51st Ukrainian National Mathematical Olympiad, 4th Round

For our convenience we denote $P(x)=x^{2011}+x^4+x^2+x+1$. It is clear that $$\max \{x^2+3x+3,P(x)\}\ge P(x)\ge \min \{1-x-x^2,P(x)\},$$ so the inequality from the problem condition can be satisfied only if $$\max \{x^2+3x+3,P(x)\}=P(x)=\min \{1-x-x^2,P(x)\},$$ and this, in turn, implies that $$x^2+3x+3\le P(x)\le 1-x-x^2.$$ So, $x^2+3x+3\le 1-x-x^2$, which yields that $2x^2+4x+2\le 0\iff (x+1{)}^2\le 0\iff x=-1$. Hence, the only possible solution is $x=-1$. A simple verification shows that this value of $x$ satisfies the original problem.