10. Let $a, b, c \in(0,1]$, $\lambda$ be a real number, such that $\frac{\sqrt{3}}{\sqrt{a+b+c}} \geqslant 1+\lambda(1-a)(1-b)(1-c)$ always holds, find the maximum value of $\lambda$.

Set $a, b, c \in(0,1]$, $\lambda$ as a real number, such that $\frac{\sqrt{3}}{\sqrt{a+b+c}} \geqslant 1+\lambda(1-a)(1-b)(1-c)$ always holds, find the maximum value of $\lambda$.
When $a=b=c=\frac{1}{4}$, $2 \geqslant 1+\lambda \cdot \frac{27}{64} \Rightarrow \lambda \leqslant \frac{64}{27}$.
$$
(1-a)(1-b)(1-c) \leqslant\left(\frac{1-a+1-b+1-c}{3}\right)^{3}=\left(1-\frac{a+b+c}{3}\right)^{3},
$$

Let $a+b+c=3 x^{2}$, then the inequality is equivalent to $\frac{1}{x} \geqslant 1+\frac{64}{27}\left(1-x^{2}\right)^{3}$
$$
\Leftrightarrow \frac{1}{x}-1 \geqslant \frac{64}{27}\left(1-x^{2}\right)^{3} \Leftrightarrow \frac{1-x}{x} \geqslant \frac{64}{27}\left(1-x^{2}\right)^{3} \Leftrightarrow 1 \geqslant \frac{64}{27} x(1-x)^{2}(1+x)^{3} \text {, }
$$

and $\frac{64}{27} x(1-x)^{2}(1+x)^{3}=64 x(1-x)^{2}\left(\frac{1+x}{3}\right)^{3}$
$$
\leqslant 64\left(\frac{x+1-x+1-x+\frac{1+x}{3}+\frac{1+x}{3}+\frac{1+x}{3}}{6}\right)^{6}=1 \text {. }
$$

Equality holds when $x=\frac{1}{2} \Rightarrow a=b=c=\frac{1}{4}$. Therefore, the maximum value of $\lambda$ is $\frac{64}{27}$.