$(x + y + z)^2 \leq 3(x^2 + y^2 + z^2)$

= $x^2 + y^2 + z^2 + 2xy + 2xz + 2yz \leq 3x^2 + 3y^2 + 3z^2$

= $x^2 + y^2 + z^2 + 2xy + 2xz + 2yz - 3x^2 - 3y^2 - 3z^2 \leq 0$

= $-2x^2 - 2y^2 - 2z^2 + 2xy + 2xz + 2yz \leq 0$

= $-2(x^2 + y^2 + z^2 - xy - xz - yz) \leq 0$

= $x^2 + y^2 + z^2 - xy - xz - yz \geq 0$

= $(x^2 - xy - xz) + (y^2 - yz) + z^2 \geq 0$

= $\frac{1}{2}(x^2 - 2xy + y^2 + x^2 - 2xz + z^2 + y^2 - 2yz + z^2) \geq 0$

= $\frac{1}{2}[(x - y)^2 + (x - z)^2 + (y - z)^2] \geq 0$

= $(x - y)^2 + (x - z)^2 + (y - z)^2 \geq 0$

Bất đẳng thức luôn đúng với mọi $x$, $y$, $z$, dấu "=" xảy ra khi $x = y = z$

@SuperSolver