Maths Olympiad Prep

Track / Stage 6 / 118 of 400 #1118 of 1964

Problem 1118

National Olympiad, first round
Algebra Difficulty 6.1 Prove it Fall Mathematical Competition · Bulgaria

Find all real numbers rr such that the inequality
r(ab+bc+ca)+(3r)(1a+1b+1c)9 r(ab + bc + ca) + (3 - r) \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \ge 9
holds true for arbitrary positive numbers a,ba, b and cc.

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.

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Official solution

Taking a=b=ca = b = c we obtain
ra2+(3r)1a3    (a1)(r(a2+a+1)3)0 ra^2 + (3 - r) \frac{1}{a} \ge 3 \iff (a - 1)(r(a^2 + a + 1) - 3) \ge 0
for any a>0a > 0. Then it easily follows that r=1r = 1.

Conversely, let r=1r = 1. Then we write the inequality as 2+abc331a+1b+1c\frac{2 + abc}{3} \ge \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}. The GM-HM inequality implies that the right hand side does not exceed abc3\sqrt[3]{abc} and, setting x=abc3>0x = \sqrt[3]{abc} > 0, it is enough to prove that
2+x33x \frac{2+x^3}{3} \ge x
which is equivalent to the obvious (x1)2(x+2)0(x-1)^2(x+2) \ge 0.

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