Example 3 Given that $a, b, c$ are positive real numbers, and $a+b+c=12, ab+bc$ $+ca=45$, try to find the maximum value of $abc$.

保留了原文的换行和格式。
Solve: From $a+b+c=12$, we get $a+c=12-b$, substitute into the transformed form of $ab+bc+ca=45$, which is $b(a+c)+ca=45$,

we get $b(12-b)+ca=45$, that is, $ca=b^{2}-12b+45$.
Thus, $b, c$ are the two roots of the following quadratic equation
$$t^{2}-(12-b)t+\left(b^{2}-12b+45\right)=0$$

Since $t$ is a real number, the equation $(*)$ has real roots, thus the discriminant $\Delta=(12-b)^{2}-4\left(b^{2}-12b+45\right) \geqslant 0$, that is, $b^{2}-8b+12 \leqslant 0$, solving this gives $2 \leqslant b \leqslant 6$. Therefore,
$$abc=\left(b^{2}-12b+45\right)b=(b-3)^{2}(b-6)+54 \leqslant 54$$

It is not difficult to see that when $a=3, b=6, c=3$, or $a=6, b=3, c=6$, or $a=3, b=3, c=6$, $abc$ achieves the maximum value of 54.