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!
 A: $$x+y=\sqrt{4z-1}\tag{1}$$
$$y+z=\sqrt{4x-1}\tag{2}$$
$$z+x=\sqrt{4y-1}\tag{3}$$
Adding $(1),(2)$ and $(3)$, we get 
$$x+y+z=\sqrt{x-\frac14}+\sqrt{y-\frac14}+\sqrt{z-\frac14}$$
Let $a=\sqrt{x-\frac14}$,$b=\sqrt{y-\frac14}$ and $c=\sqrt{z-\frac14}$
Using these substitutions, we get 
$$\left(a^2+\frac14\right)+\left(b^2+\frac14\right)+\left(c^2+\frac14\right)=a+b+c$$
$$ \Longleftrightarrow \left(a-\frac12\right)^2+\left(b-\frac12\right)^2+\left(c-\frac12\right)^2=0$$
Thus, the only possible solution is $\color{blue}{\boxed{\color{red}{(x,y,z)=\left(\frac12,\frac12,\frac12\right)}}}$
A: 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
A: We can solve the system for $x, y, z$. By the symmetry of the equations, we can say $x=y=z$, and solve any of them for this value. (all equations become the same one under that conditions: $2t = \sqrt{4t-1}$) We would get $t= \frac12$ for $t = x=y= z$. 
So, we found a faster way to one of the solutions. It is not hard to prove that it is also the only solution. For example, by squaring the first two equations and after subtraction:
$(x+y)^2 - (y+z)^2 = 4 (z-x)$
If $z>x$ then the right side is positive, and the left negative. If $z<x$ then the right side is negative, and the left positive. So $x=z$. The same way we can do it with the second and third equation to get $x=y$. (it is also clear that $x, y, z$ must be positive numbers so the roots are defined)
