Find $x+y+z$ From Equations Including Radicals Let $(x,y,z)$ be the real solution to the system of equations
\begin{align*}
x+y &= \sqrt{4z-1} ,\\
y+z &= \sqrt{4x- 1} , \\
z+x &= \sqrt{4y-1} . 
\end{align*}
Find $x+y+z.$
I could add all the equations up but that doesn't do any good.  Thanks in advance!

To initiate Arthur's hints in the comments:
$$x^2 + 2xy + y^2 = (x + y)^2 = 4z - 1$$
$$y^2 + 2yz + z^2 = (y + z)^2 = 4x - 1$$
$$z^2 + 2zx + x^2 = (z + x)^2 = 4y - 1$$
Rather than adding all three equations, we subtract them pairwise.  Then assuming $x \neq y \neq z$, we have:
$$2y(x - z) + (x + z)(x - z) = 4(z - x) \implies 2y + x + z = -4$$
$$2z(y - x) + (y + x)(y - x) = 4(x - y) \implies 2z + y + x = -4$$
$$2x(y - z) + (y + z)(y - z) = 4(z - y) \implies 2x + y + z = -4$$
Equating, we get
$$2y + x + z = 2z + y + x = 2x + y + z = -4.$$
Therefore, we obtain
$$x + y + z = -4 - y = -4 - z = -4 - x.$$
This implies that
$$x = y = z,$$
which contradicts our earlier assumption.
Therefore, the original system of equations has no real solution.
Added August 27 2016

Therefore, for the original system of equations to have a real solution, we must have $x = y = z$.

Consequently, the system is reduced to solving the lone equation
$$x + x = \sqrt{4x - 1}$$
$$(2x)^2 = 4x - 1$$
$$4x^2 - 4x + 1 = 0$$
$$x = \frac{1}{2}.$$
Thus,
$$x = y = z = \frac{1}{2}$$
is the only real solution.
QED
