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Topic: can your CAS help proving inequalities?
Replies: 19   Last Post: Mar 11, 2013 12:00 PM

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 clicliclic@freenet.de Posts: 1,245 Registered: 4/26/08
Re: can your CAS help proving inequalities?
Posted: Mar 9, 2013 10:33 AM
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Mate schrieb:
>
> On Mar 8, 8:08 pm, cliclic...@freenet.de wrote:

> >
> > Let a, b, c, d be arbitrary real numbers. Define:
> >
> > r(a, b, c, d) := (a - c)*(a - d)*(b - c)*(b - d)
> >
> > s(a, b, c, d) := (a + b)*(c + d) - 2*(a*b + c*d) - ABS((a - b)*(c - d))
> >
> > t(a, b, c, d) := (a + b)*(c + d) - 2*(a*b + c*d) + ABS((a - b)*(c - d))
> >
> > Can your CAS help proving the following inequalities?
> >
> > MIN(r(a, b, c, d), r(a, c, b, d), r(a, d, c, b)) <= 0
> >
> > MAX(s(a, b, c, d), s(a, c, b, d), s(a, d, c, b)) >= 0
> >
> > MIN(t(a, b, c, d), t(a, c, b, d), t(a, d, c, b)) <= 0
> >
> > Have fun!
> >

>
> 1.
> Denoting
> x:=r(a, b, c, d), y:=r(a, c, b, d), z:=r(a, d, c, b)
>
> ==> x*y+x*z+y*z = 0
> ==> min(x,y,z) <=0 and actually also max(x,y,z) >= 0
>
> 2,3.
> Denoting similarly x,y,z ==>
>
> y*z^3+2*y^2*z^2+y^3*z+x*z^3+4*z^2*y*x+4*z*y^2*x+y^3*x
> +2*x^2*z^2+4*y*z*x^2+2*x^2*y^2+x^3*z+y*x^3 = 0
> ==> min(x,y,z) <= 0 and also max(x,y,z) >= 0
>
> The relations in x,y,z can be easily verified with any CAS.
> I have found them using Grobner bases in Maple.
>
> I had fun indeed. Thanks for the problems.
>

Truly formidable - the relations 2,3 in particular! I had omitted the
second inequality under 1, because here one already has max(x,y) >= 0,
max(x,z) >= 0, max(y,z) >= 0.

Looks like no CAS just simplifies these inequalities to "true".

Martin.

Date Subject Author
3/8/13 clicliclic@freenet.de
3/8/13 Mate
3/9/13 clicliclic@freenet.de
3/9/13 clicliclic@freenet.de
3/8/13 Nasser Abbasi
3/9/13 Mate
3/9/13 Nasser Abbasi
3/9/13 A N Niel
3/9/13 Mate
3/10/13 A N Niel
3/10/13 Mate
3/9/13 Nasser Abbasi
3/10/13 Mate
3/10/13 Nasser Abbasi
3/10/13 Mate
3/10/13 clicliclic@freenet.de
3/10/13 Nasser Abbasi
3/11/13 Peter Pein
3/11/13 clicliclic@freenet.de
3/11/13 Peter Pein

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