Example 1 Let $f(x): \mathbf{R} \rightarrow \mathbf{R}^{+}, f\left(\frac{1}{x}\right) \geqslant x+\sqrt{x^{2}+1}$, try to find $f(x)$.
Given the function $p(x)\left(\forall p(x) \geqslant 0, x \in \mathbf{R}^{\prime}\right)$, let $f\left(\frac{1}{x}\right)=x+\sqrt{x^{2}+1}+p(x)$, then
$$
\begin{array}{l}
f\left(\frac{1}{x}\right)=x+\sqrt{x^{2}+1}+p(x)=\frac{1+\sqrt{1+\frac{1}{x^{2}}}}{\frac{1}{x}}+p\left(\frac{1}{\frac{1}{x}}\right) \text {, so the general solution of the functional inequality is } \\
f(x)=\frac{1}{x} \frac{\sqrt{1+x^{2}}}{x}+p\left(\frac{1}{x}\right) \quad\left(\forall p(x) \geqslant 0, x \in \mathbf{R}^{\prime}\right) . \\
\end{array}
$$

Check $f\left(\frac{1}{x}\right)=\frac{1+\sqrt{1+\frac{1}{x^{2}}}}{\frac{1}{x}}+p\left(\frac{1}{\frac{1}{x}}\right)=x+\sqrt{x^{2}+1}+p(x) \geqslant x+\sqrt{x^{2}+1}$, so $f(x)=\frac{1+\sqrt{1+x^{2}}}{x}+p\left(\frac{1}{x}\right) \quad\left(\forall p(x) \geqslant 0, x \in \mathbf{R}^{+}\right)$ is a solution to the given functional inequality.
Analysis (1) The verification here is an important part of solving the functional inequality. When solving for $f(x)$, we first assume that the functional inequality has a solution $f(x)$. The $f(x)$ we find only satisfies the necessity, meaning that $f(x)$ is a solution to the functional inequality only if the functional inequality has a solution. If the functional inequality has no solution, then the $f(x)$ we find is not a solution to the original functional inequality. Therefore, the sufficiency must be verified. Sometimes the solution obtained clearly satisfies the original functional inequality, so we omit the verification.
(2) When $x \in \mathbf{R}^{+}$, $p\left(\frac{1}{x}\right) \geqslant 0 \Leftrightarrow p(x) \geqslant 0$, hence the solution to the given functional inequality can also be written more concisely as:
$$
f(x) \geqslant \frac{1+\sqrt{1+\overline{x^{2}}}}{x}, x \in \mathbf{R} .
$$

However, this form of expression cannot be substituted back for verification, as inequalities do not have the property of "equality."