1. ## trigo inequality

prove that sin (a/2)sin(b/2)sin(c/2)<=1/8 where a,b,c are angles of triangle abc

2. try $\displaystyle 2.sin(\frac{A}{2})sin(\frac{B}{2}) = cos(A-B) - cos(A+B)$ and$\displaystyle A+B+C=\pi$

3. Originally Posted by prasum
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

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

4. Originally Posted by Archie Meade
$\displaystyle r=incircle\;radius$

$\displaystyle R=circumcircle\;radius$

$\displaystyle s=semiperimeter$

$\displaystyle A=triangle\;area$

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

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