16. Given that $\alpha, \beta$ are real numbers, for any real numbers $x, y, z$, we have
$$
\alpha(x y+y z+z x) \leqslant M \leqslant \beta\left(x^{2}+y^{2}+z^{2}\right),
$$

where, $M=\sum \sqrt{x^{2}+x y+y^{2}} \sqrt{y^{2}+y z+z^{2}}$, and $\sum$ denotes the cyclic sum. Find the maximum value of $\alpha$ and the minimum value of $\beta$.
16. Taking $x=y=z=1$, we have
$$
3 \alpha \leqslant 9 \leqslant 3 \beta \Rightarrow \alpha \leqslant 3, \beta \geqslant 3 \text {. }
$$
(1) Note that, $x^{2}+x y+y^{2} \geqslant \frac{3}{4}(x+y)^{2}$.
$$
\begin{array}{l}
\text { Then } M \geqslant \frac{3}{4} \sum|x+y||y+z| \\
\geqslant \frac{3}{4} \sum\left(x y+y z+z x+y^{2}\right) \\
=\frac{3}{4} \sum x^{2}+\frac{9}{4} \sum x y \\
=3 \sum x y .
\end{array}
$$
(2) From the condition, we get
$$
\begin{array}{l}
M \leqslant \sum\left(\sqrt{x^{2}+x y+y^{2}}\right)^{2} \\
=2 \sum x^{2}+\sum x y \leqslant 3 \sum x^{2} .
\end{array}
$$

Combining (1) and (2), we know that the maximum value of $\alpha$ is 3, and the minimum value of $\beta$ is also 3.