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Inequality If abc=1

  1. Apr 30, 2003 #1
    If abc=1
    prove
    1/[a3(b+c)] + 1/[b3(a+c)] + 1/[c3(b+a)] >= 3/2
    where a, b, c are positive real numbers.

    Please give me a hint.
     
    Last edited by a moderator: Feb 4, 2013
  2. jcsd
  3. May 1, 2003 #2
    I think that a>0, b>0, c>0...because if this is not true then for a=5, b=-1, c=-1/5 the left part is equal to -31.465, which is smaller than 3/2...
    So...I'll use this well known inequality (I don't know the english name)...
    (a1+a2+..+an)/n>=(a1*a2*...*an)^(1/n)...where a1,...an>0 and n>=2...
    Let n=3 and a1=1/(a^3*(b+c)), a2=1/(b^3*(a+c)), a3=1/(c^3*(a+b))...
    We obtain (a1+a2+a3)/3>(a1*a2*a3)^(1/3)...

    Let S=a1+a2+a3...
    S/3>(a1*a2*a3)^(1/3)...evidently a1*a2*a3=1/[(a+b)*(a+c)*(b+c)]...because a^3*b^3*c^3=1...
    So...S/3>(1/[(a+b)*(a+c)*(b+c)])^(1/3)...
    S>3*(1/[(a+b)*(a+c)*(b+c)])^(1/3)...
    It's simple to prove now that (1/[(a+b)*(a+c)*(b+c)])^(1/3)>1/2...the same inequality I used before...good luck...
     
    Last edited: May 1, 2003
  4. May 1, 2003 #3
    wow bogdan, thank you!

    yes, a, b, c are positive real numbers and I forgot to include that in the question.

    I only call it AM>=GM.

    prove 1/[(a+b)*(a+c)*(b+c)])^(1/3)>=1/2
    ==> Prove (a+b)*(a+c)*(b+c) <=8
    use AM>=GM again
    (a+b)/2 >=sqrt(ab)...........(1)
    (a+c)/2 >=sqrt(ac)...........(2)
    (c+b)/2 >=sqrt(cb)...........(3)
    (1)*(2)*(3)
    the result follows
     
  5. May 1, 2003 #4
    Unfortunately...it is so very wrong...
    (a+b)*(a+c)*(b+c)>=8, not <=8...
    There is a solution...use Cauchy-Buniakowsky-Schwartz inequaltity and then AM>GM...
     
  6. May 1, 2003 #5
    This problem is very strange...I strongly believe I've seen it somewhere in a contest...in the Balcaniad math contest...or something...but that sugestion (Cauchy inequality and AM>GM) is good...because I have it in a book...and that's the only hint the author gives...
     
  7. May 1, 2003 #6
    If I remember correctly, it's an IOM question, 1995.

    I think the Cauchy-Buniakowsky-Schwartz inequaltity is different from Cauchy-Schwartz inequality. I haven't heard of C-B-S inequality before. Do you have any link or some explainations about what CBS inequality is?
     
  8. May 1, 2003 #7
    No...the two are the same...
    IOM ? Are you crazy ? Do you know what a headache this inequality gave me ?
    I can't do much more simple inequalities...and I tried this...
    I have this problem in a book for preparing math contests...and the explanation is : use cauchy-...-...- inequality and AM>GM...that's all...and...what a luck...I have a book with IOMs but 1990-1994...
     
  9. May 1, 2003 #8
    well, frankly speaking, I needed to take out my textbook before I could remember what C-S inequality is!

    hehe, I'm not that good in mathematics. I just saw this question and thought it might be an interesting one. I think I got the solutions of IMO questions in 1995 somewhere in my computer but I was too lazy to find it.

    same here.

    I need to stop here now, and revise for my pure math test today!
    Perhaps we can do a little discussion on inequalities later.
     
  10. May 1, 2003 #9
    Last edited: May 1, 2003
  11. May 1, 2003 #10
    usually applying substitutions in harder inequality problems can simplify the questions a bit, but to find suitable substitions is difficult.

    yes, the solution is so nice!
     
    Last edited: May 2, 2003
  12. May 2, 2003 #11
    No...there are only a few "feasable" substitutions...x=1/a...x=a+b...x=a-b...things like these...
     
  13. May 2, 2003 #12
    what's wrong in the above steps ?
     
  14. May 3, 2003 #13
    We have to prove (a+b)*(a+c)*(b+c)<=8, but using AM>GM we obtain (a+b)*(a+c)*(b+c)>=8...exactly the opposite...I've seen another inequality like this...in a IMO romanian team selection test...where after applying AM>GM twice I obtained the opposite inequality...and took 0 points...
     
  15. May 3, 2003 #14
    I think there must be something wrong in the proof, we can't get logically correct answers which contratict each other.
     
  16. May 3, 2003 #15
    I've just shown you the mistake...(a+b)*(b+c)*(a+c)>=8, not viceversa...we wanted the opposite inequality...there's the mistake...we made the expression too small...and now the new expression is less than 3/2...got it ?
     
  17. May 4, 2003 #16
    Bogdan, I think you've misunderstood me or I wasn't explain my question clearly. I meant how come we can prove (a+b)*(b+c)*(a+c)>=8 and (a+b)*(b+c)*(a+c)<=8 at the same time ? (I know we have to prove (a+b)*(a+c)*(b+c)<=8, not the other way round.) Like we get the wrong answer if we use AM>=GM while it is ok when using other methods. So what's wrong if we use AM>=GM?


    If the question is modified a bit, say if abc=1 (a,b,c are positive real numbers)
    which of the following is correct
    1/[a3(b+c)] + 1/[b3(a+c)] + 1/[c3(b+a)] >= 3/2
    or
    1/[a3(b+c)] + 1/[b3(a+c)] + 1/[c3(b+a)] <= 3/2

    we can't have z>=3/2 and z<=3/2 at the same time while z is not necessarily equals 3/2.
     
    Last edited: May 4, 2003
  18. May 5, 2003 #17
    No...
    We took E>X, and then expected to prove X>3/2......well..this is not true for every X...got it ?
     
  19. May 5, 2003 #18
    Re: Inequality

     
  20. May 6, 2003 #19
    something must have gone wrong in our proof.

    This step is wrong, which leads the rest of our proof a contradiction. (one of the counter examples, a=1.1, b=2.2, c=3.3)
    I'm sorry,true_omega, that I didn't include that a, b, c are positive real numbers at the very beginning. I've edited it already.
     
    Last edited: May 6, 2003
  21. May 11, 2003 #20
    i did not say the only way is for two of the variables to equal -1 i said the only other way. the solution would be for each variable to equal positive 1 whereas 1*1*1 of course equals 1
     
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