5. First prove: If $a b c d=1$, then $a+b+c+d$ does not exceed the weighted average of $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$ and $\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}$.
By the AM-GM inequality, we have
$$\begin{array}{l} a=\sqrt[4]{\frac{a^{4}}{a b c d}}=\sqrt[4]{\frac{a}{b} \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{a}{d}} \\ \leqslant \frac{1}{4}\left(\frac{a}{b}+\frac{a}{b}+\frac{b}{c}+\frac{a}{d}\right) . \end{array}$$

Similarly, $b \leqslant \frac{1}{4}\left(\frac{b}{c}+\frac{b}{c}+\frac{c}{d}+\frac{b}{a}\right)$,
$$\begin{array}{l} c \leqslant \frac{1}{4}\left(\frac{c}{d}+\frac{c}{d}+\frac{d}{a}+\frac{c}{b}\right), \\ d \leqslant \frac{1}{4}\left(\frac{d}{a}+\frac{d}{a}+\frac{a}{b}+\frac{d}{c}\right) . \end{array}$$

Adding the above four inequalities, we get
$$\begin{array}{l} a+b+c+d \\ \leqslant \frac{3}{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)+\frac{1}{4}\left(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\right) \end{array}$$

If $a+b+c+d>\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$, then
$$a+b+c+d<\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d} \text {. }$$