10. (15 points) If $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=1$, find the maximum value of the real number $\lambda$ such that the inequality $a^{2}+b^{2}+c^{2}+\lambda \sqrt{a b c} \leqslant 1$ always holds.
$$
\begin{array}{l}
\text { 10. Solution: } \leqslant \frac{1-a^{2}-b^{2}-c^{2}}{\sqrt{a b c}}= \\
\frac{1-(a+b+c)^{2}+2(a b+b c+c a)}{\sqrt{a b c}} \\
=2\left(\sqrt{\frac{a b}{c}}+\sqrt{\frac{b c}{a}}+\sqrt{\frac{a c}{b}}\right), \\
\text { and }\left(\sqrt{\frac{a b}{c}}+\sqrt{\frac{b c}{a}}+\sqrt{\frac{a c}{b}}\right)^{2}=\frac{a b}{c}+\frac{b c}{a}+\frac{a c}{b}+2 \\
=\frac{1}{2}\left[\left(\frac{a b}{c}+\frac{b c}{a}\right)+\left(\frac{a b}{c}+\frac{a c}{b}\right)+\left(\frac{b c}{a}+\frac{a c}{b}\right)\right] \\
2 \geqslant 3,
\end{array}
$$

Therefore, $\lambda \leqslant 2 \sqrt{3}$.