6. Given $x, y, z \in \mathbf{R}_{+}$, and $\sqrt{x^{2}+y^{2}}+z=1$. Then the maximum value of $x y+2 x z$ is $\qquad$ .
6. $\frac{\sqrt{3}}{3}$.
$$
\begin{array}{l}
\text { Given } \sqrt{x^{2}+y^{2}}+z=1 \Rightarrow x^{2}+y^{2}=(1-z)^{2} \\
\Rightarrow x^{2}=(1-z-y)(1-z+y) .
\end{array}
$$

Then $(x y+2 x z)^{2}=x^{2}(y+2 z)^{2}$
$$
\begin{array}{l}
=\frac{1}{3}(3-3 z-3 y)(1-z+y)(y+2 z)(y+2 z) \\
\leqslant \frac{1}{3}\left(\frac{(3-3 z-3 y)+(1-z+y)+2(y+2 z)}{4}\right)^{4} \\
=\frac{1}{3} \\
\Rightarrow x y+2 x z \leqslant \frac{\sqrt{3}}{3} .
\end{array}
$$

Equality holds if and only if $x=\frac{\sqrt{3}}{3}, y=z=\frac{1}{3}$.