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2bornot2b

  • 3 years ago

If a, b, c be three positive numbers and a+b+c=1 then prove that \[(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\ge 33\frac{1}{3}\]

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  1. 2bornot2b
    • 3 years ago
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    @JamesJ Please help me out with this...

  2. JamesJ
    • 3 years ago
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    Well, by the Cauchy-Schwartz inequality, \[ (<a+1/a,b+1/b,c+1/c> . <1,1,1>)^2 \leq \] \[(1^2 + 1^2 + 1^2) (a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2 \] Play around with that for a while now.

  3. JamesJ
    • 3 years ago
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    *there should be brackets around all three terms on the right: ( (a+1/a)^2 + (b+1/b)^2 + (c+1/c)^2 )

  4. JamesJ
    • 3 years ago
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    i.e., what the Cauchy-Schwartz inequality says is that if a and b are vectors in \( \mathbb{R}^3 \), then \[ (a \cdot b)^2 \leq ||a||^2||b||^2 \]

  5. 2bornot2b
    • 3 years ago
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    I understand what you are saying. I am trying it. I will let you know if I get stuck.

  6. 2bornot2b
    • 3 years ago
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    Thanks for the help

  7. JamesJ
    • 3 years ago
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    Ok. It will come down to you being able to write a smart bound for 1/a + 1/b + 1/c

  8. mahmit2012
    • 3 years ago
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    |dw:1341502540913:dw|

  9. mahmit2012
    • 3 years ago
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    |dw:1341502627816:dw|

  10. mahmit2012
    • 3 years ago
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    |dw:1341502695269:dw|

  11. 2bornot2b
    • 3 years ago
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    @JamesJ I used A.M.>H.M. to deal with the part 1/a + 1/b + 1/c. I showed that it is greater than 9, and eventually proved the thing. I hope, this is the right way, is it?

  12. mahmit2012
    • 3 years ago
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    you can't deal with part of function !! 1/a +1/b +1/c can't consider alone.

  13. mahmit2012
    • 3 years ago
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    In minimum solution should consider all term together .

  14. JamesJ
    • 3 years ago
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    @2bornot2b, yes, sounds right. The Lagrange multiplier method that mahmit has used is also perfectly sound, but we can avoid using such a 'sophisticated' method.

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