55rd Ukrainian National Mathematical Olympiad - Third Round

Rewrite the equation as: $$2x^2{y}^2+2y^2{z}^2+2z^2{x}^2-x^4-y^4-z^4=-(x^2{y}^2+y^2{z}^2+z^2{x}^2).$$ The left-hand side can be decomposed as: $$(x+y+z)(x+y-z)(y+z-x)(z+x-y)=-(x^2{y}^2+y^2{z}^2+z^2{x}^2).$$ Hence, the left-hand side is negative, therefore, at least one of the multipliers is negative too. The first one is always positive, then, without loss of generality, we can assume the second one is negative, i.e. $x+y-z<0$, which contradicts the triangle inequality.