3. Given $x, y, z \in \mathbf{R}_{+}$ and $x+y+z=1$. Then the maximum value of $x+\sqrt{2 x y}+3 \sqrt[3]{x y z}$ is $\qquad$
3. 2 .

By the AM-GM inequality, we have
$$
\frac{x}{2}+y \geqslant \sqrt{2 x y}, \frac{x}{2}+y+2 z \geqslant 3 \sqrt[3]{x y z} \text {. }
$$

Adding the two inequalities, we get
$$
x+\sqrt{2 x y}+3 \sqrt[3]{x y z} \leqslant 2(x+y+z)=2 \text {. }
$$

The equality holds when $x: y: z=4: 2: 1$.