Prove that, for all values of \(a\) and \(b\), \[\sin a \sin b \le \sin^2\frac{1}{2}(a+b).\]

Can we find an identity for \(\sin a \sin b\)?

Show further that, if \(a\), \(b\), \(c\) and \(d\) all lie between \(0\) and \(\pi\), then \[\sin a \sin b \sin c \sin d \leq \left( \sin\frac{1}{4}(a+b+c+d)\right)^4;\]

If \(0 \leq a \leq \pi\), what does that tell us about \(\sin a\)?

Handy fact: if \(0 \leq a \leq b\), and \(0 \leq c \leq d\), then \(0 \leq ac \leq bd\).

… and, by writing \(d=\frac{1}{3}(a+b+c)\), deduce that \[\sin a \sin b \sin c \le \left( \sin\frac{1}{3}(a+b+c)\right)^3.\]

Let’s think about the case \(0 < a, b, c, d < \pi\) separately to the cases where \(a = b = c = 0\) and \(a = b = c = \pi.\)

For the first case, what do you know about \(\sin \frac{1}{3}(a+b+c)\)?