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Thread: 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$\displaystyle A/2$ + cosec$\displaystyle B/2$+ 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$\displaystyle A/2$ + cosec$\displaystyle B/2$+ cosecC/2
    Using Jensen's inequality,
    $\displaystyle cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2} \geq3cosec\frac{A+B+C}{6}= 3cosec\frac{\pi}{6}= 3*2=6$

    Thus, $\displaystyle cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2} \geq6$
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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,
    $\displaystyle cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2} \geq3cosec\frac{A+B+C}{6}$
    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 $\displaystyle P $ be a point inside $\displaystyle \Delta ABC $ , draw the feets of perpendicular of $\displaystyle P $ to the sides , namely $\displaystyle L,M,N $ .

    Then we always have :

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

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


    By applying this , we have

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

    where $\displaystyle r $ denotes the inradius and $\displaystyle I $ the incentre .

    Therefore , $\displaystyle \csc(\frac{A}{2}) + \csc(\frac{B}{2}) + \csc(\frac{C}{2}) \geq 6 $ , the equality holds when $\displaystyle 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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