16. Let $\alpha, \beta$ be real numbers. If for any real numbers $x, y, z$, we have $\alpha(x y+y z+z x) \leqslant M \leqslant \beta\left(x^{2}+y^{2}+z^{2}\right)$ always holds, where $M=\sqrt{x^{2}+x y+y^{2}} \cdot \sqrt{y^{2}+y z+z^{2}}+\sqrt{y^{2}+y z+z^{2}} \cdot \sqrt{z^{2}+z x+x^{2}}+\sqrt{z^{2}+z x+x^{2}} \cdot \sqrt{x^{2}+x y+y^{2}}$. Find the maximum value of $\alpha$ and the minimum value of $\beta$.
Take $x=y=z=1 \Rightarrow 3 \alpha \leqslant 9 \leqslant 3 \beta \Rightarrow \alpha \leqslant 3, \beta \geqslant 3$.
Next, we prove $M \geqslant 3(x y+y z+z x)$. Since $\sqrt{x^{2}+x y+y^{2}} \cdot \sqrt{y^{2}+y z+z^{2}}$
$$
=\sqrt{\left(x+\frac{1}{2} y\right)^{2}+\frac{3}{4} y^{2}} \cdot \sqrt{\left(z+\frac{1}{2} y\right)^{2}+\frac{3}{4} y^{2}} \geqslant\left|\left(x+\frac{1}{2} y\right)\left(z+\frac{1}{2} y\right)\right|+\frac{3}{4} y^{2}
$$
$\geqslant x z+\frac{1}{2} x y+\frac{1}{2} y z+y^{2}$, similarly $\sqrt{y^{2}+y z+z^{2}} \cdot \sqrt{z^{2}+z x+x^{2}} \geqslant y x+\frac{1}{2} y z+\frac{1}{2} z x+z^{2}$,
$\sqrt{z^{2}+z x+x^{2}} \cdot \sqrt{x^{2}+x y+y^{2}} \geqslant z y+\frac{1}{2} z x+\frac{1}{2} x y+x^{2}$, adding the three inequalities gives
$$
M \geqslant 2(x y+y z+z x)+\left(x^{2}+y^{2}+z^{2}\right) \geqslant 3(x y+y z+z x) \text {. }
$$

Next, we prove $M \leqslant 3(x y+y z+z x)$. Since $\sqrt{x^{2}+x y+y^{2}} \cdot \sqrt{y^{2}+y z+z^{2}}$
$$
\leqslant \frac{x^{2}+x y+y^{2}+y^{2}+y z+z^{2}}{2}=\frac{x^{2}+2 y^{2}+z^{2}+x y+y z}{2} \text {, }
$$

similarly $\sqrt{y^{2}+y z+z^{2}} \cdot \sqrt{z^{2}+z x+x^{2}} \leqslant \frac{x^{2}+y^{2}+2 z^{2}+y z+z x}{2}$,
$\sqrt{z^{2}+z x+x^{2}} \cdot \sqrt{x^{2}+x y+y^{2}} \leqslant \frac{2 x^{2}+y^{2}+2 z^{2}+z x+x y}{2}$, adding the three inequalities gives
$$
M \leqslant 2\left(x^{2}+y^{2}+z^{2}\right)+(x y+y z+z x) \leqslant 3\left(x^{2}+y^{2}+z^{2}\right) \text {. }
$$

In summary, the maximum value of $\alpha$ and the minimum value of $\beta$ are both 3.