
Re: Let Z be a complex number.
Posted:
Oct 27, 2013 4:09 AM


On Sat, 26 Oct 2013, dan.ms.chaos@gmail.com wrote:
> The original problem ... was to prove that > > (x*x + 2*x + y*y > 3) implies that > ((x^3  3*x*y*y +1)^2 + (3*x*x*yy*y*y)^2 > 1) > for x ,y real numbers . > 25  10 > 3 125 + 1
Egads, terrible notation. For all x,y in R . . x^2 + 2x + y^2 > 3 implies . . (x^3  3xy^2 + 1)^2 + (3x^2 y  y^3)^2 > 1.
> I've figured that if you write z = iy  x , you can rewrite it in a simple > way, namely:
> z  1 > 2 implies z^3  1 > 1
Are you sure of "> 2"?
z  1 = sqr((x + 1)^2 + y^2) z^3  1 = sqr((x^3 + 3xy^2  1)^2 + (3x^2 y  y^3)^2)
z^3  1 = z  1.z^2 + z + 1
If z  1 > sqr3, is z^2 + z + 1 >= 1/sqr 3?
> The question is, does this now shortened problem have a short solution?
I don't see any.

