Prove that $\left|\sin x \sin \left(\frac{\pi}{3}-x\right) \sin \left(\frac{\pi}{3}+x\right)\right| \leq \frac{1}{4}$ for all values of $x$
$\frac{\pi}{3}=60^{\circ}$
We have:
$|\sin x \sin (60-x) \sin (60+x)|$
$=\left|\sin x\left(\sin ^{2} 60-\sin ^{2} x\right)\right|$
$\left[\because \sin (A+B) \sin (A-B)=\sin ^{2} A-\sin ^{2} B\right]$
$=\left|\sin x\left(\frac{3}{4}-\sin ^{2} x\right)\right|$
$=\left|\frac{1}{4} \sin x\left(3-4 \sin ^{2} x\right)\right|$
$=\left|\frac{1}{4} 3 \sin x-4 \sin ^{3} x\right|$
$=\frac{1}{4}|\sin 3 \mathrm{x}| \quad\left(\because 3 \sin x-4 \sin ^{3} x=\sin 3 x\right)$
$\leq \frac{1}{4} \quad(\because|\sin x| \leq 1$ for all $x)$
$\therefore|\sin x \sin (60-x) \sin (60+x)| \leq \frac{1}{4}$
Hence proved.
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