Example 6 (2000 Hebei Province Competition Question) Given $a, b \in \mathbf{R}^{+}, m, n \in \mathbf{R}, m^{2} n^{2}>a^{2} m^{2}+b^{2} n^{2}$, let $M=\sqrt{m^{2}+n^{2}}$, $N=a+b$, then the relationship between $M$ and $N$ is ( ).
A. $M>N$
B. $M<N$
C. $M=N$
D. Cannot be determined
Solution: Choose A. Reason: From the given model $m^{2} n^{2}>a^{2} m^{2}+b^{2} n^{2}$, we know $m^{2} n^{2} \neq 0$. Dividing both sides of this inequality by $m^{2} n^{2}$, we get
$$
\frac{a^{2}}{n^{2}}+\frac{b^{2}}{m^{2}}\left(m^{2}+n^{2}\right)\left(\frac{a^{2}}{n^{2}}+\frac{b^{2}}{m^{2}}\right) \\
= a^{2}+b^{2}+\frac{m^{2}}{n^{2}} a^{2}+\frac{n^{2}}{m^{2}} b^{2} \\
\geqslant a^{2}+b^{2}+2 a b=(a+b)^{2} .
\end{aligned}
$$