Question

In any triangle ABC prove that identities.
$\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\le \frac{1}{8}$.

Answer 1

$\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\le \frac{1}{8}$

$C=\pi -(A+B)$

$\sin \frac{A}{2}\sin \frac{B}{2}\sin \left(\frac{\pi -(A+B)}{2}\right)$

$=\sin \frac{A}{2}\sin \frac{B}{2}\cos \left(\frac{A+B}{2}\right)$

$f(A,B)=\sin \frac{A}{2}\sin \frac{B}{2}\cos \left(\frac{A+B}{2}\right)$

$\frac{\delta F}{\delta A}=\frac{1}{2}\sin \frac{B}{2}\cos (A+\frac{B}{2})$

$\frac{\delta F}{\delta B}=\frac{1}{2}\sin \frac{A}{2}\cos (B+\frac{A}{2})$

for maxima of $F$.

$\frac{\delta F}{\delta A}=\frac{\delta F}{\delta B}=\sigma$

$A+\frac{B}{2}=B+\frac{A}{2}=\frac{\pi }{2}$

$\Rightarrow A=B=\pi /3$

$C=\pi -\frac{2\pi }{3}=\pi /3$

so max value $={\sin }^3\pi /3=\frac{1}{8}$