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Analysis Inequality

  1. Oct 1, 2008 #1
    hey guys, i came across this inequality in analysis and am not sure how to prove it. Any ideas? It's not homework, im 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
     
  2. jcsd
  3. Oct 1, 2008 #2

    morphism

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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]
     
  4. Oct 1, 2008 #3
    Thats pretty sweet. Thats hard to generalize, no?
     
  5. Oct 1, 2008 #4

    morphism

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    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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