59th Ukrainian National Mathematical Olympiad

Answer: $(x,y,z)=(t,2t,2t),(t,0,0),(0,t,0),(0,0,t)$.

Consider the given inequality as quadratic relative to the variable $x$: $$x^2(3y^2+3z^2-2yz)-2x(y^{2}z+z^{2}y)+y^2{z}^2\ge 0.$$ Since $3y^2+3z^2\ge 6|yz|\ge 2yz$, consider the case that $3y^2+3z^2=2yz$. Then both variables should be equal to zero, which gives a solution in which equality is achieved. In all other cases it is a quadratic trinomial with branches up. Let's find its discriminant: $$\frac{1}{4}D=(y^{2}z+z^{2}y{)}^2-(3y^2+3z^2-2yz)y^2{z}^2=y^2{z}^2((y+z{)}^2-(3y^2+3z^2-2yz))=-2y^2{z}^2(y-z{)}^2\le 0.$$

Thus it is proved that the left part is non-negative, because the discriminant is not positive.

Consider the conditions under which equality is possible, that is, when the discriminant is zero.

1 case. $y=0$. Then the equality transforms into $3x^2{z}^2=0$. Thus all triples $(0,0,z),(0,y,0)$ and $(x,0,0)$ are sought ones.

2 case: $y=z\ne 0$. Then $4x^2{y}^2=4x{y}^3-y^4\Rightarrow 4x^2-4xy+y^2=0\Rightarrow (2x-y{)}^2=0$. Thus the triples $(t,2t,2t)$ for any real $t$ are the answer too.