Solve for $x,y$ and $z$ when: $x+y=\sqrt{4z-1}, \ y+z=\sqrt{4x-1}\ \mathrm{and} \ z+x=\sqrt{4y-1}$ How to Solve for $x,y$ and $z$ when: $x+y=\sqrt{4z-1}, \ y+z=\sqrt{4x-1}\ \mathrm{and} \ z+x=\sqrt{4y-1}$.
I've added the equations and get: $2(x+y+z)=\sqrt{4x-1}+\sqrt{4y-1}+\sqrt{4z-1}$ and after this how I proceed further?
Even if I subtract the first two equations then I'm getting:
\begin{align*}
x-z&=\sqrt{4z-1}-\sqrt{4x-1}\\
\Rightarrow\left(x+\sqrt{4x-1}\right)^2&=\left(z+\sqrt{4z-1}\right)^2\\
\Rightarrow\left(x^2-z^2\right)+4(x-z)&=2z\sqrt{4z-1}-2x\sqrt{4x-1}
\end{align*}
after this also I can't understand how to help with this.

If there is a solution, the symmetry of the equations suggests we investigate the possibility that $x=y=z$
So assume that $x=y=z=a$.
$$2a=\sqrt{4a-1}$$
Thus $4a^2-4a+1=0$, or $(2a-1)^2=0$. Thus $x=y=z=\frac{1}{2}$.
This is a solution. But, as per the comment of dxiv below, this does not rule out solutions where $x\ne y$, $y\ne x$, $z\ne x$.
