6. Let $a, b, c > 0$, and $a+b+c=1$. Then the maximum value of the real number $\lambda$ such that $a^{2}+b^{2}+c^{2}+\lambda \sqrt{a b c} \leqslant 1$ always holds is ( ).
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{\sqrt{3}}{2}$
(C) $\sqrt{3}$
(D) $2 \sqrt{3}$
6.D.

Let $a, b, c > 0$, and $a+b+c=1$. Then
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+\lambda \sqrt{a b c} \leqslant 1 \\
\Leftrightarrow \lambda \leqslant 2\left(\sqrt{\frac{a b}{c}}+\sqrt{\frac{a c}{b}}+\sqrt{\frac{b c}{a}}\right). \\
\left(\sqrt{\frac{a b}{c}}+\sqrt{\frac{a c}{b}}+\sqrt{\frac{b c}{a}}\right)^{2}=\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}+2 \\
=\frac{1}{2}\left(\frac{a b}{c}+\frac{b c}{a}\right)+\frac{1}{2}\left(\frac{b c}{a}+\frac{a c}{b}\right)+ \\
\frac{1}{2}\left(\frac{a c}{b}+\frac{a b}{c}\right)+2 \geqslant 3.
\end{array}
$$

Therefore, the maximum value of the real number $\lambda$ that makes the inequality hold is $2 \sqrt{3}$.