Solve the system $x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$ 
Find all real numbers $x,\ y,\ z$ that satisfy $$x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$$

First natural move would be rewriting the system as:
$$x^{2}+y^{2}+2xy=4z-1$$
$$x^{2}+z^{2}+2xz=4y-1$$
$$z^{2}+y^{2}+2zy=4x-1$$
Thus,
$$2x^{2}+2y^{2}+2z^{2}+2xy+2yz+2zx-4x-4y-4z+3=0$$
And I tried to write the polynomial above as the sum of complete squares. Is this possible or do you have other ideas?

Note that these equations are defined only for $x,y,z\ge 1/4$.
Consider the first two equations
$$x+y=\sqrt{4z-1}$$
$$x+z=\sqrt{4y-1}$$
Subtracting these two equations gives us
$$y-z=\sqrt{4z-1}-\sqrt{4y-1}$$
If $y\gt z$, then the LHS is positive but the RHS is negative, so this cannot be. Similarly, if $y\lt z$, then the LHS is negative but the RHS is positive, which is also impossible. Thus, we have that
$$y-z=\sqrt{4z-1}-\sqrt{4y-1}=0$$
or 
$$y=z$$
We may show by similar reasoning that $x=y$, and so $x=y=z$.
Then we have that
$$2x=\sqrt{4x-1}$$
or
$$4x^2-4x+1=0$$
Which has one solution:
$$x=\frac{1}{2}$$
This gives us the only solution to your system:
$$x=y=z=\frac{1}{2}$$
NOTE: This problem is trivialized by the fact that both $x$ and $\sqrt{4x-1}$ are increasing functions. The following system of equations might be more interesting:
$$x+y=\frac{1}{\sqrt{4z-1}}$$
$$x+z=\frac{1}{\sqrt{4y-1}}$$
$$y+z=\frac{1}{\sqrt{4x-1}}$$
