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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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Mate

Posts: 389
Registered: 8/15/05
Re: can your CAS help proving inequalities?
Posted: Mar 9, 2013 3:38 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Mar 9, 5:04 am, "Nasser M. Abbasi" <n...@12000.org> wrote:
> On 3/8/2013 12:08 PM, cliclic...@freenet.de wrote:
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> > Hello!
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> > Let a, b, c, d be arbitrary real numbers. Define:
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> > r(a, b, c, d) := (a - c)*(a - d)*(b - c)*(b - d)
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> > s(a, b, c, d) := (a + b)*(c + d) - 2*(a*b + c*d) - ABS((a - b)*(c - d))
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> > t(a, b, c, d) := (a + b)*(c + d) - 2*(a*b + c*d) + ABS((a - b)*(c - d))
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> > Can your CAS help proving the following inequalities?
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> > MIN(r(a, b, c, d), r(a, c, b, d), r(a, d, c, b)) <= 0
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> > MAX(s(a, b, c, d), s(a, c, b, d), s(a, d, c, b)) >= 0
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> > MIN(t(a, b, c, d), t(a, c, b, d), t(a, d, c, b)) <= 0
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> > Have fun!
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> > Martin.
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> Would showing that the CAS found {} as solution for
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>     MIN(r(*)...) >0
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> but found at least one solution for
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>     MIN(r(*)...) <= 0
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> qualify?
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> --Nasser


The fact that MIN(r(*)...) <= 0 has a solution is irrelevant
(and obvious: r(a,a,a,a)=0).

If the CAS "shows" that MIN(r(*)...) >0 has no solutions,
we are done, but there are here 2 possibilities
- a solution does not exist
- the CAS was not able to find one
Which one should we consider?




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