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

Kewin Kazankow <lo@xxxxxxxx> wrote:
> there is:
[three _positive_ real numbers a,b,c with]
> ab + bc + ca = 3
> prove that:
> a^3 + b^3 + c^3 + 6abc >= 9

In response I wrote
> Somehow I guess I would want to show that (a^3 + b^3 + c^3 + 6abc - 9)
> could be written as a multiple of (ab + bc + ca - 3) plus a sum of
> (positive multiples of) squares. That is, I might hope for equations like
>
> (a^3 + b^3 + c^3 + 6abc - 9) = (ab + bc + ca - 3)(something) +
> (something)^2 + (a+b)(something)^2 + ...

(See below for a clearer example of the kind of identity I wanted.)
But I didn't solve that one.

Then in article <43879543@xxxxxxxxxxxxx>, <mautsch@xxxxxxxxxxxx> wrote:

>I don't think that breaking the symmetry of the inequality
>is such a good idea at this point.

Right.

>Instead, I would prove the homogeneous polynomial inequality
>
> (a^3 + b^3 + c^3 + 6abc)^2 >= 3 (ab + bc + ca)^3,
>
>from which the claimed inequality follows for ab + bc + ca = 3.

Clever!

>This inequality turned out to be
> *"trivial"*
>in the sense described in the thread Dave Rusin mentioned
>(see news:<slrndn4306.qcj.mautsch@xxxxxxxxxxxxxxxxxxxxx> ):
>
>One can w.l.o.g. assume that
> a <= b <= c,
>and set
> b = a + x1
> c = a + x1 + x2
>with new non-negative variables x1 and x2.
>When this substitution is applied to the difference
> (a^3 + b^3 + c^3 + 6abc)^2 - 3 (ab + bc + ca)^3,
>one obtains a polynomial in a, x1, and x2
>with only non-negative coefficients,
[deleted]
>which *obviously* cannot take on negative values for a,x1,x2>=0.

Excellent!

For comparison, let me note that the Arithmetic-Geometric Mean
inequality, (a+b+c)^3 - 27*a*b*c >= 0 (for positive real a,b,c)
also succumbs to Mautsch's technique, but cannot be proved directly
with a sum-of-squares technique. (A slightly indirect version works:
this identity proves the AGM inequality for all a,b,c > 0 :

(a+b+c)*( (a+b+c)^3-27*a*b*c ) =

4*a*b * (a-b)^2
+ 4*b*c * (b-c)^2
+ 4*c*a * (c-a)^2
+ 2*a*b * (a+b-2*c)^2
+ 2*b*c * (b+c-2*a)^2
+ 2*c*a * (c+a-2*b)^2
+ 1 * ( (a^2+b^2+c^2) - (b*a+c*a+b*c) )^2
+ 1 * ( 1/2*a*b+2*b*c-5/2*c*a)^2
+ 3 * (-3/2*a*b+ b*c+1/2*c*a)^2

which is to say that I have written a positive polynomial as a sum
of squares of _rational functions_, rather than polynomials.)

So it starts to look like Mautsch's approach may be the more powerful.
Naturally, I therefore looked for an example to defeat it :-)
Here is one:

Problem: Prove that for all real a,b we have
3 a b (2 a^2 - 3 a b + 2 b^2) < ( a^2 + b^2 )^2

Solution: RHS - LHS = (a^2+b^2 - 3*a*b)^2 .

However, when we substitute b = a + x into RHS-LHS we get a
negative coefficient: a^4 + x^4 + 2*a^3*x - a^2*x^2 - 2*a*x^3 .
So this inequality cannot be proved using the Mautsch Principle
(as useful as it is nonetheless in other cases!).

dave


PS -- In the AGM example I gave, the last two terms add up to 7 times
a^2*b^2+b^2*c^2+c^2*a^2 - a*b*c*(a+b+c)
This quantity is then a sum of squares (hence positive), and is
symmetric, but cannot be written as a sum of squares of symmetric
polynomials (which would necessarily have to be linear combinations
of s1^2 = (a+b+c)^2 and s2 = (ab+bc+ca) ) since when my expression
is written as a polynomial in the elementary symmetric polynomials s_i
it involves s3 = abc . (It's equal to s2^2 - 3 s1 s3 .)

So symmetry on the way in does not imply symmetry on the way out.
Nonetheless, symmetry can be very useful in the sum-of-squares
decompositions, I have learned; it's a very fruitful step to decompose
the vector space of polynomials of the appropriate degree into
orthogonal subspaces corresponding to the different representations
of the symmetric group. This cuts down on computations, and "explains"
why there have been repeated factors in the determinants I have
reported in other recent posts on this subject.


PPS --This inequality a^2*b^2+b^2*c^2+c^2*a^2 - a*b*c*(a+b+c) >=0
DOES follow instantly from Mautsch's technique.

.

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