5. Let non-zero real numbers $a, b, c$ be not all equal and satisfy
$$
\frac{b c}{2 a^{2}+b c}+\frac{a c}{2 b^{2}+a c}+\frac{a b}{2 c^{2}+a b}=1 \text {. }
$$

Find the value of $a+b+c$.
Let $x=\frac{2 a^{2}}{b c}, y=\frac{2 b^{2}}{a c}, z=\frac{2 c^{2}}{a b}$. Then $x y z=8$.
Therefore, the original expression can be transformed as
$$
\begin{array}{l}
\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1 \\
\Rightarrow x y z=x+y+z+2 \\
\Rightarrow x+y+z=6 .
\end{array}
$$

Substituting $a, b, c$ into the expanded form, we get
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}=3 a b c . \\
\text { Hence }(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0 \text {. }
\end{array}
$$

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

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