279 Given positive real numbers $x, y, z$ satisfying $(\sqrt{3}+1) x y + 2 \sqrt{3} y z + (\sqrt{3}+1) z x = 1$.
(1) Find the minimum value of $x + y + z$;
(2) Find the minimum value of $\frac{\sqrt{3} x y}{z} + \frac{(8-4 \sqrt{3}) y z}{x} + \frac{\sqrt{3} z x}{y}$.
(1) By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{(\sqrt{3}+1) x^{2}}{2}+(\sqrt{3}-1) y^{2} \geqslant 2 x y, \\
2 y^{2}+2 z^{2} \geqslant 4 y z, \\
(\sqrt{3}-1) z^{2}+\frac{(\sqrt{3}+1) x^{2}}{2} \geqslant 2 x z .
\end{array}
$$

Adding the above three inequalities, we get
$$
\begin{array}{l}
(\sqrt{3}+1)\left(x^{2}+y^{2}+z^{2}\right) \geqslant 2(x y+2 y z+z x) \\
\Rightarrow x^{2}+y^{2}+z^{2} \geqslant(\sqrt{3}-1)(x y+2 y z+z x) .
\end{array}
$$

Equality holds if and only if $\frac{(\sqrt{3}+1) x^{2}}{2}=(\sqrt{3}-1) y^{2}, 2 y^{2}=2 z^{2},(\sqrt{3}-1) z^{2}=\frac{(\sqrt{3}+1) x^{2}}{2}$, i.e., $y=z=\frac{(\sqrt{3}+1) x}{2}$.
$$
\begin{array}{l}
\text { Then }(x+y+z)^{2} \\
=x^{2}+y^{2}+z^{2}+2(x y+y z+z x) \\
\geqslant(\sqrt{3}-1)(x y+2 y z+z x)+2(x y+y z+z x) \\
\geqslant(\sqrt{3}+1) x y+2 \sqrt{3} y z+(\sqrt{3}+1) z x=1 .
\end{array}
$$
$$
\text { Thus, } x+y+z \geqslant 1 \text {. }
$$

Equality holds if and only if $y=z=\frac{(\sqrt{3}+1) x}{2}$ and $(\sqrt{3}+1) x y+2 \sqrt{3} y z+(\sqrt{3}+1) z x=1$, i.e., $y=z=\frac{\sqrt{3}-1}{2}$, $x=2-\sqrt{3}$.

Therefore, when $x=2-\sqrt{3}, y=z=\frac{\sqrt{3}-1}{2}$, $x+y+z$ attains its minimum value of 1.
(2) By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{y}{z}+\frac{z}{y} \geqslant 2, \\
\frac{z}{x}+\frac{(2+\sqrt{3}) x}{2 z} \geqslant \sqrt{3}+1, \\
\frac{y}{x}+\frac{(2+\sqrt{3}) x}{2 y} \geqslant \sqrt{3}+1 .
\end{array}
$$

Equality holds if and only if $y=z=\frac{(\sqrt{3}+1) x}{2}$.
Introducing positive parameters $u, v, w$, we have
$$
\begin{array}{l}
u x\left(\frac{y}{z}+\frac{z}{y}\right)+v y\left[\frac{(2+\sqrt{3}) x}{2 z}+\frac{z}{x}\right]+ \\
w z\left[\frac{(2+\sqrt{3}) x}{2 y}+\frac{y}{x}\right] \\
=\left[u+\frac{(2+\sqrt{3}) v}{2}\right] \frac{x y}{z}+(v+w) \frac{y z}{x}+ \\
{\left[u+\frac{(2+\sqrt{3}) w}{2}\right] \frac{z x}{y} .} \\
\text { Also, } u x\left(\frac{y}{z}+\frac{z}{y}\right)+v y\left[\frac{(2+\sqrt{3}) x}{2 z}+\frac{z}{x}\right]+ \\
w z\left[\frac{(2+\sqrt{3}) x}{2 y}+\frac{y}{x}\right] \\
\geqslant 2 u x+(\sqrt{3}+1) v y+(\sqrt{3}+1) w z \text {, } \\
\text { Hence, }\left[u+\frac{(2+\sqrt{3}) v}{2}\right] \frac{x y}{z}+(v+w) \frac{y z}{x}+ \\
{\left[u+\frac{(2+\sqrt{3}) w}{2}\right] \frac{z x}{y}} \\
\geqslant 2 u x+(\sqrt{3}+1) v y+(\sqrt{3}+1) w z . \\
\text { Let } u=\sqrt{3}+1, v=w=2 \text {. Substituting into the above inequality, we get } \\
(3+2 \sqrt{3}) \frac{x y}{z}+4 \times \frac{y z}{x}+(3+2 \sqrt{3}) \frac{z x}{y} \\
\geqslant 2(\sqrt{3}+1) x+2(\sqrt{3}+1) y+2(\sqrt{3}+1) z \text {, } \\
\text { Then } \frac{\sqrt{3} x y}{z}+\frac{(8-4 \sqrt{3}) y z}{x}+\frac{\sqrt{3} z x}{y} \\
\geqslant 2(\sqrt{3}-1)(x+y+z) \geqslant 2(\sqrt{3}-1) \text {. } \\
\end{array}
$$

Equality holds if and only if $x=2-\sqrt{3}, y=z=\frac{\sqrt{3}-1}{2}$.

Therefore, when $x=2-\sqrt{3}, y=z=\frac{\sqrt{3}-1}{2}$, $\frac{\sqrt{3} x y}{z}+$ $\frac{(8-4 \sqrt{3}) y z}{x}+\frac{\sqrt{3} z x}{y}$ attains its minimum value of $2(\sqrt{3}-1)$.