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Algebra Difficulty 5.4 AIME, harder Prove it Ukraine

max⁡{x2+3x+3,x2011+x4+x2+x+1}≤min⁡{1−x−x2,x2011+x4+x2+x+1}, \max\{x^2+3x+3, x^{2011}+x^4+x^2+x+1\} \le \min\{1-x-x^2, x^{2011}+x^4+x^2+x+1\},
where max⁡{a,b}={a,if a≥bb,if a<b\max\{a,b\} = \begin{cases} a, & \text{if } a \ge b \\ b, & \text{if } a < b \end{cases}, and min⁡{a,b}={b,if a≥ba,if a<b\min\{a,b\} = \begin{cases} b, & \text{if } a \ge b \\ a, & \text{if } a < b \end{cases}.

Solution

For our convenience we denote P(x)=x2011+x4+x2+x+1P(x) = x^{2011} + x^4 + x^2 + x + 1. It is clear that
max⁡{x2+3x+3,P(x)}≥P(x)≥min⁡{1−x−x2,P(x)}, \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⁡{x2+3x+3,P(x)}=P(x)=min⁡{1−x−x2,P(x)}, \max\{x^2+3x+3, P(x)\} = P(x) = \min\{1-x-x^2, P(x)\},
and this, in turn, implies that
x2+3x+3≤P(x)≤1−x−x2. x^2+3x+3 \le P(x) \le 1-x-x^2.
So, x2+3x+3≤1−x−x2x^2+3x+3 \le 1-x-x^2, which yields that 2x2+4x+2≤0⇔(x+1)2≤0⇔x=−12x^2+4x+2 \le 0 \Leftrightarrow (x+1)^2 \le 0 \Leftrightarrow x=-1. Hence, the only possible solution is x=−1x=-1. A simple verification shows that this value of xx satisfies the original problem.

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