Three. (25 points) If positive numbers $a, b, c$ satisfy
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}=3 \text {, }
$$

find the value of the algebraic expression
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$
Three, since $a, b, c$ have cyclic symmetry, without loss of generality, assume $0 < a < b < c$, then
$$
c-a>b>0, c-b>a>0 \text {. }
$$

Thus $\frac{b^{2}+c^{2}-a^{2}}{2 b c}=1+\frac{(c-b)^{2}-a^{2}}{2 b c}>1$,
$$
\begin{array}{l}
\frac{c^{2}+a^{2}-b^{2}}{2 c a}=1+\frac{(c-a)^{2}-b^{2}}{2 c a}>1, \\
\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{(a+b)^{2}-c^{2}}{2 a b}-13,
$$

Contradiction.
(2) If $c-1, \\
\frac{a^{2}+b^{2}-c^{2}}{2 a b}=1+\frac{(a-\dot{b})^{2}-c^{2}}{2 a b}<1 .
\end{array}
$$

From the above three equations, we get
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}<3,
$$

Contradiction.
Combining (1) and (2), we know $c=a+b$.

Therefore, $\frac{b^{2}+c^{2}-a^{2}}{2 b c}=1, \frac{c^{2}+a^{2}-b^{2}}{2 c a}=1$,
$$
\begin{array}{l}
\frac{a^{2}+b^{2}-c^{2}}{2 a b}=-1 . \\
\text { Hence } \frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}=1 .
\end{array}
$$