2512, 11:54 PM

P: 234

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



#17
Feb2612, 11:44 AM

P: 7

you can replace z with 1xy but can't see what you did 


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