10. Take $a=b=c=\frac{1}{4}$, then $\lambda \leqslant \frac{64}{27}$.

Below we prove: $\lambda=\frac{64}{27}$ satisfies the condition, i.e., we need to prove:
$$
\frac{\sqrt{3}}{\sqrt{a+b+c}} \geqslant 1+\frac{64}{27}(1-a)(1-b)(1-c).
$$

Notice that,
$$
(1-a)(1-b)(1-c) \leqslant\left(1-\frac{a+b+c}{3}\right)^{3}.
$$

Let $a+b+c=3 x^{2} (x>0)$. Then $0<x \leqslant 1$. We only need to prove
$$
\begin{array}{l}
\frac{1}{x} \geqslant 1+\frac{64}{27}\left(1-x^{2}\right)^{3} \\
\Leftrightarrow \frac{1-x}{x} \geqslant \frac{64}{27}(1-x)^{3}(1+x)^{3} \\
\Leftrightarrow 1 \geqslant \frac{64}{27} x(1-x)^{2}(1+x)^{3}.
\end{array}
$$

By the AM-GM inequality, we know
$$
\begin{array}{l}
x(1-x)^{2}(1+x)^{3} \\
=27 x(1-x)^{2}\left(\frac{1+x}{3}\right)^{3} \\
\leqslant 27\left(\frac{x+2(1-x)+3\left(\frac{x+1}{3}\right)}{6}\right)^{6}=\frac{27}{64}. \\
\text { Hence } \frac{64}{27} x(1-x)^{2}(1+x)^{3} \leqslant \frac{64}{27} \times \frac{27}{64}=1.
\end{array}
$$

Therefore, inequality (1) holds.
In conclusion, the maximum value of $\lambda$ is $\frac{64}{27}$.
10. (20 points) Let \( a, b, c \in (0,1], \lambda \) be a real number such that
\[
\frac{\sqrt{3}}{\sqrt{a+b+c}} \geqslant 1+\lambda(1-a)(1-b)(1-c) .
\]

Find the maximum value of \(\lambda\).