We first prove a lemma.
Lemma If $1>a, b>0$, then $a^{b} \geqslant \frac{a}{a+b}$, which is obvious when $a \geqslant 1$.
If $1>a, b>0$, by Bernoulli's inequality we have
$$a^{b}=\frac{a}{a^{1-b}} \geqslant \frac{a}{1-(1-b)(1-a)}=\frac{a}{a+b-a b} \geqslant \frac{a}{a+b}$$

Thus the lemma is proved.
Returning to the original problem, by the lemma we have
$$a^{b}+b^{c}+c^{a} \geqslant \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$$

Next, we prove
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \geqslant \frac{3}{2}$$

In fact,
$$\begin{array}{l} \left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)-\left(\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\right)= \\ \frac{(a-b)(b+c)(c+a)+(b-c)(a+b)(c+a)+(c-a)(a+b)(b+c)}{(a+b)(b+c)(c+a)}= \\ \frac{(c+a)\left(a b+a c-b^{2}-b c+a b+b^{2}-c a-c b\right)+(c-a)(a+b)(b+c)}{(a+b)(b+c)(c+a)}= \\ \frac{(a-c)\left(2 b c+2 a b-a b-a c-b^{2}-b c\right)}{(a+b)(b+c)(c+a)}= \\ \frac{(a-c)(a-b)(b-c)}{(a+b)(b+c)(c+a)} \geqslant 0 \\ \quad\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)+\left(\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\right)=3 \end{array}$$

Therefore,
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \geqslant \frac{3}{2}$$

The equality in the inequality will not hold, and we have completed the proof.