【Analysis】There are many ways to prove this problem. We might as well try constructing a function.
Notice that, $a b c>0$.
By the Pigeonhole Principle, we know that among $a, b, c$, there exist two numbers with the same sign, let's assume they are $b, c$. Hence, $b c>0$, and consequently, $a>0$.
Let $t=\sqrt{b c}>0$.
By the AM-GM inequality, we have
$$\sqrt{b c} \leqslant \frac{|b+c|}{2}=\frac{|1-a|}{2} \text {. }$$

Transform the original inequality as follows:
$$\text{a b+b c+c a0 ,}$$

where, $u=\sqrt{a} \in(0,1)$.
At this point, we only need to prove
$$g(u)=3 u^{3}-u^{2}-2 u+1>0 \text {. }$$

Since $g^{\prime}(u)=9 u^{2}-2 u-2=0$ has only one solution $\frac{1+\sqrt{19}}{9}$ in the interval $[0,1]$, which is the minimum point of $g(u)$, and
$$g\left(\frac{1+\sqrt{19}}{9}\right)=\frac{187-38 \sqrt{19}}{243}>0,$$

Therefore, $g(u)>0$ holds in the interval $(0,1)$.
(2) When $a \geqslant 1$,
$$\text { Equation (2) } \Leftrightarrow\left(3 u^{3}-2 u\right)+\left(u^{2}-1\right)>0 \text {. }$$

Since $u=\sqrt{a} \geqslant 1$, this inequality clearly holds.
Thus, we have proved that $f(x)<0$ always holds.
Substituting $x=\sqrt{b c}$, we get
$$a b+b c+c a<\frac{\sqrt{a b c}}{2}+\frac{1}{4} .$$