Track / Stage 4 / 26 of 340 #286 of 1964

Problem 286

AMC 12 late, AIME early

Algebra Difficulty 4.5 Prove it Brazilian Mathematical Olympiad · Brazil

Show that if the positive real numbers $a$ , $b$ satisfy $a^3 = a + 1$ and $b^6 = b + 3a$ , then $a > b$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

$a^3 = a + 1 > 1$ , so $a > 1$ , so $b^6 = b + 3a > 3$ , so $b > 1$ .
$a^6 - b^6 = (a+1)^2 - (b+3a) = (a-1)^2 + (a-b) > a-b$ . But $a^6 - b^6 = (a-b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)$ , so if $b > a$ , then $b^6 - a^6 \ge 6(b-a)$ and hence $a^6 - b^6 < a - b$ . Contradiction. Obviously if $a=b$ , then $a^6 = b^6$ and so $a^6 - b^6$ is not greater than $a-b$ . Hence we must have $a > b$ .

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