Example 5 Let the domain of $f$ be $\mathbf{N}^{+}$, satisfying $f(1)=1$, and for any $m, n \in \mathbf{N}^{+}$,
$$
f(m+n) \geqslant f(m)+f(n)+m n
$$

Try to find $f(n)$.
Solve the function $g(n, m)(\forall g(n, m) \geqslant 0)$, let
$$
f(m+n)=f(m)+f(n)+mn+g(n, m),
$$

Substitute $m=1$ into equation (1), we get
$$
f(n+1)=f(1)+f(n)+n+g(n, 1)=f(n)+(n+1)+g(n, 1),
$$

Substitute $n=1,2, \cdots, k-1$ into (1), we get
$$
\begin{array}{l}
f(2)=f(1)+2+g(1,1), \\
f(3)=f(2)+3+g(2,1), \\
\cdots \cdots \\
f(k)=f(k-1)+k+g(k-1,1) .
\end{array}
$$

Add these $k-1$ equations, we get
$$
\begin{aligned}
f(k) & =f(1)+2+3+4+\cdots+k+g(1,1)+g(2,1)+\cdots+g(k-1,1) \\
& =1+2+\cdots+k+\sum_{i=1}^{k-1} g(i, 1)=\frac{k(k+1)}{2}+\sum_{i=1}^{k} g(i, 1)
\end{aligned}
$$

Therefore, $f(n)=\frac{n(n+1)}{2}+\sum_{i=1}^{n-1} g(i, 1), \forall g(n, m) \geqslant 0$. It is verified that $f(x)$ satisfies the functional equation (2).
Explanation (1) It is not difficult to prove: The conclusion of this problem can be written in a more concise form as $f(n)=\frac{n(n+1)}{2}+\varphi(n), \forall \varphi(n) \geqslant 0$ and for any $m, n \in \mathbf{N}^{+}, \varphi(m+n) \geqslant \varphi(m)+\varphi(n)$.
The parameter functions in the first five examples are all in the form of adding an arbitrary function, but they can also be in the form of a product, exponential, logarithmic, etc., depending on the specific situation of the given functional inequality.
(2) Examples 4 and 5 show that it is difficult to ensure the equivalence of the transformation process when taking multi-variable parameter functions. We can only find a formal solution.