USA IMO 2003

By the Product-to-sum formulas and the Double-angle formulas, we have $$\begin{array}{rl}\sin (\alpha -\beta )\sin (\alpha +\beta )& =\frac{1}{2}[\cos 2\beta -\cos 2\alpha ]\\ & ={\sin }^2\alpha -{\sin }^2\beta .\end{array}$$ Hence, we obtain $$\begin{array}{rl}& \sin a\sin (a-b)\sin (a-c)\sin (a+b)\sin (a+c)\\ & =\sin c({\sin }^{2}a-{\sin }^{2}b)({\sin }^{2}a-{\sin }^{2}c)\end{array}$$ 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 $$z(z^2-x^2)(z^2-y^2)\ge z(y^2-x^2)(z^2-y^2)\ge y(z^2-y^2)(y^2-x^2).$$