Problem 1 Let real numbers $a, b, c \geqslant 1$, and satisfy
$$a b c+2 a^{2}+2 b^{2}+2 c^{2}+c a-b c-4 a+4 b-c=28$$

Find the maximum value of $a+b+c$.
From the transformation of the conditional equation, we get
$$\begin{aligned}
& \left(\frac{a-1}{2}\right)^{2}+\left(\frac{b+1}{2}\right)^{2}+\left(\frac{c}{2}\right)^{2}+\left(\frac{a-1}{2}\right)\left(\frac{b+1}{2}\right) \frac{c}{2} \\
= & 4
\end{aligned}$$

Let \( x=\frac{a-1}{2}, y=\frac{b+1}{2}, z=\frac{c}{2} \), then \( x \geqslant 0, y \geqslant 1 \), \( z \geqslant \frac{1}{2} \), and we have
$$ x^{2}+y^{2}+z^{2}+x y z=4 $$

Thus, \( a+b+c \leqslant 6 \).
When \( \frac{a-1}{2}=1, \frac{b+1}{2}=1, \frac{c}{2}=1 \), i.e., \( a=3, b=1, c=2 \), \( a+b+c \) achieves its maximum value of 6.