Maths Olympiad Prep

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Problem 886

AMC 12 late, AIME early
Algebra Difficulty 4.6 Prove it Saudi Arabian Mathematical Competitions · Saudi Arabia

Let aa, bb, cc be positive real numbers. Prove that
8(a+b+c)(1a+1b+1c)≤9(1+ab)(1+bc)(1+ca) 8(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq 9\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)

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

The inequality is equivalent to
8(a+b+c)(ab+bc+ca)≤9(a+b)(b+c)(c+a) 8(a+b+c)(ab+bc+ca) \leq 9(a+b)(b+c)(c+a)
Using the identity
(a+b+c)(ab+bc+ca)=(a+b)(b+c)(c+a)+abc (a+b+c)(ab+bc+ca) = (a+b)(b+c)(c+a) + abc
we get
8(a+b)(b+c)(c+a)+8abc≤9(a+b)(b+c)(c+a), 8(a+b)(b+c)(c+a) + 8abc \leq 9(a+b)(b+c)(c+a),
hence 8abc≤(a+b)(b+c)(c+a)8abc \leq (a+b)(b+c)(c+a). Applying AM-GM inequality, it follows
2ab≤a+b,2bc≤b+c,2ca≤c+a, 2\sqrt{ab} \leq a+b, \quad 2\sqrt{bc} \leq b+c, \quad 2\sqrt{ca} \leq c+a,
and by multiplication 8abc≤(a+b)(b+c)(c+a)8abc \leq (a+b)(b+c)(c+a). The equality holds if and only if a=b=ca = b = c.

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