Re: On limitations of the Mautsch Principle (was Re: ineqality)

How shocking! ;-)
After two internet-free nights, I return to this newsgroup
to see my name associated to a principle (thanks, Dave!)
which becomes increasingly useless
with every new message in this threat:
:-(
In news:<9buqo1t7kiq0stepi05gp4tet16p2pac74@xxxxxxx> schrieb quasi:
> On Wed, 30 Nov 2005 09:11:15 GMT, "Dirk Van de moortel"
><dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
>>"quasi" <quasi@xxxxxxxx> wrote:
[ ... ]
>>> >However, to defend the principle as it is, how about this ...
[ ... ]
>>> Conjecture:
>>>
>>> Let f(x_1, x_2, ..., x_n) be a
[ homogeneous ]
>>> symmetric polynomial with real coefficients such that
>>> for all x_1, x_2, ..., x_n>=0,
>>>
>>> f(x_1, x_2,..., x_n) >= 0 with equality iff x_1 = x_2 = ... = x_n.
>>>
>>> Then in the expansion of the polynomial
>>>
>>> f(y_1, y1+y_2, ..., y_1+y_2+...+y_n)
>>>
>>> there are no negative coefficients.
>>
>>Counter examples:
>>Dave Rusin's
>> ( x^2 + y^2 )^2 - 3 x y (2 x^2 - 3 x y + 2 y^2)
>> x y ( x^2 + y^2 )^2 - 3 x^2 y^2 (2 x^2 - 3 x y + 2 y^2)
>>and Thomas Mautsch's
>> (x+y-z)^2 (y+z-x)^2 (z+x-y)^2
>
> The above examples are _not_ counterexamples to my conjecture.
>
> A key requirement in my conjecture is that equality hold iff all
> variables are equal. Neither of the above examples satisfies that
> requirement.

Once you have a counterexample like Dave Rusin's from above,
you can construct very many counterexamples with special properties:

Take for example Dave's example
P(x,y) = ( x^2 + 3 - 3 x y)^2,
and add to it a constant epsilon
times an arbitrary symmetric polynomial Q(x,y)
of the same homogeneity as P(x,y)
that satisfies
Q(x,y) > 0 for all x,y>=0, except (0,0).
Take e.g.
Q(x,y) = x^6 + y^6.

Then for all small enough epsilon>0,
the resulting polynomial
P(x,y) + epsilon Q(x,y)
will also be a counter example to my original conjecture,
but with the additional property that it is only zero
at (x,y)=(0,0), i.e. for x=y.

Then this polynomial, and also the products

(x-y)^(2k) * ( P(x,y) + epsilon Q(x,y) )

will be counterexamples to your conjecture.

In addition,
the polynomials P(x,y) + epsilon Q(x,y)
from the example above are irreducible over the reals;
and I also have the strong feeling that
you can achieve similar counterexamples for
symmetric polynomials in more than two variables.


Remains the question, how I was tricked to believe
that this "symmetry-breaking trick" for symmetric inequalities
could be sort of universal. - Well, I obviously
did not check all non-negative symmetric polynomials
systematically, and concentrated on problems
from math contests instead (e.g. many inequalities
for triangle sides a,b,c - with the substitution
a=(x+y), b=(y+z), c=(z+x) and x,y,z>=0).


>>I think you need the modified requirement
>> *only* for all x_1, x_2, ..., x_n >= 0:
>> f(x_1, x_2,..., x_n) >= 0 with equality iff x_1 = x_2 = ... = x_n.
>
> No, that's far too restrictive.
>
> For example, that would make the method not applicable to any sum of
> squares.
>
> Also, the method applies perfectly to the A-M GM inequalities for all
> n,

Do you have a short proof for this fact, or just
"experimental evidence"? - That would be nice!
If the symmetry-breaking principle did not work,
then this might shatter
the idea of Bar-Natan

http://www.math.toronto.edu/~drorbn/projects/ArithGeom/

to find a geometric proof of the AGM inequality
for n numbers by filling n-dimensional hypercubes
of side length (a1 + a2 + ... + an)
with n^n disjoint boxes of dimension a1 x a2 x ... x an,
each...


> but your restriction would not allow those examples.
>
> As far as I can see, my conjecture stands.
> In fact, I have a sketch of a proof, but it's late, so I'll defer it.
> But I'm pretty confident about it (I would even bet on it).
>
> Counterexamples are welcome, but please check to make sure the
> hypothesis of the conjecture are satisfied.
>
> Also, note the latest, revised version which requires that the
> polynomial be homogeneous as well as symmetric.
.

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