<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] Solution. $\blacktriangleleft$ Subtract every pair of the equation, we would obtain something like
\begin{align*}
y-z &= (x+y) - (x+z) \\ &= \sqrt {4z-1} - \sqrt {4y-1} \\&= \frac {4z-1 - 4y+1} {\sqrt {4z-1} + \sqrt {4y-1}} \\&= 4\frac {z-y} {\sqrt {4z-1} + \sqrt {4y-1}},
\end{align*}
then $y -z$ must be zero, otherwise a sum of two square roots would be negative, which is impossible. Thus we have $x = y =z$. Solve any one of them would obtain that 
$$
x = y = z = \frac 12. \blacktriangleright
$$
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