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Math Help - in a triangle ABC

  1. #1
    Member grgrsanjay's Avatar
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    in a triangle ABC

    in a triangle ABC,the minimum value for cosec A/2 + cosec B/2+ cosecC/2
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by grgrsanjay View Post
    in a triangle ABC,the minimum value for cosec A/2 + cosec B/2+ cosecC/2
    Using Jensen's inequality,
    cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2}  \geq3cosec\frac{A+B+C}{6}= 3cosec\frac{\pi}{6}= 3*2=6

    Thus, cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2}  \geq6
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  3. #3
    Member grgrsanjay's Avatar
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    Quote Originally Posted by alexmahone View Post
    Using Jensen's inequality,
    cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2}  \geq3cosec\frac{A+B+C}{6}
    Quote Originally Posted by grgrsanjay
    can you prove Jensen's inequality?
    i don't know about Jensen's inequality..... please prove it
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    Do you know about Erdos-Mordell Inequality , it is a geometric inequality .


    Let  P be a point inside  \Delta ABC , draw the feets of perpendicular of  P to the sides , namely  L,M,N .

    Then we always have :

     AP + BP + CP \geq 2(LP + MP + NP)

    The inequality holds when it is an equilateral triangle ,  P being the centre , that means given a nonequilateral triangle , we can never find a point satisfying the equality .


    By applying this , we have

     \big{ r[ \csc(\frac{A}{2}) + \csc(\frac{B}{2}) + \csc(\frac{C}{2}) ] } = AI + BI + CI \geq 2( r + r + r ) = 6r

    where  r denotes the inradius and  I the incentre .

    Therefore ,  \csc(\frac{A}{2}) + \csc(\frac{B}{2}) + \csc(\frac{C}{2}) \geq 6 , the equality holds when  A = B = C = 60^o
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  5. #5
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    Quote Originally Posted by grgrsanjay View Post
    i don't know about Jensen's inequality..... please prove it
    It is a very useful inequality. Look here.

    For the purpose of cracking Olympiad problems, knowing Arithmetic Mean - Geometric Mean inequality, Cauchy Schwarz inequality and Jensen's inequalty helps a lot.

    Look these up in wikipedia. Various applications of Schur's inequality and Muirhead's inequality are also frequently seen in Olympiad problems.
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