Trying to solve a system of inequalities I have this kind of system of inequalities (with one equality):


*

*$x+y+z=1$

*$ax+by+cz \leq 2 $

*$a^2x+b^2y+c^2z \leq 6 $

*$a^3x+b^3y+c^3z \leq 14 $ 
and so on.. (I could continue with 5. 6. 7.... the power always increases by one and I know the value on the right).
$x,y,z$ are my unknowns and $a,b,c$ are known.
Could I get a solution (or an approximation) for $x,y$ and $z$ as the number of inequalities grows, I don't see it?
Thanks
 A: Not a full answer, but too long for a comment: let's first assume that all quantities involved are positive for convenience. Let $u=(x,y,z)$, and let $v_n=(a^n,b^n,c^n)$. The inequalities translate to the system
$$\begin{cases} u\cdot\vec{1}=1\\u\cdot v_k\leq b_k,\,1\leq k\leq n \end{cases}$$
for given constants $b_k$. By Cauchy-Schwarz, we have that $u\cdot v_k\leq |u||v_k|$, so a sufficient (but certainly not necessary) condition for the system of inequalities to hold is $|u||v_k|\leq b_k$ for each $k$, equivalently
$$x^2+y^2+z^2\leq \frac{b_k^2}{a^{2k}+b^{2k}+c^{2k}}.$$
Now, we run through $1\leq k\leq n$ and calculate
$$m=\min_{1\leq k\leq n}\left\{\frac{b_k^2}{a^{2k}+b^{2k}+c^{2k}}\right\}.$$
Once we calculate this, then we are left with the restriction $|u|<\sqrt m$ which is an open ball in $\mathbb R^3$. On the other hand $u\cdot\vec{1}=1$ represents a plane, so taking any vector $u$ which lies on their intersection is sufficient.
This is not a full solution, however: the method can fail because there might not be any $u$ in their intersection, although there does exist $u$ satisfying the original system. This is because of the use of Cauchy-Schwarz, which is not an if-and-only-if criterion.
