Can anyone prove this:
x + y + z < = (xyz)^1/3 (x/y + y/z + z/x)

Obviously we assume xyz=1 to get rid of awkward terms, but where do we go from there?
One might almost say that the trouble with this inequality is that it has the wrong kind of symmetry. It is symmetric under a cyclic permutation of the variables, which means that any asymmetric approach gets extremely messy extremely quickly, but it's not symmetric enough to do anything useful with.

I did find a "solution" by eliminating z, mutltiplying through to get a cubic in x with coefficients in y, taking the discriminant to get another polynomial in y, factoring that and taking the discriminant again, but this amounts to bludgeoning the problem to death, and would be almost impossible without a computer algebra program.
So what is the "nice" derivation I have been missing?

David Loeffler

David,

I think you need x,y,z positive.

Anyway, have you heard of the AM-GM inequality? If so then you can use this.

You have to play around a bit before it is obvious how it should be applied though. This is the only way I can see of getting it to work. First of all divide both sides by (xyz)^1/3 . Now let a = x/y, b = y/z and c = z/x. Substitute in and you're left needing to show:

(a/c)^1/3 + (b/a)^1/3 + (c/b)^1/3 < = a + b + c.

Now abc = (x/y)(y/z)(z/x) = 1. So insert a factor of (abc)^1/3 into the left hand side. (This makes the order of each side the same, so helps when applying AM-GM.) Therefore we need to show:

(a² b)^1/3 + (b² c)^1/3 + (c² a)^1/3 < = a + b + c.

Now by AM-GM (a² b)^1/3 < = 2/3a + 1/3b

Also:

(b² c)^1/3 < = 2/3b + 1/3c

and:

(c² a)^1/3 < = 2/3c + 1/3a

Adding these three inequalities together gives the required result.

Michael

Thanks.

David