Find the sum from the system of equations 
If $x,y, z$ satisfy: $$x + y = z^2 + 1, y + z = x^2 + 1, x + z = y^2 + 1 $$ Find the value of $2x  +3y + 4z$. 

This gives us (by getting $x + y + z$ that)
$z^2 + z + 1 = x^2 + x + 1 = y^2 + y + 1 \implies z^2 + z = x^2 + x = y^2 + y$.
Using the first and last, I also got:
$2x + y + z = z^2 + y^2 + 2$
But I cannot get much farther!
EDIT:
$x -z = z^2 - x^2 = (z-x)(z+x) \implies z + x = -1$.
$y - x = x^2 = y^2 = (x-y)(x+y) \implies x + y = -1$
Thus, $z - y = 0$ and $z = y$. 
$2x + 7y$ is to be found then, and $x = -1 -y $ so: $2(-1 - y) - 7y = -2 - 9y$ is to be found.
Any ideas here?
 A: You can add up the three equations to get
       $2x+2y+2z = x^2+y^2+z^2 +3
                 = (x^2+1)+ (y^2+1)+ (z^2+1)\geq 2x+2y+2z$
Hence, we know that $x=y=z=1$ 
A: Hint:
If you add up the first equation and the second times $-1$, you get:
$$x-z=z^2-x^2=-(x-z)(x+z)$$
Edit:
Therefore, $x=y=z$, because otherwise you can divide by $x-y$ to get $x+y=-1$ and, using the first set of equations, $z^2+1=-1$ and $z^2=-2$, which is not possible. The same goes for y and z. Using this in the first equation gives:
$$
x+x=x^2+1 \\
x^2-2x+1=0\\
x=y=z=1\\
2x+3y+4z=2+3+4=9
$$
A: As you've already noted, we can set $z^2 +z +1 = x^2 +x +1 = y^2 +x +1= \lambda^2$. We can do this because $x^2 +x+1>0 $ , $\forall x \in \mathbb{R}$.
Recall the law of cosines and observe that  $z^2 +z +1=\lambda^2$ can be rewritten as $\lambda^2=z^2+ 1^2 -2 \cdot z \cdot 1 \cdot \cos 120^\circ $, which is the above law for a triangle with sides of length $z$ and $\lambda$ and $1$. 
Applying this method to the other two equations we get that $x,y,z$ are the cevians of an equilateral triangle with sides of length $\lambda$ and since we've already established that one those is $1$, they're all equal to $1$, i.e $x=y=z=1$, the result then follows as mentioned above. For a visualization see this figure.
