Re: Algebriac Inequality ( Schur-type )
Robert Israel
<alexand...@ulbsibiu.ro> wrote:
>Suppose that F(x)=x^4+a_1x^3+a_2x^2+a_3x+a_4
>
>has only real roots a,b,c,d.
>
>Denote by F' the derivative of F.
>
>
>It's true that the inequality
>
>=====================================
>
> a*F'(a)+b*F'(b)+c*F'(c)+d*F'(d)+ abcd >= 0
>
>=====================================
>
>is verified ?
Yes. The left side is the symmetric polynomial
Q(a,b,c,d) = a (a-b)(a-c)(a-d) + b (b-a)(b-c)(b-d) + c(c-a)(c-b)(c-d) +
d(d-a)(d-b)(d-c) + abcd
which is homogeneous of degree 4. It's enough to show that the minimum
value is >=0 over the sphere a^2 + b^2 + c^2 + d^2 = 1, using a Lagrange
multiplier. Maple finds all the solutions to the resulting set of
polynomial equations, and all the resulting values turn out to be
nonnegative.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Dave Rusin
Robert Israel <isr...@math.ubc.ca> wrote:
>In article <1133685038.6...@g49g2000cwa.googlegroups.com>,
> <alexand...@ulbsibiu.ro> wrote:
>>Suppose that F(x)=x^4+a_1x^3+a_2x^2+a_3x+a_4
>>
>>has only real roots a,b,c,d.
>>
>>Denote by F' the derivative of F.
>>
>>
>>It's true that the inequality
>>
>>=====================================
>>
>> a*F'(a)+b*F'(b)+c*F'(c)+d*F'(d)+ abcd >= 0
>>
>>=====================================
>>
>>is verified ?
>
>Yes. The left side is the symmetric polynomial
>
>Q(a,b,c,d) = a (a-b)(a-c)(a-d) + b (b-a)(b-c)(b-d) + c(c-a)(c-b)(c-d) +
>d(d-a)(d-b)(d-c) + abcd
>
>which is homogeneous of degree 4. It's enough to show that the minimum
>value is >=0 over the sphere a^2 + b^2 + c^2 + d^2 = 1, using a Lagrange
>multiplier. Maple finds all the solutions to the resulting set of
>polynomial equations, and all the resulting values turn out to be
>nonnegative.
C'mon, Robert, get with the program! If you'll calculate
( 2*(d-c)*(b-c)*(b-d) )^2
+ ( d^3-c^3-b*c*d+b*c^2+b^2*c-b^3-a*d^2+a*c*d+a*b*d-a*b*c-a^2*d+a^3 )^2
+ ( d^3-c^3-b*d^2+b*c*d-b^2*d+b^3-a*c*d+a*c^2+a*b*d-a*b*c+a^2*c-a^3 )^2
+ ( d^3-c*d^2-c^2*d+c^3+b*c*d-b^3+a*c*d-a*b*d-a*b*c+a*b^2+a^2*b-a^3 )^2
+ ( d^3-c*d^2-c^2*d+c^3-b*d^2+3*b*c*d-b*c^2-b^2*d-b^2*c+b^3-a*d^2
+3*a*c*d-a*c^2+3*a*b*d+3*a*b*c-a*b^2-a^2*d-a^2*c-a^2*b+a^3 )^2
+(3/2)* ( (a-c)*(a*d-2*a*c+a*b-d^2-b^2+d*c+c*b) )^2
+(1/2)* ( (b-d)*(2*d*b+c*b-3*a*b-c^2+3*a^2+d*c-3*a*d) )^2
you'll get 4(a^2+b^2+c^2+d^2) Q, showing that Q is always positive.
Alternatively, you can substitute {b=a+x,c=a+x+y,d=a+x+y+z} into Q
and expand, and you will get a polynomial with only positive coefficients,
thus showing the (unfortunately weaker) result that the inequality
holds whenever a,b,c,d are all non-negative.
dave
quasi
wrote:
Haha. Sounds like a good program. Sign me up.
>Alternatively, you can substitute {b=a+x,c=a+x+y,d=a+x+y+z} into Q
>and expand, and you will get a polynomial with only positive coefficients,
>thus showing the (unfortunately weaker) result that the inequality
>holds whenever a,b,c,d are all non-negative.
So the SOS principle beats the Mautsch principle this round.
quasi
TCL
I don't think Maple can do that.
"Dave Rusin" <ru...@vesuvius.math.niu.edu> wrote in message
news:dn0aa4$i7t$1...@news.math.niu.edu...
quasi
wrote:
>In article <dmvjqu$s04$1...@nntp.itservices.ubc.ca>,
Actually, since the original function is homogeneous of degree 4,
knowing that the inequality holds for all nonnegative a,b,c,d implies
that it also holds if all a,b,c,d are nonpositive.
quasi
alexand...@ulbsibiu.ro
roots a,b,c,d,e .
I am interested to find a simple way to prove that
===============================
F'(a) +F'(b) +F'(c)+F'(d) +F'(e) >= 0 ??
================================
( F' denotes the derivative of F )
If it's possible to give such a proof, then select e=0.
======
Someone infiorm me that my posted question was proposed
as a part of a problem at IMO-71 /1.
======
However it seems that case d=e=0 is also interesting [I.Schur].
(perhaps connected to a "Polish Inequality" )
......................................
Other question of interest for me :
" Let X=(x_1,x_2,...,x_n) in R^n:=(-infty,infty)^n and
G_n(X;t):=(t-x_1)(t-x_2) ...(t-x_n) , n >=3 ,
If the inequalities
===================================
SUM_{k=1to k=n } G'_n(X;x_k) >= 0 for all X in R^n ,
===================================
are satisfied, then it's true that n= 5 ?? "
Thanking for interest,Alex
Dave Rusin
quasi <qu...@null.set> wrote:
>On 5 Dec 2005 02:57:08 GMT, ru...@vesuvius.math.niu.edu (Dave Rusin)
>wrote:
>>In article <dmvjqu$s04$1...@nntp.itservices.ubc.ca>,
>>Robert Israel <isr...@math.ubc.ca> wrote:
>>>The left side is the symmetric polynomial
>>>
>>>Q(a,b,c,d) = a (a-b)(a-c)(a-d) + b (b-a)(b-c)(b-d) + c(c-a)(c-b)(c-d) +
>>>d(d-a)(d-b)(d-c) + abcd
>>>
>>>which is homogeneous of degree 4
[...]
>>>and all the resulting values turn out to be
>>>nonnegative.
>>C'mon, Robert, get with the program! If you'll calculate
>>
[typical bizarre sum of squares deleted]
>>
>>you'll get 4(a^2+b^2+c^2+d^2) Q, showing that Q is always positive.
>>
>>Alternatively, you can substitute {b=a+x,c=a+x+y,d=a+x+y+z} into Q
>>and expand, and you will get a polynomial with only positive coefficients,
>>thus showing the (unfortunately weaker) result that the inequality
>>holds whenever a,b,c,d are all non-negative.
>Actually, since the original function is homogeneous of degree 4,
>knowing that the inequality holds for all nonnegative a,b,c,d implies
>that it also holds if all a,b,c,d are nonpositive.
I had the same instinct but of course all you get from homogeneity is
that the inequality holds along whole lines through the origin. From
a proof valid in one hexadecimant (Yes it's a word -- Google's got it)
you get a proof valid in the antipodal one. That's two for the price of
one but it does not buy the set of all sixteen hexadecimants.
To the person who asked how I concocted the sum-of-squares mess,
the answer is contained in a long posting from a few weeks ago
after I found the Matlab plugin called "SOSTools". It expresses the
everywhere-positive polynomial as a positive-definite quadratic
form on the vector space of all candidate polynomials. I then
manually computed an orthonormal basis on which the quadratic form
is a sum of squares. In this particular case, sostools reported a
quadratic form whose coefficients were all 0, +-1/2, or +-1 IIRC,
so I took it as-is; in other cases I have worked harder to find
my own SPD matrix representing the polynomial, hoping for one of
lower rank or more symmetry or something, but hey, this is finals
week and I haven't even written my final yet...
dave
quasi
wrote:
Yes, that's what I meant -- that if _all_ a,b,c,d are nonpositive,
then the inequality still holds. So yes, 2 out of 16, but 2 is better
than 1.
quasi
Thomas Mautsch
> In article <dmvjqu$s04$1...@nntp.itservices.ubc.ca>,
> Robert Israel <isr...@math.ubc.ca> wrote:
>>In article <1133685038.6...@g49g2000cwa.googlegroups.com>,
>> <alexand...@ulbsibiu.ro> wrote:
>>>Suppose that F(x)=x^4+a_1x^3+a_2x^2+a_3x+a_4
>>>has only real roots a,b,c,d.
>>>Denote by F' the derivative of F.
>>>
>>>It's true that the inequality
>>>=====================================
>>> a*F'(a)+b*F'(b)+c*F'(c)+d*F'(d)+ abcd >= 0
>>>=====================================
>>>
>>>is verified ?
>>
>>Yes. The left side is the symmetric polynomial
>>
>>Q(a,b,c,d) = a (a-b)(a-c)(a-d) + b (b-a)(b-c)(b-d) + c(c-a)(c-b)(c-d) +
>>d(d-a)(d-b)(d-c) + abcd
>>
>>which is homogeneous of degree 4.
> Alternatively, you can substitute {b=a+x,c=a+x+y,d=a+x+y+z} into Q
> and expand, and you will get a polynomial with only positive coefficients,
> thus showing the (unfortunately weaker) result that the inequality
> holds whenever a,b,c,d are all non-negative.
In de.rec.denksport, Hermann Jurksch gives a way to make this method work:
news:<9jGjy...@jurksch.focus.ping.de>
The idea is to make a case distinction
according to what values in the sequence a <= b <= c <= d are negative.
It suffices to distinguish three cases:
term:= a *(a-b)*(a-c)*(a-d) + b *(b-a)*(b-c)*(b-d)
+ c*(c-a)*(c-b)*(c-d) + d*(d-a)*(d-b)*(d-c) + a*b*c*d;
expand(eval(term, [a=a,b=a+x,c=a+x+y,d=a+x+y+z]));
2 2 2 2 2 2
2 a y z + x y z + 2 a x z + a x y + 4 x y z + a y z
2 2 2 3 2 2 3 2 2
+ 4 a x y + x y + 3 y z + 3 y z + 2 x z + x z
3 4 4 2 3 3 3
+ a z + a x y z + z + a + a x z + 3 a x + 2 a y + a z
2 2 2 2
+ 3 a x + a y
expand(eval(term, [a=-a,b=b,c=b+y,d=b+y+z]));
2 2 2 2 2 2
2 a y z + 4 a b y + b y z + 4 b y z + a b y + a y z
2 3 2 2 3 4 4
+ 2 a b z + 3 y z + 3 y z + a z + a b y z + z + a
3 3 2 2 3 2 2 2 2 2
+ 2 a y + a z + a y + 3 a b + 3 a b + b y + a b z
2 2 3
+ b z + 2 b z
expand(eval(term, [a=-b-x,b=-b,c=c,d=c+z]));
2 2 2 2 2 2 2
2 x c z + 4 b x c + b x z + x b c + 2 b x z + c z b + x c z
2 4 3 4 2 2 3
+ 4 b c z + x + x z + x b c z + z + b z + 2 b z
2 2 3 3 3 2 2 2 2
+ 3 b x + 3 b x + 2 x c + x z + x c + 3 c z
3 2 2
+ 3 c z + b c
In all cases the resulting polynomial contains
only non-negative coefficients,
hence is non-negative for non-negative values of the substituted variables;
and this proves Alexandru Lupas' claim.