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

We know that \[\cos (x+y) = \cos x\cos y-\sin x\sin y \quad \text{and} \quad \cos (x-y) = \cos x\cos y+\sin x\sin y.\]

Thus \[\sin x\sin y = \dfrac{1}{2}(\cos (x-y)-\cos (x+y)).\]

This means \[\sin \left(\dfrac{a+b}{2}\right)\sin \left(\dfrac{a+b}{2}\right) = \dfrac{1}{2}(1-\cos (a+b)).\]

We also have \[\sin a\sin b = \dfrac{1}{2}(\cos (a-b)-\cos (a+b)).\]

Since \(\cos(a-b) \leq 1\), on comparing the last two equations we see \(\sin a\sin b \leq \sin^2 \left(\dfrac{a+b}{2}\right)\).

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 \(0 \leq a, b, c, d \leq \pi,\) then

\[0 \leq \sin a, \sin b, \sin c, \sin d \leq 1.\]


\[0 \leq \sin a \sin b \leq \sin^2{\frac{1}{2}(a+b)}\]


\[0 \leq \sin c \sin d \leq \sin^2{\frac{1}{2}(a+b)}.\]

Because the left hand side of each inequality is greater than or equal to zero, we can multiply the inequalities to get

\[\begin{equation} \sin a \sin b \sin c \sin d \leq \sin^2{\frac{1}{2}(a+b)} \sin^2{\frac{1}{2}(c+d)}. \label{eq:1} \end{equation}\] Using the inequality from a), we also have \[\begin{align} \sin{\frac{1}{2}(a+b)} \sin{\frac{1}{2}(c+d)} &\leq \sin^2{\frac{1}{2}\bigl[\frac{1}{2}(a+b)+\frac{1}{2}(c+d)\bigr]} \notag \\ \quad &= \sin^2{\frac{1}{4}(a + b + c +d)}. \label{eq:2} \end{align}\]

We know

\[0 \leq \frac{1}{2}(a+ b), \frac{1}{2}(c+d) \leq \pi, \]

and so

\[0 \leq \sin \frac{1}{2}(a+b) \sin \frac{1}{2}(c+d),\]

which means we can square \(\eqref{eq:2}\) to get

\[\begin{equation} \sin^2{\frac{1}{2}(a+b)} \sin^2 \frac{1}{2}(c+d) \leq \sin^4{\frac{1}{4}(a+b+c+d)}. \label{eq:3} \end{equation}\]

Combining \(\eqref{eq:1}\) and \(\eqref{eq:3}\) we get

\[\begin{align*} \sin a \sin b \sin c \sin d &\leq \sin^4\frac{1}{4}(a+b+c+d) \\ & \quad = \bigl(\sin \frac{1}{4}(a+b+c+d)\bigr)^4, \end{align*}\]

as required.

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

Setting \(d = \frac{1}{3}(a+b+c),\) we see that

\[a + b + c + d = \frac{4}{3}(a+b+c),\]

and we can deduce that

\[\begin{equation} \sin a \sin b \sin c \sin \frac{1}{3}(a+b+c) \leq \bigl(\sin \frac{1}{3}(a+b+c)\bigr)^4. \label{eq:4} \end{equation}\]

From \(0 \leq a, b, c \leq \pi\), we deduce that \(0 \leq \frac{1}{3}(a+b+c) \leq \pi\) and, except in the cases \(a,b,c=0\) and \(a,b,c=\pi\), we have the strict inequality \(0 < \frac{1}{3}(a+b+c) < \pi\), and so

\[\sin \frac{1}{3}(a+b+c) > 0.\]

Given this, we can divide \(\eqref{eq:4}\) by \(\sin \frac{1}{3}(a+b+c)\) to get

\[\sin a \sin b \sin c \leq \bigl(\sin \frac{1}{3}(a+b+c)\bigr)^3,\]

as required.

The final task is to consider the cases \(a = b = c = 0\) and \(a = b = c = \pi.\) In both of these cases, both sides of the inequality are \(0\) (as \(\sin 0 = \sin \pi = 0\)), and so the inequality still holds.