Ta có:

x² + y² + 2z² + xy + 2yz + 2zx + x + y + 1 = 0

⇔ 2(x² + y² + 2z² + xy + 2yz + 2zx + x + y + 1) = 0

⇔ 2x² + 2y² + 4z² + 2xy + 4yz + 4zx + 2x + 2y + 2 = 0

⇔ (x² + 2xy + y²) + 4z(x + y) + 4z² + (x² + 2x + 1) + (y² + 2y + 1) = 0

⇔ (x + y)² + 4z(x + y) + 4z² + (x + 1)² + (y + 1)² = 0

⇔ (x + y + 2z)² + (x + 1)² + (y + 1)² = 0

Vì (x + y + 2z)² ≥ 0 với mọi x, y, z

(x + 1)² ≥ 0 với mọi x

(y + 1)² ≥ 0 với mọi y

Nên (x + y + 2z)² + (x + 1)² + (y + 1)²  ≥ 0 với mọi x, y, z

Suy ra  $\left\{\begin{array}{l}x+y+2\text{z}=0\\ x+1=0\\ y+1=0\end{array}\right.\iff \left\{\begin{array}{l}2\text{z}=-x-4\\ x=-1\\ y=-1\end{array}\right.\iff \left\{\begin{array}{l}\text{z}=1\\ x=-1\\ y=-1\end{array}\right.$

Vậy x = –1, y = –1, z = 1.