Maria Jose Leon-Sotelo Esteban
Posted on Saturday, 03 May, 2003 - 12:42 pm:
Let x, y and z be positive real numbers satisfying 1/3 £ xy + yz + zx £ 3. Determine the range of values for (i) xyz and (ii) x + y + z.

Thanks. Maria Jose

Peter Gyarmati
Posted on Saturday, 03 May, 2003 - 05:31 pm:
A well-known identity:

x3 +y3 +z3 -3xyz = (x+y+z)(x2 +y2 +z2 -(xy+yz+zx))

I see that the value of (x+y+z) has no upper limit. Example: we increase x ,then x+y+z is also increasing, but we can give values to y and z, for which the original condition will be true.

Peter
Brad Rodgers
Posted on Sunday, 04 May, 2003 - 12:16 am:
There's no doubt a better solution than this, but one way to do the problem is "geometrically." Interpret the set of points (x,y,z) such that

1/3 < = xy + yz + xz < = 3

as the region between the two surfaces

1/3 = xy + yz + xz .... (a)
3 = xy + yz + xz .... (b).

We can then find the minimum of xyz by minimizing it subject to constraint (a), and similarly we can maximize it by maximizing subject to (b). We can assume that the minimum/maximum lies on one of the surfaces (unless no maximum/minimum exists) because, if x can be made larger without changing the values of y and z, xyz can be made larger too. The same method works for x+y+z, for the same reason.

For (i), it's clear that xyz has no minimum; it can go arbitrarily close to 0 by making z arbitrarily close to 0 and giving x and y a value of, say, 1 each.

From finding the extrema it's also clear that the maximum of xyz occurs at either x = y = z = 1, or no maximum exists. We will prove that a maximum exists. Assume xyz can be arbitrarily high. (WLOG), if z is greater than x or y, then z must be able to be arbitrarily large. But xz < = 3, yz < = 3, or xyz2 < = 9. Hence for z arbitrarily large (i.e. z > > 1), xyz < 9, and thus xyz cannot be arbitrarily high. Contradiction. Thus the maximum of xyz is 1.

Peter has already shown that x + y + z is not bounded from above -- can you show using this method that x + y + z > = 1?

Again, I'm almost certain there's a better way to go about this problem, but this is the best way that hits me at the moment.

Brad
David Loeffler
Posted on Sunday, 04 May, 2003 - 11:02 am:
You can fill in some of the gaps in this with AM-GM and factorisation - if this is an Olympiad problem or anything like that, it is wise to avoid using calculus.
 Firstly, we want to prove that if xy + yz + xz £ 3 then xyz £ 1. We have 1 ³ (xy + yz + zx)/3 ³ (xy.yz.zx)1/3 = (xyz)3/2.

So (xyz)2/3 £ 1, and we deduce xyz £ 1.

On the other hand, we want to prove that is xy + yz + zx ³ 1/3, then x + y + z ³ 1.

We have
(x + y + z)2
=
x2 +y2 + z2 + 2xy + 2yz + 2xz
=
3(xy + yz + zx) + (x-y)2 + (y-z)2 + (z-x)2
³
3(xy + yz + zx)
³
1

To show that xyz can be made arbitrarily small, try taking y=z. If we let y=z=e, x = 1/(6e) - e/2, when we have xy + yz + zx = 1/3 so the conditions are satisfied, but xyz £ e2, so xyz can be made arbitrarily small; and x+y+z ³ 1/(6e) can be made arbitrarily large.

David

Chris Tynan
Posted on Sunday, 04 May, 2003 - 12:11 pm:
Out of interest, this problem was on the 1993 BMO.
David Loeffler
Posted on Sunday, 04 May, 2003 - 01:18 pm:
I thought it looked familiar in an Olympic sort of way.