已知$a,b,c\in\mathcal R$，且$\dfrac{1}{1+a^2}+\dfrac{1}{1+4b^2}+\dfrac{1}{1+9c^2}=1$，则$|6abc-1|$的最小值为_______．
分析与解 设$\dfrac{1}{1+a^2}=x$，$\dfrac{1}{1+4b^2}=y$，$\dfrac{1}{1+9c^2}=z$，则$x+y+z=1$，且$$\begin{split} (6abc)^2=&a^2\cdot 4b^2\cdot 9c^2\\=&\left(\dfrac 1x-1\right)\left(\dfrac 1y-1\right)\left(\dfrac 1z-1\right)\\=&\dfrac{(y+z)(z+x)(x+y)}{xyz}\geqslant 8,\end{split} $$于是$6abc$的取值范围是$$(-\infty,-2\sqrt 2]\cup[2\sqrt 2,+\infty),$$所求的最小值为$2\sqrt 2-1$．当$x=y=z=\dfrac 13$时取到等号，此时$a^2=2,b^2=\dfrac 12,c^2=\dfrac 29$，取值时保证$abc>0$即可．