Example 12 Given $a, b, c, d \in \mathbf{R}^{+}$, try to find the minimum value of $f(a, b, c, d)=\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{a+b+d}+\frac{d}{a+b+c}$.
Solution 1 Since directly eliminating the denominator is complicated, consider changing the problem.
Let $b+c+d=A, c+d+a=B, d+a+b=C, a+b+c=D$.
Adding these four equations, we get $a+b+c+d=\frac{1}{3}(A+B+C+D)$.
Thus, we can find that $a=\frac{1}{3}(B+C+D-2 A), b=\frac{1}{3}(C+D+A-2 B)$,
$$
\begin{array}{l}
c=\frac{1}{3}(D+A+B-2 C), d=\frac{1}{3}(A+B+C-2 D). \\
\text { Therefore, the original expression }= \frac{B+C+D-2 A}{3 A}+\frac{C+D+A-2 B}{3 B}+\frac{D+A+B-2 C}{3 C}+ \\
\frac{A+B+C-2 D}{3 D},
\end{array}
$$

which is $3 \cdot f(a, b, c, d)+12$
$$
\begin{array}{l}
=\frac{A+B+C+D}{A}+\frac{A+B+C+D}{B}+\frac{A+B+C+D}{C}+\frac{A+B+C+D}{D} \\
=(A+B+C+D) \cdot\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+\frac{1}{D}\right) \geqslant 16,
\end{array}
$$

i.e., $3 \cdot f(a, b, c, d)+12 \geqslant 16$, hence $f(a, b, c, d) \geqslant \frac{4}{3}$, where equality holds if and only if $A=$ $B=C=D$, i.e., $a=b=c=d$.
Therefore, the minimum value of $f(a, b, c, d)$ is $\frac{4}{3}$.
Solution 2 Noting the weighted sum inequality, we have
$$
\begin{aligned}
f(a, b, c, d) & =\frac{a^{2}}{a(b+c+d)}+\frac{b^{2}}{b(a+c+d)}+\frac{c^{2}}{c(a+b+d)}+\frac{d^{2}}{d(a+b+c)} \\
& \geqslant \frac{(a+b+c+d)^{2}}{2(a b+a c+a d+b c+b d+c d)} \\
& =\frac{(a+b+c+d)^{2}}{(a+b+c+d)^{2}-\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}
\end{aligned}
$$

and $a^{2}+b^{2}+c^{2}+d^{2}=\frac{a^{2}}{1}+\frac{b^{2}}{1}+\frac{c^{2}}{1}+\frac{d^{2}}{1} \geqslant \frac{(a+b+c+d)^{2}}{4}$, thus
$$
f(a, b, c, d) \geqslant \frac{(a+b+c+d)^{2}}{(a+b+c+d)^{2}-\frac{1}{4}(a+b+c+d)^{2}}=\frac{4}{3} \text {. }
$$

where equality holds if and only if $a=b=c=d$, hence the minimum value of $f(a, b, c, d)$ is $\frac{4}{3}$.
Solution 3 Noting the Cauchy inequality or the weighted sum inequality, we have
$$
\begin{aligned}
f(a, b, c, d) & =\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{a+c+d}+\frac{a+b+c+d}{a+b+d}+ \\
& \frac{a+b+c+d}{a+b+c}-4 \\
& \geqslant(a+b+c+d)\left(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\right. \\
& \left.\frac{1}{a+b+c}\right)-4 \\
& \geqslant(a+b+c+d) \cdot \frac{4^{2}}{3(a+b+c+d)}-4=\frac{4}{3} .
\end{aligned}
$$

where equality holds if and only if $a=b=c=d$, hence the minimum value of $f(a, b, c, d)$ is $\frac{4}{3}$.