A community for students.
Here's the question you clicked on:
 0 viewing
2bornot2b
 3 years ago
If a, b, c be three positive numbers and a+b+c=1 then prove that
\[(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\ge 33\frac{1}{3}\]
2bornot2b
 3 years ago
If a, b, c be three positive numbers and a+b+c=1 then prove that \[(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\ge 33\frac{1}{3}\]

This Question is Closed

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0@JamesJ Please help me out with this...

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Well, by the CauchySchwartz 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.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1*there should be brackets around all three terms on the right: ( (a+1/a)^2 + (b+1/b)^2 + (c+1/c)^2 )

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1i.e., what the CauchySchwartz inequality says is that if a and b are vectors in \( \mathbb{R}^3 \), then \[ (a \cdot b)^2 \leq a^2b^2 \]

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0I understand what you are saying. I am trying it. I will let you know if I get stuck.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Ok. It will come down to you being able to write a smart bound for 1/a + 1/b + 1/c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1341502540913:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1341502627816:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1341502695269:dw

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0@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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can't deal with part of function !! 1/a +1/b +1/c can't consider alone.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In minimum solution should consider all term together .

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1@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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.