Maths Olympiad Prep

Library / /56 of 264

Algebra Difficulty 5.1 AIME, harder Prove it Romania

Let aa, bb, cc be distinct positive integers.

a) Prove that a2b2+a2c2+b2c2≥9a^2b^2 + a^2c^2 + b^2c^2 \ge 9.

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

Solution

a) At least one of the numbers aa, bb, cc has modulus at least 22, whence a2b2+a2c2+b2c2≥1⋅1+1⋅4+1⋅4=9a^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
ab+ac+bc−2(a+b+c)+3≥6ab+ac+bc−3≥2(a+b+c)(ab+ac+bc−3)(ab+ac+bc+3)≥2abc(a+b+c)(ab+ac+bc)2−9≥2abc(a+b+c)a2b2+a2c2+b2c2≥9, \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).

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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