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Triangle inequality

  1. Feb 7, 2007 #1
    I am a beginner at trying to prove or disprove inequalities. In an attempt to improve on this skill I found some problems that I would like to work on. Now, I know many of you may be able to look at this and think of a solution, but please refrain from posting it, but some advice and methods would be helpful.

    Prove that in any acute triangle ABC with sides a, b, and c, the following inequality is true.


    [tex] 27 \leq (a+b+c)^2 \left( \frac{1}{a^2 + b^2 - c^2}+\frac{1}{b^2 + c^2 - a^2}+\frac{1}{c^2 + a^2 - b^2} \right) [/tex]
     
    Last edited: Feb 7, 2007
  2. jcsd
  3. Feb 7, 2007 #2

    StatusX

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    Are you trying to say:

    [tex] 27 \leq (a+b+c)^2 \left( \frac{1}{a^2 + b^2 - c^2}+\frac{1}{b^2 + c^2 - a^2}+\frac{1}{c^2 + a^2 - b^2} \right) [/tex]

    what have you tried?
     
  4. Feb 7, 2007 #3
    Well, here is where i'm at. Well, I am fairly certain that this equation is true. If we exam the smallest acute triangle I can think of, an equilateral triangle whose sides are all one, then the solutions comes out to be exactly 27. So, that is where i am at right.
     
  5. Feb 7, 2007 #4

    StatusX

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    And what makes you think the RHS will be smaller for acute triangles?

    I don't immediately see a clever way to do this, so I would start by multiplying out by a common denominator so that you're comparing polynomials, and then simplifying as much as possible.
     
  6. Feb 7, 2007 #5

    mjsd

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    mmm....problem like this is reminiscent of those you usuallly see in maths competitions or olympiad. It usually require some wit to do it elegantly. If you have absolutely no idea (like me currently :smile:), you should start by thinking about all the properties of an acute triangle and see whether that gives you other contraints relations between a, b, c. Next think about why it may not work for a triangle that is NOT acute (i guess that means it has one angle bigger than 90)

    next, look at the expression itself (change its form) ...and see if you can recognise it or part of it from somewhere.. if not, see if you can find another inequality which is easier to handle that is bigger than or equal 27 yet less than or equal to original expression.

    remember there are basic things like a + b > c, a + c> b, b + c > a. etc.
     
  7. Feb 8, 2007 #6
  8. Feb 8, 2007 #7

    Curious3141

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    Last edited: Feb 8, 2007
  9. Feb 8, 2007 #8

    mjsd

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    given the similarity of the three terms in brackets, may be 27 really means 9+9+9, and proving that each term is [tex]9 \leq[/tex] may be sufficient... just a guess.
    Certainly, acute angle means that Cos (any angle) is inside the open interval [tex](0,1)[/tex]
     
  10. Feb 8, 2007 #9
    Let a=b a=c

    27<_(a + a + a)^2((1/a^2) + (1/a^2) + (1/a^2))
    27<_(9a^2)(3/a^2)
    27<_27

    So that proves that an acute triangle that happens to be an equilateral triangle is true, that means I just have to show that the rest hold true.
     
  11. Feb 10, 2007 #10
    Yet another trivial problem you couldn't tackle ,ha Bitter ?:biggrin:
     
  12. Feb 10, 2007 #11
    I'm not really sure what your problem is with me. I'm not even sure why you said yet again. The other problem I posted was not one I couldn't solve. I merely wanted to see other people's solutions to it. I just posted it in the wrong place and forgot about it. In fact, i'm fairly certain this is my first post about solving a problem in general math section.

    So if you don't plan to give me advice that's fine, but if you feel the need to rub in the fact that you are better at math than I am, that's other thing.
     
  13. Feb 11, 2007 #12
    Really?I don't recall I've seen you ever posted your solution..But never mind that.It was another problem anyway.I got an impression your main problem isn't math or misplacing the posts,but lazyness.Again,that's just my opinion.This time you posted the problem to the right place and you will see me post the solution.But only this time.


    [tex]S=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\geq \frac{27}{(a+b+c)^2}[/tex]

    (Firstly note the right side of ineq. is always positive-> left side must be always positive too.That is ensured ,becouse [itex]a^2+b^2-c^2>0,b^2+c^2-a^2>0,a^2+c^2-b^2>0[/itex]
    is always true for acute triangles.Hence ,the restriction)

    We have:

    [tex]\frac{a^2}{a^2b^2+a^2c^2-a^4}+\frac{b^2}{a^2b^2+b^2c^2-b^4}+\frac{c^2}{a^2c^2+b^2c^2-c^4}=S \geq \frac{(a+b+c)^2}{2(a^2b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}=\frac{a+b+c}{(a+b-c)(a+c-b)(b+c-a)}[/tex]

    Now we are left to prove that for a,b,c>0 the following is true:

    [tex]\frac{a+b+c}{(a+b-c)(a+c-b)(b+c-a)}\geq \frac{27}{(a+b+c)^2}[/tex]


    It's true becouse ,by Arithmetic-Geometric ineq,we have [itex](a+b+c)^3\geq 27abc[/itex] while [itex]abc\geq (a+b-c)(a+c-b)(b+c-a)[/itex] is obvious.
    QED

    EDIT:I can't be better mathematician than you.That's becouse I'm not a mathematician (per education) at all.
    :smile:

    Best regards
     
    Last edited: Feb 11, 2007
  14. Feb 11, 2007 #13
    Sigh, I thought I said not to post an answer, but to help me out. Oh well...
     
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