16. Let $\alpha, \beta$ be real numbers. If for any real numbers $x, y, z$, we have $a(x y+y z+z x) \leq M \leq \beta\left(x^{2}+y^{2}+z^{2}\right)$ 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 $\beta$.
Let $x=y=z=1$, then $3 \alpha \leq 9 \leq 3 \beta$, so $\alpha \leq 3, \beta \geq 3$.
First, we prove: $M \geq 3(x y+y z+z x)$ for any real numbers $x, y, z$.
Since $\sqrt{x^{2}+x y+y^{2}} \cdot \sqrt{y^{2}+y z+z^{2}}=\sqrt{\left[\left(x+\frac{y}{2}\right)^{2}+\frac{3}{4} y^{2}\right] \cdot\left[\left(z+\frac{y}{2}\right)^{2}+\frac{3}{4} y^{2}\right]}$
$$
\geq\left(x+\frac{y}{2}\right)\left(z+\frac{y}{2}\right)+\frac{3}{4} y^{2}=x z+\frac{1}{2} x y+\frac{1}{2} y z+y^{2} .
$$

Therefore, $M \geq \sum\left(x z+\frac{1}{2} x y+\frac{1}{2} y z+y^{2}\right)=2(x y+y z+z x)+\left(x^{2}+y^{2}+z^{2}\right)$
$$
\geq 2(x y+y z+z x)+(x y+y z+z x)=3(x y+y z+z x) \text {. }
$$

Next, we prove: $M \leq 3\left(x^{2}+y^{2}+z^{2}\right)$ for any real numbers $x, y, z$.
Since $M \leq\left(\sqrt{x^{2}+x y+y^{2}}\right)^{2}+\left(\sqrt{y^{2}+y z+z^{2}}\right)^{2}+\left(\sqrt{z^{2}+z x+x^{2}}\right)^{2}$
$$
=2\left(x^{2}+y^{2}+z^{2}\right)+(x y+y z+z x) \leq 3\left(x^{2}+y^{2}+z^{2}\right) \text {. }
$$

In conclusion, the maximum value of $\alpha$ is $3$, and the minimum value of $\beta$ is $3$.