Example 6 Let the function $f(n)$ be defined on the set of positive integers, try to solve the functional inequality $f(n+1) \geqslant f(n)+\log _{2} \frac{n+1}{n}$.
Solution 1 Introduce a parameter function $g(n)(\forall g(n) \geqslant 0, n \in \mathbf{N})$, let $f(n+1)=f(n)+\log _{2} \frac{n+1}{n}+g(n)$. Omitted below.
Solution 2 Introduce a parameter function $g(n)\left(\forall g(n) \geqslant \log _{2} \frac{n+1}{n}, n \in \mathbf{N}^{\prime}\right)$, let $f(n+1)=f(n)+g(n)$. Omitted below.
Solution 3 Introduce a parameter function $g(n)\left(\forall g(n) \geqslant \frac{n+1}{n}, n \in \mathbf{N}^{\prime}\right)$, let $f(n+1)=f(n)+\log _{2} g(n)$. Omitted below.
Solution 4 Let $f(n)=\log _{2}[n g(n)]$. Then the function inequality $f(n+1) \geqslant f(n)+\log _{2} \frac{n+1}{n}$ becomes $\log _{2}[(n+1) g(n+1)] \geqslant \log _{2}[n g(n)]+\log _{2} \frac{n+1}{n}$, which is $g(n+1) \geqslant g(n)>0$. Therefore, the solution to the given function inequality is $f(n)=\log _{2}[n g(n)]$, where $g(n+1) \geqslant g(n)>0, n \in \mathbf{N}^{+}$. Solution 5 Convert the function inequality $f(n+1) \geqslant f(n)+\log \frac{n+1}{n}$ to
$$
f(n+1)-\log _{2}(n+1) \geqslant f(n)-\log _{2} n .
$$

Let $g(n)=f(n)-\log _{2} n$, then the above equation becomes $g(n+1) \geqslant g(n), n \in \mathbf{N}^{+}$.
Therefore, the solution to the given function inequality is $f(n)=g(n)+\log _{2} n$, where $g(n+1) \geqslant g(n), n \in \mathbf{N}^{\prime}$.