# Prove inequality

 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
 Sci Advisor P: 6,070 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.
 P: 7 You are right I did that part already and think it's not enough
Mentor
P: 18,293
Prove inequality

 Quote by JennyPA 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??
 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
Mentor
P: 18,293
 Quote by 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
But the expression reaches a maximum when x=y=z=1/3, so doesn't that tell you anything?
P: 99
 Quote by micromass ** But the expression reaches a maximum when x=y=z=1/3, so doesn't that tell you anything?

** 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.
 Sci Advisor P: 6,070 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.
 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.
P: 834
 Quote by 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.
IIRC it was a function had to have a unique global maximum.
P: 127
 Quote by pwsnafu IIRC it was a function had to have a unique global maximum.
Good point: if it's unique, then obviously x*=y*=z*
 P: 234 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.
P: 7
 Quote by 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.
I am still working on this problem and has no clue how to proceed from here.
 Emeritus Sci Advisor PF Gold 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
P: 7
 Quote by 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
The problem is I am not suppose to use calculus for this problem. This is the algebra course
 P: 234 xy/[(1-x)(1-y)] when you eliminate z, then this expression is always 1 when x+y = 1
P: 7
 Quote by coolul007 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
P: 234
 Quote by JennyPA 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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