7. For any positive numbers $a, b, c$ satisfying $a+b+c=1$, it always holds that
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+\lambda \sqrt{a b c} \leqslant 1 . \\
\text { Then } \lambda_{\text {max }}=
\end{array}
$$
7. $2 \sqrt{3}$.

From the problem, we have
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+\lambda \sqrt{a b c} \leqslant(a+b+c)^{2} \\
\Rightarrow \lambda \leqslant \frac{2(a b+b c+c a)}{\sqrt{a b c}} .
\end{array}
$$

Let $f=\frac{2(a b+b c+c a)}{\sqrt{a b c}}$. Then $\lambda_{\text {max }}=f_{\text {min }}$.
Notice that,
$$
\begin{array}{l}
f^{2}=4\left[\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}+2(a+b+c)\right] \\
=4\left(\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}+2\right) \\
\geqslant 4(a+b+c+2)=12 \\
\Rightarrow f \geqslant 2 \sqrt{3} .
\end{array}
$$

The equality holds when $a=b=c=\frac{1}{3}$, i.e.,
$$
\begin{array}{l}
f_{\min }=2 \sqrt{3} . \\
\text { Hence } \lambda_{\max }=2 \sqrt{3} \text {. }
\end{array}
$$