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Math Help - trigo inequality

  1. #1
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    trigo inequality

    prove that sin (a/2)sin(b/2)sin(c/2)<=1/8 where a,b,c are angles of triangle abc
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  2. #2
    Senior Member BAdhi's Avatar
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    try  2.sin(\frac{A}{2})sin(\frac{B}{2}) = cos(A-B) - cos(A+B) and  A+B+C=\pi
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  3. #3
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    Quote Originally Posted by prasum View Post
    prove that sin (a/2)sin(b/2)sin(c/2)<=1/8 where a,b,c are angles of triangle abc
    r=incircle\;radius

    R=circumcircle\;radius

    s=semiperimeter

    A=triangle\;area


    Arithmetic-Geometric Mean Inequality

    \displaystyle\frac{x+y}{2}\ \ge\ \sqrt{xy}


    Proof


     r=4Rsin\left[\frac{A}{2}\right]sin\left[\frac{B}{2}\right]sin\left[\frac{C}{2}\right]\Rightarrow\displaystyle\frac{r}{4R} =sin\left[\frac{A}{2}\right]sin\left[\frac{B}{2}\right]sin\left[\frac{C}{2}\right]

    A=rs

    A=\displaystyle\frac{abc}{4R}\Rightarrow\frac{rabc  s}{4R}=A^2

    \displaystyle\Rightarrow\frac{r}{4R}=\frac{A^2}{ab  cs}


    Using Heron's formula

    \displaystyle\ A^2=s(s-a)(s-b)(s-c)\Rightarrow\frac{r}{4R}=\frac{s(s-a)(s-b)(s-c)}{sabc}


    Referring to the attachment....

    \displaystyle\ s-a=\frac{b+c-a}{2}=x

    \displaystyle\ s-b=\frac{a+c-b}{2}=y

    \displaystyle\ s-c=\frac{a+b-c}{2}=z

    Therefore, utilising the "Arithmetic-Geometric Mean Inequality"

    \displaystyle\frac{r}{4R}=\frac{x\;y\;z}{abc}=\fra  c{\sqrt{xx}\;\sqrt{yy}\;\sqrt{zz}}{(y+z)(x+z)(x+y)  }

    \displaystyle\Rightarrow\frac{r}{4R}=\frac{\sqrt{x  y}\;\sqrt{xz}\;\sqrt{yz}}{(y+z)(x+z)(x+y)}=\left[\frac{1}{8}\right]\frac{\sqrt{yz}\;\sqrt{xz}\;\sqrt{xy}}{\left(\frac  {y+z}{2}\right)\left(\frac{x+z}{2}\right)\left(\fr  ac{x+y}{2}\right)}

    Hence

    sin\left[\frac{A}{2}\right]sin\left[\frac{B}{2}\right]sin\left[\frac{C}{2}\right]\ \le\ \frac{1}{8}
    Attached Thumbnails Attached Thumbnails trigo inequality-sine-product-inequality.jpg  
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  4. #4
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    Quote Originally Posted by Archie Meade View Post
    r=incircle\;radius

    R=circumcircle\;radius

    s=semiperimeter

    A=triangle\;area


    Arithmetic-Geometric Mean Inequality

    \displaystyle\frac{x+y}{2}\ \ge\ \sqrt{xy}


    Proof


     r=4Rsin\left[\frac{A}{2}\right]sin\left[\frac{B}{2}\right]sin\left[\frac{C}{2}\right]\Rightarrow\displaystyle\frac{r}{4R} =sin\left[\frac{A}{2}\right]sin\left[\frac{B}{2}\right]sin\left[\frac{C}{2}\right]

    A=rs

    A=\displaystyle\frac{abc}{4R}\Rightarrow\frac{rabc  s}{4R}=A^2

    \displaystyle\Rightarrow\frac{r}{4R}=\frac{A^2}{ab  cs}


    Using Heron's formula

    \displaystyle\ A^2=s(s-a)(s-b)(s-c)\Rightarrow\frac{r}{4R}=\frac{s(s-a)(s-b)(s-c)}{sabc}


    Referring to the attachment....

    \displaystyle\ s-a=\frac{b+c-a}{2}=x

    \displaystyle\ s-b=\frac{a+c-b}{2}=y

    \displaystyle\ s-c=\frac{a+b-c}{2}=z

    Therefore, utilising the "Arithmetic-Geometric Mean Inequality"

    \displaystyle\frac{r}{4R}=\frac{x\;y\;z}{abc}=\fra  c{\sqrt{xx}\;\sqrt{yy}\;\sqrt{zz}}{(y+z)(x+z)(x+y)  }

    \displaystyle\Rightarrow\frac{r}{4R}=\frac{\sqrt{x  y}\;\sqrt{xz}\;\sqrt{yz}}{(y+z)(x+z)(x+y)}=\left[\frac{1}{8}\right]\frac{\sqrt{yz}\;\sqrt{xz}\;\sqrt{xy}}{\left(\frac  {y+z}{2}\right)\left(\frac{x+z}{2}\right)\left(\fr  ac{x+y}{2}\right)}

    Hence

    sin\left[\frac{A}{2}\right]sin\left[\frac{B}{2}\right]sin\left[\frac{C}{2}\right]\ \le\ \frac{1}{8}
    thanks
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