Proof. Since $a b c=1$, we have
$$\begin{array}{l} \frac{1}{a^{5}(b+2 c)^{2}}=\frac{b^{3} c^{3}}{(a b+2 a c)^{2}} \\ \frac{1}{b^{5}(c+2 a)^{2}}=\frac{c^{3} a^{3}}{(b c+2 b a)^{2}} \\ \frac{1}{c^{5}(a+2 b)^{2}}=\frac{a^{3} b^{3}}{(c a+2 c b)^{2}} \end{array}$$

Therefore, we only need to prove
$$\frac{b^{3} c^{3}}{(a b+2 a c)^{2}}+\frac{c^{3} a^{3}}{(b c+2 b a)^{2}}+\frac{a^{3} b^{3}}{(c a+2 c b)^{2}} \geqslant \frac{1}{3} .$$

By the AM-GM inequality,
$$\begin{aligned} & \frac{b^{3} c^{3}}{(a b+2 a c)^{2}}+\frac{a b+2 a c}{27}+\frac{a b+2 a c}{27} \\ \geqslant & 3 \sqrt[3]{\frac{b^{3} c^{3}}{(a b+2 a c)^{2}} \cdot \frac{a b+2 a c}{27} \cdot \frac{a b+2 a c}{27}}=\frac{1}{3} b c \\ \Longrightarrow & \frac{b^{3} c^{3}}{(a b+2 a c)^{2}} \geqslant \frac{1}{3} b c-\frac{2 a b+4 a c}{27} \end{aligned}$$

Similarly, we get
$$\begin{array}{l} \frac{c^{3} a^{3}}{(b c+2 a c)^{2}} \geqslant \frac{1}{3} c a-\frac{2 b c+4 b a}{27} \\ \frac{a^{3} b^{3}}{(c a+2 c b)^{2}} \geqslant \frac{1}{3} a b-\frac{2 c a+4 c b}{27} \end{array}$$

Therefore,
$$\begin{aligned} & \frac{b^{3} c^{3}}{(a b+2 a c)^{2}}+\frac{c^{3} a^{3}}{(b c+2 b a)^{2}}+\frac{a^{3} b^{3}}{(c a+2 c b)^{2}} \\ \geqslant & \frac{1}{3}(b c+c a+a b)-\frac{2}{27}(a b+b c+c a)-\frac{4}{27}(a c+b a+c b) \\ = & \frac{1}{9}(a b+b c+c a) \geqslant \frac{1}{9} \cdot 3 \sqrt[3]{a b \cdot b c \cdot c a}=\frac{1}{3} . \end{aligned}$$