Determine all real numbers $x, y, z$ satisfying the following system of equations: $x=\sqrt{2 y+3}, y=\sqrt{2 z+3}, z=\sqrt{2 x+3}$.
It is clear that the numbers $x, y, z$ must be strictly positive. The first two equations give $x^{2}=2 y+3$ and $y^{2}=2 z+3$. By subtracting them, we obtain $x^{2}-y^{2}=2(y-z)$. From this, we deduce that if $x \leqslant y$ then $y \leqslant z$, and similarly if $y \leqslant z$ then $z \leqslant x$. Therefore, if $x \leqslant y$, we have $x \leqslant y \leqslant z \leqslant x$, which implies that $x=y=z$.

We can similarly show that if $x \geqslant y$ then $x=y=z$. Therefore, in all cases, $x, y, z$ are equal, and their common value satisfies the equation $x^{2}=2 x+3$, which can also be written as $(x-3)(x+1)=0$. Since $x$ is strictly positive, it must be that $x=y=z=3$. Conversely, it is immediately verified that $x=y=z=3$ is indeed a solution to the system.