Question

In a triangle ABC, prove that $\frac{b+c}{a}\le \csc \frac{A}{2}$ and hence show that

${\csc }^n\frac{A}{2}{\csc }^n\frac{B}{2}{\csc }^n\frac{C}{2}\ge 2^{3n}$

for all integers $n\ge 1$

Answer 1

$\frac{b+c}{a}=\frac{sinB+sinC}{sinA}=\frac{2sin\frac{B+c}{2}cos\frac{B-c}{2}}{2sin\frac{A}{2}cos\frac{A}{2}}$

$\frac{cos\frac{B-c}{2}}{sin\frac{A}{2}}<\frac{1}{sin\frac{A}{2}}(\because cos\theta \le 1)$

$\therefore cosec\frac{A}{2}\ge \frac{b+c}{a}=\frac{b}{a}+\frac{c}{a}\ge 2\sqrt{\frac{b}{a}.\frac{c}{a}}$

$\therefore cose{c}^n\frac{A}{2}\ge 2^n\left(\frac{\sqrt{bc}}{a}\right)$

write similar relations and multiply.