<s>[INST] 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!
 [/INST] $$x+y=\sqrt{4z-1}\tag{1}$$
$$y+z=\sqrt{4x-1}\tag{2}$$
$$z+x=\sqrt{4y-1}\tag{3}$$
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 $\color{blue}{\boxed{\color{red}{(x,y,z)=\left(\frac12,\frac12,\frac12\right)}}}$
</s>