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Thread: 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 $\displaystyle 2.sin(\frac{A}{2})sin(\frac{B}{2}) = cos(A-B) - cos(A+B)$ and$\displaystyle A+B+C=\pi$
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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
    $\displaystyle r=incircle\;radius$

    $\displaystyle R=circumcircle\;radius$

    $\displaystyle s=semiperimeter$

    $\displaystyle A=triangle\;area$


    Arithmetic-Geometric Mean Inequality

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


    Proof


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

    $\displaystyle A=rs$

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

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


    Using Heron's formula

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


    Referring to the attachment....

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

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

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

    Therefore, utilising the "Arithmetic-Geometric Mean Inequality"

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

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

    Hence

    $\displaystyle sin\left[\frac{A}{2}\right]sin\left[\frac{B}{2}\right]sin\left[\frac{C}{2}\right]\ \le\ \frac{1}{8}$
    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
    $\displaystyle r=incircle\;radius$

    $\displaystyle R=circumcircle\;radius$

    $\displaystyle s=semiperimeter$

    $\displaystyle A=triangle\;area$


    Arithmetic-Geometric Mean Inequality

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


    Proof


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

    $\displaystyle A=rs$

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

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


    Using Heron's formula

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


    Referring to the attachment....

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

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

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

    Therefore, utilising the "Arithmetic-Geometric Mean Inequality"

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

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

    Hence

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