Proof: Note that, if $a \geq b \geq c$, then
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)=\frac{(a-b)(a-c)(c-b)}{a b c} \leq 0$$

Therefore, we only need to consider the case $a \geq b \geq c$. Squaring both sides, we have
$$\sum_{c y c} \frac{a^{2}}{b^{2}}+\sum_{c y c} \frac{2 b}{a} \geq \frac{9\left(a^{2}+b^{2}+c^{2}\right)}{a b+b c+c a}$$

Additionally, using the following identities
$$\begin{array}{l} \frac{b}{a}+\frac{c}{b}+\frac{a}{c}-3=\frac{(b-c)^{2}}{b c}+\frac{(a-b)(a-c)}{a c} \\ \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}-3=\frac{(b-c)^{2}(b+c)}{b^{2} c^{2}}+\frac{\left(a^{2}-b^{2}\right)\left(a^{2}-c^{2}\right)}{a^{2} b^{2}} \\ a^{2}+b^{2}+c^{2}-(a b+b c+c a)=(b-c)^{2}+(a-b)(a-c) \end{array}$$

The inequality is equivalent to
$$(b-c)^{2} M+(a-b)(a-c) N \geq 0$$

where $M=\frac{2}{b c}+\frac{(b+c)^{2}}{b^{2} c^{2}}-\frac{9}{a b+b c+c a}$;
$$N=\frac{2}{a c}+\frac{(a+b)(a+c)}{a^{2} b^{2}}-\frac{9}{a b+b c+c a}$$

If $b-c \geq a-b$, then $2(b-c)^{2} \geq(a-b)(a-c)$. Thus, we have
$$\begin{array}{l} M \geq \frac{6}{b c}-\frac{9}{a b+b c+c a} \geq 0 \\ M+2 N \geq \frac{6}{b c}-\frac{18}{a b+b c+c a} \geq 0 \end{array}$$

Therefore, we have
$$(b-c)^{2} M+(a-b)(a-c) N \geq \frac{1}{2}(a-b)(a-c)(M+2 N) \geq 0 \text { . }$$

Otherwise, if $b-c \leq a-b$, then $2 b \leq a+c$, of course, $M \geq 0$ and
$$N \geq \frac{2}{a c}+\frac{a+b+c}{a b^{2}} \geq \frac{2}{a c}+\frac{3}{a b} \geq \frac{(\sqrt{2}+\sqrt{3})^{2}}{a c+a b}>\frac{9}{a b+b c+c a}$$

Equality holds if and only if $a=b=c$.