<s>[INST] 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?
 [/INST] Necessarily $x,y,z\ge{\large{\frac{1}{4}}}$.

Without loss of generality, assume $x=\min(x,y,z)$.

Then $2x \le y+z$, hence
\begin{align*}
&(y+z)^2=4x-1\\[4pt]
\implies\;&(2x)^2\le 4x-1\\[4pt]
\implies\;&4x^2-4x+1\le 0\\[4pt]
\implies\;&(2x-1)^2\le 0\\[4pt]
\implies\;&(2x-1)^2=0\\[4pt]
\implies\;&x={\small{\frac{1}{2}}}\\[4pt]
\end{align*}
Then $y,z\ge {\large{\frac{1}{2}}}$, and $4x-1=1$, hence 
\begin{align*}
&(y+z)^2=4x-1\\[4pt]
\implies\;&(y+z)^2=1\\[4pt]
\implies\;&y+z=1\\[4pt]
\implies\;&y=z={\small{\frac{1}{2}}}\\[4pt]
\end{align*}
Therefore $x=y=z={\large{\frac{1}{2}}}$.
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