## 2bornot2b Group Title If a, b, c be three positive numbers and a+b+c=1 then prove that $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\ge 33\frac{1}{3}$ 2 years ago 2 years ago

1. 2bornot2b

2. JamesJ

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

*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

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

I understand what you are saying. I am trying it. I will let you know if I get stuck.

6. 2bornot2b

Thanks for the help

7. JamesJ

Ok. It will come down to you being able to write a smart bound for 1/a + 1/b + 1/c

8. mahmit2012

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9. mahmit2012

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10. mahmit2012

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11. 2bornot2b

@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

you can't deal with part of function !! 1/a +1/b +1/c can't consider alone.

13. mahmit2012

In minimum solution should consider all term together .

14. JamesJ

@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.