6. Find all triples of positive reals $(x, y, z)$ that satisfy the system

$$
2 x \sqrt{x+1}-y(y+1)=1
$$

$$
\begin{aligned}
& 2 y \sqrt{y+1}-z(z+1)=1 \\
& 2 z \sqrt{z+1}-x(x+1)=1
\end{aligned}
$$
Solution 1. Note that, by the arithmetic and geometric mean inequality, we have

$$
x^{2}+x+1 \geq 2 \sqrt{x^{2}(x+1)}=2 x \sqrt{x+1}
$$

with equality if and only if $x^{2}=x+1$, that is, if and only if $x$ is a root of the equation $r^{2}-r-1=0$. From the first equation, we then have

$$
y^{2}+y+1=2 x \sqrt{x+1} \leq x^{2}+x+1
$$

and similarly for the other two, so

$$
x^{2}+x+1 \geq y^{2}+y+1 \geq z^{2}+z+1 \geq x^{2}+x+1
$$

which means that equality must hold in all three inequalities, that is, $x, y, z$ are solutions of the equation $r^{2}-r-1=0$. The product of the two roots of this equation is -1, so exactly one of them is negative, and $x, y, z$ are equal to each other and equal to the positive root, that is, the only solution is

$$
x=y=z=\frac{1+\sqrt{5}}{2}
$$