4.45 Let $a, b, c, d, m, n$ be positive real numbers, $P=\sqrt{a b}+\sqrt{c d}, Q=$ $\sqrt{m a+n c} \cdot \sqrt{\frac{b}{m}+\frac{d}{n}}$. Then
(A) $P \geqslant Q$.
(B) $P \leqslant Q$.
(C) $P<Q$.
(D) The size relationship between $P$ and $Q$ is uncertain and depends on the size of $m, n$.
(China High School Mathematics League, 1983)
［Solution］From the given conditions, we have
$$
\begin{array}{l}
P^{2}=a b+c d+2 \sqrt{a b c d}, \quad Q^{2}=a b+c d+\frac{m a d}{n}+\frac{n b c}{m}, \\
\because \quad \frac{m a d}{n}+\frac{n b c}{m} \geqslant 2 \sqrt{\frac{m a d}{n}} \cdot \frac{n b c}{m}=2 \sqrt{a b c d}, \\
\therefore \quad Q^{2} \geqslant P^{2} .
\end{array}
$$

Since $P, Q$ are both positive numbers, then $Q \geqslant P$.
Therefore, the correct choice is $(B)$.