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