4. Let $a, b, c$ be any real numbers, and satisfy: $a>b>c,(a-b)(b-c)(c-a)=-16$. Then the minimum value of $\frac{1}{a-b}+\frac{1}{b-c}-\frac{1}{c-a}$ is $\qquad$
4. $\frac{5}{4}$.

Notice,
$$
\begin{array}{l}
\frac{(b-c)+(a-b)}{(a-b)(b-c)}+\frac{1}{a-c} \\
=\frac{a-c}{(a-b)(b-c)}+\frac{1}{a-c} \\
=\frac{(a-c)^{2}}{16}+\frac{1}{a-c} \\
=\frac{(a-c)^{2}}{16}+\frac{1}{2(a-c)}+\frac{1}{2(a-c)} .
\end{array}
$$

Using the AM-GM inequality, it is easy to find the minimum value is $\frac{3}{4}$, when and only when $\frac{(a-c)^{2}}{16}=\frac{1}{2(a-c)}$, i.e.,
$$
a-c=2 \text {. }
$$

At this point, $(a-b)(b-c)=8, a-c=2$ has no solution.
Let $a-c=x>0$. Consider
$$
\begin{array}{l}
f(x)=\frac{x^{2}}{16}+\frac{1}{x} \\
\Rightarrow f^{\prime}(x)=\frac{x}{8}-\frac{1}{x^{2}}=\frac{x^{3}-8}{8 x^{2}} .
\end{array}
$$

By $x^{2}=(a-c)^{2}=((a-b)+(b-c))^{2}$
$$
\begin{array}{l}
\geqslant 4(a-b)(b-c)=\frac{64}{x} \\
\Rightarrow x \geqslant 4 .
\end{array}
$$

It is easy to see that $f^{\prime}(x) \geqslant 0, x \geqslant 4$.
Thus, $f(x)_{\min }=f(4)=\frac{5}{4}$.