Let there be 3 vectors, $v_1 = <2,2,1>, v_2 = <2,1,1> v_3=<1,1,1>$ and their circular permutations ($<1,2,1>$ ,.. etc).
How can I prove if these 3 can create all the vectors $<a,b,c>$ in $\mathbb{N}^3$ respecting the triangular inequality $a+b \geq c$, and in addition $a,b,c \geq 2$.
I think those I gave are not a basis, because I can not see, how $<100,100,1>$ can be created. Seems like those 3 can span the space, where the sum of any two is at least 2 times and at most 4 times that of the third. But it is not clear to me on how to start/approach a proof like this.
I hope I made the problem clear. Any proof with an elaborate explanation/hints would be greatly appreciated.