Problem 1588

According to the extreme principle, the condition for the inequality to hold with equality is $a=b=c=\frac{1}{3}$. Therefore, multiplying each term on the left side of the inequality by the constant $\sqrt{\frac{1}{3}+3}$ and combining with the Cauchy-Schwarz inequality, we get:
$$\begin{array}{l} \sqrt{\frac{1}{3}+3} \cdot \sqrt{a+\frac{1}{b}} \geqslant \sqrt{\frac{a}{3}}+\sqrt{\frac{3}{b}} \\ \sqrt{\frac{1}{3}+3} \cdot \sqrt{b+\frac{1}{c}} \geqslant \sqrt{\frac{b}{3}}+\sqrt{\frac{3}{c}} \\ \sqrt{\frac{1}{3}+3} \cdot \sqrt{c+\frac{1}{a}} \geqslant \sqrt{\frac{c}{3}}+\sqrt{\frac{3}{a}} \end{array}$$

Adding the three inequalities, we get:
$$\begin{array}{l} \sqrt{\frac{1}{3}+3}\left(\sqrt{a+\frac{1}{b}}+\sqrt{b+\frac{1}{c}}+\sqrt{c+\frac{1}{a}}\right) \\ \geqslant \sqrt{\frac{a}{3}}+\sqrt{\frac{b}{3}}+\sqrt{\frac{c}{3}}+\sqrt{\frac{3}{a}}+\sqrt{\frac{3}{b}}+\sqrt{\frac{3}{c}}, \\ \text { and } \sqrt{\frac{a}{3}}+\sqrt{\frac{b}{3}}+\sqrt{\frac{c}{3}}+\sqrt{\frac{3}{a}}+\sqrt{\frac{3}{b}}+\sqrt{\frac{3}{c}} \\ \geqslant 3 \cdot \sqrt[6]{\frac{a b c}{27}}+3 \cdot \sqrt[6]{\frac{27}{a b c}}, \text { then } \\ \sqrt{a+\frac{1}{b}}+\sqrt{b+\frac{1}{c}}+\sqrt{c+\frac{1}{a}} \\ \geqslant \frac{3}{\sqrt{10}}\left(\sqrt[6]{a b c}+3 \cdot \sqrt[6]{\frac{1}{a b c}}\right), \end{array}$$

Let $t=\sqrt[6]{a b c}$. Since the positive numbers $a, b, c$ satisfy $a+b+c=1$, we have $0<\sqrt[3]{a b c} \leqslant \frac{1}{3}$, so $0<t \leqslant \frac{1}{\sqrt{3}}$.

It is easy to see that the function $f(t)=t+\frac{3}{t}$ is monotonically decreasing in the interval $0<t \leqslant \frac{1}{\sqrt{3}}$, so $t+\frac{3}{t} \geqslant \frac{1}{\sqrt{3}}+3 \sqrt{3}=\frac{10}{3} \sqrt{3}$, with equality holding when $t=\frac{1}{\sqrt{3}}$.

Therefore,
$$\begin{array}{l} \sqrt{a+\frac{1}{b}}+\sqrt{b+\frac{1}{c}}+\sqrt{c+\frac{1}{a}} \\ \geqslant \frac{3}{\sqrt{10}} \cdot \frac{10}{3} \sqrt{3}=\sqrt{30} \end{array}$$