5. Given positive numbers $a, b, c$ satisfy $a b+b c+c a=1995$. Then the minimum value of $\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}$ is ( ).
(A) 1995
(B) $3 \sqrt{665}$
(C) $2 \sqrt{665}$
(D) $\sqrt{665}$
5. (B).
$$
\begin{array}{l}
\text { Given } a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a=1995, \text { we have } \\
\frac{a^{2} b^{2}}{c^{2}}+\frac{c^{2} a^{2}}{b^{2}}+\frac{b^{2} c^{2}}{a^{2}} \\
\geqslant \frac{a b}{c} \cdot \frac{c a}{b}+\frac{c a}{b} \cdot \frac{b c}{a}+\frac{b c}{a} \cdot \frac{a b}{c} \\
=a^{2}+b^{2}+c^{2} \geqslant 1995 . \\
\text { Also, }\left(\frac{a b}{c}+\frac{c a}{b}+\frac{b c}{a}\right)^{2} \\
=\frac{a^{2} b^{2}}{c^{2}}+\frac{c^{2} a^{2}}{b^{2}}+\frac{b^{2} c^{2}}{a^{2}}+2 a^{2}+2 b^{2}+2 c^{2} \\
\geqslant 3 \times 1995 .
\end{array}
$$

Thus, $\frac{a b}{c}+\frac{c a}{b}+\frac{b c}{a} \geqslant 3 \sqrt{665}$.
That is, the minimum value of $\frac{a b}{c}+\frac{a c}{b}+\frac{b c}{a}$ is $3 \sqrt{665}$.