3. 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
3. (A).

Obviously, $m^{2} n^{2} \neq 0$. Dividing both sides of the 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} .
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