11. Try to solve the functional inequality:
$$
n f(n)-(n-1) f(n+1) \geqslant 1 \text { and } f(2)=3, n \geqslant 2 \text {. }
$$
11. Solution 1 Introduce a parameter function $g(n)(\forall g(n) \geqslant 1)$, let
$$
\begin{array}{c}
n f(n)-(n-1) f(n+1)=g(n), \\
\frac{f(n)}{n-1}-\frac{f(n+1)}{n}=\frac{g(n)}{(n-1) n},
\end{array}
$$

Let $n=2,3, \cdots, n-1$.
$$
\begin{array}{l}
\frac{f(2)}{1}-\frac{f(3)}{2}=\frac{g(2)}{1 \cdot 2}, \frac{f(3)}{2}-\frac{f(4)}{3}=\frac{g(3)}{2 \cdot 3}, \\
\cdots \\
\frac{f(n-1)}{n-2}-\frac{f(n)}{n-1}=\frac{g(n-1)}{(n-2)(n-1)},
\end{array}
$$

Add these $(n-2)$ equations. We get
$$
\begin{array}{c}
f(2)-\frac{f(n)}{n-1}=\sum_{i=2}^{n} \frac{g(i)}{i(i+1)}, \\
f(n)=2 f(2)-(n-1) \sum_{i=1}^{n} \frac{g(i)}{i(i+1)} .
\end{array}
$$

Therefore, the solution to the given functional inequality is $f(n)=2 n-2-(n-1) \sum_{1}^{n-2} \frac{g(i)}{i(i+1)}, n \in \mathbf{N}^{+}$.
where any function $g(n)$ satisfies $g(n) \geqslant 1$.
Solution 2 Divide both sides by $n(n-1)$. We get
$$
\begin{array}{c}
\frac{f(n)}{n-1}-\frac{f(n+1)}{n} \geqslant \frac{1}{(n-1) n}=\frac{1}{n-1}-\frac{1}{n}, \\
\frac{f(n)}{n-1}-\frac{1}{n-1} \geqslant \frac{f(n+1)}{n}-\frac{1}{n},
\end{array}
$$

Introduce a parameter function $g(n)(\forall g(n) \geqslant 1)$, let $g(n)=\frac{f(n)-1}{n-1}$, then $g(2)=\frac{f(2)-1}{2-1}=2$, the general solution to the given inequality is $f(n)=1+(n-1) g(n)$
where any function $g(n)$ satisfies $g(2)=2, g(n) \geqslant g(n+1), n \in \mathbf{N}, n \geqslant 2$.