## 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 Group Title

2. JamesJ Group Title

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 Group Title

*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 Group Title

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 Group Title

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

6. 2bornot2b Group Title

Thanks for the help

7. JamesJ Group Title

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

8. mahmit2012 Group Title

|dw:1341502540913:dw|

9. mahmit2012 Group Title

|dw:1341502627816:dw|

10. mahmit2012 Group Title

|dw:1341502695269:dw|

11. 2bornot2b Group Title

@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 Group Title

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

13. mahmit2012 Group Title

In minimum solution should consider all term together .

14. JamesJ Group Title

@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.