Prove that the inequality is equivalent to
$$\sum_{c y c} \frac{a^{2}}{b^{2}}+4 \sum_{c y c} \frac{a}{b} \geq \frac{9\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a b+b c+c a}+6$$

Consider the following identities:
$$\begin{array}{l} \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{(a-b)^{2}}{a b}+\frac{(c-a)(c-b)}{a c} \\ a^{2}+b^{2}+c^{2}-a b-b c-c a=(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \end{array}$$

The inequality becomes the following form:
$$(a-b)^{2} M+(c-a)(c-b) N \geq 0$$

where $M=\frac{4}{a b}+\frac{(a+b)^{2}}{a^{2} b^{2}}-\frac{9}{a b+b c+c a}, \quad N=\frac{4}{a c}+\frac{(c+a)(c+b)}{a^{2} c^{2}}-\frac{9}{a b+b c+c a}$
Notice that, if $a \geq b \geq c$, then
$$\sum_{c y c} \frac{a}{b} \leq \sum_{c y c} \frac{b}{a} ; \quad \sum_{c y c} \frac{a^{2}}{b^{2}} \leq \sum_{c y c} \frac{b^{2}}{a^{2}}$$

Therefore, we only need to consider the case $a \geq b \geq c$, because the case $a \geq b \geq c$ after applying (*) will reduce its value
$$\begin{array}{l} N \geq \frac{5}{a c}+\frac{b}{a c^{2}}-\frac{9}{a b+b c+c a}>0 \\ M+N \geq \frac{8}{a b}+\frac{5}{a c}-\frac{18}{a b+b c+c a}>0 \end{array}$$

Let $k=\frac{1+\sqrt{5}}{2}$, consider the following cases:
(i) The case $a-b \leq k(b-c)$. Then we have $(a-b)^{2} \leq(a-c)(b-c)$, so
$$(a-b)^{2} M+(c-a)(c-b) N \geq(a-b)^{2}(M+N) \geq 0$$
(ii) The case $a-b \geq k(b-c)$. We only need to prove $M \geq 0$ or
$$\left(a^{2}+b^{2}+6 a b\right)(a b+b c+c a) \geq 9 a^{2} b^{2}$$

Since $(k+1) b-a \leq k c$ and $k+1=\frac{3+\sqrt{5}}{2}$, we have
$$a b-(a-b)^{2}=\left[\frac{(3+\sqrt{5}) b}{2}-a\right]\left[a-\frac{(3-\sqrt{5}) b}{2}\right] \leq k c\left[a+\frac{(3-\sqrt{5}) b}{2}\right] \leq 2 c(a+b)$$

Therefore
$$\begin{array}{l} \left(a^{2}+b^{2}+6 a b\right)(a b+b c+c a)-9 a^{2} b^{2} \geq a b\left[\left(a^{2}-b^{2}\right)^{2}-a b\right]+c(a+b)^{3} \\ >c(a+b)^{3}-4 a b c(a+b)=c(a+b)(a-b)^{2} \geq 0 \end{array}$$

The equality holds if and only if $a=b=c$.