14. Given real numbers $a, \mathbf{b}, \mathbf{c}$ satisfy $a^{2}+b^{2}+c^{2}=1$, find the maximum and minimum values of $M=a^{2} b c+a b^{2} c+a b c^{2}$.
【Answer】See the analysis.
【Analysis】【Detailed Explanation】
By the mean inequality and Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left|a^{2} b c+a b^{2} c+a b c^{2}\right| \leqslant|a b c|+(|a|+|b|+|c|) \\
\leqslant\left(\frac{|a|+|b|+|c|}{3}\right)^{2}(|a|+|b|+|c|) \\
\left.\frac{1}{27}\left[\sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}\right]^{4}=\frac{1}{3} . \\
\text { When } a=b=c=\frac{1}{\sqrt{3}} \text {, the equality holds, so the maximum value of } \mathrm{M} \text { is } \frac{1}{3} .
\end{array}
$$

When $a=b=c=\frac{1}{\sqrt{3}}$, the equality holds, so the maximum value of $\mathrm{M}$ is $\frac{1}{3}$.
To find the minimum value of $\mathrm{M}$, we only need to consider the case where $a, b>0, c<0$. Let $c=-t$, then $a^{2}+b^{2}+t^{2}=1$, and $a+b-t>0$. At this point, $t^{2}<\frac{2}{3}$, and $x=\frac{1}{3}$ is an extraneous root, so $x_{0}=\frac{1}{2}-\frac{\sqrt{2}}{4}$ is the maximum point of $f(x)$. Therefore, $-f\left(x_{0}\right)=-\frac{1-\sqrt{2}}{16}$ is the minimum value of $\mathrm{M}$.