Library / Ukrainian National Mathematical Olympiad, 4th Round /9 of 20

Ukrainian National Mathematical Olympiad, 4th Round

Algebra Difficulty 5.4 AIME, harder Prove it Ukraine

$\max\{x^2+3x+3, x^{2011}+x^4+x^2+x+1\} \le \min\{1-x-x^2, x^{2011}+x^4+x^2+x+1\},$
where $\max\{a,b\} = \begin{cases} a, & \text{if } a \ge b \\ b, & \text{if } a < b \end{cases}$ , and $\min\{a,b\} = \begin{cases} b, & \text{if } a \ge b \\ a, & \text{if } a < b \end{cases}$ .

Solution

For our convenience we denote $P(x) = x^{2011} + x^4 + x^2 + x + 1$ . 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 $2x^2+4x+2 \le 0 \Leftrightarrow (x+1)^2 \le 0 \Leftrightarrow x=-1$ . Hence, the only possible solution is $x=-1$ . A simple verification shows that this value of $x$ satisfies the original problem.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.