Proving Inequality: a1 + a2 + · · · + an−1 + an

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    Analysis Inequality
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Discussion Overview

The discussion revolves around proving an inequality involving strictly positive real numbers a1, a2, ..., an. Participants explore various approaches to establish the inequality, which compares the sum of the numbers to a specific expression involving their squares and ratios. The scope includes mathematical reasoning and exploration of inequalities.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents the inequality and expresses curiosity about proving it, indicating it is not a homework problem.
  • Another participant suggests that for the case n=2, the inequality can be derived using the AM-GM inequality, providing a specific mathematical expression to support this claim.
  • A third participant questions the generalizability of the approach used for n=2, implying that it may not extend easily to larger n.
  • A fourth participant proposes using the Cauchy-Schwarz inequality as an alternative method to prove the inequality, presenting a mathematical formulation that involves summation and indexing.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to prove the inequality, with multiple approaches being discussed and some uncertainty expressed regarding the generalization of the methods.

Contextual Notes

There are limitations regarding the generalization of the methods discussed, and the proofs provided are contingent on specific cases or assumptions that may not hold for all n.

hypermonkey2
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hey guys, i came across this inequality in analysis and am not sure how to prove it. Any ideas? It's not homework, I am just curious..

Let a1, a2, . . . , an be strictly positive real numbers. Show that

a1 + a2 + · · · + an−1 + an <= ((a1)^2)/a2 + ((a2)^2)/a3) +...+ ((an)^2)/a1



cheers
 
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Here's the case n=2. It follows from the AM-GM inequality.

[tex] 0 \leq \left(\frac{a_1}{\sqrt{a_2}} - \frac{a_2}{\sqrt{a_1}}\right)^2 = \frac{a_1^2}{a_2} - 2\sqrt{a_1 a_2} + \frac{a_2^2}{a_1} \leq \frac{a_1^2}{a_2} - (a_1 + a_2) + \frac{a_2^2}{a_1}.[/tex]
 
Thats pretty sweet. Thats hard to generalize, no?
 
Yeah. I thought it was an easy induction argument after that, but that's obviously not the case. The flu is making me stupid.

Anyway, we can use the Cauchy-Schwarz inequality instead.

[tex]\left(\sum a_i\right)^2 = \left(\sum \frac{a_i}{\sqrt{a_{i+1}}} \sqrt{a_{i+1}}\right)^2 \leq \sum \frac{a_i^2}{a_{i+1}} \sum a_{i+1},[/tex]

where the index of summation is taken mod n (so a_{n+1} = a_1).
 

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