Mate
Posts:
389
Registered:
8/15/05


Re: can your CAS help proving inequalities?
Posted:
Mar 9, 2013 3:38 AM


On Mar 9, 5:04 am, "Nasser M. Abbasi" <n...@12000.org> wrote: > On 3/8/2013 12:08 PM, cliclic...@freenet.de wrote: > > > > > > > > > > > > > Hello! > > > 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! > > > Martin. > > Would showing that the CAS found {} as solution for > > MIN(r(*)...) >0 > > but found at least one solution for > > MIN(r(*)...) <= 0 > > qualify? > > 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?

