I need help with the following system of equations:
$ 2y^3 +2x^2+3x+3=0 $
$ 2z^3 + 2y^2 + 3y + 3= 0 $
$2x^3 + 2z^2 + 3z + 3 = 0$
The only real solution is $x = y = z = -1$.
Claim 1: $x,y,z \ge -1$.
Proof. Suppose that $x < -1$. Then $0 = 2y^3 + 2x^3 + 3x + 3 > 2y^3 + 2$, so that $y < -1$ also. Similarly it follows that $z < -1$. Hence if one of $x,y,z$ is smaller than $-1$, all of them are. But then if for example $x<z$, we have $$0 = 2x^3 + 2z^2 + 3z + 3 < 2z^3 + 2z^2 + 3z + 3 = (z+1)(2z^2 + 3) < 0,$$ and we see that necessarily $x=y=z$, which implies that $x=y=z=-1$, contradiction.
Claim 2: $x,y,z \le -1$.
Proof. Suppose that $x > -1$ is the largest of $x,y,z$. So $z \le x$ and $$0 = 2x^3 + 2z^2 + 3z + 3 \ge 2z^3 + 2z^2 + 3z + 3 = (z+1)(2z^2 + 3),$$ which implies that $z \le - 1$. By Claim 1. $z = -1$ and hence also $x = -1$ and $y = -1$.