Example 1 Let $f: \mathbf{R} \rightarrow \mathbf{R}$, and $f(x)+f\left(\frac{x-1}{x}\right) \geqslant 1+x$, try to solve this functional inequality.
Notice that $\frac{x-1}{x} \neq 1$. Therefore, the given inequality of the function has no constraint at $x=1$.
Let $\varphi(x)=\frac{x-1}{x}$, then $\varphi^{(1)}(x)=\frac{1}{1-x} \cdot \varphi^{(3)}(x)=x$. Take $\forall p(x) \geqslant 0$. And let
$$
f(x)+f\left(\frac{x-1}{x}\right)=1+x+p(x),
$$

Substitute $\varphi(x) \rightarrow x$, we get
$$
f\left(\frac{x-1}{x}\right)+f\left(\frac{1}{1-x}\right)=\frac{2 x-1}{x}+p\left(\frac{x-1}{x}\right),
$$

Substitute $\varphi^{(2)}(x) \rightarrow x$, we get
$$
f\left(\frac{1}{1-x}\right)+f(x)=\frac{2-x}{1-x}+p\left(\frac{1}{1-x}\right) .
$$
(1) + (3) - (2) gives
$$
2 f(x)=1+x+\frac{2-x}{1-x}-\frac{2 x-1}{x}+p(x)+p\left(\frac{1}{1-x}\right)-p\left(\frac{x-1}{x}\right) .
$$

Thus, $f(x)=\frac{1+x^{2}-x^{3}}{2 x(1-x)}+\frac{1}{2}\left[p(x)+p\left(\frac{1}{1-x}\right)-p\left(\frac{x-1}{x}\right)\right]$, where $\forall p(x) \geqslant 0, x \neq 0,1$.
Take $p(x)=0, x \in(0,1)$. We get the corresponding functional equation.
$$
f(x)+f\left(\frac{x-1}{x}\right)=1+x
$$

The solution is $f(x)=\frac{x^{3}-x^{2}-1}{2 x(x-1)}, \forall x \in \mathbf{R}-\{0,1\}$.
Verification: $f(x)+f\left(\frac{x-1}{x}\right)=\frac{x^{3}-x^{2}-1}{2 x(x-1)}+\frac{\left(\frac{x-1}{x}\right)^{3}-\left(\frac{x-1}{x}\right)^{2}-1}{2 \frac{x-1}{x}\left(\frac{x-1}{x}-1\right)}=x+1$
Therefore, $f(x)=\frac{x^{3}-x^{2}-1}{2 x(x-1)}, \forall x \in \mathbf{R}-\{0,1\}$.
Analysis: Generally, for the inequality $f(x)+f[\varphi(x)] \geqslant k(x)$, as long as $\varphi(x)$ has a period of 3, its solution can be found as
$$
\begin{array}{l}
f(x)=\frac{1}{2}\left\{k(x)-k[\varphi(x)]+k\left[\varphi^{\circ}(x)\right]+p(x)-p[\varphi(x)]+p\left[\varphi^{\prime}(x)\right]\right\}, \forall p(x) \geqslant 0, \\
\text { or } f(x)=\frac{1}{2}\left\{p(x)-p[\varphi(x)]+p\left[\varphi^{\prime}(x)\right]\right\}, \forall p(x) \geqslant k(x) .
\end{array}
$$

The constraint conditions are determined by specific situations.