<s>[INST] 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.
 [/INST] Hint: $\;x,y,z \ge \frac{1}{4}$ for the square roots to be defined. Assume WLOG that $x \ge y \ge z\,$, then $\sqrt{4z-1}=x+y \ge z+z = 2z \implies 4z-1 \ge 4z^2\iff (2z-1)^2 \le 0 \,$.
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