6. Let $a, b, c, d, m, n$ all 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 relationship between $P$ and $Q$ is uncertain and depends on the sizes of $m$ and $n$.
6. (B)
$$
\begin{array}{cc}
P=\sqrt{a b}+\sqrt{c d}, Q=\sqrt{a m+n c} \cdot \sqrt{\frac{b}{m}+\frac{d}{n}} . \\
\therefore & P^{2}=a b+c d+2 \sqrt{a b c d} . \\
& Q^{2}=(a m+n c)\left(\frac{b}{m}+\frac{d}{n}\right)=a b+c d+\frac{n b c}{m}+\frac{m a d}{n} . \\
\therefore & \frac{n b c}{m}+\frac{m a d}{n} \geqslant 2 \sqrt{\frac{n b c}{m} \cdot \frac{m a d}{n}}=2 \sqrt{a b c d} . \\
\because & Q^{2} \geqslant P^{2} .
\end{array}
$$

Since $P$ and $Q$ are both positive,
$$
\therefore \quad Q \geqslant P \text {. }
$$