Example 3 (2000 Hebei Provincial 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 size relationship between $M$ and $N$ is ( ).
A. $M>N$
B. $M<N$
C. $M=N$
D. The size relationship between $M$ and $N$ cannot be determined
Solution: Choose A. Reason: Let vector $\vec{\alpha}=\left(\frac{a}{n} \cdot \frac{b}{m}\right), \beta=(n, m)$. From $M^{2}>\frac{a^{2} m^{2}+b^{2} n^{2}}{m^{2} n^{2}} \cdot M^{2}=$ $\left(\frac{a^{2}}{n^{2}}+\frac{b^{2}}{m^{2}}\right)\left(n^{2}+m^{2}\right)=|\vec{\alpha}|^{2} \cdot|\beta|^{2} \geqslant(\vec{\alpha} \cdot \vec{\beta})^{2}=(a+b)^{2}=N^{2}$, we get $M>N$.