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.
 A: Hint: $\;x,y,z \ge \frac{1}{4}$ for the square roots to be defined. Assume WLOG that $x \ge y \ge z\,$, then $\sqrt{4z-1}=x+y \ge z+z = 2z \implies 4z-1 \ge 4z^2\iff (2z-1)^2 \le 0 \,$.
A: Suppose $x,y,z$ are real numbers such that
\begin{align*}
x + y &= \sqrt{4z-1}\\[4pt]
y + z &= \sqrt{4x-1}\\[4pt]
z + x &= \sqrt{4y-1}\\[4pt]
\end{align*}
Let $s = x + y + z$. Since $x + y \ge 0,\;\;y + z \ge 0,\;\;z + x \ge 0$, we have $s \ge 0$.

Then from the original system of equations, we get
\begin{align*}
(s-z)^2 &= 4z-1\\[4pt]
(s-x)^2 &= 4x-1\\[4pt]
(s-y)^2 &= 4y-1\\[4pt]
\end{align*}
Let $f(u) = (s - u)^2 - (4u - 1) = u^2 - (2s + 4)u  + (s^2+1)$.

Then, $x,y,z$ are roots of $f$, hence, since $f$ is quadratic in $u$, at least two of $x,y,z$ must be equal.

Without loss of generality, assume $z = y$.

Suppose $x \ne y$. Then since $x,y$ are roots of $f$, Vieta's formulas yield
\begin{align*}
&x + y = 2s + 4\\[4pt]
&xy = s^2 +1\\[4pt]
\end{align*}
hence, since $s \ge 0$, we get $x,y > 0$. Then also $z > 0$, since $z=y$. But then
\begin{align*}
&x,y,z > 0\\[4pt]
\implies\; &x + y < s\\[4pt]
\implies\; &2s + 4 < s\\[4pt]
\implies\; &s < -4\\[4pt]
\end{align*}
contradiction.

It follows that $x = y$, hence $x = y = z$, so
\begin{align*}
&(s - z)^2 = 4x-1\\[4pt]
\implies\; &(2x)^2 = 4x -1\\[4pt]
\implies\; &(2x-1)^2 = 0\\[4pt]
\implies\ &x = {\small{\frac{1}{2}}}\\[4pt]
\implies\; &x = y = z = {\small{\frac{1}{2}}}\\[4pt]
\end{align*}
It's easily verfied that the triple
$$
(x,y,z) = 
{\small{
\left(
\frac{1}{2},
\frac{1}{2},
\frac{1}{2}
\right)
}}
$$
satisfies the original system of equations, hence it's the only solution.
A: 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 $x\ne y$, $y\ne x$, $z\ne x$.
