Example 2 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
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$
Notice,
$$
\begin{array}{l}
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}-1 \\
=\left(\frac{b^{2}+c^{2}-a^{2}+2 b c}{2 b c}\right)\left(\frac{b^{2}+c^{2}-a^{2}-2 b c}{2 b c}\right) \\
=\frac{(b+c+a)(b+c-a)(b-c+a)(b-c-a)}{4 b^{2} c^{2}} .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}-1 \\
=\frac{(c+a+b)(c+a-b)(c-a+b)(c-a-b)}{4 c^{2} a^{2}}, \\
\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}-1 \\
=\frac{(a+b+c)(a+b-c)(a-b+c)(a-b-c)}{4 a^{2} b^{2}} . \\
\text { Therefore, }\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}-3 \\
=-(a+b+c)(a+b-c)(b+c-a)(a+c-b) . \\
\left(\frac{1}{4 b^{2} c^{2}}+\frac{1}{4 c^{2} a^{2}}+\frac{1}{4 a^{2} b^{2}}\right) \\
=0 .
\end{array}
$$

Since \(a, b, c\) are positive numbers, therefore,
$$
\begin{array}{l}
a+b+c>0, \\
\frac{1}{4 b^{2} c^{2}}+\frac{1}{4 c^{2} a^{2}}+\frac{1}{4 a^{2} b^{2}}>0 .
\end{array}
$$

Then \((a+b-c)(b+c-a)(a+c-b)=0\).
Hence, among \(a, b, c\), there must be two numbers whose sum equals the third number.
Without loss of generality, let \(a+b=c\). Then
$$
\begin{array}{l}
b^{2}+c^{2}-a^{2}=b^{2}+c^{2}-(c-b)^{2}=2 b c, \\
c^{2}+a^{2}-b^{2}=c^{2}+a^{2}-(c-a)^{2}=2 a c, \\
a^{2}+b^{2}-c^{2}=a^{2}+b^{2}-(a+b)^{2}=-2 a b . \\
\text { Therefore, } \frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b} \\
=1+1-1=1 .
\end{array}
$$