16. Given non-negative real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}+x+2 y+3 z=\frac{13}{4}$.
(1) Find the maximum value of $x+y+z$;
16. (1) From the given equation, we have $\left(x+\frac{1}{2}\right)^{2}+(y+1)^{2}+\left(z+\frac{3}{2}\right)^{2}=\frac{27}{4}$. By the Cauchy-Schwarz inequality, we get
$$\begin{array}{l}
{\left[\left(x+\frac{1}{2}\right)+(y+F)+\left(z+\frac{3}{2}\right)\right]^{2} \leqslant} \\
3\left[\left(x+\frac{1}{2}\right)^{2}+(y+1)^{2}+\left(z+\frac{3}{2}\right)^{2}\right]=\frac{81}{4}
\end{array}$$

Therefore, $x+y+z \leqslant \frac{3}{2}$, with equality holding if and only if $x=1, y=\frac{1}{2}, z=0$.