two short doubts about the inverse function in a point the function is $F(x,y,z)=(y^2+z^2, z^2+x^2, x^2+y^2)$ the point is (-1,1,-1)
task: find the local inverse of F in that point.
I have already proved that F is actually invertible there. then i solved the system:
$$
\begin{cases}
y^2+z^2=a\\
z^2+x^2=b\\
x^2+y^2=c
\end{cases}
$$
which gave me
$$
x=\pm \sqrt{\frac{c-a+b}{2}}\\
y=\pm \sqrt{\frac{a-b+c}{2}}\\
z=\pm \sqrt{\frac{b-c+a}{2}}\\
$$
The doubt I have is: how should I choose the signs? I thought of looking at the values of x,y,z in the point, which means x,z=negative y=positive
so i picked
$$
x= -\sqrt{\frac{c-a+b}{2}}\\
y= \sqrt{\frac{a-b+c}{2}}\\
z= -\sqrt{\frac{b-c+a}{2}}\\
$$
finding the inverse G:
$$ G(a,b,c)=(-\sqrt{\frac{c-a+b}{2}},\sqrt{\frac{a-b+c}{2}},-\sqrt{\frac{b-c+a}{2}})$$
the second doubt is:
should i leave it written that way or is it better:
$G(y^2+z^2, z^2+x^2, x^2+y^2)=(-x,y,-z)$ ? (with the same doubt about the signs?)
thank you in advance
 A: First doubt: you are right
Secon doubt: the first way you have written is good, the other does not make sense.
A: The function $F:\>(x,y,z)\mapsto(a,b,c)$ is smooth, and $F(-1,1,-1)=(2,2,2)$. The Jacobian $J_F(x,y,z)=16xyz$ is $\ne 0$ at $(-1,1,-1)$; therefore "by general principles" $F$ maps any sufficiently small neighborhood $U$ of $(-1,1,-1)$ bijectively onto a neighborhood $V$ of $(2,2,2)$. The "local inverse" 
$G: \>V\to U$ is again differentiable and has $G(2,2,2)=(-1,1,-1)$.
In your case it was possible to produce via algebraic manipulations explicit expressions for the possible local inverses, and we now have to choose the right candidate. You have already done that and obtained
$$G(a,b,c)=\left(-\sqrt{\frac{c-a+b}{2}},\sqrt{\frac{a-b+c}{2}},-\sqrt{\frac{b-c+a}{2}}\right)\ .\tag{1}$$
This is the explicit expression of $G$ in terms of its inborn variables $a$, $b$, $c$; you could not hope of something better. Note that when $(a,b,c)=(2,2,2)$ all radicands in $(1)$ become $1$, which implies that $G$ is differentiable in the neighborhood of $(2,2,2)$, as guaranteed by the general implicit function theorem.
