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Inequalities #2

  1. Apr 27, 2012 #1
    1. The problem statement, all variables and given/known data

    a. Prove: If a≠b≠c are real numbers, then a2+b2+c2>ab+bc+ca
    b. Prove: If a>0, b>0and a≠b, then a/b+b/a>2

    2. Relevant equations
    (real numbers)2>0


    3. The attempt at a solution
    a.
    (a+b+c)2>0
    a2+b2+c2>-2(ab+bc+ca)
    Try to prove -2(ab+bc+ca) > ab+bc+ca but not true, -2.4 <4


    Must be wrong approach????

    b.
    Same approach with same result.
    a/b+b/a>0
    Since a>0, b>0 =>ab>0
    (a+b)2/(ab)>2

    Freeze...no idea
     
    Last edited by a moderator: Apr 27, 2012
  2. jcsd
  3. Apr 27, 2012 #2
    Mod note: Removed quoted work.
    Hello, Ahmed you're not supposed to give people the answer to the problem. You have to 'help them solve'. And by the way the first one you can do a proof by contradiction.
     
    Last edited by a moderator: Apr 27, 2012
  4. Apr 27, 2012 #3
    Oh, I'm so sorry, I am new to the forum and have only skimmed over the rules. Thank you for notifying me, I will read the entire forum rules so I don't make such mistakes again.
     
  5. Apr 27, 2012 #4
    Ok not a problem.
     
  6. Apr 27, 2012 #5
    Thank you Mr. Ali Ahmed.
    I've racking my brain(old man's brain) to max. to find solution.
     
  7. Apr 27, 2012 #6
    Lessons learned from Mr Ali.

    1.
    (a±b)>0 wrong
    instead
    (a±b)≥0

    2. Many ways of getting a2, ab, b2.....
    My only way (a+b+c)2
    but there are other ways
    (a-b)2, (b-c)2,(c-a)2....

    from lessons learn, now i can solve 3rd problems

    3. Prove: If a2 + b2=1 and c2 + d2=1,
    then, ac + bd ≤ 1

    Way of getting ac, (a-c)2 ≥ 0
    bd, (b-d)2 ≥ 0

    a2+c2-2ac ≥0
    b2+d2-2bd ≥0

    a2+c2 +b2+d2 -2(ac+bd)≥0

    -2(ac+bd)≥-2
    Changing sense of equality


    Thus, ac+bd ≤ 1
     
    Last edited: Apr 27, 2012
  8. Apr 28, 2012 #7
    Exactly azizlwl.

    The main problem in math proofs is that you sort of have to trick the solution into appearing (which is usually not obvious). But maybe for future reference, if you ever encounter such problems again, the way I start is by coming up with ways of getting what I want (necessary terms, number, etc.) in the equation and then reduce the equation to its simplest form.

    What might often happen is that the reduction turns out to be useless, so when I reach such a point in my derivation I disregard my last method of obtaining my necessary terms, and use a new one (if you are confident that you made no mistake in the reduction and you arrive at a useless conclusion, then start fresh).

    But I understand where your coming from, I often go crazy even trying to solve the simplest problems.
     
  9. Apr 28, 2012 #8
    Here difference way done by Dr. Mark.
    a/b+b/a>2
    Since both a and b are positive, multiply through by ab
    a^2+b^2>0
    a^2-2ab+b^2>0
    (a-b)^2>

    Then i asked him what is the difference and which is easier.
    Here his answer


    "Typically you start with what is given and derive a truth, but if you can start with the truth and derive what is given, then that is valid as well. Usually it is easier the first way, because you may not know what to begin with the second way."
     
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