Mate
Posts:
389
Registered:
8/15/05


Re: can your CAS help proving inequalities?
Posted:
Mar 8, 2013 6:30 PM


On Mar 8, 8:08 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.
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. Mate

