Obviously $x,y,z \ge 0$. Suppose that $x < 3$. Then $(x+1)(x-3) = x^2 - 2x - 3 < 0$, so $x^2 < 2x+3 = z^2$ and so $x < z$. However, $2x + 3 < 9$, so $z = \sqrt{2x+3} < 3$. So $x < z < 3$. By the same reasoning, we can deduce that $z < y < 3$ and $y < x < 3$. But this leads to $x < z < y < x$, contradiction. Suppose that $x > 3$. Then $(x+1)(x-3) = x^2 - 2x - 3 > 0$, so $x^2 > 2x+3 = z^2$ and so $x > z$. However, $2x + 3 > 9$, so $z = \sqrt{2x+3} > 3$. So $x > z > 3$. By the same reasoning, we can deduce that $z > y > 3$ and $y > x > 3$. But this leads to $x > z > y > x$, contradiction. Hence $x = 3$, which leads to $z = 3, y = 3$ and so the sole solution $(x,y,z) = (3,3,3)$ which satisfies.

djb86 wrote: Solve the following system of equations for real $x,y$ and $z$: \begin{eqnarray*} x &=& \sqrt{2y+3}\\ y &=& \sqrt{2z+3}\\ z &=& \sqrt{2x+3}. \end{eqnarray*} Write (squaring) the system as : $x,y,z\ge 0$ $(x-3)(x+3)=2(y-3)$ $(y-3)(y+3)=2(z-3)$ $(z-3)(z+3)=2(x-3)$ If one $x,y,z$ is $3$, we immediately get $\boxed{x=y=z=3}$ If none is $3$, multiplying the three lines, we get $(x+3)(y+3)(z+3)=8$ which is impossible since $x,y,z\ge 0$ and so $LHS\ge 27$ So no other solution. Edit : too late