I Proof Using Rearrangement Inequality

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The Rearrangement Inequality indicates that the sum of products of similarly arranged sequences is maximized. The discussion focuses on applying this theorem to prove that for any three positive whole numbers a, b, and c, the inequality $$\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} \geq \frac{b}{a} + \frac{a}{c} + \frac{c}{b}$$ holds true. Participants explore the challenge of handling denominators and consider rewriting them with negative exponents to facilitate the proof. A breakthrough occurs when a common denominator is established, leading to cancellation and simplification of the terms. This approach successfully demonstrates the validity of the inequality.
James Brady
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The Rearrangement Inequality states that for two sequences ##{a_i}## and ##{b_i}##, the sum ##S_n = \sum_{i=1}^n a_ib_i## is maximized if ##a_i## and ##b_i## are similarly arranged. That is, big numbers are paired with big numbers and small numbers are paired with small numbers.

The question given is using the above theorem to to prove that for any given three positive whole numbers a,b and c:

$$\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} \geq \frac{b}{a} + \frac{a}{c} + \frac{c}{b}$$

Thinking so far: the theorem deals with maximizing or minimizing the Cartesian product of sets, but we have denominators in each term, maybe I can rewrite the denominators using negative exponents...? But I don't see how that gets me any closer either, basically I need some kind of nudge in the right direction.​
 
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Hint:
\frac{b}{a}+\frac{a}{c} + \frac{c}{b} = \frac{b}{c}\frac{c}{a}+\frac{a}{b}\frac{b}{c}+\frac{a}{b}\frac{c}{a}
 
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Perfect, I used that to build a common denominator which then canceled out.
 
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