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Problem 1193

AIME late

Algebra Difficulty 5.1 Prove it Romanian Mathematical Olympiad · Romania

Let $a$ , $b$ , $c$ be distinct positive integers.

a) Prove that $a^2b^2 + a^2c^2 + b^2c^2 \ge 9$ .

b) If, moreover, $ab + ac + bc + 3 = abc > 0$ , show that
$(a - 1)(b - 1) + (a - 1)(c - 1) + (b - 1)(c - 1) \ge 6.$

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) At least one of the numbers $a$ , $b$ , $c$ has modulus at least $2$ , whence $a^2b^2 + a^2c^2 + b^2c^2 \ge 1 \cdot 1 + 1 \cdot 4 + 1 \cdot 4 = 9$ .

b) The required inequality can be successively written
$\begin{aligned} & ab + ac + bc - 2(a + b + c) + 3 \ge 6 \\ & ab + ac + bc - 3 \ge 2(a + b + c) \\ & (ab + ac + bc - 3)(ab + ac + bc + 3) \ge 2abc(a + b + c) \\ & (ab + ac + bc)^2 - 9 \ge 2abc(a + b + c) \\ & a^2b^2 + a^2c^2 + b^2c^2 \ge 9, \end{aligned}$
that is exactly a).

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