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

Mathematics
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@JamesJ Please help me out with this...
Well, by the Cauchy-Schwartz inequality, \[ ( . <1,1,1>)^2 \leq \] \[(1^2 + 1^2 + 1^2) (a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2 \] Play around with that for a while now.
*there should be brackets around all three terms on the right: ( (a+1/a)^2 + (b+1/b)^2 + (c+1/c)^2 )

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Other answers:

i.e., what the Cauchy-Schwartz inequality says is that if a and b are vectors in \( \mathbb{R}^3 \), then \[ (a \cdot b)^2 \leq ||a||^2||b||^2 \]
I understand what you are saying. I am trying it. I will let you know if I get stuck.
Thanks for the help
Ok. It will come down to you being able to write a smart bound for 1/a + 1/b + 1/c
|dw:1341502540913:dw|
|dw:1341502627816:dw|
|dw:1341502695269:dw|
@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?
you can't deal with part of function !! 1/a +1/b +1/c can't consider alone.
In minimum solution should consider all term together .
@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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