To prove: Without loss of generality, assume $a \geq b \geq c$. First, we will prove
$$\sqrt{\frac{a^{2}+c^{2}}{b^{2}+c^{2}}}+\sqrt{\frac{b^{2}+c^{2}}{c^{2}+a^{2}}} \geq \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}$$

In fact, this inequality is equivalent to
$$\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+\frac{b^{2}+c^{2}}{c^{2}+a^{2}} \geq \frac{a}{b}+\frac{b}{a} \Leftrightarrow \frac{(a-b)^{2}(a+b)(a b-c)^{2}}{a b\left(c^{2}+a^{2}\right)\left(b^{2}+c^{2}\right)} \geq 0$$

This is true because $a \geq b \geq c$. Using this result, we have
$$\begin{array}{l} \sqrt{\frac{a^{2}+2 b c}{b^{2}+c^{2}}}+\sqrt{\frac{b^{2}+2 c a}{c^{2}+a^{2}}}+\sqrt{\frac{c^{2}+2 a b}{a^{2}+b^{2}}} \geq \sqrt{\frac{c^{2}+a^{2}}{b^{2}+c^{2}}}+\sqrt{\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}+\sqrt{\frac{2 a b}{a^{2}+b^{2}}} \\ \geq \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}+\sqrt{\frac{2 a b}{a^{2}+b^{2}}} \end{array}$$

Let $x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}} \geq 2$. If $x \geq 3$, the inequality is obviously true. Otherwise, assume $x \leq 3$, we only need to prove
$$\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}+\sqrt{\frac{2 a b}{a^{2}+b^{2}}}=x+\sqrt{\frac{2}{x^{2}-2}} \geq 3$$

Since
$$\frac{2}{x^{2}-2}-(3-x)^{2}=\frac{(x-2)^{2}\left(-x^{2}+2 x+5\right)}{x^{2}-3} \geq 0$$

Equality holds if and only if $a=b, c=0$ and their permutations.