# Thread: in a triangle ABC

1. ## in a triangle ABC

in a triangle ABC,the minimum value for cosec$\displaystyle A/2$ + cosec$\displaystyle B/2$+ cosecC/2

2. Originally Posted by grgrsanjay
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, $\displaystyle cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2} \geq6$

3. Originally Posted by alexmahone
Using Jensen's inequality,
$\displaystyle cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2} \geq3cosec\frac{A+B+C}{6}$
Originally Posted by grgrsanjay
can you prove Jensen's inequality?

4. 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 $\displaystyle A = B = C = 60^o$

5. Originally Posted by grgrsanjay