
Re: Let Z be a complex number.
Posted:
Oct 26, 2013 8:24 AM


On Saturday, 26 October 2013 06:01:27 UTC+3, William Elliot wrote: > On Fri, 25 Oct 2013, dan.ms.chaos@gmail.com wrote: > > > > > Of course the proof boils down to elementary algebra, but I'm looking for a > > > way that avoids tedious calculations as much as possible (eg. some geometric > > > principle) ... > > > > What proof?
The proof I were to make substituting z as iy  x, and using Lagrange multipliers to solve the constraint. The original problem was not the one posted here. It 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 .
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 The question is, does this now shortened problem have a short solution?
http://www.wolframalpha.com/input/?i=%28%28x^3++3*x*y*y+%2B1%29^2+%2B+%283*x*x*yy*y*y%29^2++%3D+1%29++and+%28x*x+%2B+2*x+%2B+y*y+%3D+3%29+++and+%28x*x+%2B+y*y+%3D+2^%282%2F3%29%29%29
The shape described by z^3  1 > 1 looks like a sort of cycloid (hypotrochoid ?), other than that and the standard procedures for solving a function subject to inequality constraints, I'm stumped as to what a proper proof might look like.

