Question

In triangle ABC, $\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}\le \frac{3}{2}$ then show that $\cos \frac{\pi +A}{4}\cos \frac{\pi +B}{4}\cos \frac{\pi +C}{4}\le \frac{1}{8}$.

Answer 1

lets assume the maximum limit

$sin\left(\frac{A}{2}\right)+sin\left(\frac{B}{2}\right)+sin\left(\frac{C}{2}\right)=\frac{3}{2}$ then
$\frac{A}{2}=\frac{B}{2}=\frac{C}{2}={30}^0$
$A=B=C={60}^0=\theta$
Then
$\frac{\pi +\theta }{4}=\frac{{180}^0+{60}^0}{4}={60}^0$
Hence
$\prod cos\left(\frac{\theta +\pi }{4}\right)$
$=\prod cos\left({60}^0\right)$

$=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}$
$=\frac{1}{8}$.
Since $cos\theta$ decreases as $\theta$ increases, hence
$\prod cos\left(\frac{\theta +\pi }{4}\right)$ is maximum when $\theta ={60}^0$
($\theta$ being an angle of triangle ).
Hence
$\prod cos\left(\frac{\theta +\pi }{4}\right)\le \frac{1}{8}$ is true.