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Algebra Difficulty 5.4 AIME, harder Prove it

Let a,b,ca, b, c be lengths of triangle sides, p=ab+bc+cap=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} and q=ac+cb+baq=\frac{a}{c}+\frac{c}{b}+\frac{b}{a}.

Prove that ∣p−q∣<1|p-q|<1.

Solution

One has

abc∣p−q∣=abc∣c−ba+a−cb+b−ac∣=∣bc2−b2c+a2c−ac2+ab2−a2b∣==∣abc−ac2−a2b+a2c−b2c+bc2+ab2−abc∣==∣(b−c)(ac−a2−bc+ab)∣==∣(b−c)(c−a)(a−b)∣. \begin{aligned} a b c|p-q| & =a b c\left|\frac{c-b}{a}+\frac{a-c}{b}+\frac{b-a}{c}\right| \\ & =\left|b c^{2}-b^{2} c+a^{2} c-a c^{2}+a b^{2}-a^{2} b\right|= \\ & =\left|a b c-a c^{2}-a^{2} b+a^{2} c-b^{2} c+b c^{2}+a b^{2}-a b c\right|= \\ & =\left|(b-c)\left(a c-a^{2}-b c+a b\right)\right|= \\ & =|(b-c)(c-a)(a-b)| . \end{aligned}

Since ∣b−c∣<a|b-c|<a, ∣c−a∣<b|c-a|<b and ∣a−b∣<c|a-b|<c we infer

∣(b−c)(c−a)(a−b)∣<abc |(b-c)(c-a)(a-b)|<a b c

and

∣p−q∣=∣(b−c)(c−a)(a−b)∣abc<1 |p-q|=\frac{|(b-c)(c-a)(a-b)|}{a b c}<1

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.