System of $3$ nonlinear equations 
Find positive solutions to the following:

\begin{align*}
x^2+y^2+xy=&1\\
y^2+z^2+yz=&3\\
z^2+x^2+xz=&4
\end{align*}
I simplified and got $x+y+z=\sqrt{7}$ and $x^2+y^2+xy=1$. How do I continue?

Through the cosine theorem, the problem can be stated in the following way:

In a triangle $ABC$ with side lengths $1,\sqrt{3},2$, what are the
  distances of the Fermat point $F$ from the vertices of $ABC$?

Well, such a triangle is a right triangle, since $1^2+\sqrt{3}^2=2^2$, and $FA+FB+FC$ is the length of the Steiner net of $ABC$. It follows that:
$$ x+y+z = \sqrt{1^2+\sqrt{3}^2-2\sqrt{3}\cos 150^\circ}=\sqrt{7}$$

To find $x,y,z$, i.e. the lenghts of $FA,FB,FC$, it is enough to embed this construction in the plane by setting, for instance, $A=(0,0), B=(1,0), C=(0,2)$, then finding the coordinates of $F$ by intersecting a couple of lines. As an alternative, since the trilinear coordinates of $F$ are given by
$$ F = \left[\frac{1}{\sin(A+\pi/3)};\frac{1}{\sin(B+\pi/3)};\frac{1}{\sin(C+\pi/3)}\right]$$
with the previous assumptions we have
$$ F=\left[1;\frac{\sqrt{5}}{2+\sqrt{3}};\frac{\sqrt{5}}{1+2\sqrt{3}}\right]$$
and the tripolar coordinates of $F$ are $[1,2,4]$.
On the other hand, once we get $x+y+z=\sqrt{7}$ our job is done, since the original system of equations implies:
$$\left\{\begin{array}{rcr}(x-z)(x+y+z)&=&-2\\(y-x)(x+y+z)&=&-1\\(z-y)(x+y+z)&=&3 \end{array}\right. $$
or:
$$\left\{\begin{array}{rcr}z-x&=&\frac{2}{\sqrt{7}}\\x-y&=&\frac{1}{\sqrt{7}}\\z-y&=&\frac{3}{\sqrt{7}} \end{array}\right. $$
so that $(x,y,z)=\left(\tau,\tau-\frac{1}{\sqrt{7}},\tau+\frac{2}{\sqrt{7}}\right)$. 
By solving the original system with respect to $\tau$, we get $\tau=\frac{2}{\sqrt{7}}$. The other solutions (non-geometrical, associated with negative lengths) can be derived in a similar way.
