Conditions on Complex Inequality

[Shoelace Thm.]

Homework Statement

Find constraints on $a,b,c \in \mathbb{R}$ such that $\forall w_1,w_2,w_3 \in \mathbb{C}$,

(1) $x = |w_1|^2(1-c) + a|w_2|^2 + c|w_1+w_3|^2 + |w_3|^2(b-c) \ge 0$ and

(2) $x=0 \Rightarrow w_1=w_2=w_3=0$.

Homework Equations

The Attempt at a Solution

I believe the solution is $a>0, c = \mathbb{R_{+}} \setminus \{1\}, b>c$, but I am not sure if there are stronger bounds. How can I know for sure?

 

[Simon Bridge]

You seem to have made sure that each term is not negative - which certainly guarantees the sum is not negative. Though note: c ≤ 1 means that (1 - c) ≥ 0 doesn't it so do you need the lower bound on c?

Is it possible for some terms to be negative and still end up with x not negative?
i.e. if you allow the constraints on a,b,c, to depend on the |w_i| etc. (where i is in {1,2,3}.)

Though I'm guessing that conditions (1) and (2) have to be satisfied simultaneously ... i.e. need to see your reasoning.

note: $w_1=w_2=w_3=0$ certainly means the $x=0$ no matter what a,b,c are. But that's not what condition (2) says is it?

 

[Shoelace Thm.]

Well as you note I chose a and b so that the terms involving them are positive. c can be any positive real except 1. It can be positive and less than 1 obviously. It can't equal 1 because then w_1 can be any complex number. It can be greater than 1 because the sum of the first and third terms of x are then positive. If c is not bounded below by 0, then for large w_3, the sum of the first and third terms of x are negative.

As to your note, no, that's not what condition (2) says.

 

[Simon Bridge]

OK - if every term has to be, individually, non-negative (and that "for all" kinda suggests this) then I don't see tighter constraints. There is this feeling there is something left out isn't there - but that's the only thing that springs to my mind.