Question 79, Given $a, b, c \in R^{+}, a+b+c=3$, try to find the minimum value of $\frac{1}{a}+\frac{1}{a b}+\frac{1}{a b c}$.

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Question 79, Solution: According to the conditions and the AM-GM inequality, we have:
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc} \\
=\left(\frac{1}{a}+\frac{(3+\sqrt{5})a}{10}\right)+\left(\frac{1}{ab}+\frac{(3+\sqrt{5})a}{10}+\frac{(5+\sqrt{5})b}{10}\right)+\left(\frac{1}{abc}+\frac{(4+2\sqrt{5})a}{10}+\frac{(5+3\sqrt{5})b}{10}+\frac{(10+4\sqrt{5})c}{10}\right)-
\end{array}
$$
$\frac{(10+4\sqrt{5})}{10}(a+b+c)$
$\geq 2 \sqrt{\frac{1}{a} \cdot \frac{(3+\sqrt{5})a}{10}}+3 \sqrt[3]{\frac{1}{ab} \cdot \frac{(3+\sqrt{5})a}{10} \cdot \frac{(5+\sqrt{5})b}{10}}+4 \sqrt[4]{\frac{1}{abc} \cdot \frac{(4+2\sqrt{5})a}{10} \cdot \frac{(5+3\sqrt{5})b}{10} \cdot \frac{(10+4\sqrt{5})c}{10}}-\frac{3(10+4\sqrt{5})}{10}$ $=\frac{3+\sqrt{5}}{2}$

Equality holds when $a=\frac{5-\sqrt{5}}{2}, b=1, c=\frac{\sqrt{5}-1}{2}$.
Note 1: The AM-GM inequality and the method of undetermined coefficients can be used:
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{ab}+\frac{1}{ab} \\
=\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}+k(a+b+c-3) \\
=\left(\frac{1}{a}+xa\right)+\left(\left(\frac{1}{ab}+ya+zb\right)+\left(\frac{1}{abc}+(k-x-y)a+(k-z)b+kc\right)-3k\right. \\
\geq 2 \sqrt{x}+3 \sqrt[3]{yz}+4 \sqrt[4]{(k-x-y)(k-z)k}-3k
\end{array}
$$

The equality conditions are:
$$
\begin{array}{l}
\left\{\begin{array}{c}
\frac{1}{a}=ax \\
\frac{1}{ab}=ya=zb \\
\frac{1}{abc}=(k-x-y)a=(k-z)b=kc \\
a+b+c=3 \\
\frac{1}{a^2}=x \\
\frac{1}{a^2b}=y \\
\frac{1}{ab^2}=z \\
\frac{1}{abc^2}=k \\
\left\{\begin{array}{c}
\text { equality conditions are }
\end{array}\right. \\
\Leftrightarrow (k-y-y)a=(k-z)b \\
a+b+c=3
\end{array}\right.
\end{array}
$$

Thus, we have:
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array}{c}
b=c^2+c \\
a=c^4+c^3+c^2+c
\end{array}\right. \\
\end{array}
$$

Substituting $a+b+c=3$, we get:
$$
\begin{aligned}
& c^4+c^3+2c^2+3c=3 \\
\Rightarrow & (c^2+3)(c^2+c-1)=0 \\
\Rightarrow & c=\frac{\sqrt{5}-1}{2}
\end{aligned}
$$
$\frac{3+\sqrt{5}}{2}$.
$\left\{\begin{array}{l}x=\frac{3+\sqrt{5}}{10} \\ y=\frac{3+\sqrt{5}}{10} \\ z=\frac{5+\sqrt{5}}{10} \\ k=\frac{5+2\sqrt{5}}{5}\end{array}\right.$ . Substituting into (*), we get the minimum value of $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}$ as

Thus, we know $b=1, a=\frac{5-\sqrt{5}}{2}$. Substituting, we get $\left\{\begin{array}{l}x=\frac{3+\sqrt{5}}{10} \\ y=\frac{3+\sqrt{5}}{10} \\ z=\frac{5+\sqrt{5}}{10} \\ k=\frac{5+2\sqrt{5}}{5}\end{array}\right.$ . Substituting into (*), we get $\frac{3+\sqrt{5}}{2}$. Note 2: The method of Lagrange multipliers can also be used to find the equality conditions when the minimum value is achieved.
$$
\begin{array}{l}
\text { Let } L(a, b, c, \lambda)=\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}+\lambda(a+b+c-3) \text { . Then } \\
\left\{\begin{array}{l}
\frac{\partial L}{\partial a}=-\frac{1}{a^2}-\frac{1}{a^2b}-\frac{1}{a^2bc}+\lambda=0 \\
\frac{\partial L}{\partial b}=-\frac{1}{ab^2}-\frac{1}{ab^2c}+\lambda=0 \\
\frac{\partial L}{\partial c}=-\frac{1}{abc^2}+\lambda=0 \\
\frac{\partial L}{\partial \lambda}=a+b+c-3=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{c}
\frac{1}{a^2}+\frac{1}{a^2b}+\frac{1}{a^2bc}=\lambda \\
\frac{1}{ab^2}+\frac{1}{ab^2c}=\lambda \\
\frac{1}{abc^2}=\lambda \\
a+b+c-3=0 \\
b=c^2+c \\
a=c^4+c^3+c^2+c \\
a+b+c=3
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a=\frac{5-\sqrt{5}}{2} \\
b=1 \\
c=\frac{\sqrt{5}-1}{2}
\end{array}\right.
\end{array}
$$