18. Since $\sin (x-y) \sin (x+y)=\frac{1}{2}(\cos 2 \beta-\cos 2 \alpha)=\sin ^{2} \alpha-\sin ^{2} \beta$, we have
$$\begin{aligned} & \sin a \sin (a-b) \sin (a-c) \sin (a+b) \sin (a+c) \\ = & \sin a\left(\sin ^{2} a-\sin ^{2} b\right)\left(\sin ^{2} a-\sin ^{2} c\right) \end{aligned}$$

Let $x=\sin ^{2} a, y=\sin ^{2} b, z=\sin ^{2} c$, then the original inequality is equivalent to
$$x^{\frac{1}{2}}(x-y)(x-z)+y^{\frac{1}{2}}(y-z)(y-x)+z^{\frac{1}{2}}(z-x)(z-y) \geqslant 0,$$

which is the Schur inequality.