Example $14.13 a, b, c, x, y, z \in \mathbf{R}$ and satisfy $(x+y) c-(a+b) z=\sqrt{6}$, find the minimum value of the following expression.
$$a^{2}+b^{2}+c^{2}+x^{2}+y^{2}+z^{2}+a x+b y+c z$$
Proof: Let $a+b=2t, x+y=2s$, by Cauchy inequality, we have
$$\begin{array}{l}
a^{2}+b^{2}+x^{2}+y^{2}+a x+b y=\left(a+\frac{x}{2}\right)^{2}+\left(b+\frac{y}{2}\right)^{2}+\frac{3}{4}\left(x^{2}+y^{2}\right) \geqslant \\
\frac{1}{2}\left(a+b+\frac{x+y}{2}\right)^{2}+\frac{3}{8}(x+y)^{2}=2\left(t^{2}+t s+s^{2}\right)
\end{array}$$

Next, we prove
$$\begin{array}{l}
2\left(t^{2}+t s+s^{2}\right)+c^{2}+c z+z^{2} \geqslant \sqrt{6}(c s-t z) \Leftrightarrow \\
c^{2}+(z-\sqrt{6} s) c+2 t^{2}+(2 s+\sqrt{6} z) t+2 s^{2}+z^{2} \geqslant 0 \Leftrightarrow \\
\left(c+\frac{z-\sqrt{6} s}{2}\right)^{2}+2\left(t+\frac{2 s+\sqrt{6} z}{4}\right)^{2} \geqslant 0
\end{array}$$

The above inequality holds. From this, we easily know
$$a^{2}+b^{2}+c^{2}+x^{2}+y^{2}+z^{2}+a x+b y+c z \geqslant 3$$