USA IMO, 2003

By the Product-to-sum formulas and the Double-angle formulas, we have
$$\begin{aligned} \sin(\alpha - \beta) \sin(\alpha + \beta) &= \frac{1}{2}[\cos 2\beta - \cos 2\alpha] \\ &= \sin^2 \alpha - \sin^2 \beta. \end{aligned}$$
Hence, we obtain
$$\begin{aligned} & \sin a \sin(a-b) \sin(a-c) \sin(a+b) \sin(a+c) \\ &= \sin c(\sin^2 a - \sin^2 b)(\sin^2 a - \sin^2 c) \end{aligned}$$
and its analogous forms. Therefore, it suffices to prove that
$$x(x^2 - y^2)(x^2 - z^2) + y(y^2 - z^2)(y^2 - x^2) + z(z^2 - x^2)(z^2 - y^2) \ge 0,$$
where $x = \sin a$, $y = \sin b$, and $z = \sin c$ (hence $x, y, z > 0$). Since the last inequality is symmetric with respect to $x, y, z$, we may assume that $x \ge y \ge z > 0$. It suffices to prove that
$$x(y^2 - x^2)(z^2 - x^2) + z(z^2 - x^2)(z^2 - y^2) \ge y(z^2 - y^2)(y^2 - x^2),$$
which is evident as
$$x(y^2 - x^2)(z^2 - x^2) \ge 0$$
and
$$z(z^2 - x^2)(z^2 - y^2) \geq z(y^2 - x^2)(z^2 - y^2) \geq y(z^2 - y^2)(y^2 - x^2).$$