2. Let positive real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=1$. Then the maximum value of $s=a^{2} b c+a b^{2} c+a b c^{2}$ is $\qquad$
2. $\frac{1}{3}$.

By the AM-GM inequality and Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
a^{2} b c+a b^{2} c+a b c^{2}=a b c(a+b+c) \\
\leqslant \sqrt{\left(\frac{a^{2}+b^{2}+c^{2}}{3}\right)^{3}} \cdot \sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}=\frac{1}{3} .
\end{array}
$$

Equality holds if and only if $a=b=c=\frac{\sqrt{3}}{3}$.