Re: A family of inequalities of degree 3

Thank you "Mate"!

In news:<1172417162.321264.46310@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
schrieb Mate <mmatica@xxxxxxxxxxx>:
Thomas Mautsch wrote:
I would like to understand the set S of those triples
(X,Y,Z) in R^3 for which the inequality

(a+b+c)*(X*a^2+Y*b^2+Z*c^2)
>=
12a^2*c + 12a^2*b + 12a*b^2 + 12c*b^2 + 12a*c^2 + 12b*c^2
+ 4a^3 + 4b^3 + 4c^3 - 3abc

is true for all real a,b,c >= 0.

It is easy to see that the set S is a non-empty, convex set,
and that whenever a point (X,Y,Z) lies in S,
every point (X',Y',Z') with X'>=X, Y'>=Y, Z'>= Z also lies in S.

From computer experiments I get the impression that
the set S might be a polyhedron.
(possibly away from a neighborhood of the point (X,Y,Z) = (9,9,9)).

The set is closed, convex and it is easy to verify that it contains
(strictly) the set {(x,y,z) : x >= 9, y >= 9, z >= 9}.
(the point (11,9,8) seems to be in the set).

Well of course - because it contains the point (9,9,9).

There is an easy way to calculate two sequences of polyhedra
(with exponentially growing numbers of faces)
of which one is a decraesing sequence that contains the set S,
and the other is an increasing sequence of polyhedra that are
completely contained in S.
The decreasing sequence even converges to S. -
Plotting the first elements of these sequences,
I got the impression that S itself might be a polyhedron;
but I forgot
that *every* closed convex set can be represented
as the intersection of countably many hyperplanes...

The set can be expressd as
S = {(x,y,z) : a^2 x + b^2 y + c^2 z >= 4 - 27 abc, for each (a,b,c)
in Sigma} where Sigma is the simplex {(a,b,c): a,b,c >= 0, a+b+c=1}

Right you are! - I hadn't noticed.

S is not a polyhedron; this is obvious if you plot (e.g. with Maple}
the section z=9

It may be possible to find exactly z=f(x,y) using some nonlinear
programming
because f(x,y) = sup{ ... : a,b>0, a+b<1}

Ah! *That*'s how you think about the problem
and how you were able to plot the level set for z=9!
Nice idea! However...

but it seems that the computations are messy.

.... the formula for the boundary of S is really not nice
and can (most probably) only be given in implicit form F(x,y,z) = 0.
Even worse, it is only *one component* of the zero set
of the (symmetric) function

F(X,Y,Z) =
680244480*X*Z^3 - 156519216*X^3*Y^2 + 680244480*Y*X^3
+ 680244480*Y^3*X + 4566141072*Y^2*X^2 - 26714476188*Y^2*X
- 26714476188*Y*X^2 + 87885324*Y^2*Z*X^3 + 1492992*Y^4*X^2
+ 81960956784*Z^2 - 979776*Z^3*Y^3 + 173736565956*Y*X
- 576481687632*X - 576481687632*Y - 576481687632*Z
- 59307221277*Y*X*Z - 26714476188*Z*Y^2 + 4566141072*Y^2*Z^2
+ 10139362839*Z^2*Y*X - 345161088*Z^3*Y*X - 156519216*Z^3*Y^2
+ 1492992*Z^4*Y^2 + 2985984*Z^4*Y*X + 81960956784*X^2
+ 173736565956*Z*X - 2176782336*X^3 + 4566141072*Z^2*X^2
- 26714476188*Z*X^2 + 680244480*Z*X^3 + 81960956784*Y^2
- 156519216*Y^3*X^2 - 26714476188*Z^2*X - 2176782336*Z^3
+ 680244480*Z*Y^3 + 10139362839*Y*X^2*Z + 10139362839*Z*Y^2*X
- 1911849156*Y^2*X^2*Z - 1911849156*Y^2*X*Z^2 - 345161088*Y^3*X*Z
- 3347568*Y^3*Z*X^3 + 87885324*Y^3*Z*X^2 - 1259712*Y^4*Z*X^2
+ 2985984*Y^4*Z*X + 2985984*Y*Z*X^4 - 1259712*Y^2*Z*X^4
- 979776*Y^3*X^3 + 1492992*Y^2*X^4 - 345161088*Z*Y*X^3
+ 1492992*Z^2*Y^4 + 1492992*Z^2*X^4 - 22546512*Y^2*X^3*Z^2
- 22546512*Y^3*X^2*Z^2 - 1259712*Y*X^4*Z^2 - 1259712*Y^4*X*Z^2
+ 391630464*Z^2*Y^2*X^2 + 87885324*Z^2*Y^3*X + 87885324*Z^2*Y*X^3
- 1911849156*Z^2*Y*X^2 - 156519216*Z^2*X^3 - 156519216*Z^2*Y^3
- 979776*Z^3*X^3 + 87885324*Z^3*Y*X^2 + 87885324*Z^3*Y^2*X
- 22546512*Z^3*Y^2*X^2 - 3347568*Z^3*Y*X^3 - 3347568*Z^3*Y^3*X
- 156519216*Z^3*X^2 + 1492992*Z^4*X^2 - 1259712*Z^4*Y^2*X
- 1259712*Z^4*Y*X^2 + 680244480*Y*Z^3 - 26714476188*Y*Z^2
+ 173736565956*Y*Z + 1264464*Y^2*X^3*Z^3 + 64*Y^4*X^5*Z
- 25600*Y^4*X^2*Z^3 - 50880*Y^3*X^3*Z^3 + 1264464*Y^3*X^3*Z^2
+ 424656*Y^2*X^4*Z^2 + 640*Y^3*X^3*Z^4 - 25600*Y^3*X^2*Z^4
- 25600*Y^2*X^3*Z^4 + 1264464*Y^3*X^2*Z^3 - 1024*Y^3*X^5*Z
+ 424656*Y^2*X^2*Z^4 + 61776*Y^3*X^4*Z - 25600*Y^3*X^4*Z^2
- 1536*Y^2*X^2*Z^5 + 61776*Y^3*X*Z^4 - 25600*Y^4*Z^2*X^3
- 1024*Y^3*Z^5*X + 640*Y^4*Z^3*X^3 - 3776*Y^4*X^4*Z - 256*Y^4*X^5
+ 192*Y^3*X^5*Z^2 + 384*Y^4*X^4*Z^2 + 640*Y^3*X^4*Z^3
+ 192*Y^2*X^5*Z^3 - 1536*Y^2*X^5*Z^2 + 139968*Y^3*X^4
+ 384*Y^2*X^4*Z^4 - 25600*Y^2*X^4*Z^3 + 61776*Y*X^3*Z^4
+ 61776*Y*X^4*Z^3 - 1024*Y*X^5*Z^3 - 3776*Y*X^4*Z^4 - 256*Y^4*Z^5
+ 139968*Z^4*X^3 + 139968*Z^3*X^4 + 139968*Y^3*Z^4 - 1024*Y*X^3*Z^5
- 3776*Y^4*Z^4*X + 6912*Z^4*X^4 + 384*Y^4*Z^4*X^2 + 192*Y^2*Z^5*X^3
+ 64*Y^4*Z^5*X + 192*Y^3*Z^5*X^2 - 256*Z^5*X^4 - 256*Z^4*X^5
+ 64*Z^4*Y*X^5 + 64*Z^5*Y*X^4 + 139968*Z^3*Y^4 + 139968*Y^4*X^3
+ 61776*Y^4*Z*X^3 + 424656*Y^4*X^2*Z^2 + 61776*Z^3*Y^4*X
+ 192*Y^5*X^2*Z^3 + 192*Y^5*Z^2*X^3 + 64*Y^5*X^4*Z + 64*Y^5*Z^4*X
+ 6912*Y^4*X^4 + 6912*Y^4*Z^4 - 256*Z^4*Y^5 - 2176782336*Y^3
- 1024*X*Z^3*Y^5 - 1536*X^2*Y^5*Z^2 - 1024*X^3*Z*Y^5
+ 2008387814976 - 256*X^4*Y^5

To be precise: It's the convex connected component of F(X,Y,Z) = 0
which contains the point (9,9,9).

Sorry for wasting your time, but I found this result
only a couple of hours ago. Thanks again for your help!

Regards,
Thomas


In <news:45d48140@xxxxxxxxxxxxx> schrieb Thomas Mautsch:

========================================================================
If we knew what the (convex) set of all triples (X,Y,Z) looked like
for which the inequality

(a+b+c)*(X*a^2+Y*b^2+Z*c^2)
>=
12a^2*c+12a^2*b+12a*b^2+12c*b^2+12a*c^2+12b*c^2-3abc+4a^3+4b^3+4c^3

holds for all a,b,c >= 0,
========================================================================
then we could calculate the intersection of this set
with the surface

(X+Y+Z)^2 = XYZ,

and calculate which values (m1,m2,m3)
corresponds to these intersection points (X,Y,Z)
according to the rule:

X = m3 * (m1+m2+m3)^2/(m1*m2*m3)
Y = m2 * (m1+m2+m3)^2/(m1*m2*m3)
Z = m1 * (m1+m2+m3)^2/(m1*m2*m3)

Perhaps someone else knows how to determine this set of (X,Y,Z)'s?
.

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