Example 8 Given non-zero real numbers $a, b, c$ that are not all equal and satisfy
$$
\frac{a^{2}}{2 a^{2}+b c}+\frac{b^{2}}{2 b^{2}+c a}+\frac{c^{2}}{2 c^{2}+a b}=1 .
$$

Find the value of $a+b+c$.
Let $x=\frac{2 a^{2}}{b c}, y=\frac{2 b^{2}}{c a}, z=\frac{2 c^{2}}{a b}$.
Then, $x y z=8$.
Thus, equation (1) can be transformed into
$$
\begin{array}{l}
\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=2 \\
\Rightarrow \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1 \\
\Rightarrow x y z=x+y+z+2 \Rightarrow x+y+z=6 \\
\Rightarrow a^{3}+b^{3}+c^{3}-3 a b c=0 \\
\Rightarrow(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=0 \\
\Rightarrow \frac{1}{2}(a+b+c)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right)=0 .
\end{array}
$$

Since $a, b, c$ are not all equal, we have
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \neq 0.
$$

Therefore, $a+b+c=0$.