Proof Using Rearrangement Inequality

In summary, the Rearrangement Inequality states that when two sequences are similarly arranged, their sum is maximized. Using this theorem, it is possible to prove that for any given 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. This can be achieved by rewriting the denominators using negative exponents and using the fact \frac{b}{a}+\frac{a}{c} + \frac{
  • #1
James Brady
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4
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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  • #2
Hint:
[tex]\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}[/tex]
 
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Likes James Brady
  • #3
Perfect, I used that to build a common denominator which then canceled out.
 

Related to Proof Using Rearrangement Inequality

1. What is the rearrangement inequality?

The rearrangement inequality is a mathematical concept that states that if two sequences of real numbers are arranged in the same order, then their dot product will be maximum. This means that the sum of the products of the largest numbers in each sequence will be greater than or equal to the sum of the products of the smallest numbers in each sequence.

2. How is the rearrangement inequality used in proofs?

The rearrangement inequality is often used in proofs to show that a certain expression or equation is the maximum or minimum possible value. By rearranging the terms in a specific way, the inequality can be applied to show that the value is indeed the maximum or minimum.

3. Can the rearrangement inequality be applied to any set of numbers?

Yes, the rearrangement inequality can be applied to any set of real numbers as long as there are two sequences of numbers that can be arranged in the same order. This can be proven using basic algebra and the properties of inequalities.

4. Are there any special cases where the rearrangement inequality does not hold?

Yes, there are certain cases where the rearrangement inequality may not hold, such as when one sequence contains all negative numbers. In these cases, the inequality may need to be modified or a different approach may need to be used in the proof.

5. How does the rearrangement inequality relate to other mathematical concepts?

The rearrangement inequality is closely related to other concepts such as the Cauchy-Schwarz inequality and the AM-GM inequality. These inequalities all involve the comparison of different sequences of numbers and their products, and can often be used together in proofs to strengthen the argument.

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