A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

zepdrix

  • one year ago

Let a, b, and c be non-negative real numbers. Prove the following inequality:

  • This Question is Closed
  1. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    same here

  2. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \(\large\rm ab+bc+ca\le a^2+b^2+c^2\) So here is was I was attempting... For \(\large\rm a,~b,~c\ge0\) we have \(\large\rm (a+b+c)^2\ge0\). Expanding this out gives us: \(\large\rm a^2+b^2+c^2+2ab+2bc+2ca\ge0\)

  3. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    idk but i think you are going to need something else, since you did not really use the fact that they are non negative

  4. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \[(x+b+c)^2\geq 0\] no matter what

  5. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Oh haha good point XD

  6. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    how about thinking about it like the sum sub of 3 perfect square vs the sum of interchanging perfect squares

  7. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so like rectangles formed from mixing and matching the sides of perfect squares

  8. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1441246380649:dw|

  9. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Use Cauchy / Shwarz

  10. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    i think the trick is to write \[a^2+b^2+c^2-ab-bc-ab\] as a sum of squares

  11. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1441246436682:dw|

  12. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Doh >.< Yah teacher mentioned something about Cauchy, I thought it was for a different problem though.. hmm

  13. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    You are allowed?

  14. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    think we can do it without that maybe

  15. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hey but loook my picture works!!

  16. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I am allowed to use Cauchy Schwarz if I first first prove it lol. That's a separate problem though I suppose :)

  17. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    The squares thing? :O works?

  18. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if you arrange any set of A^2,B^2,C^2 whjere A>B>C

  19. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    it is not hard to prove for \(\mathbb{R}\).

  20. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \[a^2+b^2+c^2-ab-ac-bc=(a-b)^2+(b-c)^2+(c-a)^2\] or maybe half of that

  21. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Else, I would assume wlog a<=b<=c to get rid of so many cases.

  22. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    since the right hand side is clearly positive, you get \[a^2+b^2+c^2-ab-ac-bc\geq 0\]

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you don't need cases, do what @misty1212 said, except the algebra is not quite right

  24. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Oh so the idea I started with was kind of a mess? Where do you come up with the subtraction like that though? :d hmm

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it is a common gimmick is all i think

  26. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Ooo that works out very nicely Misty!! :O

  27. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Hmm I have a couple more like this :) Imma give them a shot though before I bug you guys.

  28. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Another great trick in these is to add 0. like +a-a ...

  29. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ya ya ya ya ya :) good call boss

  30. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I'll post the next one just in case anyone wanted to look at it :d\[\large\rm 8abc\le(a+b)(b+c)(c+a)\]

  31. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    hmm

  32. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Well, if you expand the RHS you get 8 terms, maybe use the last one?

  33. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    suppose x,y,z are positive real numbers can we say min(xyz) = min(x)*min(y)*min(z) ?

  34. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ya you get 8 terms :d two of which combine to give us some abc's. Hmm. min what? +_+

  35. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\min\{(a+b)(b+c)(c+a)\} =\min\{a+b)\} *\min\{b+c\}* \min\{c+a\} \] By AM-GM inequality we have \(a+b\ge 2\sqrt{ab}\) giving the desired result

  36. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    presuming \(a,b,c\) are positive reals which keeps the sums \(a+b, ~b+c, ~c+a\) also positive

  37. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Hmm neato :) Thanks

  38. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.