Problem 1152

Analysis: Transform each term, convert to $\frac{a_{1}}{1-q}(|q|<1)$, since $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>0$, so, $0<\frac{1}{t a} 、 \frac{1}{t b} 、 \frac{1}{t c}<1$.
Also, $a b c=1$, then
$$\begin{array}{l} \frac{1}{a^{3}(b+c)}=\frac{a b c}{a^{3}(b+c)}=\frac{b c}{a^{2}(b+c)} \\ =\frac{\frac{1}{a^{2}}}{\frac{1}{b}+\frac{1}{c}}=\frac{\frac{1}{a^{2}}}{t-\frac{1}{a}}=\frac{1}{a} \cdot \frac{\frac{1}{t a}}{1-\frac{1}{t a}} \\ =\frac{1}{a} \sum_{k=1}^{\infty}\left(\frac{1}{t a}\right)^{k} . \end{array}$$

Similarly, $\frac{1}{b^{3}(c+a)}=\frac{1}{b} \sum_{k=1}^{\infty}\left(\frac{1}{t b}\right)^{k}$,
$$\begin{aligned} & \frac{1}{c^{3}(a+b)}=\frac{1}{c} \sum_{k=1}^{\infty}\left(\frac{1}{t c}\right)^{k} . \\ \text { Hence } & \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \\ = & \frac{1}{t}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)+\frac{1}{t^{2}}\left(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\right)+ \\ & \cdots+\frac{1}{t^{n}}\left(\frac{1}{a^{n+1}}+\frac{1}{b^{n+1}}+\frac{1}{c^{n+1}}\right)+\cdots \\ \geqslant & \frac{1}{t} \cdot \frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}}{3}+\frac{1}{t^{2}} \cdot \frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{3}}{3^{2}}+ \\ & \cdots+\frac{1}{t^{n}} \cdot \frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{n+1}}{3^{n}}+\cdots \\ = & \frac{1}{t} \cdot \frac{t^{2}}{3}+\frac{1}{t^{2}} \cdot \frac{t^{3}}{3^{2}}+\cdots+\frac{1}{t^{n}} \cdot \frac{t^{n+1}}{3^{n}}+\cdots \\ = & t \sum_{k=1}^{\infty} \frac{1}{3^{k}} \\ = & t \cdot \frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2} t=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \\ \geqslant & \frac{1}{2} \times 3 \sqrt[3]{\frac{1}{a b c}}=\frac{3}{2} . \end{aligned}$$

Equality holds if and only if $a=b=c=1$.
Therefore, the original inequality holds.